MASDOC A1 Linear Partial Di erential Equationsmasdh/A1.pdf · 2010-12-09 · 4.4 Eigenfunctions as...

MASDOC A1

Linear Partial Differential Equations

James C. Robinson

Contents

1 Introduction: Hilbert spaces and PDEs page 1

1.1 Inner products, norms, and Hilbert spaces 1

1.2 Closest points in closed linear subspaces 3

1.3 Linear operators and dual spaces 5

1.4 The Riesz Representation Theorem 6

1.5 Poisson’s equation 9

1.6 ‘Variational formulation’ 12

1.7 Non-symmetric B: the Lax–Milgram Lemma 13

2 Function spaces 18

2.1 Euclidean spaces 18

2.2 The sequence spaces `p 18

2.3 Density and separability 21

2.4 Spaces of continuous and differentiable functions 22

2.4.1 Spaces of continuous functions 22

2.4.2 Multi-index notation and spaces of differentiable functions 24

iii

iv 0 Contents

2.4.3 ‘Test functions’ 24

2.4.4 Mollification 25

2.5 The Lebesgue integral on R 26

2.6 The Lebesgue spaces Lp 29

2.7 Locally integrable functions 34

2.8 Sobolev spaces 36

2.8.1 Weak derivatives 36

2.8.2 Sobolev spaces 38

2.8.3 Completions 39

2.8.4 Density results for Hk(Ω) 41

2.8.5 Extension theorems 44

2.9 The Fourier transform and Sobolev spaces 48

2.9.1 Compact embeddings 51

2.9.2 Sobolev spaces and Lp spaces 53

2.9.3 Scaling and inequalities 54

2.9.4 Boundary values and the trace operator 55

3 Elliptic PDEs 57

3.1 Elliptic regularity 60

3.1.1 Interior regularity 60

3.1.2 Boundary regularity 64

4 Eigenvalues, eigenfunctions, and bases in Hilbert spaces 66

4.1 Eigenvalues and the spectrum 66

4.2 The spectrum of a compact symmetric operator 67

Contents v

4.3 Bases in Hilbert spaces 71

4.4 Eigenfunctions as basis elements; the Hilbert–Schmidt Theorem 74

4.5 Eigenvalues and eigenfunctions for elliptic PDEs 76

5 Linear parabolic problems 78

5.1 Banach-space valued function spaces and ‘Sobolev-type’ results 79

5.2 Weak solutions of parabolic problems 83

5.3 Galerkin approximations 84

6 Nonlinear equations, weak convergence 91

6.1 Dual spaces, the Hahn–Banach Theorem, and weak convergence 91

6.2 Weak compactness and our linear parabolic PDE 96

1

Introduction: Hilbert spaces and PDEs

To begin with, we outline a ‘simple’ approach to proving the existence and

uniqueness of solutions of Poisson’s equation with Dirichlet boundary con-

ditions:

−∆u = f with u|∂Ω = 0, (1.1)

where Ω ⊂ Rn is a smooth bounded domain. Our approach uses ideas from

the theory of Hilbert spaces, so we recall some of these ideas.

A normed vector space1 is complete if every Cauchy sequence converges2.

A complete normed space is called a Banach space.

1.1 Inner products, norms, and Hilbert spaces

Examples of Banach spaces: Rn, `p, Lp(Ω), C0(Ω), C0(Ω), C0,α(Ω), W k,p(Ω),

Hk(Ω)...

An inner product on a normed vector space is a map (·, ·) : X ×X → Ksuch that

(i) (x, x) ≥ 0 for all x ∈ X, with equality iff x = 0;

1 All vector spaces will be over K = R or C, and usually over R.2 • A norm on a vector space X is a map ‖ · ‖ : X → [0,∞) such that (i) ‖x‖ = 0 iff x = 0; (ii)‖λx‖ = |λ|‖x‖ for every λ ∈ K and x ∈ X; and (iii) ‖x+ y‖ ≤ ‖x‖+ ‖y‖ for all x, y ∈ X.• A sequence xn ∈ X is Cauchy if for every ε > 0 there exists an N such that ‖xn−xm‖ < εfor all n,m ≥ N .• A sequence xn in X converges to x ∈ X if ‖xn − x‖ → 0 as n→∞.

1

2 1 Introduction: Hilbert spaces and PDEs

(ii) (x+ y, z) = (x, z) + (y, z) for all x, y, z ∈ X;

(iii) (αx, y) = α(x, y) for all α ∈ K, x, y ∈ X; and

(iv) (x, y) = (y, x).

An inner product induces a norm via

‖x‖2 = (x, x). (1.2)

Lemma 1.1 (Cauchy-Schwarz inequality) If (·, ·) is an inner product

and ‖ · ‖ is defined as in (1.2) then

|(x, y)| ≤ ‖x‖‖y‖ for all x, y ∈ X.

In particular, ‖ · ‖ defines a norm on X.

Proof Consider

‖x− λy‖2 = (x− λy, x− λy) = ‖x‖2 − λ(y, x)− λ(x, y) + |λ|2‖y‖2 ≥ 0;

setting λ = (x, y)/‖y‖2 yields

‖x‖2 − 2|(x, y)|2

‖y‖2+|(x, y)|2

‖y‖2≥ 0

and the inequality holds.

To show that ‖ · ‖ is a norm, properties (i) and (ii) are clear. To show

(iii), note that

‖x+ y‖2 = (x+ y, x+ y) = ‖x‖2 + (x, y) + (y, x) + ‖y‖2

≤ ‖x‖2 + 2|(x, y)|+ ‖y‖2 ≤ ‖x‖2 + 2‖x‖‖y‖+ ‖y‖2

≤ (‖x‖+ ‖y‖)2.

A complete inner product space is called a Hilbert space.

Every Hilbert space is a Banach space. The converse is not true, since in

a Hilbert space H the norm must satisfy the parallelogram law1

‖x+ y‖2 + ‖x− y‖2 = 2(‖x‖2 + ‖y‖2) for all x, y ∈ H. (1.3)

1 In fact this characterises those norms that can be derived from an inner product. If a normcomes from an inner product you can recover the inner product from the norm using the‘polarisation identity’ 4(x, y) = ‖x+y‖2−‖x−y‖2 in the real case (the complex case is messier;in both cases you prove the identity by expanding the norms as inner products). If a norm ona real vector space satisfies the parallelogram law and you define a map (·, ·) : X × X → Rusing this polarisation identity, you will end up with an inner product (this is quite awkward,if not difficult, to prove).

1.2 Closest points in closed linear subspaces 3

(To prove simply expand both sides using the inner product.)

1.2 Closest points in closed linear subspaces

We now use the parallelogram law to show that in a Hilbert space, there is

always a ‘closest point’ in a closed linear subspace1.

Theorem 1.2 Let U be a closed linear subspace2 of a Hilbert space H and

let x ∈ H. Then there exists a unique a ∈ U such that

‖x− a‖ = inf‖x− a‖ : a ∈ U.

Furthermore, x− a ∈ U⊥, where

U⊥ = u ∈ H : (u, x) = 0 for all x ∈ U.

The map P : H → U given by x 7→ a is the orthogonal projection of x onto

U ; P 2 = P and ‖Px‖ ≤ ‖x‖.

Two elements x and y of an inner product space are said to be orthogonal

if (x, y) = 0. (We sometimes write x ⊥ y.) U⊥ is the orthogonal complement

of U .

Proof Set δ = inf‖x− a‖ : a ∈ A and find a sequence an ∈ A such that

‖x− an‖2 ≤ δ2 +1

n. (1.4)

We will show that an is a Cauchy sequence. To this end, we use the

parallelogram law:

‖(x− an) + (x− am)‖2 + ‖(x− an)− (x− am)‖2 = 2[‖x− an‖2 + ‖x− am‖2].

Which gives

‖2x− (an + am)‖2 + ‖an − am‖2 < 4δ2 +2

m+

2

nor

‖an − am‖2 ≤ 4δ2 +2

m+

2

n− 4‖x− 1

2(an + am)‖2.

1 Linear subspaces of infinite-dimensional spaces are not always closed. Can you find an example?However, for any set U , U⊥ is always a closed linear subspace: clearly linear, and closed sinceif xn → x with xn ∈ U⊥, then (x, u) = limn→(xn, u) = 0 for every u ∈ U .

2 There also exists a closest point in any closed convex subset of H: a set U is convex if for everyu, v ∈ U , λu+ (1− λ)v ∈ U for all λ ∈ [0, 1].


Since A is convex, an + am ∈ A, and so ‖x− 12(an + am)‖2 ≥ δ2, which gives

‖an − am‖2 ≤2

m+

2

n.

It follows that an is Cauchy, and so an → a. Since A is closed, a ∈ A.

To show that a is unique, suppose that ‖u − a∗‖ = δ with a∗ 6= a. Then

‖u− 12(a∗ + a)‖ ≥ δ since A is a linear subspace, and so, using the parallel-

ogram law again,

‖a∗ − a‖2 ≤ 4γ2 − 4γ2 = 0,

i.e. a∗ = a and a is unique.

Now consider v = x− a; the claim is that v ∈ U⊥, i.e. that

(v, y) = 0 for all y ∈ U.

Consider ‖x− (a− ty)‖ = ‖v + ty‖; then

∆(t) = ‖v + ty‖2 = (v + ty, v + ty)

= ‖v‖2 + (ty, v) + (v, ty)) + ‖y‖2

= ‖v‖2 + t(y, v) + t(y, v) + |t|2‖y‖2

= ‖v‖2 + 2 Ret(y, v)+ |t|2‖y‖2.

We know from the construction of a that ‖v+ ty‖ is minimal when t = 0. If

t is real then this implies that d∆/dt(0) = 2 Re(y, v) = 0. If t = is, with

s real, then d∆(is)/ds = −2 Im(y, v) = 0. So (y, v) = 0 for any y ∈ U ,

i.e. v ∈ U⊥ is claimed.

Clearly P 2 = P , and since

‖x‖2 = ‖a‖2 + 2(a, x− a) + ‖x− a‖2,

the inequality ‖Pu‖ ≤ ‖u‖ is clear.

Note that when U is a one-dimensional subspace1, the closest point can

be given explicitly: take any u ∈ U with ‖u‖ = 1; then for the closest point

to be tu we need

(x− tu, u) = 0 ⇒ (x, u)− t‖u‖2 = 0 ⇒ t = (x, u), (1.5)

i.e. the closest point is (x, u)u.

1 This approach can be extended to any space spanned by a finite/countable collection of or-thonormal vectors ej, with the closest point being given by

∑(x, ej)ej .

