MATHEMATISCHES INSTITUT

PROF. DR. PETER MÜLLER

Summer Term 2013

Lecture Course

Functional Analysis

Typesetting by Kilian Lieret and Marcel Schaub

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Thanks!

Version of April 10, 2014

Contents

1 Topological and metric spaces 51.1 Topological spaces: basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Limits and continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3 Metric spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4 Example: sequence spaces `p . . . . . . . . . . . . . . . . . . . . . . . . . . 141.5 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.6 Example: spaces of continuous functions . . . . . . . . . . . . . . . . . . . . 191.7 Baire’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2 Banach and Hilbert spaces 262.1 Vector spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.2 Banach spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.3 Linear operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.4 Linear functionals and dual space . . . . . . . . . . . . . . . . . . . . . . . . 352.5 Hilbert spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3 Measures, integration and Lp-spaces 473.1 Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 473.2 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 493.3 Lp-spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533.4 Decomposition of Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

4 The cornerstones of functional analysis 694.1 Hahn-Banach theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 694.2 Three consequences of Baire’s theorem . . . . . . . . . . . . . . . . . . . . . 724.3 (Bi)-Dual spaces and weak topologies . . . . . . . . . . . . . . . . . . . . . . 76

5 Bounded operators 855.1 Topologies on the space of bounded linear operators . . . . . . . . . . . . . 855.2 Adjoint operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 885.3 The spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 915.4 Compact operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 955.5 Fredholm alternative for compact operators . . . . . . . . . . . . . . . . . . 98

3

Introductory remarks

Functional analysis is...

• a child of linear algebra and analysis

• a theory of infinite-dimensional vector spaces

Functional analysis has lots of applications

• partial differential equations (PDE’s)

• approximation theory

• numerical maths

• probability theory

• quantum mechanics (functional analysis is its language!)

4

1 Topological and metric spaces

1.1 Topological spaces: basics

1.1 Definition. Let X be a set. T ⊆ P(X) is a topology iff

(1) ∅, X ∈ T .

(2) T is closed under arbitrary unions (i.e. if I is an arbitrary index set and for everyα ∈ I let a set Aα ∈ T be given. Then⋃

α∈IAα ∈ T

holds.).

(3) T is closed under finite intersections (i.e. if n ∈ N and A1, . . . , An ∈ T , then

n⋂k=1

Ak ∈ T

holds.).

• (X, T ) is called topological space (often just X)

• A ∈ P(X) is open iff A ∈ T .

• Let T1, T2 be topologies on X. T1 is finer than T2 iff T1 ⊇ T2 and coarser than T2 iffT1 ⊆ T2.

1.2 Examples. (a) Indiscrete topology: T = ∅, X.

(b) Discrete topology: T = P(X).

(c) Euclidean (or standard) topology on Rn, n ∈ N: A ⊆ Rn is open iff ∀x ∈ A∃ε > 0such that Bε(x) ⊆ A, where Bε(x) := y ∈ Rn : |x − y| < ε is the Euclidean ball ofradius ε > 0 about x ∈ Rn.

Induced topology on subsets

1.3 Definition. Let (X, T ) be a topological space, A ∈ P(X) (not necessarily open!).TA ⊆ P(A) is the relative topology on A iff

TA := B ⊆ A : ∃C ∈ T with B = C ∩A.

1.4 Remark. (a) TA is topology on A.

(b) If A /∈ T and B ∈ TA then it may happen that B /∈ T .

Example. Let X = R with standard topology, A = [0, 1]. Then B := [0, 1/2[ ∈ TAbut B /∈ T .

1.5 Definition. Let X be a topological space, A ⊆ X, x ∈ X.

(a) A is closed iff Ac := X \A ∈ T .

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(b) U ⊆ X (not necessarily open) is a neighbourhood of x iff ∃A ∈ T such that x ∈ A andA ⊆ U .

(c) X is a Hausdorff space iff for all x, y ∈ X, x 6= y, there exist neighbourhoods Ux of xand Uy of y such that Ux ∩ Uy = ∅.

(d) x is a limit point of A (or accumulation point) iff for all neighbourhoods U of x

U ∩A 6= ∅.

Note: Every point of A is also a limit point according to this definition.

(e) x is an interior point of A iff there exists a neighbourhood U of x such that U ⊆ A.

(f) x is a boundary point of A iff for every neighbourhood U of x: U ∩ A 6= ∅ andU ∩Ac 6= ∅.Boundary of A: ∂A := x ∈ X : x boundary point of A.

(g) Interior of A: A := A \ ∂A closure of A: A := A ∪ ∂A = x ∈ X :x limit point of A.

(h) A is dense in X iff X = A.

1.6 Lemma. Let X be a topological space, A ⊆ X.

(a) A is open ⇐⇒ ∀x ∈ A : x is an interior point of A.

(b) A is closed ⇐⇒ A = A.

(c) A, ∂A are closed.

Proof. See Problem 1.

1.7 Definition. Let (X, T ) be a topological space, B ⊆ T a family of open sets.

(a) B is a base for T iff T consists of unions of sets from B.

(b) B is a subbase for T iff finite intersections of sets from B form a base.

(c) N ⊆ T is a neighbourhood base at x iff every N ∈ N is a neighbourhood of x and forevery neighbourhood U of x there exists N ∈ N with N ⊆ U .

1.8 Remark. (a) Let S ⊆ P(X). Then there exists a topology T on X such that S is asubbase for T and T is the coarsest topology containing S. Jargon: T is generatedby S.

(b) Example: Rn with standard topology. Let x ∈ Rn.

(i) B1/k(x) : k ∈ N is a neighbourhood base at x.

(ii) B1/k(q) : k ∈ N, q ∈ Qn is a base for the standard topology (see the proof ofThm. 1.19 later).

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1.9 Definition. Let J 6= ∅ be an arbitrary index set. For every j ∈ J let (Xj , Tj) be atopological space. T is the product topology on the Cartesian product space

×j∈J

Xj :=

f : J →

⋃j∈J

Xj with f(j) ∈ Xj

iff T has the base

×j∈J

Aj : Aj ∈ Tj ∀j ∈ J, Aj 6= Xj for at most finitely many j’s

.

1.10 Remark. If J is finite, then the condition “Aj 6= Xj for at most finitely many j’s”can be dropped.

1.11 Definition. Let X be a topological space.

(a) X is separable iff ∃A ⊆ X countable with A = X.

(b) X is 1st countable iff every x ∈ X has a countable neighbourhood base.

(c) X is 2nd countable iff there exists a countable (sub-)base for the topology.[Note: countable base ⇐⇒ countable sub base (see Problem T2).]

1.12 Theorem. Let X be a topological space. ThenX is 2ndcountable =⇒ X is 1stcountable and separable.

Proof. Let B be a countable base for the topology.

• Let x ∈ X and Nx := B ∈ B : x ∈ B. Then Nx is a neighbourhood base andcountable (hence 1st countable).

• ∀∅ 6= B ∈ B choose xB ∈ B and let A := xB : ∅ 6= B ∈ B. We claim A iscountable (trivial) and A = X. For all x ∈ X and neighbourhoods U of x thereexists C ∈ T such that x ∈ C ⊆ U . C is a union from sets in B, so there existsB ∈ B such that x ∈ B ⊆ U . On the other hand xB ∈ B means xB ∈ U and xB ∈ Aimplies A ∩ U 6= ∅.

1.2 Limits and continuity

1.13 Definition. Let X be a topological space and (xn)n∈N ⊆ X be a sequence.(xn)n converges to x ∈ X iff for every neighbourhood U of x there exists n0 ∈ N and forall n ≥ n0: xn ∈ U .Notations: limn→∞ xn = x or xn

n→∞−−−→ x.

1.14 Remark. (a) convergence is harder for finer topologies.

(b) X Hausdorff =⇒ limits are unique (see Problem T3).

1.15 Definition. Let (X, TX) and (Y, TY ) be topological spaces and let f : X −→ Y .

(a) f is sequentially continuous iff xnn→∞−−−→ x (in X) implies f(xn)

n→∞−−−→ f(x) (in Y ).

(b) f is continuous iff for every A ∈ TY : f−1(A) ∈ TX (f−1(A) is the inverse image).

(c) f is open iff ∀A ∈ TX : f(A) = y ∈ Y : ∃x ∈ A : y = f(x) ∈ TY .

7

(d) f is a homeomorphism iff f is bijective, open and continuous(bijection compatible with topological structure).

1.16 Theorem. Let X,Y be topological spaces and f : X −→ Y a map. Then

(a) f is continuous =⇒ f is sequentically continuous.

(b) f is sequentially continuous and X is first countable =⇒ f is continuous.

Proof. (a) Let xnn→∞−−−→ x inX. Let V ⊆ Y be a neighbourhood of f(x). W.l.o.g.1 assume

V open (see below). Set U := f−1(V ). U is open (because f is continuous) and x ∈ U=⇒ U is a neighbourhood of x and we can apply the definition of convergence to U=⇒ ∃n0 ∈ N : ∀n ≥ n0 : xn ∈ U . But that also means ∃n0 ∈ N : ∀n ≥ n0 : f(xn) ∈ V .

If V is not open there exists an open subset V0 ⊆ V with x ∈ V0 and one can repeatthe same argument with V0 instead of V .

(b) Proof by contradiction: Suppose f is not continuous, i.e. there exists an open subsetV ⊆ Y s.t. U := f−1(V ) is not open: ∃x ∈ U s.t. for all neighbourhoods N of x:

N * U =⇒ N ∩ UC 6= ∅. (1)

Let Nkk∈N be a countable neighbourhood base at x. Consider Nk :=⋂kj=1Nj . Then

Nk+1 ⊆ Nk ∀k and (2)

Nk is a neighbourhood of x ∀k. (3)

So Nkk∈N is a neighbourhood base of x. (1) and (3) imply that ∀k ∈ N ∃xk ∈ Nk∩UC

• ∀l ≥ k : xl ∈ Nk ∩ UC =⇒ xll→∞−−−→ x.

• f(xk) ∈ f(U)C ⊆ V C =⇒ f(xk) 6∈ V ∀k. But f(x) ∈ V so f(xk) cannot convergeto f(x) .

1.3 Metric spaces

Known notions and notations: Let X 6= ∅ be a set.

• Metric d : X ×X −→ [0,∞[ with the known properties:

– d(x, y) ≥ 0 ∀x, y ∈ X with d(x, y) = 0⇐⇒ x = y.

– d(x, y) = d(y, x) ∀x, y ∈ X.

– d(x, y) ≤ d(x, z) + d(z, y) ∀x, y, z ∈ X.

(X, d) is called a metric space.

• Y ⊆ X =⇒ d|Y×Y is the so-called induced metric on Y .

• Open metric ball of radius ε > 0 about x ∈ X:

Bε(x) := y ∈ X : d(x, y) < ε.

• Cauchy sequence, completeness2

1without loss of generality2Completeness is not a topological notion! See Problem 4.

8

Figure 1: Triangle inequality

• For A ⊆ X, x ∈ X define

– diam(A) := supa,a′∈A d(a, a′) diameter of A.

– dist(x,A) := infa∈A d(x, a) distance of x to A.

1.17 Lemma. Let X be a metric space. Then X is a topological space, which is 1st

countable and Hausdorff w.r.t.3 the metric topology, i.e. the one given by the baseB1/k(x)k∈N

x∈X.

Proof. • B1/k(x)k∈N is a neighbourhood base at x which is clearly 1stcountable.

• X is Hausdorff: Let x, y ∈ X be two distinct points. Then ε := d(x, y) > 0. Forarbitrary points u ∈ Bε/2(x) and v ∈ Bε/2(y) the triangle inequation yields (seeFigure 1):

ε = d(x, y) ≤ d(x, u) + d(u, v) + d(v, y) < ε/2 + d(u, v) + ε/2

Thus d(u, v) > 0 and Bε/2(x) ∩Bε/2(y) = ∅ and X is Hausdorff.

1.18 Corollary. All topological notions are available in a metric space X. In particularfor (xn)n ⊆ X, x ∈ X, A ⊆ X:

(a) (xn)n converges to x ⇐⇒ ∀ε > 0∃n0 ∈ N : ∀n ≥ n0 : xn ∈ Bε(x)⇐⇒ limn→∞ d(xn, x) = 0.

(b) x ∈ A ⇐⇒ ∃(yn)n ⊆ A : limn→∞ yn = x.

(c) If Y is a topological space and f : X −→ Y , thenf is continuous ⇐⇒ f is sequentially continuous.

Proof. (a), (b): simple exercises. (c): See Theorem 1.16.

1.19 Theorem. X a metric space. Then X separable implies X 2nd countable.

Proof. There is a countable and dense subsetA ⊆ X (A = X). We claim that B1/n(a)n∈Na∈A

is a base of the (metric) topology: Let C ⊆ X open. ∀x ∈ C ∃ε(x) > 0 such thatBε(x) ⊆ C. Thus

C =⋃x∈C

Bε(x)(x).

Denseness ofA inX implies: ∀x ∈ C ∃a(x) ∈ A and n(x) ∈ N such that x ∈ B1/n(x)(a(x)) ⊆Bε(x)(x) (see Figure 2). But then we can also write

C =⋃x∈C

B1/n(x)(a(x)).

3with respect to

9

Figure 2: Finding n(x)

NB. This also proves Remark 1.8 (b): B1/k(q) k∈Nq∈Qn

is a base of the Euclidean topology

in Rn.

1.20 Lemma. Let X be a complete metric space and A ⊆ X. Then: A closed ⇐⇒ Acomplete.

Proof. See Analysis II.

1.21 Example. Consider the space of all continuous functions f : [0, 1] −→ C:

C([0, 1]) := f : [0, 1] −→ C : f is continuous

with two different metrics

(a) Supremum metric

d∞(f, g) := ‖f − g‖∞, where ‖f‖∞ := supx∈[0,1]

|f(x)|

The metric space (C([0, 1]), d∞) is complete and separable (see Thms. 1.41 and 1.42later).

(b) 1-metric: d1(f, g) :=∫ 1

0 dx |f(x)− g(x)| defines a metric on C([0, 1]):

(i) d1(f, g) ≥ 0 is obvious. Suppose f 6= g. Then there exists at least one pointx0 ∈ [0, 1] with f(x0) 6= g(x0) and we find ε > 0 so small that |f(x0)−g(x0)| > ε.But f, g are continuous, so there exists δ > 0 such that for any x ∈ [0, 1] with|x− x0| < δ we have |f(x)− f(x0)| < ε/3 and |g(x)− g(x0)| < ε/3. Combiningthese inequalities we obtain

|f(x)− g(x)| ≥ |f(x0)− g(x0)|︸ ︷︷ ︸>ε

− |f(x)− f(x0)|︸ ︷︷ ︸<ε/3

− |g(x0)− g(x)|︸ ︷︷ ︸<ε/3

> ε/3

for all x ∈ [0, 1] with |x− x0| < δ. Thus

d1(f, g) =

∫ 1

0dx |f(x)− g(x)| ≥

∫ 1

0dx 1]x0−δ,x0+δ[(x) |f(x)− g(x)|︸ ︷︷ ︸

>ε/3

> ε/3 · δ > 0.

10

Figure 3: fn

(ii) The symmetry d1(f, g) = d1(g, f) is obvious.

(iii) Triangle inequality:

d1(f, h) =

∫ 1

0dx |f(x)− h(x)|︸ ︷︷ ︸≤|f(x)−g(x)|+|g(x)−h(x)|

≤ d1(f, g) + d1(g, h).

Thus, d1 is indeed a metric.

• (C([0, 1]), d1) is separable because d1(f, g) ≤ d∞(f, g) and (C([0, 1]), d∞) is sep-arable.

• (C([0, 1]), d1) is not complete: Consider fn from Figure 3.

(fn)n is a Cauchy sequence: Let n,m ∈ N, m ≥ n:

d1(fn, fm) =

∫ 1

0dx |fn(x)− fm(x)| =

∫ 1/2

1/2−1/ndx |fn(x)− fm(x)|︸ ︷︷ ︸

≤1

≤ 1

n.

But ∫ 1

0dx |fn(x)− 1[1/2,1](x)| ≤

∫ 1/2

1/2−1/ndx |fn(x)− 1[1/2,1](x)|︸ ︷︷ ︸

≤1

≤ 1

n

and 1[1/2,1](x) /∈ C([0, 1]). Suppose ∃f ∈ C([0, 1]) with d1(fn, f)→ 0 as n→∞.Then ∫ 1

0dx |f(x)− 1[1/2,1](x)| = 0

which cannot be true (consider a neighbourhood of x = 1/2) =⇒ fn does notconverge in C([0, 1])!

1.22 Definition. Let (X, dX), (Y, dY ) be metric spaces and T : X −→ Y .

• T is called an isometry iff ∀x, x′ ∈ X : dX(x, x′) = dY (T (x), T (x′)).

• X,Y are isometric iff there exists a bijective isometry T : X −→ Y .

1.23 Remark. Let T : X −→ Y be an isometry. Then

(a) T is injective and continuous,

(b) X and the range of T ,

ranT := T (X) = y ∈ Y : ∃x ∈ X with y = T (x),

are isometric.

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1.24 Theorem. Let X be a metric space. Then there exists a complete metric space Xand an isometry i : X −→ X such that i(X) is dense in X and X is unique up to isometricspaces.

Proof. Consists of 4 steps:

(a) Construct X.

(b) Construct the isometry i and W := i(X) is dense in X.

(c) X is complete

(d) Uniqueness.

(a) Define an equivalence relation ∼ on the set of all Cauchy sequences in X:

(xn)n ∼ (yn)n :⇐⇒ limn→∞

d(xn, yn) = 0

(∼ is clearly reflexive, symmetric and transitive). We look at the equivalence classeswe obtain in this way , so consider

X := equivalence classes x of Cauchy sequences in X.

We write (xn)n ∈ x for a representative (xn)n of the equivalence class x. Define ametric on X:

d(x, y) := limn→∞

d(xn, yn) x, y ∈ X, (xn)n ∈ x, (yn)n ∈ y

We now have to check that d...

(1) ...is well defined:

(i) Existence of the limit :

d(xn, yn) ≤ d(xn, xm) + d(xm, ym) + d(ym, yn)

d(xn, yn)− d(xm, ym) ≤ d(xn, xm) + d(ym, yn)

And the same inequality holds with m and n exchanged. Thus we have:

| d(xn, yn)︸ ︷︷ ︸=:αn

− d(xm, ym)︸ ︷︷ ︸=:αm

| ≤ d(xn, xm) + d(ym, yn) (∗)

Because of (xn)n and (yn)n being Cauchy sequences we have ∀ε > 0 ∃N ∈ Nsuch that ∀n,m ≥ N :

d(xn, xm) < ε/2 and d(yn, ym) < ε/2

So |αn − αm| < ε and (αn)n ⊆ R is a Cauchy sequence, thus convergent(because R is complete).

(ii) Independence of representatives: Let (xn)n ∼ (x′n)n and (yn)n ∼ (y′n)n. Wehave to show

limn→∞

d(xn, yn) = limn→∞

d(x′n, y′n)

Repeat derivation of (∗) with exchanging xm ←→ x′n and ym ←→ y′n andobtain

|d(xn, yn)− d(x′n, y′n)| ≤ d(xn, x

′n) + d(yn, y

′n)

n→∞−−−→ 0 by definition of ∼

12

(2) ...fulfills the axioms of a metric:

• d ≥ 0 X

d(x, y) = 0⇐⇒ limn→∞

d(xn, yn) = 0 for some representatives

⇐⇒ (xn)n ∼ (yn)n

⇐⇒ x = y

• Symmetry and triangle inequality are clear from the properties of d.

(b) Define

i :X −→ X

b 7−→ b

where b is the unique equivalence class with (b, b, b, . . .) ∈ b. Set W := i(X). The mapi is an isometry because

d(b, a) = d(i(a), i(b)) = limn→∞

d(a, b) = d(a, b).

It remains to show that W is dense in X with respect to d. Let x ∈ X and ε > 0.Pick any representative (xn)n ∈ x. (xn)n being a Cauchy sequence implies ∃N ∈ N∀n,m ≥ N : d(xn, xm) < ε. Let b ∈W defined by (xN , xN , . . .) ∈ b

=⇒ d(b, x) = limn→∞

d(xN , xn) < ε

So W is dense.

(c) Let (x(k))k∈N) be a Cauchy sequence in X; W is dense in X: ∀k ∈ N ∃z(k) ∈ W suchthat

d(x(k), z(k)) <1

k

Let (zk, zk, zk, . . .) ∈ z(k) be the corresponding constant representative for each k. Forevery k, l ∈ N we get (due to the isometry property of i):

d(zk, zl) = d(i(zk)︸︷︷︸z(k)

, i(zl)︸︷︷︸z(l)

) ≤ d(z(k), x(k))︸ ︷︷ ︸<1/k

+d(x(k), x(l)) + d(x(l), z(l))︸ ︷︷ ︸<1/l

Thus (z1, z2, z3, . . .) ⊆ X is a Cauchy sequence and hence a representative of somex ∈ X. We show now, that

limn→∞

d(x(k), x) = 0.

Let ε > 0, thend(x(k), x) ≤ d(x(k), z(k))︸ ︷︷ ︸

<1/k

+ d(z(k), x)︸ ︷︷ ︸=limn→∞ d(zk,zn)

< ε

for all k large enough, as (zn)n is Cauchy.

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(d) Suppose, there exists X with a dense subset V and

X

j

i

V

X

j i−1

WX

Figure 4: Uniqueness

X with a dense subset W and bijective isometries j,respectively i between X and V , respectively X andW .We show: X and X are isometric. We know: V :=j(X) and W := i(X) are isometric with isometryj i−1. According to the Problem T5, j i−1 can beuniquely extended to an isometry X −→ X. Thuswe have the uniqueness up to isometries.

1.4 Example: sequence spaces `p

1.25 Definition (`p-spaces). Set

`p :=

x = (xn)n∈N : xn ∈ C ∀n and ‖x‖p :=

(∑n∈N|xn|p

)1/p<∞

whenever p ∈ [1,∞[ and

`∞ :=x = (xn)n∈N : xn ∈ C ∀n and ‖x‖∞ := sup

n∈N|xn| <∞

(‖·‖p will be a norm for every p ∈ [1,∞]).

1.26 Lemma. For every p ∈ [1,∞], dp(x, y) := ‖x− y‖p, x, y ∈ `p defines a metric on`p.

Proof. All properties clear, except for triangle inequality, this follows from Lemma 1.27.

1.27 Lemma. Let p, q ∈ [1,∞] be conjugated exponents, i.e. 1/p+ 1/q = 1 (convention:

”1/∞ = 0“). Then

(a) Dual pairing and Holder inequality: ∀x ∈ `p, ∀y ∈ `q: 〈x, y〉 :=∑

n∈N xnyn is welldefined and ∣∣〈x, y〉∣∣ ≤∑

n∈N|xnyn| ≤ ‖x‖p ‖y‖q .

(b) Minkowski inequality: Let x, y ∈ `p. Then

‖x+ y‖p ≤ ‖x‖p + ‖y‖p .

Proof. From corresponding inequalities on CN and subsequent limitN →∞. For example,to prove (a), use

N∑n=1

|xnyn| ≤( N∑n=1

|xn|p)1/p( N∑

n=1

|yn|q)1/q

(see e.g. Forster, vol. 1) and perform limit.

1.28 Remark. Minkowski inequality does not remain true for p ∈ ]0, 1[. Therefore weconsider `p-spaces only for p ≥ 1.

1.29 Theorem. (a) `p is separable ∀p ∈ [1,∞[.

14

(b) `∞ is not separable.

Proof. (a) We use the separability of C. For n ∈ N let

Mn := (x1, . . . , xn, 0, . . .) ∈ `p with xj ∈ Q+ iQ, j = 1, . . . , n.

So Mn is countable and M :=⋃n∈NMn is also countable. Claim M = `p. Let y ∈ `p

and ε > 0, then there exists N ∈ N:

∞∑j=N+1

|yj |p <εp

2

and since Q + iQ is dense in C there exists x ∈ MN such that∑N

j=1 |xj − yj |p < εp

2 .This implies (

dp(y, x))p

= ‖x− y‖pp < εp

(b) See Problem 6.

1.30 Theorem. `p is complete for every p ∈ [1,∞].

Proof. (a) Case p ∈ [1,∞[. Let (x(n))n∈N ⊂ `p be a Cauchy sequence (x(n) = (x(n)1 , x

(n)2 , . . .)).

Let ε > 0. Then there exists N ∈ N such that for every n,m ≥ N and J1, J2 ∈ N:

J2∑j=J1

|x(n)j − x

(m)j |p < εp (∗)

Step 1: We have to show the existence of a candidate for the limit using the complete-

ness of C. Let J1 = J2 = J . (∗) implies (x(n)J )n ⊂ C is a Cauchy sequence and

since C is complete, there is some xJ ∈ C such that limn→∞ x(n)J = xJ for all

J ∈ N. So our candidate is x := (x1, x2, . . .).

Step 2: Set J1 = 1 in (∗) and use the Minkowski inequality in CJ2 :( J2∑j=1

|xj |p)1/p

≤( J2∑j=1

|xj − x(n)j |p

)1/p

+

( J2∑j=1

|x(n)j |p

)1/p

︸ ︷︷ ︸≤‖x(n)‖

p<∞

≤ ε+ ‖x(n)‖p ∀J2 ∈ N.

So we obtain ‖x‖p ≤ ε + ‖x(n)‖p < ∞ for every n ∈ N and therefore x ∈ `p.For every n ≥ N and m→∞ in (∗):

J2∑j=1

|x(n)j − xj |p < εp ∀ J2 ∈ N,

so for sending J2 →∞ in addition, we have

‖x(n) − x‖p︸ ︷︷ ︸dp(x(n),x)

< ε.

So x(n) −→ x in `p.

(b) Case p =∞: Replace “∑J2

j=J1” by “supJ1≤j≤J2” and “| · |p” by “| · |”.

15

1.5 Compactness

1.31 Definition. Let X be a topological space and A ⊆ X.

(a) A is compact iff for every open cover A ⊆ ⋃α∈I Bα with an index set I 6= ∅ andBα ⊆ X open for every α ∈ I there exists a finite open subcover, i.e. N ∈ N andα1, . . . , αN ∈ I with A ⊆ ⋃N

n=1Bαn (”Heine-Borel-property“).

(b) A is sequentially compact iff every sequence in A has a convergent subsequence withlimits in A (

”Bolzano-Weierstraß property“).

(c) A is relatively compact iff A is compact.

1.32 Remark. (a) Def. applies in particular to A = X. In this case we have”=“ instead

of”⊆“ in (a).

(b) Some books (e.g. Bourbaki) use compactness only for Hausdorff spaces, otherwise thenotion is quasi-compact.

1.33 Theorem. Let X be a topological space.

(a) If X is 1st countable, then X compact implies X sequentially compact.

(b) If X is 2nd countable, then X compact is equivalent to X sequentially compact.

1.34 Theorem (Lindelof). Let X be a 2nd countable topological space and let X =⋃α∈I Aα be an open cover. Then there exists (αn)n∈N ⊆ I such that X =

⋃n∈NAαn,

i.e. a countable subcover.

Proof. Let Bkk∈N be a countable base of the topology. Set

K := k ∈ N : ∃α ∈ I with Bk ⊆ Aα.Since every Aα is a union of Bk’s, we have⋃

k∈KBk =

⋃α∈I

Aα = X (*)

So for every k ∈ K pick αk ∈ I such that Bk ⊆ Aαk . Then we get

X(∗)=⋃k∈K

Bk︸︷︷︸⊆Aαk

⊆⋃k∈K

Aαk ⊆⋃α∈I

Aα(∗)= X

So⋃k∈K Aαk = X.

Proof of Theorem 1.33. (a) By contradiction: Assume X is compact but there exists asequence (xn)n∈N ⊂ X without convergent subsequence.

Claim: ∀x ∈ X there exists a neighbourhood U(x) such that xn ∈ U(x) for at mostfinitely many n.

Proof of the claim: Consider a countable neighbourhood base Uii∈N at x and letVm :=

⋂mi=1 Ui for m ∈ N. Suppose the claim was false. Then for every m ∈ N there

exist in finitely many k ∈ N such that xk ∈ Vm. This means there exists a subsequence(km)m ⊆ N increasing such that xkm ∈ Vm for every m ∈ N. So, for every m,m′ ∈ N,m′ ≥ m: xkm′ ∈ Vm. This implies limm→∞ xkm = x, i.e. convergent subsequence. Now X =

⋃y∈X U(y) =

⋃ni=1 U(yi) for some y1, . . . , yn ∈ X (because X is compact).

The claim implies that U(yi) contains at most finitely many sequence members xk ∈U(yi), so there are at most finitely many k such that xk ∈ X.

16

(b) “=⇒” follows from (a) and Lemma 1.17.

We now show “⇐=” by contradiction: Assume every sequence has a convergent sub-sequence, but there exists an open cover without finite subcover. Since X is 2nd

countable there exists a countable subcover X =⋃j∈NCj of this cover by Lindelof.