1.3 Linear operators and dual spaces 5

1.3 Linear operators and dual spaces

Much of functional analysis concerns operators - particularly linear operators

- between different spaces. An operator L : X → Y is linear if

L(αx1 + βx2) = αLx1 + βLx2 for all α, β ∈ K, x1, x2 ∈ X.

A linear operator L : X → Y is bounded if there exists a constant M such

that

‖Lx‖Y ≤M‖x‖X ; (1.6)

we denote the space of all bounded linear maps1 from X into Y by L (X,Y ),

and define the operator norm of L (as an operator from X into Y ) by

‖L‖L (X,Y ) = infM : ‖Lx‖Y ≤M‖x‖X. (1.7)

(When there is no ambiguity this is often abbreviated to ‖L‖.)

The space L (X,Y ) is a vector space (with the obvious definitions of

addition and scalar multiplication), and ‖ · ‖L (X,Y ) is a norm on this space.

What is a somewhat remarkable is that if Y is complete then so is L (X,Y ).

Theorem 1.3 If X is a normed space and Y is a Banach space, then

L (X,Y ) is complete (i.e. is a Banach space).

Proof Let Ln be a Cauchy sequence in L (X,Y ). We identify a candidate

L for the limit of the sequence Ln, then show that this is an element of

L (X,Y ), and that Ln → L (wrt the operator norm).

Since Ln is Cauchy, for any ε > 0 there exists an N such that

‖Ln − Lm‖L (X,Y ) < ε for all n,m ≥ N.

It follows that for any fixed x ∈ X,

‖Lnx− Lmx‖Y = ‖(Ln − Lm)x‖Y < ε‖x‖, (1.8)

i.e. that Lnx is a Cauchy sequence in Y . So for each x ∈ X we can define

Lx = limn→∞

Lnx.

We claim that (i) L ∈ L (X,Y ) and (ii) that Ln → L (wrt ‖ · ‖L (X,Y )).

1 There are many unbounded linear maps; in particular, differential operators...


To show (i), we first check that L is linear:

L(αx1 + βx2) = limn→∞

Ln(αx1 + βx2)

= limn→∞

[αLnx1 + βLnx2]

= α limn→∞

Lnx1 + β limn→∞

Lnx2

= αLx1 + βLx2,

and that L is bounded: take m→∞ in (1.8) to obtain

‖Lnx− Lx‖Y < ε‖x‖.

This shows that L = (L − Ln) + Ln is bounded, and that Ln converges to

L.

The dual space of a normed space X (over K) is the space X∗ = L (X,K).

We investigate the dual space of Hilbert space in the next section. When

not ambiguous for f ∈ X∗ we write

‖f‖∗ = ‖f‖X∗ = ‖f‖L (X,K).

1.4 The Riesz Representation Theorem

First we show that if H is a Hilbert space, from any y ∈ H we can construct

an element of H∗ with the same norm, using the inner product.

Example 1.4 Let U be a Hilbert space. Given any y ∈ H, define

ly(x) = (x, y). (1.9)

Then ly is clearly linear, and

|ly(x)| = |(x, y)| ≤ ‖x‖‖y‖

using the Cauchy-Schwarz inequality. It follows that ly ∈ H∗ with ‖ly‖ ≤‖y‖. Choosing x = y in (1.9) shows that

|ly(y)| = (y, y) = ‖y‖2

and hence ‖ly‖ = ‖y‖.

The Riesz Representation Theorem shows that this example can be ‘re-

versed’, i.e. every linear functional on H corresponds to some inner product:

1.4 The Riesz Representation Theorem 7

Theorem 1.5 (Riesz Representation Theorem) For every bounded lin-

ear functional f on a Hilbert space H there exists a unique element y ∈ Hsuch that

f(x) = (x, y) for all x ∈ H; (1.10)

furthermore ‖y‖H = ‖f‖H∗.

Proof Let

K = Ker(f) = x ∈ H : f(x) = 0,

which is a closed linear subspace of H. [Clearly a linear subspace; closed

since if xn ∈ Ker(f) and xn → x then

|f(x)| = |f(x)− f(xn)| = |f(x− xn)| ≤ ‖f‖∗‖x− xn‖;

this holds for all n, hence |f(x)| = 0.]

If f = 0 then K = H and we set y = 0.

Otherwise, if f 6= 0 then K 6= H; it follows from Theorem 1.2 that K⊥ is

non-empty. In fact, K⊥ is a one-dimensional linear subspace of H. Indeed,

given u, v ∈ K⊥ we have

f(f(u)v − f(v)u) = 0. (1.11)

Since u, v ∈ K⊥ it follows that f(u)v−f(v)u ∈ K⊥, while (1.11) shows that

f(u)v − f(v)u ∈ K. It follows1 that

l(u)v − l(v)u = 0,

and so u and v are proportional.

Therefore we can choose z ∈ K such that ‖z‖ = 1, and decompose any

x ∈ H as2

x = (x, z)z + w with w ∈ K.

(We are using Theorem 1.2 along with the observation in 1.5.) Therefore

l(x) = (x, z)l(z) = (x, l(z)z).

Setting y = l(z)z we obtain (1.10).

To show that this choice of y is unique, suppose that

(x, y) = (x, y) for all x ∈ H.

Then (x, y − y) = 0 for all x ∈ H, i.e. y − y ∈ H⊥ = 0.1 We always have U ∩ U⊥ = 0: if x ∈ U and x ∈ U⊥ then ‖x‖2 = (x, x) = 0.2 If U is a closed linear subspace of H then (U⊥)⊥ = U , which we are in fact using here...


Finally, the calculation in (1.4) shows the equality of the norms of y and

f .

The map f 7→ y is known as the Riesz mapping. Note that it is a linear

isometry (this just means that ‖f‖∗ = ‖y‖); as is its inverse. It follows that

H and H∗ are (isometrically) isomorphic (which we write as H ' H∗).

The following useful corollary is almost immediate. We say that a map

B : H ×H → R is a bilinear form if B(u, v) is linear in both u and v, i.e.

B(αu1 + βu2, v) = αB(u1, v) + βB(u2, v) for all α, β ∈ R, u1, u2, v ∈ H

and

B(u, αv1 + βv2) = αB(u, v1) + βB(u, v2) for all α, β ∈ R, u, v1, v2 ∈ H.

Corollary 1.6 Suppose that B : H ×H → R is a bilinear form that is

(i) bounded, i.e. there exists an α ≥ 0 such that

|B(u, v)| ≤ α‖u‖‖v‖ for all u, v ∈ H,

(ii) coercive, i.e. there exists a β > 0 such that

B(u, u) ≥ β‖u‖2 for all u ∈ H,

and

(iii) symmetric,

B(u, v) = B(v, u) for all u, v ∈ H.

Then for any f ∈ H∗ there exists a unique u ∈ H such that

B(u, v) = f(v) for all v ∈ H, (1.12)

and

‖u‖ ≤ ‖f‖∗β

. (1.13)

Proof B is an alternative inner product on H; from (ii) B(u, u) ≥ 0 with

equality iff u = 0; B satisfies the linearity requirements of an inner product;

and B is symmetric by assumption. We can immediately apply the Riesz

Representation Theorem to deduce that existence of a unique u satisfying

(1.12). The bound on u in (1.13) follows immediately from (ii), since

β‖u‖2 ≤ B(u, u) = f(u) ≤ ‖f‖∗‖u‖.


In a theorem below (the Lax–Milgram Lemma) we will remove the re-

quirement that B is symmetric.

1.5 Poisson’s equation

We return to Poisson’s equation

−∆u = f x ∈ Ω, u|∂Ω = 0.

Multiply this equation by a C∞ function ϕ that is zero on the boundary,

and integrate over Ω to obtain

−∫

Ωϕ(x)∆u(x) dx =

∫Ωf(x)ϕ(x) dx.

Now integrate the first term by parts1 to obtain∫Ω∇ϕ · ∇udx =

∫Ωfϕdx. (1.14)

We want to recast this in a form suitable to apply Corollary 1.6. To do

this we require an appropriate Hilbert space.

Let

L2(Ω) = f ∈ C∞(Ω) :

∫Ω|f(x)|2 dx <∞;

we define the L2(Ω) norm of f to be

‖f‖2L2 =

(∫Ω|f(x)|2 dx

)1/2

; (1.15)

this is induced by the L2-inner product

(f, g)L2 =

∫Ωf(x)g(x) dx. (1.16)

It is straightforward to check that (·, ·)L2 is indeed an inner product (and

1 This is the divergence theorem:∫Ω∇ · (ϕ∇u) dx =

∫∂Ω

ϕ∇u · n dS;

the right-hand side is zero since ϕ = 0 on ∂Ω, and the left-hand side is∫Ω∇ϕ · ∇u dx+

∫Ωϕ∆u dx.


hence that ‖·‖L2 is indeed a norm). The space L2(Ω) is in fact not complete

with respect to this norm (exercise: find a sequence in L2(Ω) that is Cauchy

with respect to this norm but does not have a limit in L2(Ω)); its completion

is the Lebesgue space L2(Ω) which we will study in some detail later.

We also let

H10 (Ω) = f ∈ C∞(Ω) : f |∂Ω = 0,

∫Ω|f(x)|2 + |∇f(x)|2 dx <∞,

and define the H1 norm of f to be given by

‖f‖2H1 =

∫Ω|f(x)|2 + |∇f(x)|2 dx = ‖f‖2L2 + ‖∇f‖2L2 ;

this can be induced by the inner product

((f, g))H1 = (f, g)L2 +

n∑j=1

(∂jf, ∂jg)L2 =: (f, g)L2 + (∇f,∇g)L2 .

Again, H10 (Ω) is not in fact complete with respect to this norm; its comple-

tion is the Sobolev space H10 (Ω).

We can now rewrite (1.14) as

(∇u,∇ϕ)L2 = (f, ϕ)L2 .

We want to consider this as an equation in the Hilbert space H10 to which

we can apply Corollary 1.6.

First we note that (f, ϕ)L2 maps H10 ⊂ L2 into R, and is clearly linear in

ϕ; it is also a bounded mapping, since

|(f, ϕ)L2 | ≤ ‖f‖L2‖ϕ‖L2 ≤ ‖f‖L2‖ϕ‖H1 ,

using the Cauchy-Schwarz inequality in L2 and the definition of the H1

norm.

Now, if we define

B(u, v) = (∇u,∇v)L2

for u, v ∈ H10 , then this is clearly a symmetric bilinear form. It is bounded

since

|B(u, v)| = |(∇u,∇v)L2 | ≤ ‖∇u‖L2‖∇v‖L2 ≤ ‖u‖H1‖v‖H1 ,

again using the Cauchy-Schwarz inequality in L2 and the definition of the

H1 norm.


However, the coercivity is more delicate. For this we require the following

Poincar’e inequality :

Lemma 1.7 (Poincare’s inequality) Suppose that Ω is bounded in one

direction, i.e. |x1| ≤ d <∞. Then for every f ∈ H10

‖f‖L2 ≤ 2d‖∇f‖L2 . (1.17)

(The result actually holds in H10 (Ω) with essentially the same proof, as we

will see; note that it follows from the proof that in fact we do not need

all the derivatives on the right-hand side, only the one derivative in the x1

direction.)

Proof The inequality is trivial if f = 0, so we assume that f 6= 0, and hence

that ‖f‖L2 6= 0. We integrate by parts in the x1 variable:

‖f‖2L2 =

∫Ω

1 · |f(x)|2 dx = −∫

Ωx1

∂

∂x1|f(x)|2 +

∫∂Ωx1|u(x)|2n1 dS,

where n1 is the component of the outward normal in the x1 direction. Using

the fact that u = 0 on ∂Ω the boundary term vanishes, and hence

‖f‖2L2 = −2

∫Ωx1f(x)

∂

∂x1f(x) dx

≤ 2

∫Ω|x1||f(x)|

∣∣∣∣ ∂∂x1f(x)

∣∣∣∣ dx

≤ 2d

∫Ω|f(x)||∇f(x)|dx

≤ 2d

(∫Ω|f(x)|2 dx

)1/2(∫Ω|∇f(x)|2 dx

)1/2

= 2d‖f‖L2‖∇f‖2L2 ,

where we have used the Cauchy-Schwarz inequality (in L2) and the fact that

|∂f/∂x1| ≤ |∇f |; (1.17) follows if we divide through by ‖f‖L2 .

Two norms ‖ · ‖1 and ‖ · ‖2 on a space X are said to be equivalent if there

exist constants c2 > c1 > 0 such that

c1‖x‖1 ≤ ‖x‖2 ≤ c2‖x‖1 for all x ∈ X.

Corollary 1.8 If Ω is bounded in one direction then the norms ‖f‖H1 and

‖∇f‖L2 are equivalent on H10 (Ω).


Proof

‖∇f‖2L2 ≤ ‖f‖2L2 + ‖∇f‖2L2︸︷︷︸‖u‖2

H1

≤ (4d2 + 1)‖∇f‖2.

We have therefore shown that B(u, u) is coercive on H10 :

B(u, u) = ‖∇u‖2L2 ≥1

1 + 4d2‖u‖2H1 ,

and hence all the requirements of Corollary 1.6 are satisfied, from which we

can deduce that there is a unique solution u ∈ H10 (Ω) of the problem

B(u, v) = f(v) for all v ∈ H10

with ‖u‖H1 ≤ c‖f‖L2 .

There are some gaps/problems in this argument, mainly arising from the

fact that we haven’t defined any function spaces properly:

(i) We derived the equation

B(u, ϕ) = (f, ϕ)

under the assumption that everything was smooth; in particular, that

ϕ was a C∞ function zero on ∂Ω.

(ii) The spaces L2 and H10 are not complete, so not Hilbert spaces.

(iii) We would really like a classical (at least C2) solution of the problem,

but this method only gives us a u ∈ H10 .

1.6 ‘Variational formulation’

PDEs are closely related to certain problems in the ‘calculus of variations’

(i.e. minimisation/maximisation problems). Here we will show that the ab-

stract formulation of our PDE,

find u ∈ H such that B(u, v) = f(v) for all v ∈ H, (1.18)

is the same as the following minimisation problem:

find u ∈ H such that J(u) ≤ J(v) for all v ∈ H, (1.19)

where J(u) = 12B(u, u)− f(u).


Theorem 1.9 Under the conditions on B and f imposed in Corollary 1.6

the two problems (1.18) and (1.19) are equivalent.

Proof First, suppose that we have a solution u ∈ H of (1.18), and for an

arbitrary v ∈ H write

J(v) = J(u+ (v − u))

=1

2B(u, u)− f(u) +B(u, v − u)− f(v − u) +

1

2B(v − u, v − u)

= J(u) + [B(u, v − u)− f(v − u)]︸︷︷︸= 0 since u solves (1.18)

+1

2B(u− v, u− v)

≥ J(u)

since B(u− v, u− v) ≥ β‖u− v‖2 ≥ 0 using coercivity of B. So u minimises

J .

On the other hand, if u minimises J then for any v ∈ H consider

J(u+ tv) =1

2B(u+ tv, u+ tv)− f(u+ tv)

=1

2B(u, u)− f(u) + t(B(u, v)− f(v)) +

1

2t2B(v, v).

Since we must have J(u+ tv) ≥ J(u) for every t, it follows that

B(u, v) = f(v).

Since v ∈ H was arbitrary this shows that u solves (1.18).

Corollary 1.10 Under the conditions of Corollary 1.6, the minimisation

problem (1.19) has a unique solution u ∈ H.

1.7 Non-symmetric B: the Lax–Milgram Lemma

We now prove the Lax–Milgram Lemma, a version of Corollary 1.6 valid

when B is not symmetric. In fact the argument is somewhat surprising in

the light of where we began, namely a linear PDE, which we could write as

Lu = f,

where L is some linear operator. By taking inner products we turned this

into an abstract problem involving a bilinear form B : H ×H → R,

B(u, v) = f(v) for all v ∈ H.


The main ‘trick’ in the proof of the Lax–Milgram Lemma is to turn this

problem back into a linear equation; properties of the bilinear form then

guarantee the existence of a solution of this linear equation.

Theorem 1.11 (Lax–Milgram Lemma) Suppose that B : H × H → Ris a bilinear form that is

(i) bounded, i.e. there exists an α ≥ 0 such that

|B(u, v)| ≤ α‖u‖‖v‖ for all u, v ∈ H

and

(ii) coercive, i.e. there exists a β > 0 such that

B(u, u) ≥ β‖u‖2 for all u ∈ H.

Then for any f ∈ H∗ there exists a unique uf ∈ H such that

B(uf , v) = f(v) for all v ∈ H. (1.20)

Furthermore

‖uf‖ ≤ β−1‖f‖∗; (1.21)

in particular uf depends continuously on f , i.e.

‖uf − ug‖ ≤ β−1‖f − g‖∗.

Proof Once we have a solution it is clearly unique: if

B(u, v) = B(u, v) = f(v)

for every v ∈ H then B(u − u, v) = 0 for every v ∈ H (since B is bilinear)

and in particular for v = u− u, whence

β‖u− u‖2 ≤ B(u− u, u− u) = 0,

i.e. u = u. The bound in (1.21) follows as before (set v = uf in (1.20)) and

the continuity result follows by considering

B(uf , v)−B(ug, v) = B(uf − ug, v) = (f − g, v)

and setting v = uf − ug. So only existence requires any work.

Fix u ∈ H, and consider the map v 7→ B(u, v). We claim that this defines

a bounded linear functional on H: it is clearly linear, since

B(u, αv1 + βv2) = αB(u, v1) + βB(u, v2)


be the linearity of B, and it is bounded since

|B(u, v)| ≤[α‖u‖

]‖v‖.

It follows from the Riesz Representation Theorem that there exists a w ∈ Hsuch that

(w, v) = B(u, v) for all v ∈ H.

We define Au = w, i.e. by definition

(Au, v) = B(u, v),

and claim that this definition yields a bounded linear operator from H into

itself.

Indeed, for every v ∈ H,

(A(αu1 + βu2), v) = B(αu1 + βu2, v)

= αB(u1, v) + βB(u2, v)

= α(Au1, v) + β(Au2, v)

= (αAu1 + βAu2, v).

Since this holds for every v ∈ H, it follows1 that

A(αu1 + βu2) = αAu1 + βu2,

i.e. A is linear. To show that A is bounded, note that

‖Au‖2 = (Au,Au) = B(u,Au) ≤ α‖u‖‖Au‖,

so that ‖Au‖ ≤ α‖u‖ and A is bounded.

Using the Riesz Representation Theorem in a standard way, we know that

there exists a ϕ ∈ H such that

(ϕ, v) = f(v) for all v ∈ H.

We can therefore rewrite our equation as

(Au, v) = (ϕ, v) for all v ∈ H.

This implies that u satisfies (1.20) iff Au = ϕ; we have regained a linear

equation from the formulation in terms of a bilinear form. We now have to

show that this equation has a solution.

1 If (u, v) = (u, v) for every v ∈ H, then (u−u, v) = 0 for every v ∈ H, in particular for v = u−u,whence u = u.


Method 1: coercivity implies that A is invertible

Now, not only is A ‘bounded above’; the coercivity implies that it is also

bounded below:

β‖u‖2 ≤ B(u, u) = (Au, u) ≤ ‖Au‖‖u‖ ⇒ β‖u‖ ≤ ‖Au‖.

As a consequence A is one-to-one (if Au = Av then u = v since ‖u − v‖ ≤β−1‖Au−Av‖) and onto. To show that A is onto, first note that

R(A) = Au : u ∈ H

is a closed linear subspace of H. It is clearly a linear subspace, and it is

closed since if vn ∈ R(A) (so that vn = Aun, un ∈ H) and vn → v, then

‖un − um‖ ≤ β−1‖Aun −Aum‖ = β−1‖vn − vm‖,

so that un is Cauchy. Since H is complete, un → u ∈ H, and since A is

bounded it is continuous: it follows that Au = v, i.e. v ∈ R(A), so R(A) is

closed.

So now suppose that R(A) 6= H. It follows that there exists a non-zero

x ∈ H with x ∈ (R(A))⊥. Thus

β‖x‖2 ≤ B(x, x) = (Ax, x) = 0,

a contradiction. Since A is one-to-one and onto, it must have an inverse, i.e.

we can find a solution of Au = ϕ for any ϕ ∈ H.

Method 2: contraction mapping argument

Clearly, for any % > 0,

Au = ϕ ⇔ u = u− %(Au− ϕ).

We use the contraction mapping theorem1, applied to the map

Tu = u− %(Au− ϕ).

We have

‖Tu− Tv‖2 = ‖(u− v)− %A(u− v)‖2

= ‖u− v‖2 − 2%(A(u− v), u− v) + %2‖A(u− v)‖2

= ‖u− v‖2 − 2%B(u− v, u− v) + %2‖A(u− v)‖2

≤ ‖u− v‖2 − 2%β‖u− v‖2 + %2α2‖u− v‖2

= (1− 2%β + %2α2)‖u− v‖2,1 Let (X, d) be a complete metric space, and T : X → X a map such that d(Tx, Ty) ≤ θd(x, y)

for some θ < 1. Then T has a unique fixed point, i.e. there exists a unique x ∈ X such thatTx = x; furthermore for any initial x0, the iterates Tkx0 converge to the fixed point x. [Toprove this, show that Tkx forms a Cauchy sequence, and use the completeness of X; thelimit is the fixed point.]


where we have used the coercivity of u and the fact that A is bounded. It

follows that if we choose % sufficiently small, T is a contraction. It therefore

has a unique fixed point, which provides our solution.