For every n ∈ N pick a point

xn ∈ X \( n⋃j=1

Cj

)(always possible ∀n ∈ N because there does not exist a finite subcover). Let (xn)n ⊂ Xbe a sequence in X. By hypothesis, it has a convergent subsequence xnk

k→∞−−−→ x ∈ X.There exists N ∈ N such that x ∈ CN , so CN is a neighbourhood of x. xnk beingconvergent means xnk ∈ CN for finally all k, but for every k such that nk ≥ N wehave by definition of xn that xnk /∈ CN .

1.35 Theorem. Let X be a compact topological space and A ⊆ X. Then

(a) A closed =⇒ A compact

(b) If X is also Hausdorff, then A compact =⇒ A closed.

Proof. (a) Let A ⊆ ⋃α∈I Uα be an open cover. Since A is closed Ac is open and

X = Ac ∪(⋃α∈I

Uα

)is an open cover. Since X is compact, there exist α1, . . . , αn ∈ I such that

X = Ac ∪( n⋃i=1

Uαi

)and A ⊆ ⋃n

i=1 Uαi is a finite subcover.

(b) See Problem 8.

1.36 Warning. Bounded and closed do not imply compact in general! Example: `p,p ∈ [1,∞] and

B1(0) = x ∈ `p : ‖x‖p ≤ 1 = x ∈ `p : ‖x‖p < 1 = B1(0)

is bounded and closed but consider e(n) := (. . . , 0, 1, 0, . . .) ∈ `p (with 1 at the nth position)n ∈ N. It fulfills

dp(e(n), e(m)) = ‖e(n) − e(m)‖p =

21/p p <∞1 p =∞

∀n,m ∈ N, n 6= m,

so there exists no convergent subsequence and B1(0) is not sequentially compact. Hence,by Theorem 1.33 (a), B1(0) is not compact.

1.37 Theorem. Let X be a metric space. Then

X is compact ks(a) +3

(c)

X is sequentially compact

X is 2ndcountable ks(b)

+3 X is separable.

17

Proof. (a) (⇒) This is Theorem 1.33 (a).

(⇐) Using the implication (c) and the equivalence (b), which will be proven below:If X is sequentially compact it is also 2ndcountable and we can apply Theo-rem 1.33 (b).

(b) For any topological space X 2nd countability implies separability (Theorem 1.12). IfX is a separable metric space, then it is also 2nd countable (Theorem 1.19).

(c) We show that sequential compactness implies separability by constructing a countable

set M with M = X. Fix n ∈ N and use the following algorithm to define points x(n)k :

• Choose an arbitrary x(n)1 ∈ X

• If R(n)1 := X \B1/n(x

(n)1 ) 6= ∅, pick x

(n)2 ∈ R(n)

1 , otherwise stop.

• Suppose x(n)1 , . . . , x

(n)k are chosen. If

R(n)k := X \

( k⋃j=1

B1/n(x(n)1 )

)6= ∅,

pick x(n)k+1 ∈ R

(n)k , otherwise stop.

Claim: This algorithm stops after finitely many steps.

True, because for k 6= l we have d(x(n)k , x

(n)l ) > 1/n. Thus, if the algorithm did

not stop after finitely many steps we would have an infinite sequence (x(n)l )l∈N ⊂ X

without a convergent subsequence which is a contradiction to X being sequentiallycompact.

The claim implies the existence of Kn ∈ N such that

X =

Kn⋃j=1

B1/n

(x

(n)j

).

SetMn := x(n)j : j = 1, . . . ,Kn andM :=

⋃k∈NMn. M is countable andM = X.

1.38 Theorem (Tychonoff). Let J 6= ∅ be an index set and Xα a compact topologicalspace for all α ∈ J . Then

×α∈J

Xα =f : J −→

⋃α∈J

Xα such that f(α) ∈ Xα

is compact (in the product topology).

Proof. See any textbook on topology, e.g. Kelley, v. Querenburg.

1.39 Definition. Let X,Y be topological spaces. Define

C(X,Y ) := f : X −→ Y such that f is continuous

in particular for Y = C set C(X) := C(X,C).

1.40 Theorem. Let X,Y be topological spaces, X compact, f ∈ C(X,Y ). Then

(a) f(X) is compact.

18

(b) If in addition X,Y are Hausdorff and f a bijection, then f is a homeomorphism.

(c) If in addition X,Y are metric spaces, then f is uniformly continuous iff

∀ε > 0 ∃δ ≡ δε : ∀x ∈ X : f(Bδ(x)) ⊆ Bε(f(x))

(Note that Bδ(x) and Bε(f(x)) are balls corresponding to potentially different metrics.However we will not use different notations for them as long as it is clear from thecontext to which metric they belong).

(d) If even Y = R then f takes on its maximum and minimum.

Proof. (b), (c), (d): See Problems 9, 10 and T7 .

(a) Let f(X) ⊆ ⋃α∈J Fα be openly covered. Then

X ⊆ f−1

(⋃α∈J

Fα

)=⋃α∈J

f−1(Fα).

Because f is continuous f−1(Fα) is open ∀α ∈ J and we have an open cover of X.But X is compact which by definitions means that we find N ∈ N and α1, . . . , αn suchthat X ⊆ ⋃N

n=1 f−1(Fαn) and, thus, f(X) ⊆ ⋃N

n=1 Fαn .

1.6 Example: spaces of continuous functions

General assumptions in this subsection:

• X is a compact Hausdorff space,

• K means R or C,

• C(X,K) is equipped with the uniform (supremum) metric

d∞(f, g) := supx∈X|f(x)− g(x)| = ‖f − g‖∞ .

(well defined by Theorem 4.40 (a)).

1.41 Theorem. C(X,K) is complete.

Proof. Follows from completeness of the bounded continuous functions Cb(X,K), seeProblem 5, and C(X,K) = Cb(X,K), which follows from compactness of X and The-orem 4.40.

1.42 Theorem. X metrisable ⇐⇒ C(X,K) separable.

Proof. For ”⇐=” see e.g. Bourbaki, Elements of Mathematics, General Topology, Part 2,Sect. X.3.3. Thm. 1.1. Here, we only show ”=⇒”:For m,n ∈ N define

Gmn :=

f ∈ C(X,K) : f(B1/m(x)) ⊆ B1/n(f(x)) ∀x ∈ X

Compactness of X implies that,

19

• . . . f ∈ C(X,K) is automatically uniformly continuous. Fix n and consider Theorem4.40(c) with ε := 1/n. f is uniformly continuous, i.e. there exists δ > 0 such that

∀x ∈ X : f(Bδ(x)) ⊆ B1/n(f(x)).

Then we can find m ∈ N such that 1/m ≤ δ and get that f ∈ Gmn. This shows

C(X,K) =⋃m∈N

Gmn ∀n ∈ N. (1)

• . . . for any m ∈ N we can find p(m) ∈ N and a1, . . . , ap(m) ∈ X such that X can bewritten a union of open balls of radius 1/m:

X =

p(m)⋃j=1

B1/m(aj). (2)

Now, K is separable, i.e. there exists a countable set κν ∈ K : ν ∈ N that is dense in K.For fixed m ∈ N and ϕ : 1, . . . , p(m) −→ N (or equivalently ϕ ∈ Np(m)) define

G(ϕ)mn := f ∈ Gmn : |f(ak)− κϕ(k)| < 1/n ∀k = 1, . . . , p(m).

We only want to consider those ϕ with G(ϕ)mn 6= ∅, so we set

Φmn := ϕ ∈ Np(m) : G(ϕ)mn 6= ∅.

Φmn is not empty (see below). For ϕ ∈ Φmn pick some gϕ ∈ G(ϕ)mn. Now define

Lmn := gϕ : ϕ ∈ Φmn and L :=⋃

m,n∈NLmn

Lmn is countable because Φmn ⊆ Np(m). Thus, L is countable as well. We want to showthat L is dense in C(X,K). Pick an arbitrary f ∈ C(X,K). Because of (1) we have:

∀n ∈ N ∃m ∈ N such that f ∈ Gmn. There also exists ϕf ∈ Φmn such that f ∈ G(ϕf )mn : We

only have to choose ϕf (k) such that |f(ak)− κϕf (k)| < 1/n. But κν : ν ∈ N is dense inK, so this is obviously possible.For each x ∈ X choose kx ∈ 1, . . . , p(m) such that x ∈ B1/m(akx) (this is possiblebecause of (2)). Now we consider gϕf (x) ∈ Lmn ⊆ L. Using that f and gϕf both lie in

G(ϕf )mn we can estimate the upper bound of the distance |f(x)− gϕf (x)|:

|f(x)− gϕf (x)| ≤ |f(x)− f(akx)|︸ ︷︷ ︸<1/n by definition of Gmn

+ |f(akx)− κϕf (kx)|︸ ︷︷ ︸<1/n by definition of G

ϕfmn

+

+ |κϕf (kx) − gϕf (akx)|︸ ︷︷ ︸<1/n by definition of G

ϕfmn

+ |gϕf (akx)− gϕf (x)|︸ ︷︷ ︸<1/n by definition of Gmn

<4

n

Thus, the distance vanishes for n → ∞ and L is countable and dense in C(X,K), i.e.C(X,K) is separable.

An alternative way to separability:

20

1.43 Definition. (a) A K-vectorspace A is a K-algebra iff there exists a multiplicationA×A −→ A which fulfils the distributive laws

(a+ b)c = ac+ bc ∀ a, b, c ∈ A,c(a+ b) = ca+ cb ∀ a, b, c ∈ A,λ(ac) = (λa)c = a(λc) ∀ a, c ∈ A, λ ∈ K.

(b) A subspace B ⊆ A is a subalgebra iff B is closed under multiplication.

(c) A subset B ⊆ C(X,K) separates points in X iff for all distinct points x, y ∈ X thereexists f ∈ B such that f(x) 6= f(y).

1.44 Theorem (Stone-Weierstraß). Let B ⊆ C(X,K) be a subalgebra with the followingproperties:

• 1 ∈ B (where 1: X −→ K, x 7−→ 1 ∀x ∈ X)4,

• B is closed with respect to d∞,

• B separates points.

• If K = C, assume further that B is closed under complex conjugation.

Then B = C(X,K).

Proof. e.g. Reed/Simon, vol. 1, Appendix to Sect. IV.3

1.45 Corollary. Let K ⊆ Rd compact, d ∈ N. Then the set of all polynomials is dense(with respect to d∞) in C(K,K). In particular, C(K,K) is separable.

Proof. Let B0 be the K-Algebra generated by the monomials K −→ K

x = (x1, . . . , xd) 7−→ xnα n ∈ N0, α ∈ 1, . . . , d

Set B := B0. Then Stone-Weierstraß gives us B = C(K,K). Now we show separability:If K = R, let BQ0 be the Q-Algebra generated by the monomials, if K = C, let BQ0 be the(Q+ iQ)-algebra. Then

• BQ0 is countable

• Since K is compact, BQ0 = B0 = C(K,K).

1.46 Definition. Let X be a metric space (not necessarily compact). Let F be a familyof continuous functions f : X −→ K.

• F is equicontinuous iff for every ε > 0 and x ∈ X there exists a δ > 0 such that forevery f ∈ F :

f(Bδ(x)) ⊆ Bε(f(x))

• F is uniformly equicontinuous iff for every ε > 0 there exists δ > 0 such that forevery x ∈ X and f ∈ F :

f(Bδ(x)) ⊆ Bε(f(x))

4This property is called “unital”

21

1.47 Remark. (a) If X is even compact, then (by Problem T7): equicontinuous ⇐⇒uniformly equicontinuous.

(b) Examples

• X = [0, 1]: x 7−→ cos(x/n) : n ∈ N is uniformly continuous

• X = [0, 1]: x 7−→ x1/n : n ∈ N is not equicontinuous.

• X = ]0,∞[: x 7−→ arctan(nx) : n ∈ N is equicontinuous, (individually)uniformly continuous, but not uniformly equicontinuous.

1.48 Theorem. Let X be a metric space (not necessarily compact) and (fn)n∈N ⊆C(X,K) an equicontinuous sequence. Suppose there exists a dense subset D ⊆ X suchthat for every x ∈ D the limit limn→∞ fn(x) exists. Then limn→∞ fn(x) =: f(x) exists∀x ∈ X and the limit function f ∈ C(X,K).

Proof. Firstly, we show convergence everywhere. Let x ∈ X be arbitrary but fixed and letε > 0. By assumption there exists δ > 0 such that for every n ∈ N

fn(Bδ(x)) ⊆ Bε(fn(x)). (1)

Since D = X, there is some y ∈ D ∩Bδ(x). At y, the sequence (fn(y))n converges and istherefore a Cauchy sequence, so there exists N ∈ N such that ∀n,m ≥ N

|fn(y)− fm(y)| < ε. (2)

So we have

|fn(x)− fm(x)| ≤ |fn(x)− fn(y)|︸ ︷︷ ︸<ε by (1)

+ |fn(y)− fm(y)|︸ ︷︷ ︸<ε by (2)

+ |fm(y)− fm(x)|︸ ︷︷ ︸<ε by (1)

< 3ε

So limn→∞ fn(x) =: f(x) exists for every x ∈ X.Secondly, we show the continuity of f . (1) is equivalent to

∃δ > 0 ∀n ∈ N ∀x′ ∈ X with d(x, x′) < δ =⇒ |fn(x)− fn(x′)| < ε,

so for n → ∞ there exists δ > 0, such that for every x′ ∈ X with d(x, x′) < δ we have|f(x)− f(x′)| < ε.

1.49 Lemma. In addition to the assumptions in Theorem 1.48 suppose that X is compact.Then we have even limn→∞ d∞(fn, f) = 0.

Proof. We need to show uniform convergence. Let ε > 0. Since X is compact, there existsδ > 0 such that for every x ∈ X and n ∈ N:

fn(Bδ(x)) ⊆ Bε(f(x)) (1)

(see Remark 1.47) and there is l ∈ N and a1, . . . , al ∈ X:

X =

l⋃j=1

Bδ(aj). (2)

We have

|fn(x)− f(x)| ≤ |fn(x)− fn(ajx)|︸ ︷︷ ︸=:T1

+ |fn(ajx)− f(ajx)|︸ ︷︷ ︸=:T2

+ |f(ajx)− f(x)|︸ ︷︷ ︸T3

where ajx is the centre of some Ball Bδ such that x ∈ Bδ(ajx).

22

• T1 < ε for every n ∈ N by (1) (and the definition of ajx)

• T2 < ε for all n sufficiently large (there are only finitely many j’s and pointwiseconvergence)

• T3 < ε by taking n→∞ in (1) (compare the end of the proof of Thm. 1.48).

So finally we have

∀ε > 0 ∃N ∈ N ∀n ≥ N ∀x ∈ X : |fn(x)− f(x)| < 3ε.

1.50 Theorem (Arzela-Ascoli). Let X be a compact metric space and (fn)n∈N ⊂ C(X,K)an equicontinuous and pointwise bounded sequence, i.e.

supn∈N|fn(x)| <∞ ∀x ∈ X.

Then there exists a uniformly convergent subsequence (fnj )j∈N. Equivalently, every equicon-tinuous and pointwise bounded subset F ⊆ C(X,K) is relatively compact.

Proof. Since C(X,K) is a metric space the equivalence of the 2 statements follows fromTheorem 1.37. We prove the “sequence version”: SinceX is compact, due to Theorem 1.37,X is also separable, so there exists a dense subset al ∈ X : l ∈ N ⊆ X. Pointwiseboundedness gives

supn∈N|fn(al)| <∞ ∀l ∈ N,

so by the Bolzano-Weierstraß theorem for every l there exists a subsequence (n(l)j )j∈N ⊆ N

such thatlimj→∞

fn(l)j

(al)

exists. W.l.o.g. assume (n(l+1)j )j ⊆ (n

(l)j )j . Define nj := n

(j)j (diagonal sequence trick),

then (nj)j≥l ⊆ (n(l)j )j and limj→∞ fnj (al) exists for every l ∈ N. Since the set of the al’s

is dense, Lemma 1.49 gives the claim.

1.51 Remark. Both assumptions, compactness and equicontinuity are essential for The-orem 1.50 to turn uniform boundedness into convergence of a subsequence.

1.7 Baire’s Theorem

. . . the mother of 3 (out of 4) fundamental theorems of functional analysis.

1.52 Remark. Let X be a metric space and A1, A2 ⊆ X be open and dense. Then A1∩A2

is also open and dense.

Completeness gives more:

1.53 Theorem (Baire). Let X be a complete metric space and ∀n ∈ N let An ⊆ X beopen and dense. Then

⋂n∈NAn is dense in X.

1.54 Remark. (a) Openness of⋂n∈NAn is false in general.

(b) Completeness is essential for the theorem: Consider Q with the metric inherited fromR. Let qn ∈ Q : n ∈ N be an enumeration of Q. Define An := Q \ qn. This isopen and dense in Q but

⋂n∈NAn = ∅.

23

Figure 5: Constructing the sequence (xn).

Proof. Define D :=⋂n∈NAn. Let x0 ∈ X be arbitrary and fix and ε > 0. To prove the

denseness of D in X we have to show that D ∩Bε(x0) 6= ∅. We do this by constructing asequence (xn) that converges against x ∈ D ∩Bε(x0) (see Figure 5).A1 dense implies that A1 ∩ Bε(x0) is non-empty and we can pick x1 ∈ A1 ∩ Bε(x0) andε1 > 0 with ε1 ∈]0, ε/2[ such that

Bε1(x1) ⊆ A1 ∩Bε(x0).

Proceeding analogously with A2 and Bε1(x1) we pick x2 ∈ A2 ∩Bε1(x1) and ε2 ∈ ]0, ε1/2[such that

Bε2(x2) ⊆ A2 ∩Bε1(x1) ⊆ A1 ∩A2 ∩Bε(x0).

We proceed in the same fashion and get two sequences:

(a) (εn)n with 0 < εn < 2−nε ∀n ∈ N,

(b) (xn)n ⊂ X with Bεn+1(xn+1) ⊆ An+1 ∩Bεn(xn) ⊆ A1 ∩ · · · ∩An ∩Bε(x0)

In particular we have:

∀N ∈ N ∀n ≥ N : xn ∈ BεN (xN ). (∗)

That is, (xn)n is Cauchy and since X is complete, (xn) converges to some x ∈ X. Butbecause of (∗) we have x ∈ BεN (xN ) ∀N ∈ N and (b) yields x ∈ ⋂n∈NBεn+1(xn+1) ⊆D ∩Bε(x0). Thus, D ∩Bε(x0) 6= ∅.

1.55 Definition. Let X be a topological space and A ⊆ X.

1. A is a Gδ(-set) iff A is a countable intersection of open sets.

2. A is nowhere dense iff A has no interior points.

3. A is meagre (or of 1st category) iff A is a countable union of nowhere dense sets.

4. A is non-meagre (or of 2nd category) iff A is not meagre.

24

1.56 Example. Q is meager in R (just consider the countable union Q =⋃q∈Qq).

1.57 Lemma. Let X be a topological space and A ⊆ X. Then

(a) A is nowhere dense ⇐⇒ (A)c dense,

(b) A is meagre and B ⊆ A =⇒ B is meagre,

(c) An ⊆ X meagre ∀n ∈ N =⇒ ⋂n∈NAn meagre.

Proof. (b) and (c) are clear by the very definitions. Statement (a) follows from the equiv-alence

B has no interior points⇐⇒ Bc dense.

To show ”=⇒”, consider x ∈ X. If x 6∈ B then x ∈ BC . If x ∈ B it is by hypothesis nointerior point of B and thus must be a limit point of BC .

We now show 3 equivalent reformulations of Baire’s theorem:

1.58 Lemma. Let X be a topological space. Then the statements (i)–(iv) are equivalent:

(i) An ⊆ X open and dense ∀n ∈ N =⇒ ⋂n∈NAn dense.

(ii) An ⊆ X is a dense Gδ ∀n ∈ N =⇒ ⋂n∈NAn is a dense Gδ.

(iii) A ⊆ X,A 6= ∅ and A is open =⇒ A non-meagre.

(iv) A ⊆ X meagre =⇒ AC dense.

1.59 Corollary. Let X be a complete metric space. Then (i)–(iv) in Lemma 1.58 hold.In particular if X 6= ∅ then X is non-meagre.

Proof of Lemma 1.58. (i) ⇔ (ii) by definition of Gδ.

(i) ⇔ (iii) Let ∅ 6= A ⊆ X be open. Suppose A is meagre, that is A =⋃n∈NAn with

An nowhere dense ∀n ∈ N. According to 1.57 (a), this is equivalent to (An)C denseand open ∀n ∈ N and because of implication (i), the intersection

⋂n∈N(An)C =: B

is dense as well. But then BC =⋃n∈NAn ⊇ A has no interior points. Thus A has

no interior points either and this is clearly a contradiction to A being open.

(iii) ⇒ (iv) Problem 13.

(iv) ⇒ (i) Problem 13.

25

2 Banach and Hilbert spaces

2.1 Vector spaces

General assumption: X 6= 0 is a K-vector space, where K = R or C.

2.1 Definition. Let ∅ 6= M ⊆ X.

• M is linearly independent iff all non-empty finite subsets F ⊆ M are linearly inde-pendent, i.e. the following implication holds:∑

f∈Fαf︸︷︷︸∈K

f = 0 =⇒ αf = 0 ∀f ∈ F.

• M is linearly dependent iff M is not linearly independent.

• B ⊆ X is a Hamel basis (or algebraic basis) iff

– B is linearly independent

– every x ∈ X can be represented as a finite linear combination from elements inB (B ”spans” X).

• X has finite dimension if there exists a Hamel basis B with |B| <∞. dimX := |B|is called the dimension of X.

• X has infinite dimension iff X does not have finite dimension.

2.2 Remark. The dimension is well defined: |B| is the same for every Hamel basis in agiven space.

2.3 Example. Consider

cc := x = (xj)j∈N : xj ∈ C ∀j ∈ N and xj 6= 0 for only finitely many j’s.

The index c stands for “compact support”. Let em := (. . . , 0, 1, 0, . . . ) with a 1 at the n-thposition. Claim: B := en : n ∈ N is a Hamel basis for cc.

Remark. Even though `1 is separable there exists no countable hamel basis for `1.

2.4 Theorem. Every vector space X 6= 0 has a Hamel basis.

Proof. Uses Zorn’s lemma. See later.

2.5 Corollary. X has infinite dimension iff for every n ∈ N there exists Mn ⊆ X suchthat |Mn| = n and Mn is linearly independent.

Proof. Existence of a Hamel basis B with |B| =∞.

2.6 Example. Infinite dimensional vector spaces: cc, `p, C(X) (where ∅ 6= X ⊆ Rd

open).

26

2.2 Banach spaces

2.7 Definition. Let X be a vector space. A mapping X −→ [0,∞[, x 7−→ ‖x‖ is a normiff

(1) ‖x‖ > 0 ∀0 6= x ∈ X,

(2) ‖λx‖ = |λ|‖x‖ ∀λ ∈ K ∀x ∈ X,

(3) ‖x+ y‖ ≤ ‖x‖+ ‖y‖ ∀x, y ∈ X

(X, ‖ · ‖) is called a normed space. If only (2) and (3) hold, ‖ · ‖ is called a seminorm.

2.8 Remark. Let X be a normed space. Then d(x, y) := ‖x− y‖ is a metric on X. Thusall topological notions and results from the theory of metric spaces are available.

2.9 Example. • `p is a normed space ∀p ∈ [1,∞].

• C(X,K) (where X a compact Hausdorff space) is a normed space with

‖f‖ := ‖f‖∞ := supx∈X|f(x)|

• Rn, Cn are normed spaces for n ∈ N (Euclidian norm or p-norm).

2.10 Lemma. Let X be a normed space. Then

• the mapping X 3 x 7−→ ‖x‖ is continuous.

Addition and multiplication are continuous as well:

• Let xkk→∞−−−→ x and yk

k→∞−−−→ y. Then xk + ykk→∞−−−→ x+ y.

• Let αkk→∞−−−→ α in K and xk

k→∞−−−→ x. Then αkxkk→∞−−−→ αx.

Proof. • Let (xk)k ⊆ X with xkk→∞−−−→ x. This is equivalent to ‖xk − x‖ x→∞−−−→ 0

(Corollary 1.18). To show that ‖ · ‖ is sequentially continuous, we have to prove that∣∣‖xk‖ − ‖x‖∣∣ k→∞−−−→ 0. This follows from the following inequalities:

‖xk‖ − ‖x− xk‖ ≤ ‖x− xk + xk‖ = ‖x‖ ≤ ‖xk‖+ ‖x− xk‖=⇒ lim sup

k→∞‖xk‖ ≤‖x‖ ≤ lim inf

k→∞‖xk‖

The following sets are a base of the metric topology of X:

x+B1/k(0) : x ∈ X, k ∈ N = B1/k(x) = y ∈ X : ‖x− y‖ ≤ 1/k,

Here we used the following notation for sets A,B: A + B = a + b : a ∈ A, b ∈ B anda+B := a+B. Warning: In general not every metric comes from a norm.

• Addition is continuous:

‖xk + yk − x− y‖ ≤ ‖xk − x‖︸ ︷︷ ︸k→∞−−−→0

+ ‖yk − y‖︸ ︷︷ ︸k→∞−−−→0

k→∞−−−→ 0.

27

• Multiplication is continuous:

‖αkxk − αx‖ = ‖αkxk − αx+ αkx− αkx‖ =

= ‖αk(xk − x) + x(αk − α)‖ ≤ ‖αk(xk − x)‖+ ‖(αk − α)x‖ ≤≤ |αk|︸︷︷︸

bounded in k

‖xk − x‖︸ ︷︷ ︸k→∞−−−→0

+ |αk − α|︸ ︷︷ ︸k→∞−−−→0

‖x‖ k→∞−−−→ 0.

2.11 Definition. A complete normed space is called a Banach space.

2.12 Examples. • All spaces in Example 2.9 are Banach spaces.

• Consider C([0, 1]) with L1 norm: ‖f‖1 :=∫ 1

0 |f(x)| dx is not a Banach space. Thiswas already discussed in Example 1.21 (b).

2.13 Theorem. Every normed space X can be completed, so that X is isometric to a denselinear subspace W of the Banach space X, which is unique up to isometric isomorphisms.

Proof. Analogous to the proof of Theorem 1.24. Note that the isometry is even a linearbijection (thus a isomorphism) in this case (see Section 2.3).

2.14 Definition. Let X be a normed space. A set en ∈ X : n ∈ N is a (Schauder) basisin X iff for all x ∈ X there exists a sequence (xn)n ⊆ KN such that

limN→∞

∥∥∥∥x− N∑n=1

xnen

∥∥∥∥ = 0.

Notation: x =∑

n∈N xnen ”infinite linear combinations”, ”convergent series”.

2.15 Example. Let p ∈ [1,∞[. Then (en)n∈N with en := (0, . . . , 1, 0, . . . ) (where the 1 isat the nth position) is a (Schauder) basis of `p: For x = (xn)n∈N ∈ `p we have∥∥∥∥x− N∑

n=1

xnen

∥∥∥∥pp

=

∞∑n=N+1

|xn|p N→∞−−−−→ 0.

Note that this construction fails for `∞!

2.16 Lemma. Let X be a normed space. Then

X has a (Schauder) basis =⇒ X is separable.

Proof. Let K0 := Q for K = R, respectively K0 := Q+ iQ for K = C. Define

AN :=

N∑n=1

xnen : xn ∈ K0

.Then the union A :=

⋃N∈NAN is dense in X and countable.

2.17 Remark. The implication ”⇐=” in Lemma 2.16 does not hold (Enflo, 1973).

2.18 Lemma. Let X be a Banach space, A ⊆ X a (linear) subspace. Then

A is closed ⇐⇒ A is complete.

28

Proof. See Analysis II, like proof of Lemma 1.20.

2.19 Theorem. Let X be a normed space and F ⊆ X a finite-dimensional subspace.Then F is complete and closed.

Proof. Choose a basis e1, . . . , en in F . (F, ‖ · ‖) is a normed space and isometric to(Kn, ||| · |||) with

|||α||| :=∥∥∥∥ n∑j=1

αjej

∥∥∥∥ ∀α = (α1, . . . , αn) ∈ Kn.

Kn is complete with respect to the Euclidean norm. But all norms in finite-dimensionalspaces are equivalent5. Thus (Kn, ||| · |||) is complete and because of the isometry, (F, ‖ · ‖)is complete as well. F is closed by Lemma 2.18 with X = A = F .

As a preparation for Theorem 2.21 we prove the following lemma:

2.20 Lemma (Riesz, 1918). Let X be a normed space and U ( X a closed subspace.Then for all λ ∈]0, 1[ there exists xλ ∈ X \ U such that

‖xλ‖ = 1 and ‖xλ − u‖ ≥ λ ∀u ∈ U.Proof. Since U is closed, we have d := dist(x, U) = infu∈U d(x, u) > 0 for all x ∈ X \ U(see Problem 9(c)!). Since λ < 1 there exists uλ ∈ U such that

d ≤ ‖x− uλ‖ ≤d

λ, hence γ :=

1

‖x− uλ‖≥ λ

d.