2

Function spaces

We now need to introduce various function spaces so that we can perform

our analysis rigorously.

2.1 Euclidean spaces

Rn consists of all n-tuples x = (x1, . . . , xn). We use the standard Euclidean

norm

|x| =

n∑j=1

|xj |21/2

and inner product

x · y =n∑j=1

xjyj .

Rn is complete.

2.2 The sequence spaces `p

The sequence spaces `p, 1 ≤ p <∞ consist of all sequences x = (x1, x2, . . .)

for which the `p norm

‖x‖`p :=

∞∑j=1

|xj |p1/p

18

2.2 The sequence spaces `p 19

is finite. In `2 the norm can be derived from the inner product

(x, y) =∞∑j=1

xjyj .

The space `∞ consists of all bounded sequences, with norm

‖x‖`∞ := supj|xj |; (2.1)

this space is slightly ‘odd’ (as, in some ways, is `1). Some work is required

to show that these are norms.

We say that two indices 1 ≤ p, q ≤ ∞ are conjugate if

1

p+

1

q= 1. (2.2)

The following simple inequality is fundamental.

Lemma 2.1 (Young’s inequality) Let a, b > 0 and let (p, q) be conjugate

indices with 1 < p, q <∞. Then

ab ≤ ap

p+bq

q. (2.3)

Proof The function ex is convex1, i.e. eλx+(1−λ)y ≤ λex + (1− λ)ey, and so

ab = exp(log a+ log b) = exp

(1

plog ap +

1

qlog bq

)≤ 1

pelog(ap) +

1

qelog(bq) =

ap

p+bq

q.

Lemma 2.2 (Holder’s inequality in `p spaces) Let x ∈ `p and x ∈ `qwith p, q conjugate, 1 ≤ p, q ≤ ∞. Then if z = (x1y1, x2y2, . . .), z ∈ `1 with

‖z‖`1 =∞∑j=1

|xjyj | ≤ ‖x‖`p‖y‖`q . (2.4)

1 A twice differentiable function on an interval (a, b) is convex on (a, b) iff its second derivativeis non-negative, see examples.

20 2 Function spaces

Proof For 1 < p <∞, consider

n∑j=1

|xj |‖x‖`p

|yj |‖y‖`q

≤n∑j=1

1

p

|xj |p

‖x‖p`p+

1

q

|yj |q

‖y‖q`q≤ 1.

So for each n ∈ Nn∑j=1

|xjyj | ≤ ‖x‖`p‖y‖`q

and (2.4) follows. For p = 1, q =∞,

n∑j=1

|xjyj | ≤n

maxj=1|yj |

n∑j=1

|xj |

≤ ‖x‖`1‖y‖`∞ .

Lemma 2.3 (Minkowski’s inequality in `p spaces) If x, y ∈ `p then

x+ y ∈ `p and

‖x+ y‖`p ≤ ‖x‖`p + ‖y‖`p .

Proof The cases p = 1,∞ are easy. For 1 ≤ p < ∞ let q be the conjugate

exponent to p, and note that (p−1)q = p. Since |xj +yj |p ≤ 2p(|xj |p+ |yj |p)clearly x+ y ∈ `p; now use Holder’s inequality to write

n∑j=1

|xj + yj |p ≤n∑j=1

|xj + yj |p−1|xj |+n∑j=1

|xj + yj |p−1|yj |

≤

∞∑j=1

|xj + yj |(p−1)q

1/q n∑j=1

|xj |p1/p

+

n∑j=1

|yj |p1/p

≤

n∑j=1

|xj + yj |p1/q

(‖x‖`p + ‖y‖`p),

and so n∑j=1

|xj + yj |p1−(1/q)

≤ ‖x‖`p + ‖y‖`q ,

from whence the triangle inequality in `p follows.

2.3 Density and separability 21

Proposition 2.4 For each 1 ≤ p ≤ ∞, the sequence space `p (equipped with

its standard norm) is complete.

For a proof see Problems 2.

Thus every `p is a Banach space, and `2 is a Hilbert space. Every (infinite-

dimensional) separable Hilbert space is isometrically isomorphic to `2.

Since the norm on `2 is the natural generalisation of the norm on Rn,

and since it is complete, it is tempting to think that `2 will behave just like

Rn. However, it does not have the ‘Bolzano-Weierstrass property’ (bounded

sequences have a convergent subsequence) as we can see easily by considering

the sequence ej∞j=1, where ej consists entirely of zeros apart from a 1 in

the jth position. Then clearly ‖ej‖`2 = 1 for all j; but if i 6= j then

‖ei − ej‖2`2 = 2,

i.e. any two elements of the sequence are always√

2 away from each other.

It follows that no subsequence of the ej can form a Cauchy sequence, and

so there cannot be a convergent subsequence.

This is really the first time we have seen a significant difference between Rnand the abstract normed vector spaces that we have been considering. The

failure of the Bolzano-Weierstrass property is in fact a defining characteristic

of infinite-dimensional spaces.

2.3 Density and separability

A set A is dense in a normed space (X, ‖ · ‖) if every x ∈ X can be approx-

imated arbitrarily closely by an element of A: given x ∈ X and ε > 0 there

exists an a ∈ A such that

‖x− a‖ < ε.

When X is complete, this is equivalent to requiring that X = A, the closure

of A (wrt the norm ‖ · ‖).

If a space has a countable dense subset, it is called separable.

Rn and `p, 1 ≤ p <∞ are separable; however, `∞ is not separable.

Proposition 2.5 `p, 1 ≤ p <∞, is separable.


Proof Let Qσ be the collection of sequence that have only a finite number of

non-zero terms, each of which is an element of Q. Then Qσ is countable (it

is a countable union of countable sets, i.e. those with ≤ n non-zero terms),

and dense in `p: given x ∈ `p and ε > 0, choose N such that

∞∑j=N+1

|xj |p < εp/2,

and then choose y ∈ Qσ such that yj = 0 for j ≥ N+1 and |yj−xj | < εp/2N

for j < N . It follows that ‖x− y‖`p < ε.

Proposition 2.6 `∞ is not separable.

Proof Suppose that A is dense in `∞. Note that the uncountable set L =

0, 1∞ is a subset of `∞. Choose ε < 1/2; then we claim that for every

element of L there exists a distinct element of A; indeed, if x, y ∈ L with

x 6= y then there exists an index j such that xj 6= yj ; wlog xj = 1 and

yj = 0. It follows that if a, b ∈ A with

‖a− x‖`∞ < ε and and ‖b− y‖`∞ < ε

then aj > 1/2 and bj < 1/2, i.e. aj 6= bj and hence a 6= b. It follows that A

is uncountable, and hence `∞ cannot be separable.

The subspace c0 of `∞, consists of all sequences that converge to zero,

c0 = x : xj → 0

equipped with the `∞ norm is still complete, and is separable.

2.4 Spaces of continuous and differentiable functions

Let Ω be an open subset of Rn (which could be Rn itself). We will be

interested in a variety of spaces of continuous functions (and functions with

continuous derivatives). Note that not all of these spaces are complete.

2.4.1 Spaces of continuous functions

The space C0(Ω) consists of all (real-valued) continuous functions on Ω.


The space C0(Ω) consists of all real-valued continuous functions on Ω, the

closure of Ω.

Theorem 2.7 The space C0(Ω) is complete when equipped with the supre-

mum norm,

‖f‖∞ = supx∈Ω

|f(x)| = supx∈Ω|f(x)|.

Proof Let fn be a Cauchy sequence wrt the sup norm. Then given ε > 0,

there exists an N such that

supx∈Ω

|fn(x)− fm(x)| < ε for all n,m ≥ N.

It follows that for each x ∈ Ω,

|fn(x)− fm(x)| < ε for all n,m ≥ N, (2.5)

i.e. fn(x) is a Cauchy sequence (but note that N does not depend on x).

Define a function f : Ω→ R by

f(x) = limn→∞

fn(x).

Then f is in fact the uniform limit on the fk; letting m→∞ in (2.5) shows

that for each fixed x ∈ Ω,

|fn(x)− f(x)| < ε for all n ≥ N,

where again N does not depend on x, whence

supx∈Ω

|fn(x)− f(x)| < ε for all n ≥ N.

It is then standard that f (the uniform limit of continuous functions) is

continuous itself.

The space C0(Ω) is not complete with the sup norm.

Using the Weierstrass Approximation Theorem (polynomials are dense in

the space of continuous functions on a compact set, see Problems sheet), it

follows immediately that:

Theorem 2.8 The space C0(Ω) is separable.


2.4.2 Multi-index notation and spaces of differentiable functions

A n-dimensional multi-index is an n-tuple α = (α1, . . . , αn) of integers. We

write

|α| = α1 + α2 + · · ·+ αn.

If x ∈ Rn is a vector; or ∂ = (∂1, ∂2, · · · , ∂n) is a vector of operators (we

abbreviate ∂/∂xj to ∂j) then

xα = xα11 · · ·x

αnn , ∂α = ∂α1

1 · · · ∂αnn =

∂|α|

∂xα11 · · · ∂x

αnn.

For example, in R2,∑|α|=2

∂αf =∂2f

∂x21

+ 2∂2f

∂x1∂x2+∂2f

∂x22

;

and Leibniz’s rule becomes

∂α(fg) =∑β≤α

(α

β

)∂α−βf ∂βγg,

where (α

β

)=

α!

(α− β)!β!,

α! = α1! · · ·αn!, and β ≤ α if βi ≤ αi for all i = 1, . . . , n.

The space Ck(Ω) consists of all real-valued functions

f : Ω→ R| ∂αf ∈ C0(Ω), |α| ≤ k,

i.e. all derivatives up to order k exist and are continuous on Ω.

With a similar definition for Ck(Ω), it is relatively straightforward to show

that Ck(Ω) is complete wrt the Ck norm

‖f‖Ck =∑|α|≤k

‖∂αf‖∞.

2.4.3 ‘Test functions’

Extremely useful are functions that are infinitely differentiable and have

compact support. The support of a function f : Ω → R is the smallest


closed set containing

x ∈ Ω : f(x) 6= 0.

We write C∞(Ω) = ∩k≥0Ck(Ω) [and similarly for Ω]. Then space C∞c (Ω)

of ‘test functions in Ω (beware, this is not a terminology that is used in any

way consistently) consists of all C∞ functions with compact support in Ω –

since Ω is open, the support of f will be bounded away from ∂Ω (and this

is often extremely helpful).

One could make similar definitions with less differentiability, e.g. C0c (Ω),

continuous functions with compact support in Ω. This space is not complete

with the sup norm.

2.4.4 Mollification

We now introduce the important idea of mollification. Essentially this is

a way of obtaining a very smooth function from a less smooth function by

averaging. We choose a non-negative function ρ ∈ C∞c (Rn) whose support

is contained in B(0, 1) = x : |x| ≤ 1 and which integrates to one, for

example

ρ(x) =c exp

(1

|x|2−1

), |x| ≤ 1, 0, |x| ≥ 1,

,

where c is chosen such that ∫Rn

ρ(x) dx = 1.

We write ρh(x) = h−nρ(x/h) (the support of ρh is contained in a ball about

the origin of radius h, and it still integrates to 1), and define the mollification

of f , fh, as the convolution of ρh with f ,

fh(x) = (ρh ? f)(x) = h−n∫Rn

ρ

(x− yh

)f(y) dy.

[If we start with a function u ∈ C0(Ω), say, then we simply extend u by zero

outside Ω; equivalently, we restrict the range of integration to Ω.]

Proposition 2.9 (Mollification of functions in C0c (Ω).) Let f ∈ C0

c (Ω).

Then fh ∈ C∞c (Ω) if h < dist(suppu, ∂Ω), and fh → f uniformly in Ω as

h→ 0.


Proof If h < dist(supp f, ∂Ω) then (i) we can differentiate under the integral

sign with respect to x to show that fh is in C∞, and (ii) the support of fhmust be contained within an h-neighbourhood of the support of f , and hence

within a compact subset of Ω.

To show convergence note that

fh(x)− f(x) = h−n∫ρ

(x− yh

)[f(y)− f(x)] dy,

since∫ρh = 1. Then clearly

|fh(x)− f(x)| ≤ sup|y−x|≤h

|f(y)− f(x)|.

Since the support of f is compact, f is uniformly continuous on Ω; given an

ε > 0 there exists a δ > 0 such that |x−y| < δ implies that |f(x)−f(y)| < ε,

and hence for h < δ, |fh(x)− f(x)| < ε, uniformly over Ω.

2.5 The Lebesgue integral on R

Here is a very cursory outline of the construction of the Lebesgue integral

on R. For more details see Priestley’s book Introduction to Integration or

my Functional Analysis I notes.

A subset U of R has zero measure if for every ε > 0 there exists a collection

of intervals (aj , bj) such that

U ⊂∞⋃j=1

(aj , bj)

and∞∑j=1

|bj − aj | < ε.

The class Lstep(R) of step functions consists of all those functions s(x) that

are non-negative and piecewise constant on a finite number of intervals, i.e.

s(x) =n∑j=1

cjχ[Ij ](x),

where cj ≥ 0, each Ij is an interval, and χ[A] denotes the characteristic

2.5 The Lebesgue integral on R 27

function of A,

χ[A](x) =

1 x ∈ A0 x /∈ A.

We define the integral of s(x) by∫s =

n∑j=1

cj |Ij |,

where |(aj , bj)| = (bj − aj). It is tedious but elementary to check that this

gives a well-defined integral on Lstep(R).

Now, if sn is a monotonically increasing sequence of functions in Lstep(R),

i.e. sn+1(x) ≥ sn(x) for each x ∈ R then the sequence∫sn

is also monotonically increasing. So provided that these integrals are bounded

above,

limn→∞

∫sn

must exist. A key point – whose proof is fairly technical – is that under

these conditions, the functions sn converge almost everywhere, i.e. except

on a set of measure zero, to a function f .

Denote the set of all functions that can be produced in this way (the

almost everywhere limit of an increasing sequence of step functions whose

integrals are uniformly bounded above) by Linc(R); for every f ∈ Linc(R),

with f(x) = limn→∞ sn(x) almost everywhere, we define∫f = lim

n→∞

∫sn.

Again, one has to show that this definition is unambiguous (and this is not

straightforward).

Finally, we let L1(R) be the collection of all functions f = g − h, where

g, h ∈ Linc(R), and for such an f define∫f =

∫g −

∫h.

We have to check that this is well-defined (this time it’s easy), and can then

show that this integral – the Lebesgue intergal – has the following desirable

properties.


(L) (Linearity) If f1, f2 ∈ L1(R) and λ ∈ R then f1 + λf2 ∈ L1(R) and∫f1 + λf2 =

∫f1 + λ

∫f2.

(P) (Positivity) If f ∈ L1(R) and f ≥ 0 almost everywhere then∫f ≥ 0.

(M) (Modulus property) If f ∈ L1(R) then1 |f | ∈ L1(R) and∣∣∣∣∫ f

∣∣∣∣ ≤ ∫ |f |.(T) (Translation invariance) If f ∈ L1(R) and fd(x) = f(x + d) then fd ∈

L1(R) and∫R f =

∫R fd.

Theorem 2.10 (Monotone Convergence Theorem) If fn ∈ L1(R) with

0 ≤ f1(x) ≤ f2(x) ≤ · · · for a.e. x ∈ R

and ∫fn ≤M

for some M > 0 then there exists an f ∈ L1(R) such that fn(x) converges

to f(x) almost everywhere and∫f = lim

n→∞

∫fn.

Suppose that f ≥ 0 and∫f = 0. Consider fn(x) = nf(x); then fn+1 ≥

fn, and∫fn = n

∫f = 0, so by the MCT fn(x)→ g(x) almost everywhere,

with g(x) ∈ L1. Since g(x) = +∞ where f 6= 0, and g is defined almost

everywhere (since it is in L1), it follows that f = 0 almost everywhere.

This shows that if∫|f | = 0 then one can only deduce that f = 0 almost

everywhere.

Theorem 2.11 ((Lebesgue’s) Dominated Convergence theorem) If

fn ∈ L1(R) converge pointwise almost everywhere to f , and there is a func-

tion g ∈ L1(R) such that |fn(x)| ≤ g(x) for every n and almost every x ∈ R,

then f ∈ L1(R) and ∫f = lim

n→∞

∫fn.

1 This does imply that there are some functions whose ‘integral’ exists in some sense, e.g.∫∞1

1x

sinxdx, which are not in L1(R).


We say that f ∈ L1(a, b) if f |(a,b) = fχ[(a, b)] ∈ L1(R).

Note that any integrable function f ∈ L1(R) is given as f = g − h where

g, h ∈ Linc(R). We have g = limn→∞ gn and h = limn→∞, where gnand hn are monotonic sequences of functions in Lstep, converging almost

everywhere. This implies, trivially, that f is the limit almost everywhere of

the (not necessarily monotonic) sequence of step functions fn − gn. Such

a function (the almost everywhere limit of a sequence of step functions) is

called measurable.

2.6 The Lebesgue spaces Lp

The Lebesgue space Lp(Ω) consists of all those functions that are measurable

and whose pth power is Lebesgue integrable;

Lp(Ω) =

f : f is measurable and

∫Ω|f |p <∞

,

and is a Banach space when equipped with the Lp norm

‖f‖Lp =

(∫Ω|f |p

)1/p

. (2.6)

We now show (i) that this is a norm and (ii) that Lp is complete wrt this

norm.

The space L∞(Ω) consists of all measurable functions on Ω such that

‖f‖L∞ := ess supx∈Ω|f(x)| <∞;

where the essential supremum is

infM : |f(x)| ≤M for all x ∈ Ω \ E, where E has measure zero.

Lemma 2.12 (Holder’s inequality) Let 1 ≤ p, q ≤ ∞ be conjugate in-

dices. Suppose that f ∈ Lp and g ∈ Lq. Then fg ∈ L1 and

‖fg‖L1 ≤ ‖f‖Lp‖g‖Lq . (2.7)

Proof For 1 < p, q <∞,∫|f(x)|‖f‖Lp

|g(x)|‖g‖Lq

≤ 1

p

∫|f(x)|p

‖f‖pLp

+1

q

∫|g(x)|q

‖g‖qLq

=1

p+

1

q= 1.


For p = 1, q =∞ the inequality is clear:∫|f(x)||g(x)| ≤ ‖g‖L∞

∫|f(x)| = ‖f‖L1‖g‖L∞ .

One can generalise Holder’s inequality to treat the product of three or

more functions. The three-term version is sometimes useful: if f ∈ Lp,

g ∈ Lq, and h ∈ Lr with p−1 + q−1 + r−1 = 1, then fgh ∈ L1 with∫|fgh| ≤ ‖f‖Lp‖g‖Lq‖h‖Lr .

For a proof see the examples sheet.

The Lp interpolation inequality is also useful: take p < r < q with

1

r=α

p+

1− αq

;

then if f ∈ Lp and f ∈ Lq,

‖f‖Lr ≤ ‖f‖αLp‖f‖1−αLq .

Again, the proof is a simple application of Holder’s inequality (see examples).

Lemma 2.13 (Minkowski’s inequality) For 1 ≤ p ≤ ∞, f ∈ Lp and

g ∈ Lp then f + g ∈ Lp and

‖f + g‖Lp ≤ ‖f‖Lp + ‖g‖Lp .

Proof First note that |f(x)+g(x)|p ≤ 2p(|f(x)|p+ |g(x)|p), since f+g ∈ Lp.Also, if (p, q) are conjugate then (p − 1)q = p; so we can apply Holder’s

inequality to obtain

‖f + g‖pLp =

∫|f + g|p ≤

∫|f ||f + g|p−1 +

∫|g||f + g|p−1

≤ ‖f‖Lp‖f + g‖p/qLp + ‖g‖Lp‖f + g‖p/qLp

≤ (‖f‖Lp + ‖g‖Lp) ‖f + g‖p/qLp ;

dividing both sides by ‖f + g‖p/qLp yields the result, since

p− p

q= p

(1− 1

q

)= 1.


We now show that Lp is complete.

Theorem 2.14 Lp(R) is complete.

Proof Suppose that fn is a Cauchy sequence in Lp. Choose n1 < n2 < · · ·such that

‖fn − fm‖Lp < 2−k for n,m ≥ nk;

then the subsequence fnj satisfies

‖fnj+1 − fnj‖Lp < 2−j .

It follows that∥∥∥∥∥∥k∑j=1

|fnj+1 − fnj |

∥∥∥∥∥∥Lp

≤k∑j=1

‖fnj+1 − fnj‖Lp ≤k∑j=1

2−j ≤ 1.