Define xλ := γ(x − uλ) ∈ X \ U . The first property ‖xλ‖ = |γ| · ‖x − uλ‖ = 1 holds bydefinition of γ and

‖xλ − u‖ = ‖γ(x− uλ)− u‖ = ‖γx− (u+ γuλ)‖ =

= |γ| ·∥∥∥x− uλ +

u

γ︸ ︷︷ ︸∈U

∥∥∥ ≥ γd ≥ λ ∀u ∈ U.

Warning 1.36 illustrates the more general

2.21 Theorem. Let X be a normed space. Then

B1(0) = x ∈ X : ‖x‖ ≤ 1 compact ⇐⇒ dimX <∞Proof. (⇐) Let dimX <∞. Proof of Theorem 2.19: X is isometric to (Kn, ||| · |||). The

statement then follows by Heine-Borel and the equivalence of norms.

(⇒) Suppose that dimX =∞. We show that this implies that B1(0) is not sequentiallycompact by constructing a sequence (xn)n in B1(0) without a convergent subse-quence:

• Pick an arbitrary x1 ∈ X with ‖x1‖ = 1. Let U1 := spanx1 be the subspacespanned by x1. U1 is closed in X.

• Riesz Lemma, applied with λ = 1/2 shows the existence of x2 ∈ X \ U1 suchthat ‖x2‖ = 1 and ‖x2 − x1‖ ≥ 1/2. Let U2 := spanx1, x2.• If we continue this procedure we get a sequence (xn)n∈N ⊆ B1(0) that satisfies‖xn − xm‖ ≥ 1/2 for all n 6= m.

(xn)n clearly has no convergent subsequence. Thus B1(0) is not compact.

5 This is a theorem from linear algebra: For norms ‖·‖ and |||·||| on Kn we can find constants c, c ∈]0,∞[such that c · ‖ · ‖ ≤ ||| · ||| ≤ c · ‖ · ‖.

29

2.3 Linear operators

2.22 Definition. Let X,Y be vector spaces (over the same field), X0 ⊆ X a subspaceand T : X0 −→ Y .

• T is linear (a linear operator) iff T (αx+βy) = αT (x)+βT (y) ∀α, β ∈ K, x, y ∈ X0.Notation: Tx := T (x). T is a (vector space) homomorphism.

• dom(T ) := X0 is the domain of T .

• ranT := T (X0) is the range of T .

• ker(T ) := x ∈ X0 : Tx = 0 is the kernel of T .

2.23 Examples. (a) The identity operator 1 := 1X : X −→ X, x 7−→ x is obviously alinear operator.

(b) Consider X = Y = C([0, 1]).

• Then the differentiation operator T : X0 = C1([0, 1]) −→ C([0, 1]) with f 7−→ f ′,so (Tf)(x) = f ′(x) and this is a linear operator.

• (Tf)(x) :=x∫0

f(t)dt has domT = X and is called the anti-derivative.

• (Tf)(x) := xf(x) is the multiplication by x.

2.24 Lemma. Let T be a linear operator. Then

(a) ran(T ) and ker(T ) are vector subspaces

(b) dim ran(T ) ≤ dim domT

(c) ker(T ) = 0 ⇐⇒ there exists an inverse T−1 of T such that T−1 : ran(T ) −→dom(T ) and T−1T = 1dom(T ).

Proof. Copy from linear algebra.

2.25 Remark. • If T−1 exists, it is linear.

• Even if we have T : X −→ X with ker(T ) = 0 (then T−1 exists) this does notimply TT−1 = 1, but only TT−1 = 1ran(T ). In other words, ker(T ) = 0 does notimply T is bijective (only injective).

2.26 Example (illustrating Remark 2.25). Let X = `∞ and x = (x1, x2, . . .) ∈ X.Consider the shift operator defined by Tx := (0, x1, x2, . . .), i.e.

(Tx)n :=

0 n = 1,

xn−1 n ≥ 2.

Then we have domT = `∞ and ranT = y = (y1, y2, . . .) ∈ `∞ : y1 = 0. Define(T−1x)n := (x2, x3, . . .). Then domT−1 = `∞ and T−1T = 1. But TT−1x = x holds onlyfor x ∈ ran(T ). Even though dom(T ) = X and ker(T ) = 0, we have ran(T ) X. Thisis not possible if Y = X and dimX <∞.

30

2.27 Definition. Let X,Y be normed spaces, T : X ⊇ dom(T ) −→ Y . T is bounded iffits operator norm is finite, i.e.

‖T‖dom(T )−→Y ≡ ‖T‖ := supx∈dom(T )

x6=0

‖Tx‖‖x‖ = sup

x∈dom(T )‖x‖=1

‖Tx‖ <∞.

2.28 Examples. (cf. Example 2.23)

(a) 1 : (X, ‖ · ‖) −→ (X, ‖ · ‖) is bounded with ‖1‖ = 1.

(b) Let X = C([0, 1]) with ‖·‖∞.

• T = ddx is unbounded. For n ≥ 2 take fn : x 7−→ sin(nx). Then ‖fn‖∞ = 1 and

‖Tfn‖ = supx∈[0,1]

n| cos(nx)| = n

so that

‖T‖ ≥ ‖f′n‖∞‖fn‖∞

= n.

• The anti-derivative fulfills

‖Tf‖∞ = supx∈[0,1]

∣∣∣∣x∫

0

f(t)dt

∣∣∣∣ ≤ 1 ‖f‖∞

for every f ∈ C([0, 1]) and f 6= 0. Because of

‖Tf‖‖f‖ ≤ 1

we obtain ‖T‖ ≤ 1. Moreover,

‖T · 1‖∞ = supx∈[0,1]

x∫0

dt = supx∈[0,1]

x = 1,

and since ‖1‖∞ = 1 we have ‖T‖ = 1.

• The multiplication operator by x has ‖T‖ = 1.

2.29 Theorem. Let X,Y be normed spaces, T : X ⊇ dom(T ) −→ Y a linear operator.Then the following statements are equivalent.

(a) T is continuous.

(b) T is continuous in some x0 ∈ dom(T )

(c) There exists c ∈]0,∞[ such that ‖Tx‖ ≤ c ‖x‖ for every x ∈ dom(T ).

(d) T is bounded.

Proof. (a) ⇒ (b) obvious.

31

(b) ⇒ (c) Suppose T is continuous at x0. Claim: It follows continuity at 0. This is truebecause let (xn)n ⊂ dom(T ) be a sequence which converges to 0. Then

(xn + x0)nn→∞−−−→ x0,

and by linearity and continuity in x0 we obtain

Txn + Tx0 = T (xn + x0)n→∞−−−→ Tx0.

So Txnn→∞−−−→ 0. From this we know that for ε = 1 there is a δ > 0 such that

whenever ‖x‖ ≤ δ for x ∈ dom(T ) we have ‖T x‖ ≤ 1. For general 0 6= x ∈ dom(T ),let x := x

‖x‖δ, so that ‖x‖ ≤ δ. Then we have

δ

‖x‖ ‖Tx‖ = ‖T x‖ ≤ 1 =⇒ ‖Tx‖ ≤ δ−1 ‖x‖ ,

i.e. c = 1/δ.

(c) ⇒ (d) We know ‖Tx‖ ≤ c ‖x‖ and so

supx∈dom(T )

x 6=0

‖Tx‖‖x‖ ≤ c <∞.

(d) ⇒ (a) Let (xn)n ⊆ dom(T ) with xnn→∞−−−→ x then

‖Txn − Tx‖ = ‖T (xn − x)‖ ≤ ‖T‖ · ‖xn − x‖ n→∞−−−→ 0

2.30 Lemma. Let T be linear and dim dom(T ) <∞. Then T is bounded.

Proof. See linear algebra.

2.31 Definition. Let X,Y be vector spaces and T : X ⊇ dom(T ) −→ Y . Let U ⊆ dom(T )and W ⊇ dom(T ).

Restriction of T to U : T |U : U −→ Y, x 7−→ T |Ux := Tx

Extension of T to W : T : W −→ Y with T x = Tx ∀x ∈ dom(T ), i.e. T |dom(T ) = T .

2.32 Theorem (Bounded linear extension). Let X be a normed space and Y a Banachspace. Let T : X ⊇ dom(T ) −→ Y be a bounded linear operator. Let Z be the completion ofdom(T ). Then there exists a bounded linear extension T : Z −→ Y of T with ‖T‖ = ‖T‖.If X is identified with a subspace of its completion X, i.e. X = W in Theorem 2.13, thenT is unique.

NB. (a) If X is a Banach space, then the uniqueness holds anyway.

(b) If dom(T ) ⊆ X, this gives extension to closure as a special case.

Proof. Let x ∈ Z. Then there exists a sequence (xn)n ⊆ dom(T ) with xnn→∞−−−→ x in X.

Since (xn)n is a Cauchy sequence, (Txn)n is a Cauchy sequence in Y because

‖Txn − Txm‖ = ‖T (xn − xm)‖ ≤ ‖T‖ · ‖xn − xm‖ .

Using that Y is a Banach space, there exists y ∈ Y such that Txnn→∞−−−→ y in Y . Define

T x := y (then T |dom(T ) = T ). We have to check several things:

32

• Well-definedness, i.e. independence of approximating sequence: Let xmm→∞−−−−→ x in

X. Then we have‖Txn − T xm‖ ≤ ‖T‖ · ‖xn − xm‖ .

In the limit n,m→∞ we get ‖y − limm→∞ T xm‖ = 0.

• Linearity: Let xnn→∞−−−→ x, x′n

n→∞−−−→ x′. Then

T (αx+ α′x) = limn→∞

T (αxn + α′x′n) = limn→∞

αTxn + limn→∞

α′Tx′n = αTx+ αTx′.

• Norm: as T is an extension, we have ‖T‖ ≥ ‖T‖, because

‖T‖ = supx∈dom(T )

x6=0

‖T x‖‖x‖ ≥ sup

x∈dom(T )x 6=0

‖T x‖‖x‖ = ‖T‖ .

On the other hand, since ‖·‖ is continuous

‖T x‖ = limn→∞

xn∈dom(T )

‖T xn‖

= limn→∞

‖Txn‖ ≤ limn→∞

‖T‖ · ‖xn‖= ‖T‖ lim

n→∞‖xn‖

= ‖T‖ ‖x‖ .

So ‖T x‖‖x‖ ≤ ‖T‖ for every x ∈ dom(T ) and therefore ‖T‖ ≤ ‖T‖. So ‖T‖ = ‖T‖.

• Uniqueness: The fact that we defined T x := y is necessary to ensure the continuityin x.

2.33 Definition. Let X,Y be normed spaces. Define

BL(X,Y ) := T : X −→ Y such that T is linear and bounded

and set BL(X) := BL(X,X).

2.34 Theorem.(

BL(X,Y ), ‖·‖X→Y)

is a normed space. If Y is complete, so is BL(X,Y ).

Proof. • First of all, BL(X,Y ) is a K-vector space with zero element

0: X −→ Y

x 7−→ 0.

For T1, T2 ∈ BL(X,Y ) and α ∈ K define T1 + T2 and αT1 by

(T1 + T2)x := T1x+ T2x

(αT1)x := αT1x

for every x ∈ X.

• ‖·‖X→Y is a norm on BL(X,Y ).

– ‖T‖X→Y ≥ 0 X and if ‖T‖X→Y = 0 then ‖Tx‖ = 0 for every x ∈ X, so Tx = 0for every x ∈ X and so T = 0.

33

– We have

‖αT‖X→Y = sup06=x∈X

‖(αT )x‖‖x‖ = sup

06=x∈X

‖αTx‖‖x‖ = sup

0 6=x∈X|α| · ‖Tx‖‖x‖ = |α| · ‖T‖.

– We have ‖(T1 + T2)x‖ = ‖T1x+ T2x‖ ≤ ‖T1x‖+ ‖T2x‖. Then

‖T‖ ≤ sup06=x∈X

(‖T1x‖‖x‖ +

‖T2x‖‖x‖

)≤ ‖T1‖+ ‖T2‖.

• Let Y be complete. We show completeness of BL(X,Y ). Let (Tk)k∈N ⊆ BL(X,Y )be a Cauchy sequence. For every x ∈ X and k, l ∈ N:

‖Tkx− Tlx‖ ≤ ‖Tk − Tl‖ · ‖x‖. (∗)

This implies that (Tkx)k∈N ⊆ Y is a Cauchy sequence for every x ∈ X and since Yis complete, there is some limit

limk→∞

Tkx =: Tx ∈ Y.

This defines a map

T : X −→ Y

x 7−→ Tx := limk→∞

Tkx.

• T is linear, because let α, α′ ∈ K and x, x′ ∈ X. Then

T (αx+ α′x′) = limk→∞

Tk(αx+ α′x′)

= limk→∞

(αTkx+ α′Tx′)

= αTx+ α′Tx′.

• We show T is bounded and that it is the norm limit of the Tk’s. According to (∗)we have that for every ε > 0 there exists an N ∈ N such that ∀k, l ≥ N and x ∈ X:

‖Tkx− Tlx‖ ≤ ε‖x‖.

So applying the limit l → ∞ gives ‖Tkx − Tx‖ ≤ ε‖x‖ and so ‖Tk − T‖ ≤ ε. Thisgives two things:

1. T is bounded because Tk ∈ BL(X,Y ) and (Tk − T ) ∈ BL(X,Y ) and sinceBL(X,Y ) is a vector space:

T = Tk − (Tk − T ) ∈ BL(X,Y ).

2. Tkk→∞−−−→ T in ‖·‖X→Y .

34

2.4 Linear functionals and dual space

2.35 Definition. Let X be a normed space.A linear functional (on X) is a linear operator l : dom(l) −→ K.The dual space of X is X∗ := BL(X,K)6. Notations for the norm on X∗: ‖·‖X→K =:‖·‖X∗ =: ‖·‖∗ =: ‖·‖.

2.36 Corollary (from Theorem 2.34). (X∗, ‖·‖X∗) is a Banach space (no matter whetherX is complete).

2.37 Examples. Let X = C([a, b]) (where a < b ∈ R) equipped with ‖·‖∞.

(a) For f ∈ X let

I(f) :=

b∫a

dx f(x)

This is clearly a linear functional I : X −→ K with

|I(f)| ≤b∫a

dx |f(x)|︸ ︷︷ ︸≤‖f‖∞

≤ ‖f‖∞ (b− a)

so ‖I‖X∗ ≤ (b−a) and the function f ≡ 1 yields equalities above and so ‖I‖X∗ = b−a.

(b) For f ∈ X and t ∈ [a, b] let δt(f) := f(t). This gives a linear functional δt : X −→ Kwhich is called the Dirac-δ functional with |δt(f)| = |f(t)| ≤ ‖f‖∞ and equality againby considering f ≡ 1. So ‖δt‖X∗ = 1.

NB. Pay attention that the boundedness of δt is heavily dependent on the norm onX. For example it’s not bounded, if X is equipped with the 1-norm ‖ · ‖1 =

∫ 10 dx | · |.

Notation. Let X,Y be metric spaces. We write X ∼= Y iff X is isometrically isomorphicto Y .

2.38 Theorem. Let p ∈ [1,∞[ and 1p + 1

q = 1. Then (`p)∗ ∼= `q.

Proof. • Case 1 < p < ∞: Let x = (xn)n∈N ∈ `p, f ∈ (`p)∗. We work with thecanonical (Schauder) basis enn∈N for `p (introduced in Example 2.15). Thereforex can be written as a ‖·‖p-convergent series x =

∑n∈N xnen with xn ∈ K. Since f is

continuous and linear:

f(x) =∑n∈N

f(xnen) =∑n∈N

xnf(en). (∗)

For N ∈ N (fixed), define x := (xn)n∈N ∈ `p by

xn :=

|f(en)|qf(en) if n ≤ N ∧ f(en) 6= 0,

0 otherwise.

6Note: These are only bounded (or equivalently continuous) linear functionals, hence the name topo-logical dual – not to be confused with the algebraic dual f : X −→ K : f linear that is common in linearalgebra!

35

Apply (∗) to x:

f(x) =

N∑n=1

|f(en)|q (∗∗)

so that 0 ≤ f(x) ≤ ‖f‖(`p)∗ ‖x‖p where

‖x‖p =

( N∑n=1

|f(en)|(q−1)p

)1/p

.

Since 1p = 1− 1

q ⇐⇒ p = qq−1 , we obtain p(q − 1) = q and

(∗∗)=⇒

N∑n=1

|f(en)|q ≤ ‖f‖(`p)∗

( N∑n=1

|f(en)|q)1/p

=⇒( N∑n=1

|f(en)|q)1/q

≤ ‖f‖(`p)∗ . (∗∗∗)

Thus, the map

I :(`p)∗ −→ `q

f 7−→(f(en)

)n∈N

is well defined and also

– I is linear,

– ‖If‖q =∥∥(f(en))n

∥∥q≤ ‖f‖(`p)∗ by (∗∗∗),

– I is onto7, because if y = (yn)n ∈ `q, define

fy(x) :=∑n∈N

xnyn

for x = (xn)n∈N ∈ `p. This is well defined because Holder’s inequality yields|fy(x)| ≤ ‖x‖p ‖y‖q so that ∥∥fy∥∥(`p)∗

≤ ‖y‖q

and fy ∈ (`p)∗. But fy(en) = yn for all n ∈ N.

– ‖f‖(`p)∗ ≤∥∥(f(en))n

∥∥q, due to (∗) and Holder as in the previous argument.

So I is an isometric isomorphism.

• The case p = 1 is analogous, but instead of defining xn, use

|f(en)| ≤ ‖f‖(`1)∗ ‖en‖1︸ ︷︷ ︸=1

.

This implies∥∥(f(en))n

∥∥∞ ≤ ‖f‖(`1)∗ , which replaces (∗∗∗), and the properties of I

follow as above.

2.39 Remarks. (a) (c0)∗ ∼= `1 (see problem sheet).

7surjective

36

(b) The map `1 −→ (`∞)∗, x 7−→ fx, defined by

fx(y) =∑n∈N

xnyn, y ∈ `∞, x ∈ `1,

is well-defined (Holder!), linear, isometric but not onto! In other words, (`∞)∗ isstrictly “bigger” than `1. (proof later)

2.5 Hilbert spaces

The main new feature is the “geometry” from the scalar product.

2.40 Definition. Let X be a (K-)vector space. A mapping 〈·, ·〉 : X×X −→ X is a scalarproduct (or inner product) iff

(i) 〈x, x〉 ≥ 0 for every x ∈ X and if 〈x, x〉 = 0 then x = 0 (non-degenerate).

(ii) 〈x, αy + βz〉 = α 〈x, y〉+ β 〈x, z〉 for every α, β ∈ K and x, y, z ∈ X.

(iii) 〈x, y〉 = 〈y, x〉 for every x, y ∈ X.

(X, 〈·, ·〉) is an inner product space (or pre-Hilbert space).

2.41 Lemma (Cauchy-Schwarz (-Bunjakowski) inequality). Let X be an inner productspace, let x, y ∈ X. Then

| 〈x, y〉 | ≤ 〈x, x〉1/2 〈y, y〉1/2

with equality if and only if x, y are linear dependent.

Proof. See linear algebra!

2.42 Lemma. Let X be an inner product space. Then X is a normed space with norm‖x‖ := 〈x, x〉1/2. All notions from topological, metric and normed spaces are available. Inparticular, the scalar product 〈·, ·〉 : X ×X −→ K is continuous.

Proof. • ‖ · ‖ fulfills all axioms of a norm, see linear algebra.

• Continuity of 〈·, ·〉: Let xnn→∞−−−→ x, yn

n→∞−−−→ y, then

| 〈xn, ym〉 − 〈x, y〉 | ≤ | 〈xn, ym〉 − 〈xn, y〉 |+ | 〈xn, y〉 − 〈x, y〉 | =

= | 〈xn, ym − y〉 |+ | 〈xn − x, y〉 |2.41≤

≤ ‖xn‖ · ‖ym − y‖︸ ︷︷ ︸m→∞−−−−→0

+ ‖y‖︸︷︷︸<∞

· ‖xn − x‖︸ ︷︷ ︸n→∞−−−→0

n,m→∞−−−−−→ 0

because supn∈N ‖xn‖ <∞ (as (xn)n converges!).

2.43 Definition. A complete inner product space is called Hilbert space.

2.44 Theorem. Let X be an inner product space. Then there exists a Hilbert space H,a dense subset W ⊆ H and a unitary map U : X −→ W (i.e. U is an isomorphism with〈x, y〉X = 〈Ux,Uy〉H for all x, y ∈ X).

37

Idea of the proof. See the proofs of Theorem 2.13 and 1.24. Additional aspect here: Definea scalar product in H := equivalence classes x of Cauchy sequences in X:

〈x, y〉H := limn→∞

〈xn, yn〉X

with (xn)n, (yn)n ∈ X being representatives of the equivalence classes x, y, and

U :X −→Wx 7−→ x

where x is the equivalence class of the constant representative (x, x, x, . . . ).

2.45 Examples. (a) `2 is a Hilbert space with the scalar product 〈x, y〉 :=∑

n∈N xnyn,where x = (xn)n, y = (yn)n.

(b) C([0, 1]) is an inner product space with the scalar product

〈f, g〉 :=

∫ 1

0dx f(x)g(x)

but not a Hilbert space (proof analogous to Example 1.21 (b)).

(c) `p for p 6= 2 is not an inner product space, because of the following theorem:

2.46 Theorem (Jordan-von Neumann). Let (X, ‖ · ‖) be a normed space. Then X is aninner product space with ‖·‖ = 〈·, ·〉1/2 if and only if ‖·‖ satisfies the parallelogram identity

‖x+ y‖2 + ‖x− y‖2 = 2(‖x‖2 + ‖y‖2)

for all x, y ∈ X.

Proof. Idea: ”=⇒” elementary computation.”⇐=” Define inner product by polarisation

〈x, y〉 :=

14

(‖x+ y‖2 − ‖x− y‖2

)K = R,

14

(‖x+ y‖2 − ‖x− y‖2

)+ i

4

(‖x+ iy‖2 − ‖x− iy‖2

)K = C.

Verify that this definition satisfies axioms of an inner product:

1. Symmetry and definiteness is obvious

2. bi-/sesquilinearity follows from parallelogram identity: (here only K = R)

〈x, y〉+ 〈x, z〉 =1

4

(‖x+ y‖2 − ‖x− y‖2 + ‖x+ z‖2 − ‖x− z‖2

)=

1

2

(∥∥∥∥x+y + z

2

∥∥∥∥2

−∥∥∥∥x− y + z

2

∥∥∥∥2)

= 2

⟨x,y + z

2

⟩(1)

z = 0 in (1)⇒〈x, y〉 = 2⟨x, y/2

⟩(2)

(1) and (2)⇒〈x, y〉+ 〈x, z〉 = 〈x, y + z〉 (3)

(2) with 2y instead of y : 〈x, 2y〉 = 2 〈x, y〉 (4)

by induction from (3) and (4) : 〈x, ny〉 = n 〈x, y〉 ∀n ∈ N0 (5)

38

z = −y in (3) ⇒ 〈x, y〉 = −〈x,−y〉 ⇒ (5) holds for all n ∈ Z.Let m ∈ Z \ 0 and use (5) with y

m instead of y (writing λ := nm):

〈x, λy〉 = n

⟨x,y

m

⟩= λm

⟨x,y

m

⟩= λ 〈x, y〉 .

This holds for all λ ∈ Q and thus for all λ ∈ R by continuity.

2.47 Definition. Let X be an inner product space and A ⊆ X.

(a) x, y ∈ X are orthogonal iff 〈x, y〉 = 0. In symbols: x ⊥ y.

(b) The Orthogonal complement of A is A⊥ := x ∈ X : 〈x, a〉 = 0 ∀a ∈ A.

(c) Let J 6= ∅ be an index set and for all α ∈ J let eα ∈ X. eαα∈J is an orthonormalset iff

• ‖eα‖ = 1 ∀α ∈ J•⟨eα, eβ

⟩= 0 ∀α, β ∈ J, α 6= β.

(d) eαα∈J is an orthonormal basis (ONB) or complete orthonormal set iff

• eαα∈J is an orthonormal set

• if 〈eα, x〉 = 0 for some x ∈ X and all α ∈ J , then x = 0 (i.e. the zero vector isthe only vector orthogonal to all eα’s!).

2.48 Lemma. Let X be an inner product space, A ⊆ X a subspace.

(a) A⊥ is a closed subspace in X.

(b) Every orthonormal set eαα∈J is linearly independent.

Proof. (a) Tutorial sheet.

(b) Suppose∑n

j=1 λjeαj = 0 for some λ1, . . . , λn ∈ K, some α1, . . . , αn ∈ J with αj 6=αk ∀j 6= k, and n ∈ N. Then we have

0 =⟨eαk , 0

⟩=⟨eαk ,

n∑j=1

eαjλj

⟩=

n∑j=1

λj 〈eαk , eαj 〉︸ ︷︷ ︸δjk

= λk ∀k = 1, . . . , n.

2.49 Lemma. Let X be an inner product space, x ∈ X and eαα∈J an orthonormal set.Then:

‖x‖2 ≥ supJ0⊆J|J0|<∞

∑α∈J0| 〈eα, x〉 |2 =:

∑α∈J| 〈eα, x〉 |2. (Bessel’s inequality)

If X is even a Hilbert space and if eαα∈J is even an ONB, then x =∑

α∈J 〈eα, x〉 eα,where at most countable many terms are 6= 0, and

‖x‖2 =∑α∈J| 〈eα, x〉 |2. (Parseval’s equality)

39

Proof. Let J0 ⊆ J be a finite subset. Then we can write x as follows:

x =∑α∈J0〈eα, x〉 eα︸ ︷︷ ︸=:u

+

(x−

∑α∈J0〈eα, x〉 eα

)︸ ︷︷ ︸

=:v

.

Using that eαα∈J0 is an orthonormal set, we get:

〈u, u〉 =

⟨∑α∈J0〈eα, x〉 eα,

∑β∈J0

⟨eβ, x

⟩eβ

⟩=

∑α,β∈J0

〈eα, x〉⟨eβ, x

⟩ ⟨eα, eβ

⟩=∑α∈J0〈eα, x〉 〈eα, x〉 =

∑α∈J0| 〈eα, x〉 |2,

〈u, v〉 = 〈u, x− u〉 = 〈u, x〉 − 〈u, u〉 =∑α∈J0〈eα, x〉 〈eα, x〉 − 〈u, u〉 = 0.

Thus we can estimate ‖x‖2:

‖x‖2 = 〈u+ v, u+ v〉 = 〈u, u〉+ 〈v, v〉︸ ︷︷ ︸≥0

≥∑α∈J0| 〈eα, x〉 |2 ∀J0 ⊆ J, |J0| <∞,

and taking the supremum over all finite subsets J0 ⊆ J , we get Bessel’s inequality.

Digression on uncountable sums: Suppose∑α∈J

cα := supJ0⊆J|J0|<∞

∑α∈J0

cα <∞ where 0 ≤ cα <∞.

Then cα = 0 for all but countable many α. To see that, we define the following sets:

S0 := α ∈ J : cα ≥ 1Sn := α ∈ J : 1/n > cα ≥ 1/(n+ 1), n ∈ N.

Claim: |Sn| <∞ ∀n ∈ N0.True, because if this was violated for some k ∈ N, then

∑α∈J cα ≥

∑α∈Sk cα =∞.

But thenα ∈ J : cα > 0

=⋃n∈N0

Sn is a countable union of finite sets and thus

countable. End of digression

This means that there exists a sequence (αn)n∈N ⊆ J such that∑α∈J| 〈eα, x〉 |2 =

∑n∈N| 〈eαn , x〉 |2 ≤ ‖x‖ <∞, (∗)

where we have used Bessel’s inequality.

Digression on series in Banach and Hilbert spaces:

• Let Z be a Banach space and (zn)n∈N ⊆ Z. Assume that∑

n∈N ‖zn‖ < ∞. Then∑n∈N zn exists in Z.

For, let Sn :=∑n

j=1 zj for n ∈ N. Since Z is a Banach space, it suffices to show that(Sn)n is Cauchy: Let m ≥ n. Then

‖Sm − Sn‖ =∥∥∥ m∑j=n+1

zj

∥∥∥ ≤ m∑j=n+1

‖zj‖ = σm − σn

with σn :=∑n

j=1 ‖zj‖. But by hypothesis (σn)n∈N converges in R and, hence, isCauchy.

40

• Let Z be a Hilbert space and let (zn)n∈N ⊆ Z be a sequence of pairwise orthogonalvectors. Then:

∑n∈N ‖zn‖2 <∞ ⇐⇒ ∑

n∈N zn exists in Z.