If we define

vk(x) =k∑j=1

|fnj+1(x)− fnj (x)|,

then it follows from the Monotone Convergence Theorem that vk → v almost

everywhere for some v ∈ Lp; in particular the series

k∑j=1

fnj+1(x)− fnj (x)

converges absolutely almost everywhere. Since the partial sums of this series

are just fnk+1(x)−fn1(x), it follows that fnj (x) converges almost everywhere;

at these points we set

f(x) = limj→∞

fnj (x)

and define f however we like elsewhere. All that remains is to show that

f ∈ Lp(R) and that ‖fn − f‖Lp → 0.


Now consider

|f − fnk+1|p =

∣∣∣∣∣∣(f − fn1)−k∑j=1

(fnj+1 − fnj )

∣∣∣∣∣∣p

=

∣∣∣∣∣∣∞∑

j=k+1

(fnj+1 − fnj )

∣∣∣∣∣∣p

≤

∞∑j=k+1

|fnj+1 − fnj |

p

≤ |v|p,

where v ∈ Lp. It follows that f − fnk+1∈ Lp and hence, since fnk+1

∈ Lp,that f ∈ Lp.

Finally, note that we can use the dominated convergence theorem to ob-

tain

‖f − fnk+1‖Lp =

∥∥∥∥∥∥f − fn1 −k∑j=1

(fnj+1 − fnj )

∥∥∥∥∥∥Lp

=

∥∥∥∥∥∥∞∑

j=k+1

(fnj+1 − fnj )

∥∥∥∥∥∥Lp

≤∞∑

j=k+1

‖fnj+1 − fnj‖Lp

≤ 2−(k−1),

so that f ∈ Lp (since f = (f − fnk+1) + fnk+1

and both the terms on the

right-hand side are in Lp) and fnk→ f in Lp. We have already seen (in the

first problems sheet) that if a subsequence of a Cauchy sequence converges

then the entire sequence converges, and the proof is complete.

The following useful corollary is contained in the proof.

Corollary 2.15 Suppose that fn → f in Lp. Then there is a subsequence

fnj such that fnj → f almost everywhere.

The following observation, which follows from the definition of L1, is fun-

damental.


Lemma 2.16 The space C0c (Ω) is dense in Lp(Ω) for 1 ≤ p <∞.

Proof (Sketch) Given f ∈ Lp(Ω) and ε > 0 find a compact subset K of Ω

such that

‖f − fχK‖Lp < ε/2.

The problem reduces to approximating fχK ∈ Lp(K). Write fχK = f1−f2

with f1, f2 ∈ Linc(K). The problem reduces to approximating fj to within

ε/4.

If f ∈ Linc then it is the limit of an increasing sequence sn of step func-

tions. Since sn(x) ≤ f(x) it follows that sn ∈ Lp, and since (u(x)−sn(x))p ≤u(x)p it follows that sn → u in Lp using the DCT.

Any step functions can be approximated arbitrarily closely in the Lp norm

by continuous functions, which completes the proof.

By mollification we immediately obtain:

Corollary 2.17 The space C∞c (Ω) is dense in Lp(Ω) for 1 ≤ p <∞.

Proof (Sketch) The uniform norm on any compact set K dominates the Lp

norm on K: ∫K|f(x)|p dx ≤ |K|‖f‖p∞.

So approximate f to within ε/2 by a function g ∈ C0c (Ω); then mollify.

Furthermore it follows from the separability of C0(Ω) that the ‘nice’ Lp

spaces are separable:

Theorem 2.18 Lp(Ω) is separable for 1 ≤ p <∞.

Proof (Sketch) Write

Ω =

∞⋃j=1

Ωj ,

where

Ωj = x ∈ Ω : dist(x, ∂Ω) > 1/j and |x| < j.

Use the facts that (i) C0(Ωj) is separable for each j, (ii) the sup norm on Ωj


dominates the Lp(Ωj) norm, and (iii) a countable union of countable sets is

still countable.

2.7 Locally integrable functions

We say that a function f is locally in Lp in Ω, or f ∈ Lploc(Ω), if f ∈ Lp(K)

for every compact subset K of Ω. We say that fn → f ‘locally in Lp(Ω)’ if

fn → f in Lp(K) for every compact subset K of Ω [usually Ω = Rn for such

local spaces].

Recall that the mollification fh of f is given by

fh(x) = (ρh ? f)(x) =

∫Rn

ρh(x− y)f(y) dy = h−n∫Rn

ρ

(x− yh

)f(y) dy.

(2.8)

Lemma 2.19 (Mollification of functions in Lploc) If f ∈ Lploc(Ω), 1 ≤p < ∞, then fh ∈ Lploc(Ω) and fh → f in Lploc(Ω). Furthermore, fh ∈C∞(Ω), and if f has compact support in Ω and h < dist(supp f, ∂Ω) then

fh ∈ C∞c (Ω).

Why can this result not hold for p =∞?

Proof First we show that if K is a compact subset of Ω, then there is another

compact subset K ′ of Ω such that

‖fh‖Lp(K) ≤ ‖f‖Lp(K′).

Take h < 12 dist(K, ∂Ω), and then change variables in (2.8) with z = x − y

to give

fh(x) =

∫|z|≤h

ρh(z)f(x− z) dz.

2.7 Locally integrable functions 35

Since∫ρh = 1, using Holder’s inequality

|fh(x)|p =

(∫|z|≤h

ρh(z)f(x− z) dz

)p

=

(∫|z|≤h

ρh(z)1−(1/p)ρh(z)1/pf(x− z) dz

)p

≤

(∫|z|≤h

ρh(z)q(1−(1/p)) dz

)p/q (∫|z|≤h

ρh(z)|f(x− z)|p dz

)

=

∫|z|≤h

ρh(z)|f(x− z)|p dz.

So ∫K|fh(x)|p dx ≤

∫K

∫|z|≤h

ρh(z)|f(x− z)|p dz dx

=

∫|z|≤h

ρh(z)

∫K|f(x− z)|p dx dz

≤∫|z|≤h

ρh(z)

(∫K′|f(x)|p dx

)=

∫K′|f(x)|p dx,

where

K ′ = y = x+ z : x ∈ K, |z| ≤ h.

To complete the proof we use the density of C0c (Ω) in Lp(Ω), and mollifi-

cation of functions in C0c (Ω): given ε > 0, choose φ ∈ C0

c (Ω) such that

‖f − φ‖Lp(Ω) < ε/3.

Choose h small enough that ‖φ − φh‖Lp(K′) < ε/3 (the sup norm on K ′

dominates the Lp(K ′) norm); then

‖f − fh‖Lp(K) ≤ ‖f − φ‖Lp(K) + ‖φ− φh‖Lp(K) + ‖φh − fh‖Lp(K)

≤ 2ε/3 + ‖φ− f‖Lp(K′) ≤ ε.


2.8 Sobolev spaces

2.8.1 Weak derivatives

Suppose that f is differentiable on Ω. Then for any test function φ ∈ C∞c (Ω),

one can integrate by parts to obtain∫Ω

∂f

∂xjφ dx = −

∫Ωf∂φ

∂xjdx, (2.9)

using the fact that φ has compact support in Ω (so the boundary term

vanishes). Repeating this process |α| times yields∫Ω

(∂αf)φ dx = (−1)|α|∫

Ωf ∂αφ dx

for any multi-index α.

Now, while the left-hand side of (2.9) makes sense only if ∂f/∂xj exists,

the right-hand side makes sense for any f ∈ L1, since we always have ∂jφ ∈L∞. For a function f ∈ L1

loc(Ω), we say that g is the weak derivative of f

with respect to xj , and write ∂jf = g, if g ∈ L1loc(Ω) and∫

Ωg φdx = −

∫Ωf ∂jφ dx

for every φ ∈ C∞c (Ω). Similarly, f has weak derivative ∂αf = g if∫Ωg φdx = (−1)|α|

∫Ωf ∂αφ dx.

The following lemma is fundamental.

Lemma 2.20 When they exist, weak derivatives are unique.

Proof It suffices to show that if g ∈ L1loc(Ω) and∫

Ωgφdx = 0

for every φ ∈ C∞c (Ω) then g = 0 almost everywhere.

Fix Ω′ ⊂⊂ Ω and ε > 0; then g ∈ L1(Ω′)) and we can approximate g in

L1(Ω′) by a smooth function g ∈ C∞c (Ω′) with

‖g − g‖L1(Ω′) < ε/3.

Write ψ = sgn(g); then |ψ| ≤ 1, and so ψ ∈ L1(Ω′); since g has compact


support in Ω′, so does ψ. Approximate ψ by its mollification ψh, with h

chosen small enough that supp(ψh) ⊂ Ω′ and

‖ψ − ψh‖L1 <ε

3‖g‖∞.

Note that

|ψh(x)| =∣∣∣∣∫ ρh(y − x)ψ(x) dy

∣∣∣∣ ≤ ∫ ρh(y−x)|ψ(x)| dy ≤∫ρh(y−x) dy = 1.

Then – noting that the support of ψh is a subset of Ω′, and that gψ = |g|,

0 =

∫Ωgψh =

∫Ω′gψh

=

∫Ω′gψ − g(ψ − ψh) + (g − g)ψh

≥∫

Ω′|g| −

∫Ω′|g||ψ − ψh| −

∫Ω′|g − g||ψh|

= ‖g‖L1 − ‖g‖∞‖ψ − ψh‖L1 − ‖g − g‖L1

≥ ‖g‖L1 − ‖g − g‖L1 −ε

3− ε

3,

and so for any ε > 0,

‖g‖L1(Ω′) ≤ ε

i.e. ‖g‖L1(Ω′) = 0, g = 0 almost everywhere in Ω′; since Ω′ ⊂⊂ Ω was

arbitrary, this shows that g = 0 almost everywhere in Ω.

As an example, consider first the function

f1(x) =

x, 0 < x ≤ 1

1 1 < x < 2.

Then for any φ ∈ C∞c (0, 2)

−∫ 2

0f1(x)φ′(x) dx = −

∫ 1

0xφ′(x) dx−

∫ 2

1φ′(x)

= − [xφ(x)]10 +

∫ 1

0φ(x) dx− φ(2) + φ(1)

=

∫ 1

0φ(x) dx,

since φ(2) = 0 (as φ ∈ C∞c (0, 2)). So f1 has weak derivative

v(x) =

1, 0 < x ≤ 1

0 1 < x < 2.


However, if we perform the same calculation starting with the function

f2(x) =

x, 0 < x ≤ 1

2 1 < x < 2

we end up with

−∫ 2

0f2(x)φ′(x) dx = −

∫ 1

0xφ′(x) dx− 2

∫ 2

1φ′(x)

= − [xφ(x)]10 +

∫ 1

0φ(x) dx− 2φ(2) + 2φ(1)

=

∫ 1

0φ(x) dx+ φ(1),

There is no v ∈ L1loc(0, 2) such that∫ 2

0v(x)φ(x) dx = φ(1).