For, consider again the sequence of partial sums (Sn)n. Let m ≥ n, then

‖Sm − Sn‖2 =∥∥∥ m∑j=n+1

zj

∥∥∥2=

m∑j,k=n+1

〈zj , zk〉 =m∑

j=n+1

‖zj‖2 = σm − σn

with σn :=∑n

j=1 ‖zj‖2. Thus, (Sn)n is Cauchy if and only if (σn)n∈N is Cauchy.

End of digression

(∗) and the above convergence criterion for sequences in Hilbert spaces implies that

x′ :=∑n∈N〈eαn , x〉 eαn

exists in X. We need to show x = x′:We have x− x′ ⊥ eαn ∀n ∈ N:⟨eαm , x− x′

⟩= 〈eαm , x〉 −

⟨eαm , x

′⟩ = 〈eαm , x〉 −∑n∈N〈eαn , x〉 〈eαm , eαn〉︸ ︷︷ ︸

δmn

= 0 ∀m ∈ N.

If α 6∈ αnn∈N we have:⟨eα, x− x′

⟩= 〈eα, x〉 −

∑n∈N〈eαn , x〉 〈eα, eαn〉︸ ︷︷ ︸

=0︸ ︷︷ ︸=0

= 〈eα, x〉(∗)= 0.

Taken together, we have x− x′ ⊥ eα ∀α ∈ J ONB=⇒ x = x′.

Moreover,

‖x‖2 =

∥∥∥∥∥ limn→∞

n∑j=1

〈eαj , x〉eαj

∥∥∥∥∥2

=

=

⟨limn→∞

n∑j=1

〈eαj , x〉eαj , limn′→∞

n′∑k=1

〈eαk , x〉eαk

⟩=

= limn→∞n′→∞

⟨n∑j=1

〈eαj , x〉eαj ,n′∑k=1

⟨eαk , x

⟩eαk

⟩=

= limn→∞n′→∞

n∑j=1

n′∑k=1

〈eαj , x〉〈eαk , x〉 〈eαj , eαk〉︸ ︷︷ ︸δjk

=

=∑j∈N|〈eαj , x〉|2

(∗)=∑α∈J|〈eα, x〉|2.

The above proof showed

2.50 Corollary. Let X be a Hilbert space, eαα∈J an ONB.

(a) If (cα)α∈J ⊂ K with∑

α∈J |cα|2 <∞, then∑

α∈J cαeα is well defined in X.

41

(b) If J is countable, then eαα∈J is a Schauder basis.

2.51 Theorem. Every Hilbert space X 6= 0 has an ONB. Moreover,X separable ⇐⇒ there exists a countable ONB.

Proof. ”⇐=” Suppose there exists a countable ONB. By Corollary 2.50 (b), this is aSchauder basis. Thus, X is separable by Lemma 2.16.

”=⇒” Let xn : n ∈ N be dense in X. W.l.o.g. assume xn 6= 0 ∀n ∈ N. We construct anONB using the the Gram-Schmidt procedure:

• Define e1 := x1‖x1‖ .

• Throw away all xn’s that are linearly dependent of e1. Let n1 > 1 be the smallestindex of the remaining elements. Set

e2 := xn1 −⟨e1, xn1

⟩e1 and e1 :=

e2

‖e2‖.

• Throw away all xn’s that are linearly dependent of span(e1, e2). Let n2 > n1 be thesmallest index of the remaining elements. Set

e3 := xn2 −⟨e1, xn2

⟩e1 −

⟨e2, xn2

⟩e2 and e3 :=

e3

‖e3‖.

• Continue this procedure.

This terminates if and only if dimX <∞. It is clear, that enn∈N is an orthonormal set.spanenn is dense in X because each xn is a finite linear combination of en’s. It remainsto show that this is a basis: Assume there is y ∈ X such that y ⊥ en ∀n. Denseness

=⇒ ∃ (yn)n ⊆ X such that yk ∈ spanen : n ∈ N for all k ∈ N and ykk→∞−→ y in X.

Thus 〈y, yk〉 = 0 ∀k and

‖y‖2 = 〈y, y〉 = limk→∞

〈y, yk〉︸ ︷︷ ︸=0

= 0.

The existence of an ONB in the non-separable case will be proven later (when Zorn’sLemma is available).

2.52 Theorem. Let X be a Hilbert space. If dimX = n, then X is isomorphic to Kn. IfdimX =∞ and X is separable, then X is (unitarily equivalent) isomorphic to `2.

Proof. We show only the 2nd case here: Fix a countable ONB enn∈N (exists by Theo-rem 2.51 because X is separable) and consider the linear operator

U :X −→ `2

x 7−→ (〈en, x〉)n∈N.

U is

• well defined by Parseval’s equality.

• surjective by Corollary 2.50 (a),

• preserves the scalar product, because Parseval implies ‖x‖2 =∑

n | 〈en, x〉 |2 =‖Ux‖`2 ∀x ∈ X. This implies 〈x, y〉X = 〈Ux,Uy〉`2 ∀x, y ∈ X by polarisation(see proof of Thm. 2.46).

42

=⇒ U is unitary.

2.53 Definition (Direct sum). Let (X1, 〈·, ·〉1), (X2, 〈·, ·〉2) be inner product spaces. Then

X1 ⊕X2 :=

(x1

x2

): x1 ∈ X1, x2 ∈ X2

is an inner product space with the scalar product⟨(

x1

x2

),

(y1

y2

)⟩⊕

:= 〈x1, y1〉1 + 〈x2, y2〉2 .

(Note: If X1, X2 are Hilbert spaces, so is X1 ⊕X2)

2.54 Examples. (a) X1 = X2 = C([0, 1]) with 〈f, g〉 =∫ 1

0 dx fg. Then

X1 ⊕X2 =

(f1

f2

): fj ∈ C([0, 1]), j = 1, 2

,

is the space of the the 2-vector-valued continuous functions with⟨(f1

f2

),

(g1

g2

)⟩⊕

=

∫ 1

0dx(f1g1 + f2g2

).

(b) Orthogonal decomposition. Let X be a Hilbert space and A ⊆ X a closed subspace.Then (A, 〈·, ·〉A) is a Hilbert space (w.r.t. to inherited scalar product from X) and thesame is true for A⊥. Direct sum:

A⊕A⊥ =

(ab

): a ∈ A, b ∈ A⊥

with the scalar product⟨(

a1

b1

),

(a2

b2

)⟩⊕

= 〈a1, a2〉A + 〈b1, b2〉A⊥ =

= 〈a1, a2〉+ 〈b1, b2〉 = 〈a1 + b1, a2 + b2〉 .

The spaces A ⊕ A⊥ and X can be identified as Hilbert spaces by(ab

)←→ x = a + b,

if the orthogonal decomposition of x is unique. Indeed, this holds because of

2.55 Theorem (Projection theorem). Let X be a Hilbert space and A ⊆ X a closedsubspace. Then for all x ∈ X there exists a unique z ∈ A and a unique w ∈ A⊥ such thatx = z + w.

The proof relies on

2.56 Lemma. For every x ∈ X there exists a unique z ∈ A such that dist(x,A) = ‖x−z‖.z is the ”closest element” to x in A.

43

Proof. Existence: Let d := dist(x,A) = infy∈A ‖x− y‖. There exists a sequence (yk)k∈N ⊂A such that

d = limk→∞

‖x− yk‖︸ ︷︷ ︸≥d

. (∗)

Using the parallelogram identity ‖u+ v‖2 = 2(‖u‖2 + ‖v‖2

)− ‖u− v‖2, we get:

‖yn − ym‖2 = ‖(yn − x) + (x− ym)‖2 =

= 2(‖yn − x‖2 + ‖ym − x‖2

)− ‖2x− ym − yn‖2︸ ︷︷ ︸

4 ·∥∥∥∥x− ym − yn

2︸ ︷︷ ︸∈A

∥∥∥∥2

︸ ︷︷ ︸≥4d2

≤ 2(‖yn − x‖2 + ‖ym − x‖2

)− 4d2 m,n→∞−−−−−→ 0. (∗∗)

Thus, (yn)n is a Cauchy sequence. A is complete: ynn→∞−−−→ z ∈ A and d = ‖x− z‖ by (∗)

and continuity of the norm.Uniqueness: assume that there exists (y′n)n ⊆ A with the same properties, i.e.

limn→∞

‖y′n − x‖ = d and y′ = limn→∞

y′n.

Replacing ym by y′m in (∗∗), yields

‖z − y′‖2 = limn→∞

limm→∞

‖yn − y′m‖2

≤ limn→∞

limm→∞

2(‖yn − x‖2 + ‖y′m − x‖2

)− 4d2 ≤ 0,

thus z = y′.

Proof of Theorem 2.55. Let z ∈ A be as in Lemma 2.56 and set w := x − z. Let ξ ∈ Cand y ∈ A. Then

‖x− z‖2︸ ︷︷ ︸=:d2

≤ ‖x− (z + ξy)︸ ︷︷ ︸∈A︸ ︷︷ ︸

w−ξy

‖2

= ‖w‖2︸ ︷︷ ︸=d2

+‖ξy‖2 − 2 Re 〈w, ξy〉

So for y 6= 0 we have

|ξ|2 − 2 Re 〈w, ξy〉‖y‖2 ≥ 0

• For ξ = t ∈ R we obtain therefore

t2 − 2 Re 〈w, y〉‖y‖2 t ≥ 0 ∀t ∈ R

This implies 〈w, y〉 = 0.

44

• For ξ = i, t ∈ R:

t2 − 2 Im 〈x, y〉‖y‖2 t ≥ 0 ∀t ∈ R

So 〈w, y〉 = 0 for every y ∈ A. So w ∈ A⊥.

Uniqueness: Assume there are z, z′ ∈ A and w,w′ ∈ A⊥ with z + w = x = z′ + w′. Then

z − z′︸ ︷︷ ︸∈A

= w′ − w︸ ︷︷ ︸∈A⊥

So z = z′ and w = w′ because A ∩A⊥ = 0.

2.57 Theorem (Riesz representation). Let X be a Hilbert space and let ` ∈ X∗. Thenthere exists a unique y` ∈ X such that

`(x) = 〈y`, x〉 x ∈ X (∗)

and ‖`‖∗ = ‖y`‖.

Proof. Note ker ` = `−1(0) is closed by continuity.

• If ker ` = X then the theorem is true with y` = 0.

• Suppose now ker ` ( X. Then theorem 2.55 tells that (ker `)⊥ ) 0 and this allowsto choose 0 6= x0 ∈ (ker `)⊥. Define

y` :=`(x0)

‖x0‖2· x0 ∈ X

– so (∗) holds for every x ∈ ker ` (x0 ⊥ ker `).

– Let x = αx0, α ∈ K. Then `(x) = α`(x0) and

〈y`, x〉 =

⟨`(x0)

‖x0‖2· x0, αx0

⟩

=`(x0)

‖x0‖2α 〈x0, x0〉

= α`(x0)

So ` and 〈y`, ·〉 agree on span(ker `, x0).

But span(ker `, x0) = X, because for every x ∈ X we have

x =

(x− `(x)

`(x0)· x0

)︸ ︷︷ ︸

∈ker `

+`(x)

`(x0)· x0︸ ︷︷ ︸

∈span(x0)

because

`

(x− `(x)

`(x0)· x0

)= `(x)− `(x)

`(x0)· `(x0)

= 0

so ` = 〈yl, ·〉 on X.

45

• Uniqueness of y`: Assume there exists y′ ∈ X with ` =⟨y′, ·⟩. Then for every x ∈ X:

0 = `(x)︸︷︷︸〈y`,x〉

− `(x)︸︷︷︸〈y′,x〉

=⟨y` − y′, x

⟩Choose x = y` − y′ so that 0 = ‖y` − y′‖2, so y` = y′.

• Norm: We have

‖`‖∗ = sup06=x∈X

|`(x)|‖x‖︸ ︷︷ ︸|〈y`,x〉|‖x‖

≤ ‖y`‖

Note y` 6= 0, because ker ` ( X in the case considered it is allowed to choose x = y`in the sup above:

‖`‖∗ ≥|`(x0)|‖y`‖

=| 〈y`, y`〉 |‖y`‖

= ‖y`‖

So ‖`‖∗ = ‖y`‖.

2.58 Corollary. Let X be a Hilbert space. Then the map

A : X∗ −→ X

` 7−→ y`

defined as in the theorem 2.57 is an anti-linear, isometric bijection.

Proof. Let α, β ∈ K and `1, `2 ∈ X∗ with `j =⟨yj , ·

⟩, j = 1, 2. Then

α`1 + β`2 = α 〈y1, ·〉+ β 〈y2, ·〉= 〈αy1, ·〉+

⟨βy2, ·

⟩=⟨αy1 + βy2, ·

⟩So A(α`1 +β`2) = αA(`1)+βA(`2). This means A is anti-linear. The property of A beingisometric was proved in Theorem 2.57 (so A is also injective). Futher, A is onto, becausefor y ∈ X, ` := 〈y, ·〉 ∈ X∗ (use Cauchy-Schwarz for continuity) and A(`y) = y.

46

3 Measures, integration and Lp-spaces8

3.1 Measures9

Measures cannot be defined satisfactorily on the power set, thus:

3.1 Definition. Let X be a set and A ⊆ P(X).A is a σ-algebra (or σ-field) :⇐⇒

• X ∈ A

• A ∈ A =⇒ Ac ⊂ A

• An ∈ A, ∀n ∈ N =⇒ ⋃n∈NAn ∈ A

thus ∅ ∈ A, ⋂n∈NAn ∈ A; A \B ∈ A if A,B ∈ A. We say A measurable :⇐⇒ A ∈ A

3.2 Lemma. Let G ⊆ P(X), then there exists a smallest σ-algebra σ(G) : G ⊆ σ(G)(σ-algebra generated by G)

3.3 Definition. Let (X, T ) be a topological space. Then B ≡ B(X) := σ(T ) Borelσ-algebra

3.4 Definition. (a) Let A be a σ-algebra on X. A mapping µ : A → [0,∞] is a (positive)measure :⇐⇒

• µ(∅) = 0

• An ∈ A, n ∈ N, An ∩ Am = ∅∀n 6= m. =⇒ µ(⋃n∈NAn) =

∑n∈N µ(An)

(σ-additivity)

(X,A, µ) is called a measure space.

(b) If µ(X) = 1 then µ is called a probability measure,a finite measure, if µ(X) <∞ anda σ-finite measure, if X =

⋃nAn, An ∈ A, ∀n ∈ N and µ(An) <∞.

(c) If X is Hausdorff, A = B(X) and µ(K) <∞ for all compacts sets K ⊆ X then µ is aBorel measure.

3.5 Definition. (a) H ⊆ P(X) is a semi-ring :⇐⇒

• ∅ ∈ H• A,B ∈ H =⇒ A ∩B ∈ H• A,B ∈ H =⇒ ∃C1, · · · , Cn ∈ H, Cj ∩ Ck = ∅ ∀j 6= k : A \B =

⋃nk=1Ck

(b) A mapping µ : H → [0,∞] is a pre-measure :⇐⇒

• µ(∅) = 0

• An ∈ H ∀n ∈ N, An ∩Am = ∅∀n 6= m and⋃n∈NAn ∈ H

=⇒ µ

( ⋃n∈N

An

)=∑n∈N

µ(An)

8Parts of this section were typewritten by Max Klinger, to whom I express my gratitude.9Literature: H. Bauer, Measure and integration theory ; J. Elstrodt, Maß- und Integrationstheorie; P. R.

Halmos, Measure theory.

47

3.6 Theorem (Kolmogorov extension of pre-measures).Let µ be a σ-finite pre-measure on a semi-ring H, then it has a unique extension to a

measure on σ(H).

3.7 Example. X = R, H :=

]a, b] : a, b ∈ R, a ≤ b

is a semi-ring.

• µ(]a, b]

):= b− a defines a σ-finite pre-measure on H

• σ(H) = B(R)

=⇒ ∃ extension λ of µ to B(R): λ is called the Lebesgue (-Borel) measure. It is a Borelmeasure in the sense of Def. 3.4 (c).

3.8 Definition. Let (X, T ) be a topological Hausdorff space, B its Borel σ-algebra andµ a Borel measure on X.

(a) µ is inner regular :⇐⇒ ∀B ∈ B µ(B) = supK⊆B

K compact

µ(K)

(b) µ is outer regular :⇐⇒ ∀B ∈ B µ(B) = infU⊇BU open

µ(U)

(c) µ is regular :⇐⇒ µ is inner regular and outer regular

3.9 Theorem. Let X be a locally compact10, 2nd countable Hausdorff space. Then everyBorel measure on X is regular.

Proof. See e.g. H. Bauer, Maß- und Integrationstheorie, Thm. 29.12.

This theorem (or the more much elementary Thm. 11.42 in P. Muller, Analysis III ) yieldthe following special case

3.10 Corollary. The Lebesgue (-Borel) measure on Rd is regular.

3.11 Definition. Let (X,A, µ) be a measure space

• A ⊆ X is a µ-null set :⇐⇒ A ∈ A and µ(A) = 0.

• a statement P (x) holds for µ-almost all x ∈ X :⇐⇒ ∃ a µ-null set N ⊂ X:P (x) holds ∀x ∈ X \ N

• (X,A, µ) is complete :⇐⇒ if A is a µ-null set and B ⊂ A, then so is B. (The pointis B ∈ A!)

3.12 Theorem (Completion). Every measure space (X,A, µ) has a completion (X,A∗, µ∗),that is, C ∈ A∗ ⇐⇒ ∃A,B ∈ A : A ⊆ C ⊆ B and µ(A) = µ(B) =: µ∗(C)

3.13 Definition (Lebesgue measure). The completion of(R,B(R), λ

)is(R,B(R)∗, λ∗

)where λ∗ is the Lebesgue measure.

3.14 Definition. Let A′ be a σ-algebra in X ′ and f : X → X ′. Then f is (A,A′)-measurable :⇐⇒ ∀A′ ∈ A′ : f−1(A′) ∈ A.

3.15 Lemma. Let A′ = σ(G′). Then: ∀G′ ∈ σ

(G′)

: f−1(G′) ∈ A =⇒ f isA-A′-measurable.

10locally compact :⇐⇒ ∀ x ∈ X ∃ a compact neighbourhood of x

48

3.16 Corollary. Let (X, T ) and (X ′, T ′) be topological spaces and f : X → X ′ continuous

=⇒ f is(B(X),B(X ′)

)-measurable.

3.17 Theorem. Let f, g, fn : X → K be measurable. Then the following are measurable,too: Re f, Im f, f + g, f · g, sup fn, inf fn, lim sup fn, lim inf fn, limn fn (if it exists).

3.18 Remark. The theorem remains true with K = R := R ∪ ±∞ and B := B(R) =σ(B ∪ +∞ ∪ −∞).

3.2 Integration

3.19 Definition. Let (X,A, µ) be a measure space.

(a) f : X → [0,∞[ is a simple function (or step function) :⇐⇒ ∃N ∈ N, α1, · · · , αN >0 and A1, · · · , AN ∈ A such that

f =N∑n=1

αn1AN , where 1A(x) :=

1, x ∈ A0, x 6∈ A

is the indicator function of A. (Note: 1A measurable ⇐⇒ A measurable)Without loss of generality one can assume that α1, · · · , αN and A1, . . . , AN are pair-wise disjoint.

(b) (µ-) integral of a simple function:∫Xµ(dx)f(x) :=

N∑n=1

αNµ(An) (can be +∞).

Other notations:∫X dµ(x)f(x),

∫X dµf ,

∫X f(x) dµ(x).

3.20 Lemma. Let f : X → [0,∞]. Then

f is (A,B)-measurable ⇐⇒

∃(fn)n, a sequence of simple functionsX → [0,∞[

with fn ≤ fn−1 ∀n and f = limn→∞

fn = supnfn

3.21 Definition. Let (X,A, µ) be a measure space.

(a) For f : X → [0,∞] measurable let∫X

dµ f := limn→∞

∫X

dµ fn,

be the (µ-) integral of f , where (fn)n is any sequence as in Lemma 3.20.

(b) For f : X → R let f+ := maxf, 0, f− := max0,−f such that f = f+ − f−(decomposition into negative and positive parts). Then

f (µ-) integrable :⇐⇒ f measurable and

∫X

dµ f± <∞.

In this case ∫X

dµ f := dµ f+ −∫X

dµ f−.

49

(c) For f : X → C measurable let

f integrable :⇐⇒ Re f, Im f integrable.

In this case ∫X

dµ f :=

∫X

dµ Re f + i

∫X

dµ Im f.

3.22 Remark. (a) Definition 3.21 (a) does not depend on the choice of (fn)n.

(b) For A ⊆ X measurable ∫A

dµ f :=

∫X

dµ f1A,

in particular for f = 1, ∫A

dµ =

∫X

dµ 1A = µ(A).

(c) For f , g integrable and α, β ∈ C∫X

dµ (αf + βg) = α

∫X

dµ f + β

∫X

dµ g (linearity).

For f , g integrable with f ≤ g∫X

dµ f ≤∫X

dµ g (monotonicity).

3.23 Example. (a) Lebesgue-Borel measure λd on Rd is defined by

λd

(d×

α=1

]aα, bα]

):=

d∏α=1

(bα − aα)

on rectangles and extended to Bd := B(Rd) by Theorem 3.6, compare Example 3.7. Iff : Rd → C is Riemann-integrable =⇒ f is λd-integrable and∫

Rdddx f(x) =

∫Rdλd(dx) f(x).

Justifies notation λd(dx) ≡ ddx.

(b) Dirac measure on Rd: for x0 ∈ Rd let

δx0(A) :=

1, x0 ∈ A0, x0 6∈ A

∀A ∈ Bd. Hence, every measurable f : Rd → R is δx0-integrable and∫Rdδx0(dx) f(x) = f(x0).

3.24 Lemma. Let (X,A, µ) be a measure space and f : X → R measurable. Then thefollowing are equivalent:

(a) f integrable

50

(b) f+, f− integrable

(c) |f | integrable

(d) ∃u, v : X → R integrable: f+(x) ≤ u(x), f−(x) ≤ v(x) for µ-almost all x ∈ X.

(e) ∃g : X → R integrable:∣∣f(x)

∣∣ ≤ g(x) for µ-almost all x ∈ X3.25 Theorem. Let (X,A, µ) be a measure space and f : X → R measurable. Then

(a) if f integrable and∫

dµ f <∞ then µ(x ∈ X : f(x) = ±∞

)= 0

(b)∫

dµ |f | = 0 ⇐⇒ f(x) = 0 for µ-almost all x ∈ XThe benefit of general integration theory is the interchangeability of integration and limits.

3.26 Theorem (Beppo Levi – monotone convergence). Let

(fn)n∈N ⊂ L1(X,µ) := g : X → R : g is µ-integrable ≡ L1(X) ≡ L1

with 0 ≤ fn ≤ fn+1 ∀n ∈ N. Then

limn→∞

∫dµ fn =

∫dµ lim

n→∞fn.

3.27 Theorem (Fatou’s Lemma). Let (fn)n ⊂ L1 with fn ≥ 0 ∀n ∈ N. Then∫dµ lim inf

n→∞fn ≤ lim inf

n→∞

∫dµ fn.

3.28 Theorem (Lebesgue – dominated convergence). Let (fn)n ⊂ L1, f : X → R measur-able, limn→∞ fn(x) = f(x) for µ-almost all x ∈ X and ∃g ∈ L1 :

∣∣fn(x)∣∣ ≤ g(x), ∀x ∈ X.

Then f ∈ L1 and

limn→∞

∫dµ fn =

∫dµ f.

Useful notions of convergence in

3.29 Definition. Let f, fn : X → R measurable ∀n ∈ N.

(a) Convergence (µ-) almost everywhere:

fna.e.−−→ f (for µ-a.a. x) :⇐⇒

∃N ∈ A, µ(N ) = 0, such that

limn→∞

fn(x) = f(x), ∀x ∈ X\N .

(b) Stochastic Convergence (convergence in measure):

fnµ−→ f :⇐⇒

∀ε > 0 ∀A ∈ A with µ(A) <∞limn→∞

µ(x ∈ A :

∣∣fn(x)− f(x)∣∣ > ε

)= 0

(c) Convergence in L1-norm:

fn‖·‖1−−→ f :⇐⇒ lim

n→∞‖fn − f‖1 = 0,

where

‖f‖1 :=

∫X

dµ |f | L1-norm.

51

3.30 Theorem. (a) convergence almost everywhere =⇒ stochastic convergence

(b) convergence in L1-norm =⇒ stochastic convergence

A partial converse:

3.31 Theorem. Let µ be σ-finite. Thenfn

µ−→ f =⇒ ∀ subsequences (fnk)k of (fn)n∃ a subsequence (fnkl )l such that fnkla.e.−−→ f

as l→∞

Next we turn to product spaces. Here only the product of two spaces. The generalisationto finitely many factors is obvious.

Notation. (Xj ,Aj , µj), j = 1, 2, are measure spaces in the following.

3.32 Definition. Product σ-algebra on Cartesian product X1 ×X2

A1 ⊗A2 := σ(A1 ×A2 : A1 ∈ A1, A2 ∈ A2

)⊆ P (X1 ×X2)

3.33 Theorem (Existence and uniqueness of σ-finite product measures). Let µ1, µ2, beσ-finite. Then ∃1 measure on A1 ⊗A2, the product measure µ1 ⊗ µ2, with

(µ1 ⊗ µ2)(A1 ×A2) = µ1(A1)µ(A2) ∀Aj ∈ Aj , j = 1, 2.

3.34 Example. Lebesgue-Borel measure on Rd = R × · · · × R. We have B(Rd) = Bd =⊗di=1 B and λd =

⊗di=1 λ (Lebesgue Borel-measure on R).

3.35 Theorem (Fubini-Tonelli). Let (X1,A1, µ1), (X2,A2, µ2) be σ-finite measure spacesand let f : X1 ×X2 → R be (A1 ⊗A2,B)-measurable. Then

x1 7→∫X2

µ2(dx2) f(x1, x2) is (A1,B)-measurable,

x2 7→∫X1

µ1(dx1) f(x1, x2) is (A2,B)-measurable.

If one of the three integrals∫X1×X2

(µ1 ⊗ µ2)(dx1dx2)∣∣f(x1, x2)

∣∣∫X1

µ1(dx1)

∫X2

µ2(dx2)∣∣f(x1, x2)

∣∣∫X2

µ2(dx2)

∫X1

µ1(dx1)∣∣f(x1, x2)

∣∣is finite, then so are the other two and∫

X1×X2

(µ1 ⊗ µ2)(dx1dx2) f(x1, x2) =

∫X1

µ1(dx1)

∫X2

µ2(dx2) f(x1, x2)

=

∫X2

µ2(dx2)

∫X1

µ1(dx1) f(x1, x2).

52

3.3 Lp-spaces

Notation. In this section: (X,A, µ) a measure space, K = R or C, f : X → K measurable.

3.36 Definition (Lp-norm). (a) For p ∈]0,∞[ set

‖f‖p :=

(∫dµ |f |p

) 1p

(possibly ∞),

‖f‖∞ := inf

α > 0: µ

(x ∈ X :

∣∣f(x)∣∣ > α

)= 0

= infN∈Aµ(A)=0

supx∈X\N

∣∣f(x)∣∣

=: ess supx∈X

∣∣f(x)∣∣ µ-essential supremum

(b) For p ∈]0,∞] introduce vector space of p-integrable functions w.r.t. µ

Lp ≡ Lp(µ) ≡ Lp(X,µ) :=f : X → K : f measurable and ‖f‖p <∞

= f : X −→ K : |f |p is µ-integrable

(c) Equivalence relation ∼ on Lp

f ∼ g :⇐⇒ f = g µ-a.e.

Vector space of equivalence classes of p-integrable functions w.r.t. µ: Lp := Lp/ ∼.

3.37 Lemma. Let f, g : X → K be measurable. Then

(a) ∀r, p, q ∈ [1,∞] with 1p + 1

q = 1r : ‖fg‖r ≤ ‖f‖p ‖g‖q (gen. Holder’s inequality)

(b) ∀p ∈ [1,∞] : ‖f + g‖p ≤ ‖f‖p + ‖g‖p (Minkowski’s inequality)

Proof. (a) Case q =∞: follows from |fg| ≤ |f | ‖g‖∞.Remains to treatCase q <∞, 1 ≤ p <∞: Without loss of generality assume f, g ≥ 0 and ‖f‖p, ‖g‖q >0.Claim:

xrp︸︷︷︸

u(x)

≤ 1 +r

p(x− 1) =

r

px+

r

q︸ ︷︷ ︸v(x)

∀x ≥ 1, p ≥ r

true, because u(1) = v(1) = 1 and u′(x) = rpx

rp−1 ≤ r

p = v′(x).Now, write x = α

β in claim and multiply inequality by β =⇒

αrpβ

rq ≤ r

pα+

r

qβ ∀α ≥ β > 0 (∗)

• also true with p↔ q =⇒ condition α ≥ β can be dropped.

• also true, if α = 0 or β = 0 =⇒ (*) holds ∀α, β ≥ 0.

apply (∗) with α := fp

‖f‖pp, β := gp

‖g‖qq=⇒ (fg)r

‖f‖rp‖g‖rq≤ r

pfp

‖f‖pp+ r

qgq

‖g‖qq∫dµ

==⇒ ‖fg‖rr‖f‖rp ‖g‖rq

≤ r

p+r

q= 1.

53

(b) Without loss of generality assume

• f, g ≥ 0 because ‖f + g‖p ≤∥∥|f |+ |g|∥∥

p

• ‖f + g‖p > 0 (otherwise nothing to prove)

Case p =∞: note that for α, β > 0 and γ = α+ β

µ(f > α

)= 0 and µ

(g > β

)= 0 =⇒ µ

(f + g > γ

)= 0.

Thus

‖f + g‖∞ = infγ > 0: µ

(f + g > γ

)= 0

≤ infα > 0: µ

(f > α

)= 0

+ infβ > 0: µ

(g > β

)= 0

= ‖f‖∞ + ‖g‖∞ .

Case q ≤ p <∞:

‖f + g‖pp =

∫X

dµ f(f + g)p−1 +

∫X

dµ g(f + g)p−1

Holder≤ ‖f‖p

∥∥∥(f + g)p−1∥∥∥q

+ ‖g‖p∥∥∥(f + g)p−1

∥∥∥q︸ ︷︷ ︸(∫

Xdµ (f + g)

q(p−1)︸ ︷︷ ︸p

) p−1p

︸ ︷︷ ︸‖f+g‖p−1

≤ ‖f + g‖p−1p

(‖f‖p + ‖g‖p

).

3.38 Lemma. (Lp, ‖·‖p) is a normed space ∀p ∈ [1,∞]. Moreover, L2 is an inner productspace with scalar product

〈f, g〉 :=

∫X

dµ(x) f(x) g(x) ∀f, g ∈ L2.

Proof. ‖f‖p ≥ 0 clear.Let ‖f‖p = 0 =⇒ f(x) = 0 for µ-a.e. x ∈ X by Thm. 3.25 (b), i.e. f = 0 in Lp.‖αf‖p = |α| ‖f‖p clear.Triangle inequality given by Thm. 3.37 (b) – needs p ≥ 1!〈·, ·〉 is a scalar product with 〈f, f〉 = ‖f‖22.

3.39 Warning. Notation does not distinguish between equivalence classes and represen-tatives!

3.40 Lemma. If µ(X) < ∞, then Lq is dense in Lp ∀1 ≤ p ≤ q ≤ ∞ and ‖·‖p ≤[µ(X)

] 1p− 1q ‖·‖q.

Proof. From 3.37 (a) with g = 1, we get ‖f‖r ≤ ‖f‖p ‖1‖q︸︷︷︸µ(X)

1r−

1p

, and cyclic permutation

(r, p, q) → (p, q, r). Thus, it suffices to prove L∞ dense in Lp, ∀p ≥ 1. Let f ∈ Lp.Without loss of generality assume f ≥ 0 (otherwise decompose Re f, Im f in positive and

54

negative parts).Now fn := minf, n ∈ L∞ ∀n ∈ N =⇒

‖f − fn‖pp =

∫X

dµ(f −minf, n

)p ≤ ∫f≥n

dµ fp

=

∫X

dµ fp −∫f<n

dµ fp︸ ︷︷ ︸n→∞−−−→∫X dµ fp

n→∞−−−→ 0,

where we used dominated convergence or monotone convergence for the limit in the laststep.

3.41 Theorem (Riesz-Fischer). Let (X,A, µ) be a measure space. Then Lp is a Banachspace ∀p ∈ [1,∞]. In particular, L2 is a Hilbert space.

Proof. All that remains to be shown is completeness ∀p ∈ [1,∞].

Case p =∞: Let (fn)n∈N ⊂ L∞ be Cauchy. Since

∀n,m ∈ N ∃ null set Nn,m ∈ A, µ(Nn,m) = 0:∣∣fn(x)− fm(x)

∣∣ ≤ ‖fn − fm‖∞ , ∀x ∈ N cn,m

=⇒ N :=⋃n,m∈NNn,m is a null set and

(fn(x)

)n

is Cauchy in K, uniformly ∀x ∈ N c. (∗)K complete

=⇒ X 3 x 7→ f(x) :=

limn→∞ fn(x), x ∈ N c

0, x ∈ Nwell defined, measurable and

‖f − fn‖∞ ≤ supx∈Nc

∣∣f(x)− fn(x)∣∣

= supx∈Nc

limm→∞

∣∣fm(x)− fn(x)∣∣

≤ supx∈Nc

supm≥n

∣∣fm(x)− fn(x)∣∣︸ ︷︷ ︸

≤‖fm−fn‖∞

(∗)=⇒ lim

n→∞‖f − fn‖∞ = 0.

Case 1 ≤ p <∞: Let (fn)n ⊂ Lp be Cauchy.

=⇒ ∃ subsequence (nk)k ⊂ N: ‖fnk − fnk+1‖p < 2−k ∀k ∈ N.

Now, h := |fn1 |+∞∑k=1

|fnk+1− fnk | ∈ Lp because ‖h‖p ≤︸︷︷︸

triangle

∥∥fn1

∥∥p

+∞∑k=1

2−k︸ ︷︷ ︸1

<∞.

In particular,∞∑k=1

|fnk+1−fnk | ∈ Lp

Thm. 1.25======⇒

∞∑k=1

|fnk+1(x)−fnk(x)| <∞ for µ-a.a. x ∈ X

=⇒(fnk(x)

)k⊂ K is Cauchy for µ-a.a. x ∈ X, because

∣∣∣fnk2 (x) − fnk1 (x)∣∣∣ ≤ k2−1∑

j=k1

∣∣∣fnj+1(x)− fnj (x)∣∣∣ ≤ ∞∑

j=k1

|. . . | k1→∞−−−−→ 0

K complete======⇒ ∃f : X → K measurable with f(x) = limk→∞ fnk(x) for µ-a.a. x ∈ X.

Moreover,∣∣fnk ∣∣ ≤ h ∀k ∈ N since fnk = fn1 +

∑k−1j=1

(fnj+1 − fnj

)=⇒ |f | ≤ h a.e.

55

=⇒∣∣fnk ∣∣p , |f |p ∈ L1.

Let gk :=∣∣fnk − f ∣∣p =⇒ gk

k→∞−−−→ 0 a.e. and

0 ≤ gk ≤(∣∣fnk ∣∣+ |f |

)p≤ 2p |h| ∈ L1 ∀k ∈ N

dom. cvg.=⇒ limk→∞

∫X dµgk = 0 =⇒ subsequence (fnk)k converges to f in ‖·‖p.

(fn)n Cauchy =⇒ (fn)n converges to f in ‖·‖p.

3.42 Theorem (Clarkson inequalities). Let p ∈]1, 2[, let f, g ∈ Lp and 1p + 1

q = 1. Then:∥∥∥∥f + g

2

∥∥∥∥qp

+

∥∥∥∥f − g2

∥∥∥∥qp

≤(

1

2‖f‖pp +

1

2‖g‖pp

)q−1

, (1<)∥∥∥∥f + g

2

∥∥∥∥pp

+

∥∥∥∥f − g2

∥∥∥∥pp

≥ 1

2‖f‖pp +

1

2‖g‖pp . (2<)

For p ∈ [2,∞[, the inequalities are reversed. We refer to them as (1>) and (2>).

Proof. The case p = 2 reduces to the parallelogram identity. For the case of general p, seee.g. R. A. Adams, Sobolev spaces, or Hirzebruch/Scharlau.

3.43 Theorem. Let p ∈]1,∞[. Then Lp is uniformly convex, i.e. ∀ε > 0 ∃δ > 0 ∀f, g ∈ Lpwith ‖f‖p = ‖g‖p = 1

‖f − g‖p ≥ ε =⇒∥∥∥∥f + g

2

∥∥∥∥p

≤ 1− δ.

Proof. Case 1 < p ≤ 2: From (1<) we get∥∥∥∥f + g

2

∥∥∥∥qp

≤ 1− 2−qεq. X

Case 2 ≤ p <∞: From (2>) we get∥∥∥∥f + g

2

∥∥∥∥pp

≤ 1− 2−pεp. X

‖·‖p = 1 f

g

f + g

2

3.44 Theorem (Riesz representation for (Lp)∗). Let (X,A, µ) be a measure space. Letp ∈ [1,∞[ and 1/p + 1/q = 1. If p = 1, assume in addition that µ is σ-finite. Then themapping

J :Lq −→ (Lp)∗

f 7−→ `fwhere `f : Lp 3 g 7→

∫X

dµ fg

is an isometric isomorphism, Lq ∼= (Lp)∗.

Proof. J is clearly linear because of the linearity of the integral. Now let f ∈ Lq. `f is well-defined because fg ∈ L1 by Holder’s inequality for every g ∈ Lp and |`f (g)| ≤ ‖f‖q ‖g‖p.So `f ∈ (Lp)∗ and ∥∥`f∥∥(Lp)∗

≤ ‖f‖q . (∗)Thus, J is well defined and bounded.

Claim:∥∥`f∥∥(Lp)∗

= ‖f‖q (=⇒ J is isometric and hence injective).

Proof of the claim. W.l.o.g. assume f 6= 0 (the case f = 0 is clear!).

56

• Case 1 < p <∞: Let

g := f |f |q−2 (∗∗)with the convention g(x) := 0 if f(x) := 0. Then

|g|p = |f |(q−1)p = |f |q(1−1/q)p = |f |q ∈ L1.

So g ∈ Lp and ‖g‖p = ‖f‖q/pq . Furthermore

`f (g)

‖g‖p=

∫dµ|f |q‖g‖p

=‖f‖qq‖f‖q/pq

= ‖f‖q(1−1/p)q = ‖f‖q .

This gives ∥∥`f∥∥(Lp)∗≥ ‖f‖q ,

and the claim follows with (∗).

• Case p = 1: Since µ is σ-finite we have

X =⋃n∈N

Xn, Xn ⊆ Xn+1 ∀n ∈ N

with µ(Xn) <∞ for every n ∈ N. Let ε > 0 and define

M := x ∈ X : |f(x)| ≥ ‖f‖∞ − ε.

By definition of the essential supremum, µ(M) > 0. Then Mn := M ∩Xn obeys

0 < µ(Mn) <∞

for all sufficiently large n. Let

g :=f

|f |1

µ(Mn)1Mn

(again with the convention g(x) := 0 if f(x) := 0). Then g ∈ L1 with ‖g‖1 = 1 and

`f (g) =

∫X

dµ fg =1

µ(Mn)

∫Mn

dµ |f | ≥ ‖f‖∞ − ε.

Since this is true for every ε > 0, this gives∥∥`f∥∥(L1)∗≥ ‖f‖∞ ,

and the claim follows with (∗).

It remains to prove that J is surjective. Let ξ ∈ (Lp)∗. We have to show that there issome f ∈ Lq such that ξ = `f . W.l.o.g. assume ‖ξ‖(Lp)∗ = 1.Case p > 1: Idea: “Determine maximiser of ξ, i.e. find g ∈ Lp such that ξ(g) = 1 =‖ξ‖(Lp)∗ and determine f from g as in (∗∗).”By definition of the sup ∃ a sequence (gk)k ⊆ Lp, ‖gk‖p = 1 ∀k ∈ N,with |ξ(gk)| k→∞−→ 1.W.l.o.g. assume ξ(gk) > 0 for finally every k (otherwise multiply gk by an appropriateoverall phase factor).

Claim: (gk)k is a Cauchy sequence.

57

Proof by contradiction (needs uniform convexity): assume this is not the case. Then∃ ε > 0 and subsequences (kj)j ⊆ N and (k′j)j ⊆ N with

‖gkj − gk′j‖p ≥ ε ∀j ∈ N.

Uniform convexity (Thm. 3.43) =⇒ ∃ δ > 0 such that∥∥∥∥gkj + gk′j2

∥∥∥∥p

≤ 1− δ ∀j ∈ N.

On the other hand, we also have

1‖ξ‖∗=1

≥ ξ

(gkj + gk′j‖gkj + gk′j‖p

)=

1

‖gkj + gk′j‖p

(ξ(gkj

)+ ξ

(gk′j

))≥ 1

2(1− δ)(ξ(gkj + ξ(gk′j )

)︸ ︷︷ ︸

j→∞−−−→2

=⇒ 1 ≥ 11−δ , proving the claim.

So (gk)k is a Cauchy sequence and by completeness of Lp, there exists g ∈ Lp, such that

gkk→∞−→ g in Lp. Also ‖g‖p = 1 and

ξ(g) = limk→∞

ξ(gk) = 1.

Define f := g |g|p−2 (with the convention f(x) := 0, if g(x) = 0). Note |f |q = |g|(p−1)q =

|g|p ∈ L1 =⇒ f ∈ Lq with ‖f‖q = ‖g‖p/qp = 1. Thus,

`f : Lp 3 g 7→∫X

dµ fg ∈ (Lp)∗ and

`f (g) =

∫X

dµ fg︸︷︷︸=|g|p

= 1 = ‖f‖q =∥∥`f∥∥(Lp)∗

(see the claim after (∗) for the last equality). Now, ξ = `f follows from

Claim: Let ξ1, ξ2 ∈ (Lp)∗ with ‖ξ1‖(Lp)∗ = 1 = ‖ξ2‖(Lp)∗ . Suppose ∃ g ∈ Lp, ‖g‖p = 1,such that ξ1(g) = ξ2(g) = 1. Then ξ1 = ξ2.

Proof by contradiction (needs Clarkson inequalities). Suppose the claim is wrong =⇒∃ h ∈ Lp such that ξ1(h) 6= ξ2(h). Then the vectors(

11

)=

(ξ1(g)ξ2(g)

)=: u and

(ξ1(h)

ξ2(h)

)=: v

are linearly independent in K2. So they span K2. Thus ∃α, β ∈ K such that(1−1

)= αu+ βv =

(ξ1(h)ξ2(h)

)where h := αg + βh ∈ Lp.

=⇒ ξ1(g + th) = 1 + t = ξ2(g − th) ∀t ≥ 0. Since∥∥ξj∥∥(Lp)∗

= 1, j = 1, 2, we must have

1 + t

‖g ± th‖p≤ 1.

58

• Case p ∈]1, 2]: Clarkson inequality =⇒

α(t) := 1 + tp ‖h‖pp =

∥∥∥∥(g + th) + (g − th)

2

∥∥∥∥pp

+

∥∥∥∥(g + th)− (g − th)

2

∥∥∥∥pp

(2<)

≥ 1

2‖g + th‖pp +

1

2‖g − th‖pp

≥ (1 + t)p =: β(t). (∗∗∗)

We know

– α(0) = 1 = β(0),

– α′(t) = ptp−1 ‖h‖pp,– β′(t) = p(1 + t)p−1.

Since p > 1 we have α′(t) < β′(t) ∀ t ≥ 0 sufficiently small. So α(t) < β(t) ∀ t > 0sufficiently small, which contradicts (∗∗∗).

• Case p ∈ [2,∞[: Clarkson inequality =⇒

1 + tq ‖h‖qp =

∥∥∥∥(g + th) + (g − th)

2

∥∥∥∥qp

+

∥∥∥∥(g + th)− (g − th)

2

∥∥∥∥qp

(1>)

≥(

1

2‖g + th‖pp +

1

2‖g − th‖pp

)q−1

≥ (1 + t)q

Contradiction as in the case p ∈]1, 2] above. =⇒ Claim is true =⇒ J surjective forp > 1.

Case p = 1: Idea: “Step 1. reduce to the above case p > 1, if µ is finite. Step 2.generalise to σ-finite µ.”

1. Assume first µ(X) < ∞. Let p > 1 and 1/p + 1/q = 1. Recall from Lemma 3.40that L∞ ⊆ Lp ⊆ L1 (the inclusions being dense)

=⇒∣∣ξ(g)

∣∣ ≤ ‖ξ‖(L1)∗︸ ︷︷ ︸1

‖g‖1 ≤[µ(X)

]1−1/p︸ ︷︷ ︸=:Mp

‖g‖p ∀ g ∈ Lp =⇒ ξ ∈ (Lp)∗.

From the already proved Riesz representation for p > 1 =⇒ ∃1fp ∈ Lq such thatξ|Lp = `fp and

∥∥fp∥∥q ≤Mp. Thus ∀p1, p2 > 1, ∀g ∈ L∞

`fp1 (g) = `fp2 (g) i.e.

∫X

dµ (fp1 − fp2)g = 0.

Choose g := (fp1 − fp2)/|fp1 − fp2 | =⇒ fp1 = fp2 =: f is independent of the index.

=⇒ f ∈⋂q>1

Lq and ‖f‖q ≤Mp = [µ(X)]1/q︸ ︷︷ ︸bounded in q

∀ q ∈]1,∞[

Problem 30=⇒ f ∈ L∞ and ‖f‖∞ ≤ lim

q→∞[µ(X)]1/q = 1

=⇒ `f ∈ (L1)∗. Since Lp dense in L1 =⇒ ξ and `f agree on a dense subspaceThm. 2.32

=⇒ ξ = `f on L1.

59

2. Relax finiteness assumption: assume µ is σ-finite. =⇒ X =⋃n∈NXn with µ(Xn) <

∞ ∀ n ∈ N. Let ξ ∈ L1(X)∗ =⇒

ξ ∈ L1(Xn)∗ ∀n ∈ N 1.=⇒ ∃1fn ∈ L∞(Xn) : ξ = `fn on L1(Xn)

and ‖ξ‖L1(Xn)∗ = ‖fn‖L∞(Xn) ∀ n ∈ N.

Define

f :=∑n∈N

fn, fn(x) :=

fn(x) x ∈ Xn

0 x 6∈ Xn

=⇒ f ∈ L∞(X) because ‖f‖L∞(X) ≤ ‖ξ‖L1(X)∗ .

For g ∈ L1(X) let gN :=∑N

n=1 g 1Xn =⇒

‖g − gN‖1 =

∫X

dµ |g|∞∑

n=N+1

1XnN→∞−−−−−−→

dom. cvg.0

=⇒ ξ(g) = limN→∞

ξ(gN ) =∑n∈N

ξ(g 1Xn)︸ ︷︷ ︸`fn (g)

=∑n∈N

∫Xn

dµ fng =∑n∈N

∫X

dµ fng.

Since∣∣∣∑N

n=1 fng∣∣∣ ≤ |fg| ∈ L1 ∀N ∈ N (Holder!), dominated convergence gives

ξ(g) =

∫X

dµ fg = `f (g).

3.45 Definition. Let X be a topological space. Then

(a) X is locally compact :⇐⇒ For all x ∈ X there exists a compact neighbourhood

(b) X is σ-compact :⇐⇒ There exist Xn ⊆ X compact ∀n ∈ N such that X =⋃n∈NXn

(c) space of continuous functions with compact support supp(f) := x ∈ X : f(x) 6= 0

Cc(X) :=f ∈ C(X) : supp(f) compact

(d) space of continuous functions ”converging to 0 outside compacts”

C0(X) :=f ∈ C(X) : ∀ε > 0 ∃Kε ⊆ X compact with |f(x)| ≤ ε ∀x ∈ X \Kε

3.46 Example. • X compact =⇒ X locally compact and σ-compact.

• Rd is locally compact and σ-compact

• C0(Rd) is the set of all continuous functions vanishing at ∞.

3.47 Theorem. Let X be a locally compact and σ-compact Hausdorff space. Let µ be aregular Borel measure on X and p ∈ [1,∞[. Then Lp(X,µ) is separable.

Idea of the proof: 1. Show Cc(X) dense in Lp (see e.g. Rudin, Real and complex anal-ysis, Thm. 3.14). Needs regularity of Borel measures on X!

2. Show that Cc(X) is separable by reducing it to separability of C(Xn) known!

3.48 Corollary. Let p ∈ [1,∞[. Then Lp(Rd) ≡ Lp(Rd, λd) is separable. Here, λd is theLebesgue-Borel measure on Rd.

60

From now on we only consider X = Rd and µ = λd.

3.49 Theorem (Young’s inequality). Let p, q, r ∈ [1,∞] such that

1

p+

1

q+

1

r= 2. (∗)

Let f ∈ Lp(Rd), g ∈ Lq(Rd), h ∈ Lr(Rd). Then

I :=

∣∣∣∣∣∫Rd×Rd

dx dy f(x)g(x− y)h(y)

∣∣∣∣∣ ≤∫Rd×Rd

dx dy |f(x)||g(x− y)||h(y)|

≤ ‖f‖p‖g‖q‖h‖r.Proof. A smart application of Holder. W.l.o.g. assume f ≥ 0, g ≥ 0, h ≥ 0. Introduceconjugate exponents

1

p+

1

p′= 1,

1

q+

1

q′= 1,

1

r+

1

r′= 1

Then

1

p′+

1

q′+

1

r′= 3− 1

p− 1

q− 1

r

(∗)= 1. (∗∗)

(a) Case p, q, r ∈]1,∞[: Let

α(x, y) := f(x)p/r′g(x− y)q/r

′,

β(x, y) := g(x− y)q/p′h(y)r/p

′,

γ(x, y) := f(x)p/q′h(y)r/q

′.

We have

p

r′+p

q′= p

(1

r′+

1

q′

)= p

(1− 1

p′

)= 1,

q

r′+q

p′= · · · = 1,

r

p′+r

q′= · · · = 1.

Thus,

α(x, y)β(x, y)γ(x, y) = f(x)g(x− y)h(y).

Apply the (generalised) Holder inequality twice for functions on Rd × Rd, using thatr and r′ are conjugate, respectively (∗∗),

I = ‖αβγ‖1 ≤ ‖α‖r′‖βγ‖r ≤ ‖α‖r′‖β‖p′‖γ‖q′ , (∗∗∗)where

‖α‖r′ =

(∫Rd×Rd

dx dy f(x)pg(x− y)q

)1/r′

=

=

((∫Rd

dx f(x)p)(∫

Rddy g(y)q

))1/r′

= ‖f‖p/r′p ‖g‖q/p′q ,

‖β‖p′ = · · · = ‖g‖q/p′q ‖h‖r/p′r ,

‖γ‖q′ = · · · = ‖f‖p/q′p ‖h‖r/q′r

so that ‖α‖r′‖β‖p′‖γ‖q′ = ‖f‖p‖g‖q‖h‖r, and the claim follows with (∗∗∗).

61

(b) Case p =∞ (analogous to q =∞, r =∞): =⇒ q = r = 1 by (∗).Take out f from I by ‖f‖∞ and separate the remaining integral.

(c) Case p = 1 (analogous to q = 1 or r = 1): =⇒ 1q + 1

r = 1 by (∗).Apply Holder to obtain

∫Rd dy g(x− y)h(y) ≤ ‖g‖q‖h‖r.

3.50 Theorem. Let 1 ≤ 1p + 1

q ≤ 2 and f ∈ Lp(Rd), g ∈ Lq(Rd). Then the convolutionof f and g,

(f ∗ g)(x) :=

∫Rd

dy f(x− y)g(y),

exists for Lebesgue almost every x ∈ Rd, is commutative f ∗ g = g ∗ f and (the equivalenceclass) obeys

f ∗ g ∈ Lr(Rd) with ‖f ∗ g‖r ≤ ‖f‖p‖g‖q,where 1

r := −1 + 1p + 1

q .

Proof. Rename r → r′ and let 1r := 1− 1

r′ . Let h ∈ Lr(Rd). Because of Young’s inequalityand 1− 1

r = 1r′ = −1 + 1

p + 1q , we have∫

Rd×Rddx dy |f(y)||g(x− y)||h(x)| ≤ ‖f‖p‖g‖q‖h‖r <∞,

so Fubini gives (x 7−→ |h(x)|

∫Rd

dy |f(y)||g(x− y)|)∈ L1.

(a) Case r =∞: choose h = 1 (allowed!) =⇒ f ∗ g ∈ L1 and ‖f ∗ g‖1 ≤ ‖f‖p‖g‖q.

(b) Case r <∞: define `(h) :=∫Rd dx (f ∗ g)(x)h(x) for h ∈ Lr(Rd).

• ` is linear in h

• ` is bounded, because |`(h)| ≤ ‖f‖p‖g‖q‖h‖r.

=⇒ ` ∈ Lr(Rd)∗ with ‖`‖(Lr)∗ ≤ ‖f‖p‖g‖q. By Theorem 3.44 (Riesz representation)

∃1 ξ ∈ Lr′(Rd) such that `(h) =∫Rd dx ξ(x)h(x) =⇒∫

Rddx [(f ∗ g)(x)− ξ(x)]h(x) = 0 ∀ h ∈ Lr(Rd). (∗)

Now, choose h = 1B (f ∗ g − ξ)/|f ∗ g − ξ|, where B ∈ Bd is of finite Lebesguemeasure, λd(B) < ∞, but otherwise arbitrary. (Here, we use the convention thath(x) = 0, whenever f ∗ g(x)− ξ(x) = 0.) Thus, h ∈ Lr(Rd) and (∗) gives∫

Bdx |(f ∗ g)(x)− ξ(x)| = 0 ∀ B ∈ Bd with λd(B) <∞

=⇒ ξ = f ∗ g and ‖f ∗ g‖r′ = ‖ξ‖r′ Riesz= ‖`‖(Lr)∗ ≤ ‖f‖p‖g‖q.

Commutativity:

(f ∗ g)(x) =

∫Rd

dy f(x− y)g(y)y=:x−z

=

∫Rd

dz f(z)g(x− z) = (g ∗ f)(x).

62

3.51 Definition. Let ∅ 6= Ω ⊆ Rd be open.

• For k ∈ N: space of k-times continuously differentiable functions

Ck(Ω) :=

f ∈ C(Ω) such that ∂α1

x1 · . . . · ∂αdxd f ∈ C(Ω),

where α1, . . . , αd ∈ N0 with∑d

j=1 αd = k

space of arbitrarily often differentiable functions

C∞(Ω) :=⋂n∈N

Ck(Ω)

• For k ∈ N ∪ ∞: space of k-times continuously differentiable functions withcompact support

Ckc (Ω) := Ck(Ω) ∩ Cc(Ω)

and space of k-times continuously differentiable functions vanishing at ∞

Ck0 (Ω) := Ck(Ω) ∩ C0(Ω).

The main result:

3.52 Theorem. Let ∅ 6= Ω ⊆ Rd be open and let p ∈ [1,∞[. Then C∞c (Ω) is dense inLp(Ω).

3.53 Remark. Thus

(a) C∞c (Ω)‖·‖p

= Lp(Ω).

(b) Lp(Ω) is separable, since C∞c (Ω) ⊂ Cc(Ω) is separable.

(c) C∞c (Ω)‖·‖∞

= C∞0 (Ω) (exercise).

An important technical result: Lp functions are made arbitrarily often differentiable by aconvolution with a mollifier.

3.54 Lemma. Let p ∈ [1,∞], f ∈ Lp(Ω) with compact support in Ω. Let j ∈ C∞c (Rd),j ≥ 0,

∫Rd dx j(x) = 1 and for ε > 0 define the mollifier

jε :Rd −→ R≥0

x 7−→ ε−dj(x/ε).

Then for all sufficiently small ε > 0: fε := f ∗ jε ∈ C∞c (Ω) and ‖fε‖p ≤ ‖f‖p. (Here, fmay be trivially extended to Rd).

Proof.Norm: use Theorem 3.50 with r = p. Then q = 1, because 1

r = −1 + 1p + 1

q there. Thus,‖fε‖p ≤ ‖f‖p‖jε‖1 = ‖f‖p.Support: let δ := dist(supp f, ∂Ω). δ > 0 by Problem 9c, since supp f is compactand ∂Ω is closed. By definition of the mollifier ∃ ε0 > 0 such that for all ε ∈]0, ε0]:supp jε ⊂ Bδ/2(0). Since

fε(x) =

∫Rd

dy jε(x− y)f(y),

63

we infersupp fε ⊆

⋃x∈supp f

Bδ/2(x) ⊂ Ω

and supp f is compact.

Derivatives: let e ∈ Rd, |e| = 1 be a unit vector and λ ∈ R \ 0. Then ∀x ∈ Ω

Fλ(x) :=1

λ

(fε(x+ λe)− fε(x)

)=

1

λ

∫Rd

dy

(jε(x− y + λe)− jε(x− y)

)f(y)

=1

λ

∫Rd

dy

∫ 1

0dt

d

dtjε(x− y + tλe)︸ ︷︷ ︸

λe·(∇jε)(x−y+tλe)

f(y)

=

∫Rd

dy

(∫ 1

0dt e · (∇jε)(x− y + tλe)

)︸ ︷︷ ︸

D

f(y).

Since |D| ≤ ‖|∇jε|‖∞ < ∞ and f ∈ L1 (because f ∈ Lp and of compact support), adouble application of dominated convergence shows that

e · (∇fε)(x) = limλ→0

Fλ(x) =

∫Rd

dy e · (∇jε)(x− y) f(y)

exists ∀x ∈ Ω. The existence of higher derivatives follows by induction in the samemanner.