Indeed, choose a sequence of φn ∈ C∞c (0, 2) such that 0 ≤ φn(x) ≤ 1,

φn(1) = 1, and φn(x)→ 0 for all x 6= 1. Then, using the DCT,

1 = limn→∞

φn(1) = limn→∞

∫ 2

0v(x)φn(x) dx = 0

(to use the DCT: v(x)φn(x) → 0 almost everywhere, |v(x)φn(x)| ≤ |v(x)|,and v ∈ L1

loc(0, 2)).

2.8.2 Sobolev spaces

Definition 2.21 The Sobolev space W k,p(Ω) is defined as

W k,p(Ω) = f : ∂αf ∈ Lp(Ω), 0 ≤ |α| ≤ k,

with norm

‖f‖Wk,p =

∑0≤|α|≤k

‖∂αf‖pLp

1/p

.

Lemma 2.22 W k,p(Ω) is a Banach space, and separable if 1 ≤ p <∞.

Proof If fn is a Cauchy sequence in W k,p(Ω) then ∂αfn is Cauchy in

Lp(Ω) for every multi-index α with 0 ≤ |α| ≤ k. So for each such α, there


exists an fα ∈ Lp such that ∂αf → fα in Lp. It remains to show that

∂αf0 = fα.

To do this, take φ ∈ C∞c (Ω); then∫Ωfαφ = lim

n→∞

∫Ω

(∂αfn)φ = − limn→∞

∫Ωfn(∂αφ) = −

∫Ωf(∂αφ)

=

∫Ω

(∂αf)φ,

i.e. fα = ∂αf , and f ∈ W k,p(Ω). [To justify swapping the sum and the

integral, note that if fn → f in Lp and g ∈ Lq then∣∣∣∣∫ fng −∫fg

∣∣∣∣ =

∣∣∣∣∫ (fn − f)g

∣∣∣∣ ≤ ‖fn − f‖Lp‖g‖Lq ,

i.e.∫fng →

∫fg.]

For separability, see the problems sheet.

We write Hk(Ω) = W k,2(Ω). This is a Hilbert space with the inner

product

(f, g)Hk =∑

0≤|α|≤k

(∂αf, ∂αg)L2 .

This inner product gives rise to the Hk-norm defined by

‖f‖2Hk =∑

0≤|α|≤k

‖∂αf‖2L2

(as above).

2.8.3 Completions

Given any normed space (X, ‖·‖), there is an abstract method of constructing

its ‘completion’ (X, ‖·‖X) – essentially, the ‘smallest’ complete normed space

that contains (X, ‖ · ‖) (or, in fact, an isometrically isomorphic copy) as a

subset. One would like to ‘add limits of Cauchy sequences to X’, but if we

only have X we don’t have these limits (unless they’re already in X). So

instead, elements of X consist of equivalence classes of Cauchy sequences in

X (where X ∼ Y if limn→∞ ‖Xn−Yn‖ = 0). With some care – much of the

work is in coping with the notation – you can show that the resulting space

is complete, and that there’s a copy of X (consisting of (equivalence classes

of) the constant sequences with Xj = x for every j) inside X.


However, the situation is much more straightforward when we already

have a space (X, ‖ · ‖), and we want to find the completion of a subspace in

X. Suppose that Y is a subspace of X; then the completion of Y in X is

simply the closure of Y in X, i.e.

Y = x ∈ X : x = limn→∞

yn, for some yn ∈ Y .

In this case we really can ‘add the limit of any Cauchy sequence’, since if

yn is Cauchy we know that the limit exists and lies in X. It is simple to

show that (Y, ‖ · ‖) is complete.

Note that if Y is dense in X, then the completion of Y in X is all of X,

by definition; we have already seen, for example, that C∞c (Ω) is dense in

L2(Ω), and so the completion of C∞c (Ω) in L2(Ω) is L2(Ω).

However, C∞c (Ω) is not dense in Hk(Ω) if k ≥ 1. Suppose that f is a

smooth non-zero solution of∑|α|≤k

(−1)|α|∂2αf = 0

on Ω; in particular f ∈ Hk. [Solve the equation using separation of variables

on a parallelepiped containing Ω, then restrict this to Ω.]

Since f 6= 0, ‖f‖2Hk 6= 0. Now suppose that there is a sequence φj ∈

C∞c (Ω) such that φj → f in Hk. Then

(f, f)Hk = limj→∞

(f, φj)Hk

= limj→∞

∑|k|≤α

(∂αf, ∂αφj)L2

= limj→∞

∑|k|≤α

((−1)|α|∂2αf, φj)L2

= limj→∞

∑|k|≤α

(−1)|α|∂2αf, φj

L2

= 0,

a contradiction. [Note: since (f, ·)Hk defines a bounded linear functional on

Hk, this shows that there are non-zero bounded linear functionals on Hk,

that nevertheless vanish on all test functions.]

The completion of C∞c (Ω) in Hk is very useful, and we denote this space


by Hk0 (Ω). Heuristically, this space consists of functions in Hk with ∂αf = 0

on ∂Ω for |α| ≤ k − 1.

In light of this the following version of the Poincare inequality is perhaps

unsurprising.

Lemma 2.23 Suppose that Ω is bounded in one direction in a strip of width

d. Then for every f ∈ H10 (Ω),

‖f‖L2 ≤ d‖∇f‖L2 . (2.10)

Proof We have already seen that if f ∈ C∞c (Ω) then the inequality holds.

Now take f ∈ H10 (Ω), and write f = limn→∞ fk, with fk ∈ C∞c (Ω) and the

limit taken in the H1 norm. This means in particular, that fk → f in L2

and ∂jfk → ∂jf in L2; so

‖f‖L2 = limk→∞

‖fn‖L2

and

‖∇f‖2L2 :=n∑j=1

‖∂jf‖2L2 = limk→∞

n∑j=1

‖∂jfn‖2L2 =: ‖∇fn‖2L2 ,

whence (2.10).

The space H−k(Ω) is the dual space of Hk0 (Ω) [i.e. the space of all bounded

linear functionals on Hk0 (Ω)]. (We do not want to have non-zero elements

of the dual that vanish on all test functions, as we did above.)

2.8.4 Density results for Hk(Ω)

However, C∞(Ω) is dense inHk, which will follow from the following pleasing

result – mollification and derivatives commute.

Lemma 2.24 Suppose that f ∈ L1loc(Ω) and that the weak derivative ∂αf

exists. Then provided that h < dist(x, ∂Ω),

∂αfh(x) = (∂αf)h(x).


Proof Using the definition of fh and differentiating under the integral sign,

we have, integrating by parts

∂αfh(x) =

∫Ω∂αx ρh(x− y)f(y) dy

= (−1)|α|∫

Ω∂αy ρh(x− y)f(y) dy

=

∫Ωρh(x− y)∂αf(y) dy

= (∂αf)h.

We define the spaceHkloc(Ω) in the obvious way: f ∈ Hk

loc(Ω) if f ∈ Hk(Ω′)

for every Ω′ ⊂⊂ Ω, and fn → f in Hkloc(Ω) if fn → f in Hk(Ω′) for every

Ω′ ⊂⊂ Ω.

Corollary 2.25 If f ∈ Hkloc(Ω) then fh ∈ C∞(Ω) ∩Hk

loc(Ω) and fh → f in

Hkloc(Ω).

Given this, we require one other technical device, and the proof that

C∞(Ω) is dense in Hk(Ω) is then ‘easy’. This device is known as a ‘partition

of unity’.

Theorem 2.26 Let Ω be an open subset of Rn, and Uj a countable collec-

tion of bounded open subsets of Rn that cover Ω, with the additional property

that every compact subset of Ω intersects only finitely many of the Uj (the

cover is ‘locally finite’). Then there exists a partition of unity subordinate

to the covering Uj, that is a set of functions ψj ∈ C∞c (Rn) such that

(i) 0 ≤ ψj ≤ 1,

(ii) suppψj ⊂ Uj, and

(iii)∑∞

j=1 ψj = 1 on a neighbourhood of Ω.

For a proof see Theorem 5.14 in [JCR].

Theorem 2.27 C∞(Ω) ∩Hk(Ω) is dense in Hk(Ω).

Proof Set

Ωj = x ∈ Ω : dist(x, ∂Ω) > 1/j and |x| < j


Then Ωj ⊂⊂ Ωj+1 and Ω = ∪∞j=1Ωj . Let ψj be a partition of unity

subordinate to the covering Ωj+1 \ Ωj .

Fix f ∈ Hk(Ω) and ε > 0. For each j choose

hj < dist(Ωj , ∂Ωj+1)

such that

‖(ψju)hj − ψju‖Hk(Ω) ≤ 2−jε.

By the choice of hj and ψj , only a finite number of the functions (ψju)hjare non-zero on any compact subdomain of Ω; it follows that the function

u =∞∑j=1

(ψju)hj

is in C∞(Ω). Furthermore

‖u− u‖Hk(Ω) =∞∑j=1

‖

∞∑j=1

ψj

u− u‖

≤∞∑j=1

‖(ψju)hj − ψju‖

≤∞∑j=1

2−jε = ε.

Note that there are minimal assumptions here on Ω.

We can use this result to prove the following ‘approximation up to the

boundary’ in a half space.

Theorem 2.28 The space C∞(Rn+) is dense in Hk(Rn+).

We denote by τhf the ‘shifted function’ τhf(x) = f(x′, xn + h) (note that

this shifts the values of the functions ‘downwards’).

Proof Take f ∈ Hk(Rn+). Then we know that there exist fn ∈ C∞(Rn+)

such that fn → f in Hk(Rn+). Now consider the shifted functions

τ1/nfn(x) := fn(x′, xn +1

n),


where x′ = (x1, . . . , xn−1). Clearly τ1/nfn is defined for all xn > 1/n, and

∂α(τ1/nfn) = τ1/n∂αfn.

Denote by fn the restriction of τ1/nfn to Rn+. Then fn ∈ Hk(Rn+)∩C∞(Rn+),

and for any α with 0 ≤ |α| ≤ k,

‖∂α(f − fn)‖L2(Rn+) ≤ ‖∂α(f − fn)‖L2(Rn

+) + ‖∂α(fn − fn)‖L2(Rn+).

The first term tends to zero by our choice of fn; the second tends to zero

since for any g ∈ L2(Rn+),

‖τhg − g‖L2(Rn+) → 0 as h→ 0.

(You can prove this using step functions, see examples.)

2.8.5 Extension theorems

First, extending functions in Hk(Rn+), where

Rn+ = x ∈ Rn : xn > 0.

We show that if u ∈ Hk(Rn+) then the is a linear operator Ek : Hk(Rn+) →Hk(Rn) such that

‖Ek[u]‖Hk(Rn) ≤ ck‖u‖Hk(Rn+).