3.55 Lemma. Let p ∈ [1,∞[, let A ∈ B(Rd) be bounded. Then

limε0‖1A − jε ∗ 1A‖p = 0.

Proof of Theorem 3.52.Step 1: let ε > 0, f ∈ Lp(Ω). For n ∈ N define

Kn :=

x ∈ Ω : dist(x, ∂Ω) ≥ 1

nand |x| ≤ n

.

Ω is open, thus Ω =⋃n∈NKn. Also Kn ⊆ Kn+1 for all n ∈ N. By dominated convergence

limn−→∞

∥∥∥f − f 1Kn

∥∥∥p

= limn−→∞

∥∥∥f(1− 1Kn)∥∥∥p

= 0.

This implies that there exists n ∈ N such that ‖f − f 1Kn‖p < ε/3.

Step 2: we know from Problem T14 that step functions are dense in Lp: there exists

L ∈ N, c1, . . . , cL ∈ K and A1, . . . , AL ∈ B(Rd) with Al ⊂ Kn ∀l = 1, . . . , L such that forg :=

∑Ll=1 cl 1Al we have

supp g ⊆ supp(f1Kn) and ‖g − f 1Kn‖p < ε/3.

Step 3: mollify step function: ∃ δ > 0 such that

• jδ ∗ g ∈ C∞c (Ω) by Lemma 3.54.

• ‖jδ ∗ g − g‖p < ε/3 by Lemma 3.55, because g has finitely many steps.

64

In summary we have: ‖f − jδ ∗ g︸ ︷︷ ︸∈C∞c (Ω)

‖p < ε.

Proof of Lemma 3.55.Step 1: we claim it suffices to prove the lemma for p = 2.

For, consider g ∈ L∞(Rd) with bounded support (=⇒ g ∈ Lp(Rd) ∀p ∈ [1,∞[).

• p ≥ 2: ‖g‖pp ≤ ‖g‖p−2∞︸ ︷︷ ︸

<∞

‖g‖22.

• p ∈ [1, 2]: by applying Holder with p′ = 2/p ≥ 1 and q′ conjugated to p′

‖g‖pp =

∫|g|p 1supp g ≤

(∫|g|2) 1p′

︸ ︷︷ ︸‖g‖2/p′2 =‖g‖p2

(λd(supp g)︸ ︷︷ ︸

<∞

)1− 1p′.

Step 2:

∣∣1A(x)− (jε ∗ 1A)(x)∣∣ =

∣∣∣∣∫Rd

dy jε(y)︸ ︷︷ ︸√jε(y√jε(y)

[1A(x)− 1A(y − x)

]∣∣∣∣≤ ‖jε‖

121︸ ︷︷ ︸

=1

(∫Rd

dy jε(y)[1A(x)− 1A(x− y)︸ ︷︷ ︸

=:G(x,y)

]2) 1

2

.

Thus, ∥∥∥1A − jε ∗ 1A

∥∥∥2

2≤∫Rd

dx

∫Rd

dy jε(y)G(x, y)2

=

∫Rd

dx

∫Rd

dy[1A(x) + 1A(x− y)− 2 · 1A(x)1A(x− y)

]= 2

∫Rd

dy jε(y)

∫A

dxG(x, y)

z:=y/ε= 2

∫Rd

dz j(z)

∫A

dxG(x, εz)︸ ︷︷ ︸Iε(z)

.

Since |Iε(z)| ≤ 2λd(A) and j ∈ L1, dominated convergence gives

limε0‖1A − jε ∗ 1A‖22 ≤ 2

∫Rd

dz j(z) limε0

∫A

dx[1A(x)− 1A(x− εz)

].

Step 3: We need to prove

J(z) :=

∫A

dx 1A(x− z) z→0−−−→∫A

dx.

Since J(z) ≤∫A dx ∀z it is enough to show

lim infz→0

J(z) ≥∫A

dx. (∗)

65

Lebesgue measure is outer regular (Def. 3.8 (b) and Cor. 3.10):

∀δ > 0 ∃B ⊇ A, B open and

∫B\A

dx < δ.

Now

J(z) =

∫A

dx 1B(x− z)−∫A

dx 1B\A(x− z)︸ ︷︷ ︸≤∫Rd dx 1B\A(x−z)<δ

≥∫A

dx 1B(x− z)− δ.

B open=⇒ limz→0 1B(x− z) = 1B(x) for all x ∈ B Fatou

=⇒

lim infz→0

J(z) ≥∫A

dx lim infz→0

1B(x− z)− δ A⊆B=

∫A

dx − δ

which holds for all δ > 0. This proves (∗).

3.4 Decomposition of Measures11

Notation. A is a σ-Algebra on a space X; µ, ν are measures on A.

3.56 Definition. (a)

µ is absolutely continuous w.r.t.ν,in symbols, µ ν

:⇐⇒ν(A) = 0 for A ∈ A =⇒ µ(A) = 0(“every ν-null set is a µ-null set”)

(b)µ has a density h w.r.t. ν(or Radon-Nikodym deriva-tive),in symbols, h = dµ/dν

:⇐⇒

∃ h : X → [0,∞] measur-able and unique up to modifica-tions on ν-null sets: ∀A ∈ A

µ(A) =

∫A

dν h.

(c) µ and ν are mutually singular,in symbols, µ ⊥ ν

:⇐⇒

∃A ∈ A : µ(A) = 0 and ν(X \A) = 0(“µ is concentrated on X \A, ν on A”)

(d) sum of measures

µ+ ν :A → [0,∞],A 7→ µ(A) + ν(A)

is a measure.

3.57 Theorem (Radon-Nikodym). Let µ be a finite measure and let ν be σ-finite. Then∃1 measure µac and ∃1 measure µs such that

µ = µac + µs,

where µac ν and µs ⊥ ν (hence also µac ⊥ µs). Moreover, µac has a density w.r.t. νwhich is ν-integrable.

11optional reading

66

In the special case µs = 0, we get

3.58 Corollary. Let µ be a finite measure and let ν be σ-finite. Thenµ ν ⇐⇒ µ has a density w.r.t. ν which is ν-integrable.

3.59 Remark. Thm. 3.57 and Cor. 3.58 can be extended to σ-finite measures µ, but thenthe density will only be ν-integrable over sets of finite µ-measure.

Proof of Theorem 3.57 (after von Neumann). First act: assume ν finite.Define ϕ := µ+ ν =⇒

∫X dϕf =

∫X dµ f +

∫X dν f ∀f ≥ 0 measurable.

Now, let f ∈ L2(X,ϕ)

CSI=⇒

∣∣∣∣∫X

dµ f

∣∣∣∣ ≤ ∫X

dµ |f | ≤∫X

dϕ |f | · 1 ≤ ‖f‖2;ϕ

(ϕ(X)

)1/2<∞

=⇒ f 7→∫X dµ f ∈

(L2(ϕ)

)∗.

Riesz=⇒ ∃g ∈ L2(ϕ) : ∀f ∈ L2(ϕ)

∫X

dµ f =

∫X

dϕgf. (∗)

Choose f = 1A in (∗), where A ∈ A =⇒ µ(A) =∫A dϕg, and since 0 ≤ µ(A) ≤ ϕ(A)

=⇒ 0 ≤ 1

ϕ(A)

∫A

dϕg ≤ 1 ∀A ∈ A with ϕ(A) > 0

=⇒ g(x) ∈ [0, 1] for ϕ-a.a. x ∈ X

(because otherwise choose A := x ∈ X : g(x) > 1 or A := x ∈ X : g(x) < 0 and geta contradiction). Now, modify g on ϕ-null set such that g(x) ∈ [0, 1] ∀x ∈ X. Note thisdoes not affect (∗).

(∗)=⇒

∫X

dµ f(1− g) =

∫X

dν fg ∀f ∈ L2(ϕ). (∗∗)

Let

Aac :=x ∈ X : g(x) ∈ [0, 1[

∈ A

As :=x ∈ X : g(x) = 1

= X \Aac ∈ A

and define two finite measures

µκ :A → [0,∞[B 7→ µ(B ∩Aκ)

for κ = ac, s.

Claim: µs ⊥ νFor, choose f = 1As in (∗∗) =⇒ 0 =

∫As

dν g︸︷︷︸1

= ν(As) and µs(X \As︸ ︷︷ ︸Aac

) = 0.

Claim: µac has a density w.r.t. ν (=⇒ µac ν)For, choose f = (1 + g + g2 + . . . + gn)1B ∈ L∞(ϕ) ⊆ L2(ϕ) in (∗∗), where B ∈ A andn ∈ N.

=⇒∫B

dµ (1− gn+1) =

∫B

dµ

n+1∑j=1

gj︸ ︷︷ ︸=:hn

.

67

(hn)n is monotone increasing sequence with limit h := limn→∞ hn ≥ 0 (possibly ∞).

monot. cvg.=⇒

∫B

dν h = limn→∞

∫B

dν hn = limn→∞

∫B

dµ (1− gn+1)︸ ︷︷ ︸1Aac (1−gn+1︸︷︷︸

∈[0,1[

)

dom. cvg.=

∫B∩Aac

dµ 1 = µac(B) <∞

∀B ∈ A, i.e. µac ν. In particular, for B = X =⇒ h ∈ L1(X, ν).Uniqueness: Assume µ = µac + µs = µac + µs.

=⇒ µac − µac︸ ︷︷ ︸ν

= µs − µs︸ ︷︷ ︸⊥ν

=⇒ µac − µac = 0 = µs − µs.

Second act: extend to ν σ-finite.Decompose X =

⋃n∈NXn, Xn pairwise disjoint, ν(Xn) <∞, use the first act on each Xn

and assemble everything.

Towards a finer decomposition of measures:

3.60 Definition. Let (X,A, µ) be a measure space. Assume x ∈ A ∀x ∈ X. Let

App :=x ∈ X : µ(x) > 0

set of atoms of µ.

(a) pure point part of µ

µpp :A → [0,∞]

B 7→ µ(B ∩App

)=

∑x∈B∩App

µ(x)

(b) continuous part of µ

µc :A → [0,∞]

B 7→ µ(B ∩ (X \App)

)(c) µ is only pure point :⇐⇒ µ = µpp (i.e. µc = 0)

(d) µ is continuous :⇐⇒ µ = µc (i.e. µpp = 0)

(e) µ is singular continuous w.r.t. another measure ν :⇐⇒ µpp = 0 and µ ⊥ ν3.61 Remark. µ is σ-finite =⇒ App is at most countably infinite. (See digression onuncountable series in the proof of Lemma 2.49)

Collecting facts, we get

3.62 Theorem (Lebesgue decomposition). Let µ, ν be σ-finite measures on A, x ∈A ∀x ∈ X and assume νpp = 0. Then ∃1 decomposition

µ = µac + µsc + µpp = µac + µs = µc + µpp ,

where µac ν, µsc ⊥ ν and µpp ⊥ ν.

3.63 Remark. (a) Theorem 3.62 mostly used with ν = λd (Lebesgue-Borel).

(b) Condition νpp = 0 needed, because otherwise ∃ ambiguities in distinguishing µpp andµsc .

68

4 The cornerstones of functional analysis

4.1 Hahn-Banach theorem

4.1 Definition. Let M be a set and D ⊆M×M . Let ≺ be the associated binary relation:x ≺ y :⇐⇒ (x, y) ∈ D.

(a) ≺ is a partial ordering (or M a partially ordered set) iff for all x, y, z ∈M :

• x ≺ x (reflexive)

• if x ≺ y and y ≺ x, then x = y, (antisymmetry)

• if x ≺ y and y ≺ z, then x ≺ z (transitive).

(b) x, y ∈ M are comparable iff x ≺ y or y ≺ x. x, y are incomparable iff x, y are notcomparable.

(c) ≺ is a total ordering iff

• ≺ is a partial ordering

• x, y are comparable for all x, y ∈M .

(d) Let W ⊆M . u ∈M is an upper bound of W iff w ≺ u ∀w ∈W .

(e) m ∈W is a maximal element of W iff the implication

m ≺ w for w ∈M =⇒ m = w

holds. (Note: a maximal element need not be an upper bound and vice versa.)

4.2 Example. • “≤” is a total ordering on R.

• “⊆” is a partial ordering on P(X), but not a total ordering.

The following axiom is equivalent to the axiom of choice.

4.3 Axiom (Zorn’s Lemma). Let M 6= ∅ be a partially ordered set. Suppose every totallyordered subset of M has an upper bound. Then M hat a maximal element.

The next Theorem finishes the proof of Theorem 2.51.

4.4 Theorem. Every Hilbert space X 6= 0 has an orthonormal basis.

Proof. Let M := E ⊆ X | E orthonormal =⇒

• M 6= ∅.

•”⊆“ is a partial ordering on M .

• Let W be a totally ordered subset of M . Then⋃E∈W E ∈M (because it is orthonor-

mal due to W being totally ordered) is an upper bound for W .

Zorn=⇒ M has a maximal element M.

Claim: M is an orthonormal basis

• Orthonormal is clear since M∈M .

69

• Suppose: M is not complete. Then there is some 0 6= x ∈ X such that x ⊥ m forevery m ∈M. Thus

M′ :=

x

‖x‖

∪M ∈M

(since M′ is orthonormal). So M &M′ which contradicts M being maximal.

4.5 Remark. Similar arguments prove Theorem 2.4 (existence of a Hamel basis; seeexercise).

4.6 Theorem (Hahn-Banach; real version). Let X be an R-vector space, let p : X → Rbe convex, i.e. for every x, y ∈ X and α ∈ [0, 1]

p(αx+ (1− α)y) ≤ αp(x) + (1− α)p(y).

Let Y ⊆ X be a subspace and let λ : Y → R be linear with λ(x) ≤ p(x) for every x ∈ Y .Then there exists Λ: X → R linear such that

• Λ|Y = λ

• Λ(x) ≤ p(x) for every x ∈ X.

Proof. • Step 1: (A preparation for the main step). W.l.o.g. assume Y & X. Thenthere is an element z ∈ X \ Y (⇒ z 6= 0). Let Y := span(Y, z). For every y ∈ Ythere is a unique decomposition y = y + αz (for y ∈ Y and α ∈ R). The extensionλ of λ to Y is uniquely specified by λ(z) because

λ(y) = λ(y) + αλ(z)

must hold because of linearity. To ensure

λ(y) ≤ p(y) ∀ y ∈ Y , (∗)

we consider β1, β2 > 0, y1, y2 ∈ Y :

β1λ(y1) + β2λ(y2) = λ(β1y2 + β2y2)

= (β1 + β2)λ

(β1

β1 + β2(y1 − β2z) +

β2

β1 + β2(y2 + β1z)

)︸ ︷︷ ︸

≤p(

β1β1+β2

(y1−β2z)+ β2β1+β2

(y2+β1z))

p convex≤ β1p(y1 − β2z) + β2p(y2 + β1z)

=⇒ 1

β2(λ(y1)− p(y1 − β2z)) ≤

1

β1(p(y2 + β1z)− λ(y2)).

Thus, there exists a ∈ R:

supβ>0y∈Y

[1

β

(λ(y)− p(y − βz)

)]≤ a ≤ inf

β>0y∈Y

[1

β

(p(y + βz)− λ(y)

)]. (∗∗)

Set λ(z) := a. Check that this choice ensures (∗). (To do so, use the right inequalityin (∗∗), if y = y + αz with α ≥ 0 there, otherwise use the left inequality.)

70

• Step 2: Use Zorn to construct the extension. Let

E :=

linear extensions e of λ with e ≤ p on dom(e).

=⇒ E 6= ∅ because λ ∈ E . Define partial ordering ≺ on E

e1 ≺ e2 :⇐⇒ dom(e2) ⊇ dom(e1) ∧ e2|dom(e1) = e1.

Let W ⊆ E be totally ordered. Write W = eαα∈I . Consider

e :

⋃α∈I dom(eα) −→ R

x 7−→ eβ(x)

where β is any element of I such that x ∈ dom(eβ)

– e is well-defined (because W is totally ordered)

– e is linear

– e is an upper bound for W

– e(x) ≤ p(x) for every x ∈ dom(e) =⋃α∈I dom(eα)

Zorn=⇒ E has a maximal element Λ. Claim: dom(Λ) = X (then the Theorem follows).Suppose not. Then Y := dom(Λ) and there is some 0 6= z ∈ X \ Y . Let Y :=span(Y, z. Then by step 1 there is some linear extension Λ ∈ E of Λ to Y whichcontradicts Λ being maximal.

4.7 Theorem (Hahn-Banach; complex version). Let X be a C-vector space. Let p : X → Rwith

p(αx+ βy) ≤ |α|p(x) + |β|p(y)

for every x, y ∈ X and α, β ∈ C with |α|+|β| = 1. Let Y ⊆ X be a subspace and λ : Y → Clinear with |λ(x)| ≤ p(x) for every x ∈ Y . Then there exists a linear extention Λ: X → Cof λ (i.e. Λ|Y = λ ) with

|Λ(x)| ≤ p(x) ∀x ∈ X.

Proof. Define `(x) := Reλ(x) for every x ∈ Y . Then ` : X → R is R-linear and allhypotheses of Theorem 4.6 are fulfilled. Then, by Theorem 4.6, there is an R-linearfunctional L : X → R with L|Y = ` and L ≤ p on X. Note: λ(x) = `(x) − i`(ix)12 forevery x ∈ X. Define Λ(x) := L(x)− iL(ix), x ∈ X. Then

• Λ is R-linear on X by Theorem 4.6

• Λ|Y = λ

• Λ is C-linear, because Λ is R-linear and

Λ(ix) = L(ix)− iL(−x) = i(L(x)− iL(ix)

)= iΛ(x)

• Claim: |Λ(x)| ≤ p(x) for every x ∈ X. For, fix x ∈ X and let θ ≡ θ(x) := arg(Λ(x))13.Then

|Λ(x)| = e−iθΛ(x)C-linear

= Λ(e−iθx)L=Re Λ

= L(e−iθx) ≤ p(e−iθx)p convex≤ p(x).

12because `(ix) = Reλ(ix) and λ(ix) = iλ(x), so Reλ(ix) = − Imλ(x)13This means θ should be the angle of the polar representation of Λ(x) = |Λ(x)| · eiθ

71

4.8 Corollary. Let X be a normed space, Y ⊆ X be a subspace and f ∈ Y ∗. Then thereis F ∈ X∗ with F |Y = f and ‖F‖X∗ = ‖f‖Y ∗.Proof. If K = C, apply Theorem 4.7 with p(x) := ‖f‖Y ∗ · ‖x‖ for x ∈ X (fulfills assump-tions) to λ = f . If K = R, apply Theorem 4.6 with the same p to λ = ±f .=⇒ ∃ F : X → K linear with |F (x)| ≤ ‖f‖Y ∗ ‖x‖ =⇒

‖F‖X∗ ≤ ‖f‖Y ∗ .But

‖F‖X∗ = sup0 6=x∈X

|F (x)|‖x‖ ≥ sup

06=x∈Y

|f(x)|‖x‖ = ‖f‖Y ∗

4.9 Corollary. Let X be a normed space and let 0 6= x0 ∈ X. Then there exists somef ∈ X∗ with f(x0) = ‖x0‖ and ‖f‖X∗ = 1.

Proof. Let Y = spanx0. If y ∈ Y then y = αx0 for some unique α ∈ K. Define ϕ onY by ϕ(y) := α · ‖x0‖. This implies ϕ(x0) = ‖x0‖ and ϕ ∈ Y ∗ with ‖ϕ‖Y ∗ = 1 because|ϕ(y)| = ‖y‖.Cor. 4.8=⇒ ∃ f : X → K linear with f |Y = ϕ (in particular f(x0) = ‖x0‖) and ‖f‖X∗ =‖ϕ‖Y ∗ = 1.

4.10 Corollary. Let X be a normed space, Z ⊆ X be a subspace and x0 ∈ X \ Z with0 < dist(x0, Z) =: d. Then there exists some f ∈ X∗ with f |Z = 0, f(x0) = d and‖f‖X∗ = 1.

Proof. Exercise.

4.2 Three consequences of Baire’s theorem

4.11 Theorem (Banach-Steinhaus, uniform boundedness principle). Let X be a Banachspace, Y a normed space and F ⊆ BL(X,Y ). If supT∈F ‖Tx‖ < ∞ for all x ∈ X, thensupT∈F ‖T‖ <∞.

Proof. For n ∈ N, define

An := x ∈ X : ‖Tx‖ ≤ n ∀T ∈ F.Then X =

⋃n∈NAn by hypothesis. Also An = An for all n ∈ N (since ‖ · ‖ and ‖T‖

continuous). By Cor. 1.59 (a reformulation of Baire’s theorem) there exists n0 ∈ N suchthat An0 has non-empty interior, i.e. there exists x0 ∈ An0 and r > 0 such that Br(x0) ⊆An0 .Now let z ∈ Br(0) and T ∈ F . Using that z+x0 ∈ Br(x0) and therefore ‖T (z+x0)‖ ≤ n0,we can estimate:

‖Tz‖ ≤ ‖T (z + x0)‖+ ‖Tx0‖ ≤ n0 + ‖Tx0‖.Finally, let 0 6= x ∈ X. Define z := r

2‖x‖x ∈ Br(0). Plugging this into the last inequality,we get:

‖Tx‖ =2‖x‖r· ‖Tz‖ ≤ 2‖x‖

r·(n0 + ‖Tx0‖

),

and by taking the supremum over all 0 6= x ∈ X:

‖T‖ = sup06=x∈X

‖Tx‖‖x‖ ≤

2

r·(n0 + ‖Tx0‖

).

Taking the supremum over all T ∈ F proves the claim.

72

Figure 6: Claim 1

4.12 Theorem (Open mapping theorem). Let X,Y be Banach spaces. Let T ∈ BL(X,Y )be onto. Then T is open.

Proof. We need to show the implication

A ⊆ X open =⇒ T (A) open in Y.

This will follow from 3 subsequent claims. For clarity, we include the ambient space inthe notation of a ball, e.g. BX

r (x).

Claim 1: It suffices to show that there exists r > 0 such that T (BXr (0)) has non-empty

interior.

Proof: Suppose T (BXr (0)) has non-empty interior. Let y = Tx be an interior point of

T (BXr (0)), i.e. there exists ry such that BY

ry(y) ⊆ T (BXr (0)) (see Figure 4.2). By continuity

of T ,Nx := T−1(BY

ry(y)) ∩BXr (0) ⊆ BX

r (0)

is an open neighbourhood of x (the intersection with BXr (0) is necessary because T is not

necessarily injective). Thus, there exists ρ > 0 such that

Nx ⊆ BXρ (x) and T (BX

ρ (x)) ⊇ BYry(y).

By scaling, translations and linearity, we get ∀ r′ > 0

T(BXr′ (x)

)= T

(BXr′ (0) + x

)= T

(r′

ρBXρ (0) + x

)=r′

ρT(BXρ (0)

)+ Tx =

=r′

ρT(BXρ (x)− x

)+ Tx ⊇ r′

ρ

(BYry(y)− Tx︸︷︷︸

=y

)+ Tx =

=r′

ρBYry(0) + Tx = BY

r′ryρ

(0) + Tx = BYr′ryρ

(y).

Now, let A ⊆ X be open and let y′ := Tx′ ∈ T (A) with x′ ∈ A arbitrary. Since A is open,there exists r′ > 0 such that BX

r′ (x′) ⊆ A. Using the inclusion above:

T (A) ⊇ T (BXr′ (x

′)) = T(BXr′ (x) + x′ − x

)⊇ BY

r′ryρ

(y) + y′ − y = BYr′ryρ

(y′),

73

i.e. T (A) is open. This proves Claim 1.

From now on all balls will be centred about 0 (unless otherwise specified), and we dropthe centre from the notation.

Claim 2: There exists ε > 0 such that

BYε ⊆ T (BX

1 ). (∗)

Proof: Since T is onto we can write Y =⋃n∈N T (BX

n ). Y is complete and we can apply

Baire’ Theorem in the form of Cor. 1.59: There exists n ∈ N such that T (BXn ) has non-

empty interior, i.e. there exists y ∈ T (BXn ) and ε′ > 0 such that BY

ε′ (y) ⊆ T (BXn ), or

equivalently, BYε′ ⊆ T (BX

n )− y.We have

• y = limk→∞ Txk with xk ∈ BXn ∀k ∈ N

• T (BXn )− Txk = T (BX

n − xk) ⊆ T (BX2n),

thus BYε′ ⊆ T (BX

2n) and, by scaling, the claim holds with ε := ε′/2n.

Claim 3: T (BX1 ) ⊆ T (BX

2 ).

Proof: Let ε be as in Claim 2. Let y ∈ T (BX1 ). Then there exists x1 ∈ BX

1 such that

y − Tx1 ∈ BYε/2

(∗)⊆ T (BX

1/2).

Analogously there exists x2 ∈ BX1/2 such that

y − Tx1 − Tx2 ∈ BYε/4

(∗)⊆ T (BX

1/4).

Inductively we get: For all n ∈ N there exists xn ∈ BX2−(n−1) such that

y −n∑j=1

Txj ∈ BYε2−n . (∗∗)

Since ∑n∈N

‖xn‖︸ ︷︷ ︸<2−(n−1)

< 2 <∞

and X is a Banach space,∑

j∈N xj =: x ∈ X exists (see digression of series in Banachspaces). T is continuous =⇒ Tx =

∑j∈N Txj . Using (∗∗) and the continuity of ‖ · ‖,

‖y − Tx‖ = limn−→∞

∥∥∥y − n∑j=1

Txj

∥∥∥ = 0,

thus y = Tx and ‖x‖ < 2, i.e. y ∈ T (BX2 ).

4.13 Corollary (Inverse mapping theorem). Let X,Y be Banach spaces and T ∈ BL(X,Y )a bijection. Then T−1 ∈ BL(Y,X).

Proof. T is open by Thm. 4.12, thus T−1 is continuous.

74

4.14 Definition. Let X,Y be normed spaces and T : X ⊇ domT −→ Y a linear operator.

(a) Graph of T :

G(T ) :=

(x, Tx) ∈ X × Y : x ∈ dom(T ).

We equip X × Y with the norm ‖(x, y)‖X×Y := ‖x‖X + ‖y‖Y . Recall: X,Y complete=⇒ X × Y complete.

(b) T closed iff G(T ) closed in X × Y .

4.15 Remark.

(a) T is closed if and only if the following implication holds:

(xn)n ⊆ domT with xnn−→∞−−−−→ x ∈ T and Txn

n−→∞−−−−→ y ∈ Y=⇒ x ∈ domT and y = Tx.

(b) Compare (a) with the definition of T continuous, where convergence of (Txn)n mustbe shown. Here, it is given!

4.16 Theorem (Closed graph theorem). Let X,Y be Banach spaces and

T : X ⊇ domT −→ Y

a closed linear operator. Then domT is closed (in X) if and only if T is bounded.

Proof. “⇐” Let (xn)n ⊆ domT with xnn→∞−−−→ x ∈ X. Then (xn)n is Cauchy in X.

By hypothesis: T is bounded, thus (Txn)n is Cauchy in Y . But Y is complete, i.e.there exists y ∈ Y such that Txn

n→∞−−−→ y and since T is closed limn→∞ xn ∈ domT(and Tx = y).

“⇒” Define the projection

Px :G(T ) −→ dom(T )

(x, Tx) 7−→ x.

– G(T ) is closed in X × Y , therefore G(T ) is itself a Banach space.

– dom(T ) is closed in X (by hypothesis), therefore dom(T ) is itself a Banachspace.

– Px is a bijection

– Px is bounded because let z = (x, Tx) ∈ G(T ), then

‖Pxz‖X = ‖x‖X ≤ ‖x‖X + ‖Tx‖Y = ‖z‖X×Y .

So ‖Px‖ ≤ 1.

By the inverse mapping theorem P−1x : dom(T ) −→ G(T ), x 7−→ (x, Tx), is also

bounded. =⇒ ∃ c <∞ : ‖P−1x x‖X×Y ≤ c ‖x‖X . Since ‖P−1

x x‖X×Y = ‖x‖X+‖Tx‖Y ,

=⇒ ‖Tx‖Y ≤ (c− 1) ‖x‖X .

So T is bounded.

75

4.17 Example. Let X = Y = Cc(R) with supremum norm. Consider

T :=d

dx:C1c (R) −→ Cc(R)

f 7−→ f ′

Claim: T is closed. Let (fn)n∈N ⊆ dom(T ) be a sequence with the properties:

• ∃ g ∈ Cc(R) such that ‖fn − g‖∞n→∞−−−→ 0,

• ∃h ∈ Cc(R) such that∥∥f ′n − h∥∥∞ n→∞−−−→ 0,

i.e. (fn, f′n) ⊆ G(T ) and we need to show: (g, h) ∈ G(T ) (equivalently g ∈ C1

c (R) andh = g′). By uniform convergence, we can exchange limits:∫ x

0dt h(t) =

∫ x

0dt lim

n→∞f ′n(t) = lim

n→∞

∫ x

0dt f ′n(x)︸ ︷︷ ︸

fn(x)−fn(0)

= g(x)− g(0)

where the last equality follows from pointwise convergence of fn to g.

=⇒ g(x) = g(0) +

∫ x

0dt h(t).

So by the fundamental theorem of calculus: g ∈ C1(R) with g′ = h. Since g ∈ Cc(R) wehave g ∈ C1

c (R). So T is closed.

4.3 (Bi)-Dual spaces and weak topologies

4.18 Theorem. Let X be a normed space. Then for every x ∈ X

‖x‖ = sup06=f∈X∗

|f(x)|‖f‖∗

.

Proof. Let S := sup06=f∈X∗ |f(x)|/ ‖f‖∗.

• Since |f(x)| ≤ ‖f‖∗ ‖x‖ we have S ≤ ‖x‖.

• Corollary 4.9 gives the existence of f ∈ X∗ such that ‖f‖∗ = 1 and f(x) = ‖x‖. So

S ≥ |f(x)|‖f‖∗

= ‖x‖ .

4.19 Definition. Let X be a normed space. X∗∗ := (X∗)∗ is the bidual space of X.More generally, we introduce the n-fold dual space

X

n-fold︷ ︸︸ ︷∗ · · · ∗ :=

(X

(n−1)-fold︷ ︸︸ ︷∗ · · · ∗)∗

for n ∈ N in a recursive way.

76

4.20 Theorem. Let X be a normed space. The canonical embedding

J :X −→ X∗∗

x 7−→ Jx

with

Jx :X∗ −→ Kf 7−→ f(x)

is well-defined, linear and isometric. If J is surjective, X is called reflexive.

4.21 Remarks. • Hilbert spaces are reflexive (Riesz!).

• Every finite dimensional normed space is reflexive (use the dual basis!).

• `p, Lp are reflexive for p ∈]1,∞[.

• X reflexive =⇒ X complete.

Proof of Theorem 4.20. J is well-defined (⇐⇒ Jx ∈ (X∗)∗ ∀x ∈ X). Let α, β ∈ K andf, g ∈ X∗:

(Jx)(αf + βg) = (αf + βg)(x)

= αf(x) + βg(x)

= α(Jx)(f) + β(Jx)(g).

So Jx is linear. J is bounded, because

|(Jx)(f)| = |f(x)| ≤ ‖f‖∗ ‖x‖ .

This implies|(Jx)(f)|‖f‖∗

≤ ‖x‖ .

So ‖Jx‖∗∗ ≤ ‖x‖ and Jx ∈ X∗∗. J is isometric, because

‖Jx‖∗∗ = sup0 6=f∈X∗

|f(x)|‖f‖∗

= ‖x‖

by Theorem 4.18. J is linear, because let α, β ∈ K, f ∈ X∗ and x, y ∈ X. Then(J(αx+ βy)

)(f) = f(αx+ βy)

= αf(x) + βf(y)

= α(Jx)(f) + β(Jy)(f)

Since this is true for every f ∈ X∗, we have J(αx+ βy) = αJ(x) + βJ(y).

4.22 Theorem. Let X be a Banach space. Then

X reflexive ⇐⇒ X∗ reflexive.

Proof. “⇒” (X∗)∗∗ = ((X∗)∗)∗ = (X∗∗)∗ = X∗

“⇐” See e.g. Theorem III.3.4 in Werner, Funktionalanalysis.

77

4.23 Theorem. Let X be a normed space. Then

X∗ separable =⇒ X separable.

4.24 Remark. X = `1 shows that “⇐=” does not hold, since `∞ ∼= (`1)∗ which is notseparable.

Proof of Thm. 4.23. Let A := fn ∈ X∗ : n ∈ N be dense (exists by hypothesis). Forn ∈ N choose xn ∈ X such that ‖xn‖ = 1 and |fn(xn)| ≥ ‖fn‖∗ /2. Let

D := spanKxn : n ∈ N ⊆ X

Claim: D is dense in X. (Note: D is not countable, but a K-linear subspace). SupposeD is not dense. Then there exists z ∈ X such that dist(z,D) > 0. By Corollary 4.10 thereexists a linear functional f ∈ X∗ with f |D = 0 and f(z) > 0. Now:

• A dense in X∗ implies, there exists a sequence (fnk)k ⊆ A such that∥∥fnk − f∥∥∗ k→∞−−−→ 0.

• We have ∥∥fnk − f∥∥∗ ≥ |fnk(xnk)− f(xnk)| = |fnk(xnk)| ≥ 1

2

∥∥fnk∥∥∗ .Then

∥∥fnk∥∥∗ k→∞−−−→ 0 =⇒ fnkk→∞−−−→ 0 in X∗ =⇒ f = 0 . So D is dense.

Define K := Q if K = R, resp. K := Q+ iQ if K = C and set

D := spanKxn : n ∈ N.

Thus, D is countable and dense in X.

4.25 Definition. Let X be a K-vector space, let pαα∈I be a separating family of semi-norms on X, i.e.

∀0 6= x ∈ X ∃α ∈ I : pα(x) > 0.

For given x ∈ X, the collectionUα,r(x) := x+ Uα,r : α ∈ I, r > 0

,

where Uα,r := y ∈ X : pα(y) < r defines the neighbourhood subbase14 of the locallyconvex topology associated with pαα∈I .

4.26 Remarks. (a) Uα,r(x) =y ∈ X : pα(y − x) < r

(b) pα separating implies that the locally convex topology is Hausdorff.

(c) The elements of the neighbourhood subbase are convex sets, i.e. for y1, y2 ∈ Uα,r andλ ∈ [0, 1] we have λy1 + (1− λ)y2 ∈ Uα,r, because

pα(λy1 + (1− λ)y2) ≤ pα(λy1) + pα((1− λ)y2)

= λ pα(y1)︸ ︷︷ ︸<r

+(1− λ) pα(y2)︸ ︷︷ ︸<r

< r.

14i.e. finite intersections give a base.

78

4.27 Definition. Let X be a normed space. The weak topology on X is the locally convextopology on X induced by the family of seminorms x 7→ |f(x)|f∈X∗ .4.28 Lemma. Let X be a normed space. Then the weak topology is

(a) Hausdorff,

(b) the coarsest topology T on X such that every f ∈ X∗ is continuous with respect to T .In particular, the weak topology is coarser than the norm (≡ strong) topology.

(c) identical to the norm topology, if dimX <∞,

(d) such that (xn)n ⊂ X converges to x ∈ X in the weak topology (or “weakly”) if andonly if

limn→∞

f(xn) = f(x) ∀f ∈ X∗.

Notation: xnw−→ x or xn

n→∞−−− x.

Proof. (a) By Remark 4.26 (b), it suffices to check that the family of seminorms is sepa-rating. This follows from Corollary 4.9: for every x ∈ X ∃ fx ∈ X∗ : f(x) = ‖x‖.

(b) • Show the implication: f ∈ X∗ =⇒ f weakly continuous:

Let G ⊆ K be open. Then G =⋃γ∈GBrγ (γ).

=⇒ f−1(G) =⋃γ∈G

f−1(Brγ ).

Using that the union over γ ∈ G can be restricted to γ ∈ G ∩ ran(f) =⇒

f−1(G) =⋃

γ∈G∩ran(f)

x ∈ X : |f(x)− γ| < rγ

=⋃

γ∈G∩ran(f)

⋃xγ∈X:f(xγ)=γ

x ∈ X : |f(x− xγ)| < rγ

︸ ︷︷ ︸

Uf,rγ (xγ)

,

where Uf,rγ (xγ) is weakly open by definition. Thus f−1(G) is weakly open.

• Let f ∈ X∗ be continuous with respect to some topology T on X. Since | · | : K→[0,∞[ is continuous (with respect to the standard topologies on K and [0,∞[)the mapping

(X, T ) −→ [0,∞[x 7−→ |f(x)|

is continuous =⇒ Uf,r := y ∈ X : |f(x)| < r ∈ T , and since T is a topology,the weak topology is coarser than T .

(c) See Problem 43.

(d) (xn)n converges to x in the weak topology

⇐⇒ ∀K ∈ N ∀f1, . . . , fK ∈ X∗ ∀r1, . . . , rK > 0 : xn ∈K⋂k=1

Ufk,rk(x) for finally all n

⇐⇒ ∀f ∈ X∗ ∀r > 0 : xn ∈ Uf,r(x)︸ ︷︷ ︸⇐⇒ |f(xn)−f(x)|<r

for finally all n

︸ ︷︷ ︸⇐⇒ lim

n→∞f(xn)=f(x)

.

79

4.29 Remarks. (a) Weak limits are unique by Lemma 4.28 (a).

(b) If dimX =∞, then the weak topology is not 1st countable (hence not metrisable)[for a proof, see Problem 44].

(c) Strong convergenceLemma 4.28 (b)

=⇒ weak convergence (but in general not viceversa).

(d) In a Hilbert space X (by Lemma 4.28 (d) and Riesz):

xnw−→ x ⇐⇒ 〈y, xn〉 n→∞−−−→ 〈y, x〉 ∀y ∈ X

(e) In `1:xn

w−→ x ⇐⇒ ‖xn − x‖ n→∞−−−→ 0

[I. Schur, J. Reine Angew. Math. 151, 79–111 (1921); see alsoJ. B. Conway, A course in functional analysis, 2nd ed., Springer, New York, 1990,Prop. 5.2]

4.30 Example. Let X = `p and p ∈ [1,∞], let (en)n∈N be the canonical basis withen = (ekn)k∈N, ekn = δnk (for p =∞ just the sequence itself). Then

• (en)n has no ‖·‖p-convergent subsequence.

• If p = 1, then (en)n is not weakly convergent. For, (`1)∗ ∼= `∞ and choosing f ↔(−1, 1,−1, 1, · · · ) ∈ `∞ =⇒ f(en) = (−1)n ∀n ∈ N =⇒ not convergent asn→∞. [Also consistent with Remark 4.29 (e)]

• If 1 < p ≤ ∞, then (en)n is weakly convergent to 0. For, in the case 1 < p <∞ wehave (`p)∗ ∼= `q by Theorem 2.38 with 1 < q <∞ =⇒ for f ↔ y = (yk)k ∈ `q :f(en) = yn

n→∞−→ 0.

In the case p =∞ use also Problems 23 & 24.

4.31 Lemma. Let X be a normed space and xnw−→ x. Then

‖x‖(a)

≤ lim infn→∞

‖xn‖ ≤ supn∈N‖xn‖

(b)< ∞.

Proof. Inequality (b): let xnw−→ x, so f(xn)

n→∞−−−→ f(x) for every f ∈ X∗.

=⇒ ∀f ∈ X∗ (fixed) : supn∈N| f(xn)︸ ︷︷ ︸

(Jxn)(f)

| <∞,

with the canonical embedding J : X −→ X∗∗. Since X∗ is Banach =⇒ (uniform bound-edness principle Thm. 4.12)

supn∈N‖Jxn‖∗∗︸ ︷︷ ︸‖xn‖

<∞.

Inequality (a):Cor. 4.9=⇒ ∃ fx ∈ X∗: ‖fx‖∗ = 1 and fx(x) = ‖x‖.

=⇒ ‖x‖ = |fx(x)| = limn→∞

|fx(x)| ≤ limn→∞

‖fx‖∗︸ ︷︷ ︸=1

‖xn‖ ≤ lim infn→∞

‖xn‖ .

80

4.32 Theorem. Let X be a normed space. Then xnw−→ x if and only if the following

two statements hold:

• supn∈N ‖xn‖ <∞.

• ∃ F ⊆ X∗ with span(F ) is (‖ ··· ‖∗-) dense in X∗ such that for every f ∈ F :

limn→∞

f(xn) = f(x).

Proof. “⇒” from Lemma 4.31 and by definition of weak convergence.

“⇐” use an ε/3-argument. Let ε > 0 and g ∈ X∗. Let K :=(‖x‖+ supn∈N ‖xn‖

)/2 <

∞.

– Since span(F ) is dense in X∗, there exists f ∈ span(F ): ‖f − g‖∗ < ε3K .

– For f ∈ span(F ) there exists N ∈ N such that for every n ≥ N :

|f(xn)− f(x)| < ε

3.

(Note: convergence f(xn) −→ f(x) holds not only for f ∈ F but also forf ∈ span(F ), because of finite linear combinations.)

=⇒

|g(x)− g(xn)| ≤ |g(x)− f(x)|+ |f(x)− f(xn)|+ |f(xn)− g(xn)|≤ ‖g − f‖∗︸ ︷︷ ︸

< ε3K

(‖x‖+ ‖xn‖︸ ︷︷ ︸

≤2K

)+ |f(x)− f(xn)|︸ ︷︷ ︸

< ε3

< ε.

4.33 Theorem (Eberlein-Smulian). Let X be a Banach space and A ⊆ X. Then

A weakly compact ⇐⇒ A weakly sequentially compact.

Proof. R. Whitley, Math. Ann. 172, 116–118 (1967).

4.34 Definition. Let X be a normed space. Then the weak* topology on X∗ is the locallyconvex topology on X∗ induced by the family of seminorms f 7→ |f(x)|x∈X .

4.35 Lemma. Let X be a normed space. Then the weak* topology is

(a) Hausdorff,

(b) the coarsest topology on X∗ such that ∀x ∈ X the map X∗ → K, f 7→ f(x), iscontinuous,

(c) coarser than the weak topology on X∗, and the two coincide if and only if X is reflexive,

(d) such that (fn)n ⊆ X∗ converges to f ∈ X∗ in the weak* topology if and only if

fn(x)n→∞−−−→ f(x) ∀x ∈ X.

Notation: fnw∗−−→ f .

81

Proof. (a) f 7−→ |f(x)|x∈X is a separating family of seminorms since for f 6= 0 ∃x ∈ Xwith f(x) 6= 0.

(b) Analogous to Lemma 4.28 (b).

(c) J(X) ⊆ X∗∗ with equality iff X is reflexive (J : canonical embedding of X in X∗∗).

(d) Analogous to Lemma 4.28 (d).

The next lemma is the weak* analogue of Lemma 4.31 and Theorem 4.32. But notice thathere X must be complete in order to apply the uniform boundedness principle in (a).

4.36 Lemma. Let X be a Banach space, f ∈ X∗, (fn)n ⊆ X∗. Then

(a) If fnw∗−→ f , then

‖f‖X∗ ≤ lim infn→∞

‖fn‖X∗ ≤ supn∈N‖fn‖X∗ <∞.

(b) fnw∗−→ f if and only if

• supn∈N ‖fn‖X∗ <∞• ∃A ⊆ X with span(A) (norm-) dense in X such that ∀x ∈ A

fn(x)n→∞−−−→ f(x).

Proof. Exercise.

4.37 Theorem (Banach-Alaoglu). Let X be a Banach space. Then the closed unit ballin X∗

B∗1 :=f ∈ X∗ : ‖f‖∗ ≤ 1

is compact in the weak* topology.

Proof. Define A :=×x∈X Ax, where Ax := z ∈ K : |z‖ ≤ ‖x‖ compact in K. Theelements of A are of the form

f :X −→ Kx 7−→ f(x)

with f(x) ∈ Ax, i.e. |f(x)| ≤ ‖x‖. By Tychonoff (Thm. 1.38), A is compact in the product-space topology (the coarsest topology such that for all x ∈ X the map Πx : A −→ Ax,f 7−→ f(x), is continuous). Now

• B∗1 =f ∈ A : f is linear

• The restriction of the product-space topology to B∗1 coincides with the weak* topol-

ogy.

82

We will show B∗1 closed in A (=⇒ B∗1 compact, because A is compact). For x, y ∈ X,α, β ∈ K, let

Kx,y,α,β :=f ∈ A : f(αx+ βy)− αf(x)− βf(y)︸ ︷︷ ︸

(Παx+βy−αΠx−βΠy)(f)

= 0

=(

Παx+βy − αΠx − βΠy

)−1(0).

But Παx+βy − αΠx − βΠy is continuous and 0 is closed in K, thus Kx,y,α,β is closed inA and therefore also

B∗1 =⋂

x,y∈Xα,β∈K

Kx,y,α,β.

4.38 Theorem (Helly, version 2 of Banach-Alaoglu). Let X be a separable Banach space.Then B∗1 is weak* sequentially compact.

Proof. Let xk : k ∈ N ⊆ X be (countable) dense in X. We need to show that anysequence (fn)n ⊆ B∗1 has a weak* convergent subsequence. Fix k ∈ N arbitrary. Considerthe sequence (fn(xk))n ⊆ K. This sequence is bounded, because

|fn(x)| ≤ ‖fn‖∗︸ ︷︷ ︸≤1

‖xk‖ ≤ ‖xk‖.

Thus, for all fixed k, (fn(xk))n has a convergent subsequence.

Claim: There exists a common subsequence (mj)j ⊆ N : ∀k ∈ N, (fmj (xk))j is convergent.

Proof: Use Cantor’s diagonal sequence trick: There exists (n(1)j )j ⊆ N such that (f

n(1)j

(x1))j

converges. Then there exists (nj)(2) ⊆ (n

(1)j )j such that (f

n(2)j

(x2))j converges. Continuing

this procedure, there exists (n(l+1)j )j ⊆ (n

(l)j )j such that (f

n(l+1)j

(xl+1))j converges.

The claim then holds with mj := n(j)j .

Now, define g(x) := limj→∞ fmj (x) ∀x ∈ span(xk : k ∈ N) =: dom(g).

• dom(g) is a dense subspace of X with respect to ‖ · ‖

• g : dom(g) −→ K is linear

• g is bounded:

|g(x)| = limj→∞

|fmj (x)|︸ ︷︷ ︸≤‖fmj‖∗︸ ︷︷ ︸

≤1

‖x‖

≤ ‖x‖

Since K is complete, we can apply the bounded linear extension theorem 2.32: There existsg : X → K such that g|dom(g) = g and ‖g‖∗ = ‖g‖ ≤ 1. Thus g ∈ B∗1 . Then by Lemma

4.36 (b), fmjw∗−−→ g as j →∞.

4.39 Theorem (version 3 of Banach-Alaoglu). Let X be Banach space. Then

X reflexive ⇐⇒ B1 :=x ∈ X : ‖x‖ ≤ 1

is weakly compact.

83

Proof. ”=⇒”: Use that ”weak topology on X” = ”weak* topology on (X∗)∗”, because Xis reflexive and Thm. 4.37.

”⇐=”: See e.g. Dunford/Schwartz: Linear Operators, vol. I, Interscience, 1966, Thm.V.4.7.

4.40 Example. Compactness of B1 in different spaces. Here p, q ∈]1,∞[ are Holderconjugate.

weak weak* seq. weak seq. weak*

`1 (∼= c∗0) no yes no yes

`p (∼= (`q)∗) yes yes yes yes

`∞ (∼= (`1)∗) no yes no yes

L1 (not a dual!) no — no —

Lp (∼= (Lq)∗) yes yes yes yes

L∞ (∼= (L1)∗) no yes no yes

84

5 Bounded operators

5.1 Topologies on the space of bounded linear operators

5.1 Definition. Let X,Y be normed spaces.

(a) uniform (operator) topology on BL(X,Y ) := norm topology with respect to ‖ · ‖X→Y .

(b) strong (operator) topology on BL(X,Y ) := locally convex topology induced by theseminorms T 7−→ ‖Tx‖Y x∈X .

(c) weak (operator) topology on BL(X,Y ) := locally convex topology induced by theseminorms T 7−→ |`(Tx)|x∈X

`∈Y ∗.

5.2 Remark. (a) The families of seminorms in Definition 5.1 (b) and (c) are separating(check!) =⇒ Hausdorff topologies.

(b) Strong and weak operator topologies are not first countable if dimX =∞.

5.3 Lemma. (a) The strong topology is the coarsest topology on BL(X,Y ) such that allmaps

Mx :BL(X,Y ) −→ Y

T 7−→ Tx, x ∈ X

are continuous. The weak topology is the coarsest topology on BL(X,Y ) such that allmaps

M`,x :BL(X,Y ) −→ K

T 7−→ `(Tx), x ∈ X, ` ∈ Y ∗

are continuous.

(b)

(c) Let T, Tn ∈ BL(X,Y ), n ∈ N. Then

• (Tn)n converges to T in the strong operator topology, in symbols, Tns−−→ T

(“strongly”)⇐⇒ lim

n→∞Tnx = Tx ∀x ∈ X.

• (Tn)n converges to T in the weak operator topology, in symbols, Tnw−−→ T

(“weakly”)

⇐⇒ limn→∞

`(Tnx) = `(Tx) ∀x ∈ X ∀` ∈ Y ∗

Proof. (a) analogous to the proof of Lemma 4.28 (b).

(c) analogous to the proof of Lemma 4.28 (d).

85

(b) Upper branch: follows from M`,x ∈(

BL(X,Y ))∗∀x ∈ X ∀` ∈ Y ∗, because

|M`,xT | = |`(Tx)| ≤ ‖`‖Y ∗ ‖Tx‖Y︸ ︷︷ ︸≤‖T‖‖x‖

=⇒ |M`,xT |‖T‖ ≤ ‖`‖Y ∗‖x‖ ∀T 6= 0

lower branch:

• Mx : BL(X,Y )→ Y is linear ∀x ∈ X. Also bounded:

‖MxT‖Y = ‖Tx‖Y ≤ ‖T‖‖x‖ =⇒ sup0 6=T∈BL(X,Y )

‖MxT‖‖T‖ ≤ ‖x‖,

i.e. Mx is continuous for all x. So from (a): strong operator topology is coarserthan the uniform topology.

• M`,x = ` Mx =⇒ continuity of Mx implies continuity of M`,x, and the strongoperator topology must be finer than the weak operator topology by (a).

5.4 Lemma. Let H be a Hilbert space and (Tn)n ⊆ BL(H).

(a) If (Tnx)n Cauchy in H ∀x ∈ H then there exists T ∈ BL(H) such that Tns−→ T .

(b) If (〈y, Tnx〉)n is Cauchy in K for all x, y ∈ H, then there exists T ∈ BL(H) such thatTn

w−−→ T .

Proof. (a) Define

T :H −→ Hx 7−→ lim

n→∞Tnx

.

The map T is

• well defined (limit exists!)

• linear

• bounded: We have supn ‖Tnx‖ <∞ for all x ∈ H (due to convergence). Thus bythe uniform boundedness principle: supn ‖Tn‖ <∞ =⇒

‖T‖ = supx∈H : ‖x‖=1

‖Tx‖︸ ︷︷ ︸limn→∞

‖Tnx‖︸ ︷︷ ︸≤ ‖Tn‖ ‖x‖︸︷︷︸

=1

≤ supn∈N‖Tn‖ <∞.

Hence T ∈ BL(H) and Tns−−→ T .

(b) Define the form

Q :H×H −→ K

(x, y) 7−→ limn→∞

〈Tnx, y〉

• Q is well-defined (i.e. the limit exists)

• Q is sesquilinear

86

• Q is bounded, i.e. for every x, y ∈ H:

|Q(x, y)| ≤(

supn∈N‖Tn‖︸ ︷︷ ︸

=:S<∞

)‖x‖ ‖y‖

where S <∞ will be shown below.

By Problem T13 (a corollary of the Riesz representation) there exists a unique T ∈BL(H) such that Q(x, y) = 〈Tx, y〉 which means Tn

w−−→ T .

Proof of S <∞: Let x, y ∈ H be fixed. Then we know:

supn∈N| 〈Tnx, y〉 |︸ ︷︷ ︸

=:`n(y)

<∞

with `n ∈ H∗. By the uniform boundedness principle ?? and the Riesz-representationTheorem we infer that

supn∈N‖`n‖∗ = sup

n∈N‖Tnx‖ <∞ ∀x ∈ H,

and again by the uniform boundedness principle ?? we have

supn∈N‖Tn‖ <∞.

5.5 Examples. Let (Tn) ⊆ BL(`2).

(a) For x = (x1, x2, . . .) ∈ `2 and n ∈ N let

Tnx :=

(1

nx1,

1

nx2, . . .

)=

1

nx.

Then ‖Tn‖ = 1n

n→∞−−−→ 0, i.e. uniform convergence to zero (operator).

(b) Let

Tnx :=(

0, . . . , 0︸ ︷︷ ︸n times

, xn+1, xn+2, . . .).

• Tn s−−→ 0, because

‖Tnx‖2 =

∞∑j=n+1

|xj |2 n→∞−−−→x∈`2

0.

• (Tn)n does not converge uniformly to 0, because

Tnen+1 = en+1 =⇒ ‖Tn‖ ≥ 1.

(c) Let

Tnx :=(

0, . . . , 0︸ ︷︷ ︸n times

, x1, x2, . . .)

“n-times iterated right shift”.

87

• Tn w−−→ 0, because for y ∈ `2 by the Cauchy-Schwarz inequality

| 〈y, Tnx〉 | =∣∣∣∣ ∞∑j=1

yn+jxj

∣∣∣∣ ≤ ‖x‖ ·√√√√ ∞∑

j=1

|yn+j |2

︸ ︷︷ ︸n→∞−−−→y∈`2

0

.

• (Tn)n does not converge strongly to 0, because

‖Tnx‖ = ‖x‖ ∀x ∈ `2.

5.2 Adjoint operators

5.6 Missing. For the amusement of the reader.

5.7 Definition. (a) Let X,Y be normed spaces and T ∈ BL(X,Y ). Then

T× :Y ∗ −→ X∗

` 7−→ T×`

with (T×`)(x) := `(Tx) is a linear operator, the adjoint of T .

(b) Let H1,H2 be Hilbert spaces and T ∈ BL(H1,H2). Then

T ∗ :H2 −→ H1

y 7−→ T ∗y

where (by Riesz!) T ∗y is the unique z ∈ H1 such that for every x ∈ H1

〈y, Tx〉H2= `(x) ∈ H∗1= 〈z, x〉H1

=:⟨T ∗y, x

⟩H1

is a linear operator, the (Hilbert) adjoint of T .

5.8 Remarks. (a) In the Hilbert space case we have that

T ∗ = A1T×A−1

2

with Aj : H∗j −→ Hj is antilinear, isometric and bijective (Corollary 2.58).

(b) T ∗ is linear, because⟨T ∗(αy + βz), x

⟩H1

= 〈αy + βz, Tx〉H2

= α 〈y, Tx〉H2︸ ︷︷ ︸〈T ∗y,x〉H1

+β 〈z, Tx〉H2︸ ︷︷ ︸〈T ∗z,x〉H1

=⟨αT ∗y + βT ∗z, x

⟩H1,

and since this is true for every x ∈ H1 we have linearity (analogous for T×).

5.9 Examples. (a) Let X = Y = `1 and Tx := (0, x1, x2, . . .), the right-shift operator.Then T ∈ BL(`1). T× : (`1)∗ −→ (`1)∗ (we know f ∈ (`1)∗ acts as f(x) =

∑n∈N ξnxn

for every x ∈ `1 and some unique ξ ∈ `∞ by Riesz). So

(T×f)(x) = f(Tx) =∑n∈N

ξn(Tx)n =∑n∈N

ξn+1xn

i.e. T× corresponds to (ξ2, ξ3, ξ4, . . .), which is a left shift.

88

(b) For H1 = H2 = H and ϕ,ψ ∈ H let P := ψ 〈ϕ, ·〉, i.e. Px = ψ 〈ϕ, x〉 ∀x ∈ H.

Then P ∗ = ϕ 〈ψ, ·〉 because ∀x, y ∈ H

〈y, Px〉 = 〈y, ψ〉 〈ϕ, x〉=⟨〈ψ, y〉ϕ, x

⟩=⟨P ∗y, x

⟩.

5.10 Theorem. Let X,Y be normed spaces and T ∈ BL(X,Y ). Then

‖T×‖Y ∗→X∗ = ‖T‖X→Y

Proof. By definition and Theorem 4.18 we have

‖T‖X→Y = supx∈X‖x‖=1

‖Tx‖

= supx∈X‖x‖=1

sup`∈Y ∗‖`‖Y ∗=1

|`(Tx)|

15

= sup`∈Y ∗‖`‖Y ∗=1

supx∈X‖x‖=1

|(T×`)(x)|

︸ ︷︷ ︸‖T×`‖

X∗

= ‖T×‖Y ∗→X∗ .

5.11 Corollary. Let H1,H2 be Hilbert spaces and T ∈ BL(H1,H2). Then ‖T ∗‖H2→H1=

‖T‖H1→H2.

Proof. This rests on

• T ∗ = A1T×A−1

2

• Theorem 5.10

• Aj , j = 1, 2, are isometric and bijective.

5.12 Theorem. Let H be a Hilbert space and T, S ∈ BL(H). Then

(a) The map

∗ :BL(H) −→ BL(H)

T 7−→ T ∗

is antilinear, isometric and bijective.

(b) (TS)∗ = S∗T ∗

(c) (T ∗)∗ = T .

(d) If T has an inverse T−1 ∈ BL(H), then (T ∗)−1 = (T−1)∗. In particular (T ∗)−1 ∈BL(H).

15interchange of suprema:

supa

supb

Aab︸︷︷︸≤supa Aab

≤ supb

supa

Aab︸︷︷︸≤sup

bA

ab

≤ supa

supb

Aab =⇒ supa

supbAab = sup

bsupaAab.

89

(e) ‖TT ∗‖ = ‖T‖2.

Proof. (a) Antilinear by definition of T ∗, isometric by Cor. 5.10, surjective by part (c) ofthis theorem.

(b) See tutorial sheet.

(c) See tutorial sheet.

(d) TT−1 = 1 = T−1T together with (b) gives

T ∗(T−1)∗ = 1 = 1∗ = (T−1)∗T ∗.

(e) On the one hand, we have ‖TT ∗‖ ≤ ‖T‖ ‖T ∗‖ = ‖T‖2 by Corollary 5.11. On the otherhand

∥∥TT ∗∥∥ = sup06=x,y∈H

∣∣∣ 〈y, TT ∗x〉 ∣∣∣‖x‖ ‖y‖

≥ sup06=x∈H

‖T ∗x‖2

‖x‖2

=∥∥T ∗∥∥2

= ‖T‖2 .

5.13 Definition. Let H be a Hilbert space and T ∈ BL(H).

• T is unitary iff T is bijective and T−1 = T ∗ (i.e. TT ∗ = 1 = T ∗T ).

• T is self-adjoint iff T = T ∗.

• T is normal iff TT ∗ = T ∗T .

5.14 Remarks. (a) Self-adjoint or unitary =⇒ normal.

(b) If T is unitary, then 〈Tx, Ty〉 = 〈x, y〉 = 〈T ∗x, T ∗y〉 ∀x, y ∈ H.

(c) If T is self-adjoint, then

〈x, Ty〉 = 〈Tx, y〉 ∀ x, y ∈ H. (∗)

Setting x = y gives 〈x, Tx〉 ∈ R for every x ∈ H.

[In Problem 41, the property (∗) was called symmetry; the notions symmetric andself-adjoint agree for bounded linear operators]

(d) T normal =⇒ ‖Tx‖ = ‖T ∗x‖ for every x ∈ H. In particular ker(T ) = ker(T ∗).

5.15 Examples. (a) T = z1, z ∈ K. Then T ∗ = z1. So T is self-adjoint if and only ifz ∈ R.

(b) Let H = L2([0, 1]), k ∈ C([0, 1]2) and

(Tf)(x) :=

∫dy k(x, y)f(y).

If k(x, y) = k(x, y) ∀x, y ∈ H =⇒ T is self-adjoint.

90

5.3 The spectrum

In this subsection: X is a Banach space.

5.16 Definition. Let T ∈ BL(X).

• resolvent set (of T ):

ρ(T ) :=z ∈ C : T − z · 1 bijective

• resolvent (Green function) of T :

Rz := (T − z)−1

– if z ∈ ρ(T ), then Rz ∈ BL(X) (in particular this inverse exists),

– need not exist for z 6∈ ρ(T ).

• spectrum of T :spec(T ) := σ(T ) := C \ ρ(T )

• if there exists 0 6= x ∈ X such that Tx = λx for some λ ∈ C then λ is an eigenvalueof T and x the corresponding eigenvector.

• point spectrum (of T ):

specp(T ) := σp(T ) :=z ∈ C : T − z not injective

=

eigenvalues of T

• continuous spectrum (of T ):

specc(T ) := σc(T ) :=z ∈ C : T − z injective and ran(T − z) 6= X, but dense

• residual spectrum (of T )

specr(T ) := σr(T ) :=z ∈ C : T − z injective and ran(T − z) not dense

[specr(T ) = ∅ for most T of interest]

5.17 Lemma. Let T ∈ BL(X). Then

(a) • C = σ(T ) ∪ ρ(T )

• σ(T ) = σp(T ) ∪σc(T ) ∪σr(T )

(b) If dimX <∞, then σc(T ) = ∅ = σr(T ).

Proof. (a) clear by definition.(b) follows from linear algebra: T − z injective =⇒ dim ker(T − z) = 0 =⇒ dim ran(T −z) = dim(X) =⇒ ran(T − z) = X.

5.18 Definition. Let D ⊆ C be a region and

x :D −→ Xz 7−→ x(z)

.

91

• x is strongly differentiable (strongly analytic) in z0 ∈ D iff

limC3h→0

x(z0 + h)− x(z0)

h

exists in X.

• x is weakly differentiable (weakly analytic) in z0 ∈ D iff z 7−→ `(x(z)) is differentiablein C in z0 for all ` ∈ X∗.

• x is strongly (weakly) analytic in D iff x is strongly (weakly) differentiable in z0 forall z0 ∈ D.

5.19 Remark. (a) X-valued strongly analytic functions have analogous properties to C-valued analytic functions (e.g. a power-series expansion converging w.r.t. ‖ · ‖, etc.)

Moral: Replace | · | (absolute value in C) by ‖ · ‖.16

(b) strongly analytic ⇐⇒ weakly analytic [see e.g. Reed/Simon Thm. VI.4 for ”⇐=”]

5.20 Theorem. Let T ∈ BL(X). Then ρ(T ) is open in C and the map

ρ(T ) −→ BL(X)

z 7−→ Rz

is strongly analytic. For λ, µ ∈ ρ(T ) the first resolvent identity holds

Rλ −Rµ = (λ− µ)RλRµ, (1)

in particular RλRµ = RµRλ (they commute!).

The proof uses:

5.21 Lemma (Neumann series). Let T ∈ BL(X) with ‖T‖ < 1. Then

(1− T )−1 ∈ BL(X) and (1− T )−1 =

∞∑j=0

T j .

Proof. From ‖T j‖ ≤ ‖T‖j , ‖T‖ < 1 and the digression on series in Banach spaces (see theproof of Lemma 2.49) =⇒ S :=

∑∞j=0 T

j ∈ BL(X). Now, for N ∈ N consider

(1− T )

N∑j=0

T j = 1− TN+1 =

N∑j=0

T j(1− T ).

Note that∑N

j=0 Tj N→∞−−−−→ S and TN+1 N→∞−−−−→ 0 (since ‖TN+1‖ ≤ ‖T‖N+1 N→∞−−−−→ 0).

Thus (1− T )S = 1 = S(1− T ), i.e. S = (1− T )−1.

5.22 Corollary. Let T ∈ BL(X). Then σ(T ) ⊆ z ∈ C : |z| ≤ ‖T‖.Proof. For z 6= 0, consider

Rz = (T − z)−1 =1

z

(T

z− 1

)−1

.

For |z| > ‖T‖ we have ‖T/z‖ < 1 and get Rz ∈ BL(X) by Lemma 5.21. Thus z ∈ ρ(T ) =C \ σ(T ).

16For a discussion of Banach-space valued functions of a complex variable, see e.g. Hille, Phillips, Func-tional analysis and semigroups, AMS, 1957; or Dunford, Schwartz, Linear operators, Vol. 1, Sect. III.14.

92

Proof of Thm. 5.20. • Equation (1) follows from

(T − λ)(Rλ −Rµ)(T − µ) =(1− (T − λ)Rµ

)(T − µ) =

= T − µ− (T − λ) = λ− µ

and by multiplication by Rλ from the left and by Rµ from the right.

• Commutativity: Exchange λ←→ µ in (1) and equate.

• Openness of ρ(T ): Let λ ∈ ρ(T ) (6= ∅ by Cor. 5.22) and z ∈ C such that |z − λ| <1‖Rλ‖ . Then

T − z = T − λ− (z − λ) = (T − λ)(1− (z − λ)Rλ︸ ︷︷ ︸

=:V

).

By definition ‖V ‖ < 1 and therefore by Lemma 5.21: (1− V )−1 ∈ BL(X). Thus

Rz = (T − z)−1 = (1− V )−1Rλ ∈ BL(X) (∗)

and z ∈ ρ(T ), i.e. ρ(T ) is open.

• From (∗) and Lemma 5.21:

Rz =

∞∑j=0

V jRλ =

∞∑j=0

(z − λ)jRj+1λ for z ∈ B1/‖Rλ‖(λ).

Thus B1/‖Rλ‖(λ) 3 z 7→ Rz is strongly differentiable, i.e. Rz is analytic in B1/‖Rλ‖(λ).But since λ ∈ ρ(T ) was arbitrary this shows the claim.

5.23 Lemma. Let T ∈ BL(X). Then σ(T ) 6= ∅.

Proof. Let |z| > ‖T‖. Then (see the proof of Cor. 5.22)

Rz = (T − z)−1 =1

z

(T

z− 1

)−1

= −1

z

∞∑j=0

T j

zj.

This expansion shows that

lim|z|→∞

‖Rz‖ = 0. (∗)

Assume σ(T ) = ∅. Then ρ(T ) = C and by Thm. 5.20,

C −→ BL(X)

z 7−→ Rz

is entire (i.e. analytic on all of C). Also C 3 z 7−→ ‖Rz‖ is bounded. By Liouville’stheorem17 z 7−→ Rz is constant and by (∗) this constant has to be 0, i.e. Rz = 0 for allz ∈ C.

17For C-valued functions, see e.g. J. B. Conway, Functions of one complex variable, 2nd ed., Springer,New York, 1978, Thm. IV.3.4.

93

5.24 Definition. For T ∈ BL(X)

Spectral radius of T :⇐⇒ r(T ) := supλ∈σ(T )

|λ|.

5.25 Theorem. Let T ∈ BL(X). Then

(a) r(T ) = limn→∞

‖Tn‖1/n = infn∈N‖Tn‖1/n, in particular, r(T ) ≤ ‖T‖.

(b) If, in addition, X is a Hilbert space and T is normal, then limn→∞

‖Tn‖1/n = ‖T‖ =⇒r(T ) = ‖T‖.

Proof. (a) • We show first limn→∞

‖Tn‖1/n = infn∈N‖Tn‖1/n.

W.l.o.g. assume Tn 6= 0 ∀n ∈ N (otherwise the claim is clear). Set an :=ln ‖Tn‖ =⇒ an1+n2 ≤ an1 + an2 ∀ n1, n2 ∈ N. Fix m ∈ N arbitrary.Write n ∈ N, n ≥ m as n = qm+ r with q ∈ N and r ∈ 0, . . . ,m− 1. Then

ann≤ qam + ar

qm+ r≤ am

m+

arqm+ r

q→∞=⇒ lim sup

n→∞

ann≤ am

m∀ m ∈ N

=⇒ lim supn→∞

ann≤ inf

m∈Namm.

But (trivially) lim infn→∞

ann≥ inf

m∈Namm

. Taken together, limn→∞ an/n exists and

limn→∞

ean/n = infn∈N

ean/n.

• Thm. 5.20 =⇒ z 7−→ Rz is strongly analytic in D := C\z ∈ C : |z| ≤ r(T )and therefore has a Laurent series expansion about 018

Rz =∑j∈Z

zjAj , Aj ∈ BL(X) for j ∈ Z,

which is norm-convergent in BL(X) for all z ∈ D. For z ∈ C : |z| > ‖T‖ ⊆ D,the expansion

Rz = −1

z

∞∑j=0

T j

zj(1)

holds, see the proof of Lemma 5.23. Thus, by uniqueness of the Laurent series,Aj = 0 for j ∈ N0 and A−j = −T j−1 for j ∈ N. That is, the series (1) is normconvergent ∀ z ∈ D.

On the other hand: Given any ε > 0, the series (1) is not norm-convergent onC\z ∈ C : |z| ≤ r(T )−ε, because then we would get convergence for some z ∈spec(T ). Thus 1

r(T ) is the radius of convergence of the series ξ 7→ −∑j∈N ξjT j−1,

and Hadamard’s root criterion gives

r(T ) = lim supj→∞

‖T j‖1/j = limj→∞

‖T j‖1/j ,

as we have already shown.

18For C-valued functions, see e.g. J. B. Conway, Functions of one complex variable, 2nd ed., Springer,New York, 1978, Thm. V.1.11.

94

(b) Using the property ‖TT ∗‖ = ‖T‖2 from Thm. 5.12 (e):

‖T 2‖2 = ‖T 2(T 2)∗‖ T normal= ‖(TT ∗)(TT ∗)∗‖ = ‖TT ∗‖2 5.12(e)

= ‖T‖4,

i.e. ‖T 2‖ = ‖T‖2. By induction on k ∈ N:

‖T 2k‖ = ‖T‖2k . (2)

Thus

limn→∞

‖Tn‖1/n = limk→∞

‖T 2k‖1/2k (2)= ‖T‖.

5.4 Compact operators

Throughout we assume that X and Y are Banach spaces and H is a Hilbert space.

5.26 Definition. An operator T ∈ BL(X,Y ) is compact iff for every bounded subsetA ⊆ X, the closure T (A) ⊆ Y is compact, i.e. T (A) is relatively compact in Y .

5.27 Remark. T is compact if and only if for every bounded sequence (xn)n ⊆ X, thesequence (Txn)n has a convergent subsequence (in metric spaces, compactness is equivalentto sequential compactness)

5.28 Examples. (a) For k ∈ C([0, 1]2) let T : L2([0, 1]) −→ L2([0, 1]),

(Tf)(x) :=

∫ x

0dy k(x, y)f(y).

Then, T is compact (exercise, use Arzela-Ascoli theorem 1.50).

(b) Finite-rank operators. T ∈ BL(X,Y ) is of finite rank iff dim ran(T ) <∞, i.e. ∃ J ∈ Nsuch that

Tx =J∑j=1

αj(x)fj ,

where the fj ∈ Y are linearly independent and αj ∈ X∗ for j = 1, . . . , J . T is compactbecause Corollary 4.10 yields the existence of `1, . . . , `J ∈ Y ∗ such that

`k(fj) = δkj .

We have

|αj(x)| = |`j(Tx)| ≤∥∥`j∥∥ ‖T‖ ‖x‖

≤(

maxj∈J

∥∥`j∥∥) ‖T‖ ‖x‖ .Assuming (xn)n ⊂ X is bounded, the vector of coefficients (αj(xn))j=1,...,J is boundedin KJ =⇒ Heine-Borel tells the existence of xnk such that for every j = 1, . . . , J we

have αj(xnk)k→∞−−−→ βj with some βj ∈ K. Hence

Txnk =

J∑j=1

αj(xnk)fjk→∞−−−→

J∑j=1

βjfj ∈ Y.

95

5.29 Theorem. Let T ∈ BL(X,Y ) be compact, (xn)n ⊆ X such that xnw−−→ x ∈ X as

n→∞. Then (Txn)n ⊆ Y is strongly convergent.

Proof. Weak convergence xnw−−→ x and Lemma 4.32 give supn∈N ‖xn‖ < ∞. Let yn :=

Txn, n ∈ N and y := Tx. Let ` ∈ Y ∗. Then, using the adjoint T× : Y ∗ −→ X∗,

`(yn)− `(y) = `(Txn)− `(Tx) = (T×`)(xn − x)n→∞−→ 0.

Hence ynw−−→ y. Suppose yn

n→∞−−−→

sy. Then there is some ε > 0 and (ynk)k∈N such that∥∥ynk − y∥∥ ≥ ε for every k ∈ N. But (xnk)k∈N is bounded. By the compactness of T , there

is a subsequence Txnkll→∞−−−→ z ∈ Y , z 6= y. Since ynkl

w−−−→l→∞

y 6= z, this is a contradiction,

so ynn→∞−−−→ y.

5.30 Theorem. Let T ∈ BL(X,Y ).

(a) Assume (Tn)n∈N ⊆ BL(X,Y ), Tn compact for each n ∈ N and

‖Tn − T‖ n→∞−−−→ 0.

Then T is compact.

(b) T is compact if and only if T× is compact (Schauder’s Theorem).

(c) Let Z be a Banach space and S ∈ BL(Y, Z). If S or T is compact, then ST ∈ BL(X,Z)is compact.

Proof. (a) Let (xm)m ∈ X bounded, w.l.o.g. ‖xm‖ ≤ 1 for m ∈ N. Then Tnxm hasa convergent subsequence as m → ∞ with the limit yn ∈ Y . Now by (Cantor’s)diagonal sequence argument, there is a common convergent subsequence (xmk)k∈Nwith Tnxmk

k→∞−−−→ yn, n ∈ N. Let ε > 0 and N ∈ N such that ‖Tn − Tn′‖ ≤ ε forevery n, n′ ≥ N . Then

‖yn − yn′‖ ≤∥∥yn − Tnxmk∥∥︸ ︷︷ ︸

(1)

+∥∥Tnxmk − Tn′xmk∥∥︸ ︷︷ ︸≤‖Tn−Tn′‖≤ε

+∥∥Tn′xmk − yn′∥∥︸ ︷︷ ︸

(2)

< ε+ ε+ ε

because (1) and (2) tend to 0 as k → ∞. So (yn)n∈N is a Cauchy sequence, henceconvergent, i.e. yn

n→∞−−−→ y ∈ Y .

Claim:∥∥Txmk − y∥∥ k→∞−−−→ 0. Indeed, let ε > 0. There is some n ∈ N such that

‖T − Tn‖ ≤ ε/3 and ‖yn − y‖ ≤ ε/3.

This leads to∥∥Txmk − y∥∥ ≤ ∥∥Txmk − Tnxmk∥∥+∥∥Tnxmk − yn∥∥+ ‖yn − y‖

≤ ε/3 +∥∥Tnxmk − yn∥∥︸ ︷︷ ︸

≤ε/3

+ε/3

for k big enough. So T is compact.

96

(b) “⇒” Let (`n)n∈K be a bounded sequence in Y ∗. Then K := T (BX1 (0)) ⊆ Y is a com-

pact metric space. The sequence of restrictions fn := `n|K ∈ C(K) is boundedand equicontinuous, because

|fn(y1)− fn(y2)| ≤(

supk∈N‖`k‖

)︸ ︷︷ ︸

<∞

‖y1 − y2‖Y .

By the theorem of Arzela-Ascoli 1.50 there exists a uniformly convergent subse-quence (fnk)k∈N. Hence∥∥∥T×`nk − T×`nl∥∥∥

X∗= sup

x∈BX1 (0)

|T×`nkx− T×`nlx|K

= supx∈BX1 (0)

|`nk(Tx)− `nl(Tx)|

=∥∥fnk − fnl∥∥∞,K k,l→∞−−−−→ 0,

where the last equality is due to the fact that T (BX1 (0)) is dense in K. So

(T×`nk)k converges in X∗ and T× : Y ∗ −→ X∗ is compact.

“⇐” If T× compact, then T×× : X∗∗ −→ Y ∗∗ is compact. So T××JX is compact,where JX : X −→ X∗∗ is the canonical embedding (JXx)(f) = f(x) for everyx ∈ X and f ∈ X∗. Now, for x ∈ X and ` ∈ Y ∗ we calculate

(T××JX(x))(`) = JX(x)(T×`)

= (T×`)(x)

= `(Tx)

= (JY (Tx))(`)

This implies T××JX = JY T being compact. Since Y is closed in Y ∗∗, T iscompact aswell.

(c) is clear, since bounded linear operators preserve convergence and boundedness.

5.31 Theorem. Let H be a separable Hilbert space. Then every compact T ∈ BL(H) isthe uniform limit of a sequence of operators (Tn)n ⊆ BL(X,Y ) of finite rank.

Proof. Let ϕjj∈N ⊆ H be an orthonormal basis and for n ∈ N let

λn := supψ⊥span(ϕ1,...,ϕn)

‖ψ‖=1

‖Tψ‖

(λn)n∈N is non-negative and decreasing, so there is some limit

limn→∞

λn =: λ ≥ 0

Claim: λ = 0: True, because for n ∈ N ∃ ψn ⊥ span(ϕ1, . . . , ϕn) with ‖ψn‖ = 1 such

97

that ‖Tψn‖ ≥ λn/2 ≥ λ/2. Now, for every y ∈ H

| 〈y, ψn〉 |2 =

∣∣∣∣∣∞∑j=1

⟨y, ϕj

⟩ ⟨ϕj , ψn

⟩∣∣∣∣∣2

≤( ∞∑j=1

|⟨y, ϕj

⟩ ⟨ϕj , ψn

⟩|)2

≤ ‖ψn‖2∞∑

j=n+1

|⟨y, ϕj

⟩|2

≤∞∑

j=n+1

|⟨y, ϕj

⟩|2 n→∞−−−→ 0.

So ψnw−−→ 0 and by Theorem 5.29 Tψn

n→∞−−−→ 0. So λ = 0. Let

Rn :=n∑j=1

⟨ϕj , ·

⟩Tϕj ∈ BL(X,Y )

with dim ranRn ≤ n. Then, by Parseval’s Equality 2.49

(T −Rn)ψ = (T −Rn)∞∑l=1

〈ϕl, ψ〉ϕl

=∞∑l=1

〈ϕl, ψ〉Tϕl −n∑l=1

〈ϕl, ψ〉Tϕl

=∞∑

l=n+1

〈ϕl, ψ〉Tϕl

= T

( ∞∑l=n+1

〈ϕl, ψ〉ϕl︸ ︷︷ ︸⊥span(ϕ1,...,ϕn)

)

and ∥∥∥∥ ∞∑l=n+1

〈ϕl, ψ〉ϕl∥∥∥∥ ≤ ‖ψ‖ .

So

‖T −Rn‖ = supψ∈H‖ψ‖=1

∥∥(T −Rn)ψ∥∥ ≤ λn∥∥∥∥ ∞∑

l=n+1

〈ϕl, ψ〉ϕl∥∥∥∥ ≤ λn n→∞−−−→ 0.

5.5 Fredholm alternative for compact operators

5.32 Motivation.

• Let M be a N ×N -matrix for N ∈ N and let z ∈ C. Then

either: MΨ = zΨ has a solution 0 6= Ψ ∈ CN ,

or: (M − z)−1 exists.

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• Let H = L2([0, 2]) and M ∈ BL(L2([0, 2])) be defined by

(MΨ)(x) := xΨ(x) for a.e. x ∈ [0, 2] ∀Ψ ∈ L2([0, 2]).

– MΨ = Ψ has no solutions 0 6= Ψ ∈ L2 because xΨ(x) = Ψ(x) for Lebesguealmost every x ∈ [0, 2] =⇒ Ψ(x) = 0 for almost every x ∈ [0, 2].

– (M − 1) is not surjective, because

∗(

(M − 1)Ψ)

(x) = (x− 1)Ψ(x)

∗ The mappingϕ : x 7−→ |x− 1|−1/3

is an element of L2([0, 2]) but ϕ 6∈ ran(M − 1): Suppose ϕ ∈ ran(M − 1).Then there exists Ψ ∈ L2([0, 2]) such that (M −1)Ψ = ϕ and thus Ψ(x) =

1x−1 |x− 1|−1/3 which is not an element of L2([0, 2]).

• For compact operators the above either/or statement still holds.

In the following H is a separable Hilbert space.

5.33 Theorem (Analytic Fredholm theorem). Let D ⊆ C be open and connected, letf : D −→ BL(H) be analytic and f(z) compact for all z ∈ D. Then

either (A1): (f(z)− 1)−1 does not exist for any z ∈ D.

or (A2): there exists a discrete subset S ⊆ D without accumulation point in D such thatthe mapping

D \ S −→ BL(H)

z 7−→ (f(z)− 1)−1

is well-defined and analytic. Moreover, z ∈ S iff f(z)Ψ = Ψ has a solution 0 6= Ψ ∈H.

5.34 Corollary. Let T ∈ BL(H) be compact and z ∈ C \ 0. Then

either: TΨ = zΨ has a solution 0 6= Ψ ∈ H

or: (T − z)−1 ∈ BL(H) exists.

Proof. Choose D := C \ 0 and f(z) := Tz . Alternative (A1) does not hold in Thm. 5.33

because Tz − 1 has an inverse in BL(H) ∀|z| > ‖T‖ (Neumann series!) =⇒ Alternative

(A2) holds.

Proof of Thm. 5.33. It suffices to prove the following: For every z0 ∈ D exists a neighbour-hood N (z0) such that eiher (A1) or (A2) holds on N (z0). [Indeed, color a neighbourhoodN (z0) red or blue, depending on which alternative holds. D is connected, i.e. if both colorsappear then there exists z0 such that N (z0) cannot be uniquely coloured. Contradiction.]Fix z0 ∈ D. Since f is continuous there exists r > 0 such that ‖f(z) − f(z0)‖ < 1

2 forall z ∈ Br(z0) = Dr. By Thm. 5.31 (separability!) there exists a finite-rank operatorF ∈ BL(H) such that ‖f(z0) − F‖ < 1

2 . Let N := rank(F ). By Lemma 5.21 (Neumannseries)

Dr −→ BL(H)

z 7−→ (1− f(z) + F )−1 (1)

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is well defined and analytic. Ex. 5.28 (b) and Riesz =⇒ ∃ Ψ1, . . . ,ΨN ∈ H linearlyindependent and ϕ1, . . . , ϕN ∈ H such that F =

∑Nk=1 〈ϕn, ·〉Ψn. For n = 1, . . . , N define

the analytic H-valued function

γn :Dr −→ Hz 7−→

((1− f(z) + F )−1

)∗ϕn.

Then

g(z) := F (1− f(z) + F )−1 =N∑n=1

⟨γn(z), ·

⟩Ψn, (2)

and

f(z)− 1 = (g(z)− 1)(1− f(z) + F ). (3)

For all z ∈ Dr we have the equivalences

f(z)− 1 is not invertible ⇐⇒ g(z)− 1 is not invertible~ww(∗) [proof below]

∃ solution 0 6= Ψ ∈ H of f(z)Ψ = Ψ ⇐⇒ ∃ solution 0 6= ϕ ∈ H of g(z)ϕ = ϕ

(2)

~ww ϕ=∑Nn=1 βnΨn

∃β1, . . . ,βN ∈ C (not all = 0) : βn =N∑m=1

〈γn(z), ψm〉βm~wwd(z) := det

(A(z)− 1N×N

)= 0, where A := (Anm)1≤n,m≤N , Anm(z) :=

⟨γn(z),Ψm

⟩But d : Dr → C is analytic (because Anm is so ∀m,n)

identity theorem=⇒

either Sr := z ∈ Dr : d(z) = 0 has no accumulation point in Dr or Sr = Dr =⇒claim

It remains to prove (∗), i.e. the problem is now reduced to the finite-rank case: “⇑”clear“⇓” By contradiction. Assume d(z) 6= 0 =⇒ ∀ξ ∈ CN :

(A(z) − 1N×N

)β = ξ has

a solution β ∈ CN . Choose ξn :=⟨γn(z), ψ

⟩for ψ ∈ H fixed but arbitrary =⇒

ϕ := −ψ +∑N

n=1 βnψn solves(g(z)− 1

)ϕ = ψ, because

(g(z)− 1

)ϕ = ψ − g(z)ψ +

(g(z)− 1

) N∑m=1

βmψm

(2)= ψ +

N∑n=1

ψn

[ N∑m=1

(Anm(z)− δnm

)βm −

⟨γn(z), ψ

⟩ ]︸ ︷︷ ︸

=0

=⇒ g(z)− 1 is invertible, a contradcition.

100

5.35 Corollary (Riesz-Schauder). Let T ∈ BL(H) be compact. Then σ(T ) has no accu-mulation point in C, except possibly at zero. Moreover, any 0 6= λ ∈ σ(T ) is an eigenvalueof finite multiplicity (i.e. the corresponding eigenspace is finite-dimensional).

Proof. Choose f(z) := Tz and D := C \ 0 in Thm. 5.33. Then C \ (0∪S) ⊆ ρ(T ), also

S ⊆ σp(T ) ⊆ σ(T ) ⊆ 0 ∪ S. Thus S \ 0 = σp(T ) \ 0. Also zero is the only possibleaccumulation point of S.Finite multiplicity : Suppose not (only possible for dim(H) = ∞). Then there exists aninfinite ONB Ψnn∈N of the eigenspace corresponding to some eigenvalue λ 6= 0 andTΨn = λΨn for all n. Compactness of T implies that (TΨn)n = (λΨn)n has a convergentsubsequence, which is a contradiction since λ 6= 0.

5.36 Remark. (a) If dimH = ∞ =⇒ 0 ∈ σ(T ). Proof : If not, then 0 ∈ ρ(T ), i.e.T−1 ∈ BL(H) and 1 = TT−1. Let Ψnn∈N be an ONB. Then Ψn = T (T−1Ψn) ∀n.Since T−1ψnn ⊆ H is bounded and T is compact, Ψn has a convergent subsequence.Contradiction.

(b) Corollaries 5.34 and 5.35 have generalisations to Banach spaces and to K = R.

Outlook: Spectral theorem for compact self-adjoint operators

5.37 Theorem (Hilbert-Schmidt theorem). Let T ∈ BL(H) be self-adjoint and compact.Then there exists an ONB Ψnn of H and to each Ψn and eigenvalue λn ∈ R such thatTΨn = λnΨn for all n and T =

∑n λn 〈Ψn, ·〉Ψn (convergence in operator norm).

101