We then use a change of coordinates to treat sufficiently smooth bounded

domains.

Theorem 2.29 Let k ≥ 0 be an integer. Then there exists a bounded linear

mapping Ek : Hk(Rn+)→ Hk(Rn) such that

E[f ]|Rn+

= f for all f ∈ Hk(Rn+)

and

‖Ek[f ]‖Hj(Rn) ≤ ck‖f‖Hj(Rn+) for all j = 1, . . . , k.

Proof First we consider how to extend functions in Hk(Rn+)∩Ck(Rn+), and

then we use a density argument to treat any f ∈ Hk(Rn+).

To make f continuous across xn = 0 we could simply define

f(x′, xn) = f(x′,−xn) for all xn > 0,


i.e. ‘reflect’ in xn = 0. But this would cause problems with the first deriva-

tive; to get round these we could try

f(x′, xn) = 3f(x′,−xn)− 2f(x′,−2xn),

which now matches by f and f ′ across xn = 0. To match derivatives up to

order k, try a combination of the form

f(x′, xn) =k+1∑j=1

ajf(x′,−jxn).

For the first k derivatives to agree requires

k+1∑j=1

(−j)iaj = 1 for all i = 0, . . . , k.

This equation for aj always has a solution: to see this, consider the matrix

equation

Ja = 1,

where J is the k+ 1×k+ 1 matrix with Jij = (−j)i−1. Now, J is invertible:

if not there must be some non-zero d such that Jd = 0. But this implies

that the degree k polynomial

k∑i=0

di+1xi

has k + 1 roots, −1, . . . ,−(k + 1).

That’s it.

Let U and V be open subsets of Rn. A function Φ : U → V is a Ck-

diffeomorphism if

(i) Φ and Φ−1, and their derivatives of order up to k, are bounded and

continuous on U and V respectively, and

(ii) there are positive constants k1 and k2 such that

k1 ≤ |det∇Φ(x)| ≤ k2

for all x ∈ U .


Suppose that we have a function f : U → R. If we change coordinates we

produce a new function f∗ : V → R, given by

f∗(y) = f(Φ−1(y)).

This is the ‘pullback of f under Φ−1’, (Φ−1)∗f = f Φ−1. Note that Φ−1

acts on points in V ; (Φ−1)∗ acts on functions defined on U . Similarly, a

function defined on V gives rise to a function defined on U via the pullback

operator Φ∗,

(Φ∗g)(x) = g(Φ(x)).

The following lemma – whose proof we omit – shows that if f ∈ Hk(U)

then (Φ−1)∗f ∈ Hk(V ), and similarly if g ∈ Hk(V ) then Φ∗g ∈ Hk(U).

Lemma 2.30 If Φ : U → V is a Ck-diffeomorphism then (Φ−1)∗ and Φ∗

are bounded linear maps from Hk(U) to Hk(V ), and vice versa, i.e. for

f ∈ Hk(U) and g ∈ Hk(V ),

‖(Φ−1)∗f‖Hk(V ) ≤ c‖f‖Hk(U) and ‖Φ∗g‖Hk(U) ≤ c‖g‖Hk(V ).

Now we suppose that Ω is a Ck domain: this means that at each point

x0 ∈ ∂Ω there exists an ε > 0 and a Ck diffeomorphism Φ of B(x0, ε) onto

a subset B of Rn such that

(i) Φ(x0) = 0,

(ii) Φ(B(x0, ε) ∩ Ω) ⊂ Rn+, and

(iii) Φ(B(x0, ε) ∩ ∂Ω) ⊂ ∂Rn+.

Theorem 2.31 If Ω is a bounded Ck domain, then for each open set Ω∗

with Ω ⊂⊂ Ω∗ there exists a bounded linear extension operator E such that

if f ∈ Hk(Ω) then E[f ] ∈ Hk0 (Ω∗) and

‖E[f ]‖Hk(Ω∗) ≤ Ck,Ω∗‖f‖Hk(Ω). (2.11)

(In fact (2.11) holds for each Hj with 0 ≤ j ≤ k.)

Proof At each x ∈ ∂Ω find an εx such that there exists a Ck diffeomorphism

of B(x, εx) onto a subset of Rn as above, and such that B(x, εx) ⊂⊂ Ω∗.

Since Ω is bounded so is ∂Ω, which is therefore compact: choose a finite

collection of the balls B(x, εx), call them Bjj = 1m, that cover an open


neighbourhood of ∂Ω, and corresponding functions Φj . Write Uj = Ω ∩ Bjand Vj = Φj(Ωj)

Let ψj be a partition of unity subordinate to the Bj. Then

m∑j=1

ψjf

is equal to f in a neighbourhood of ∂Ω, and the function defined within Ω

as

f := f −m∑j=1

ψjf

has compact support within Ω. Clearly within Ω we have

f = f +m∑j=1

ψjf.

For each j consider the function f∗j ∈ Hk(Rn+) defined by

f∗j = (Φ−1j )∗(ψjf),

which is zero outside Vj . By Lemma 2.30,

‖f∗j ‖Hk(Rn+) ≤ cj‖ψjf‖Hk(Uj),

and since for any ψ ∈ C∞c (Rn), U ⊂ Rn and f ∈ Hk(U) we have

‖ψf‖Hk(U) ≤ C(ψ)‖f‖Hk(U),

it follows that

‖f∗j ‖Hk(Rn+) ≤ c′j‖f‖Hk(Uj) ≤ c′j‖f‖Hk(Ω).

Now, we know that the function f∗j can be extended to a function Ek[f∗j ] =

fj ∈ Hk(Rn), such that

‖fj‖Hk(Rn) ≤ Ck‖f∗j ‖Hk(Rn+) ≤ c′′j ‖f‖Hk(Ω).

Now multiply fj by a C∞c (Rn) cutoff function θj , that is equal to 1 on

Φj(supp(ψj)) and has compact support within Φj(Bj). Then the pullback

function

Φ∗j (θjfj)

is equal to ψju on Uj , has compact support within Bj (and so within Ω∗),

and satisfies

‖Φ∗j (θjfj)‖Hk(Ω∗) ≤ Cj‖f‖Hk(Ω). (2.12)


Since f has compact support within Ω, its extension by zero – which we

still write as f is in Hk(Ω).

So now consider the composite function

E[f ] = f +

m∑j=1

Φ∗j (θj fj).

This extends u to a function in Hk0 (Ω∗), and on summing the estimates in

(2.12) has

‖E[f ]‖Hk(Ω∗) ≤ Ck‖f‖Hk(Ω).

Now we can prove...

Theorem 2.32 If Ω is of class Ck and ∂Ω is compact, then C∞(Ω) is dense

in Hk(Ω).

Proof Take Ω∗ such that Ω ⊂⊂ Ω∗, and given f ∈ Hk(Ω) extend to a

function E[f ] ∈ Hk0 (Ω∗). Then there exists a sequence of functions gn ∈

C∞(Ω∗) ∩ Hk(Ω∗) such that gn → E[f ] in Hk(Ω∗). Set fn = gn|Ω; then

fn ∈ C∞(Ω) and

‖fn − f‖Hk(Ω) ≤ ‖gn − E[f ]‖Hk(Ω∗) → 0

as n→∞.

2.9 The Fourier transform and Sobolev spaces

Before proving various Sobolev embedding theorems we give a (very) brief

treatment of the Fourier transform (FT).

Let S (Rn) be the class of all complex-valued C∞ functions on Rn such

that |x|k|∂αf(x)| is bounded for every k ∈ N and every multi-index α.

The Fourier transform of any f ∈ S (Rn) is defined as

f(ξ) = (2π)−n/2∫Rn

e−iξ·xf(x) dx.

It is relatively easy to show that if f ∈ S (Rn) then f ∈ S (Rn).


If g ∈ S (Rn) then there exists a unique f ∈ S (Rn) such that f = g, and

f(x) = (2π)−n/2∫Rn

eiξ·xg(x) dx,

i.e. f = (g) , where g(x) = g(−x).

The FT is particularly useful because it ‘converts’ derivatives into ‘mul-

tipliers’:

(∂αf ) = (iξ)αf .

Also, the FT preserves the L2 inner product: if f, g ∈ S (Rn) then

(f , g) = (f, g);

in particular

‖f‖L2 = ‖f‖L2 .

Since the FT is linear, this enables us to define the FT for functions

f ∈ L2(Rn) using a limit procedure. Take f ∈ L2(Rn); then since S (Rn) is

dense in L2(Rn) (we know that C∞c (Rn) functions are dense in L2(Rn) and

these are a subset of functions in S (Rn)), we can find fn ∈ S (Rn) such

that f = limn→∞ fn (where the limit is taken in L2), and then define

f = limn→∞

fn,

where again the limit is taken in L2(Rn). [So if f ∈ L2, f ∈ L2.]

If f ∈ Hk(Rn), then the derivatives ∂αf are in L2(Rn) for 0 ≤ |α| ≤ k; it

follows that

(iξ)αf ∈ L2(Rn) for all 0 ≤ |α| ≤ k,

which we can combine to give a characterisation of Hk in terms of the FT,

Hk(Rn) = f ∈ L2(Rn) : (1 + |ξ|2)k/2f ∈ L2(Rn), (2.13)

and it is straightforward to show that the norm

‖(1 + |ξ|2)k/2f‖L2(Rn) =

∫Rn

(1 + |ξ|2)k|f(ξ)|2 dξ

is equivalent to the norm in Hk(Rn). [Note that there is no reason why k

should be an integer in (2.13), and this gives a way to define Sobolev spaces

with non-integral exponents.]


Theorem 2.33 Let f ∈ Hk(Rn) with k > n/2. Then there exists a constant

Ck,n such that

‖f‖∞ ≤ Ck,n‖f‖Hk(Rn),

and f is equal almost everywhere to a function in C0(Rn).

Actually the argument here shows also that the FT maps L1 to L∞.

Proof First take f ∈ S (Rn) ∩Hk(Rn); then

‖f‖∞ = supx∈Rn

(2π)−n/2∣∣∣∣∫

Rn

eiξ·xf(ξ) dξ

∣∣∣∣≤ (2π)−n/2

∫|f(ξ)| dξ

= (2π)−n/2∫

1

(1 + |ξ|2)k/2(1 + |ξ|2)k/2|f(ξ)|dξ

≤ (2π)−n/2(∫

1

(1 + |ξ|2)kdξ

)1/2(∫(1 + |ξ|2)k|f(ξ)|2 dξ

)= Ck,n‖f‖Hk(Rn),

using the fact (see exercises) that∫Rn

1

(1 + |ξ|2)kdξ <∞

if and only if k > n/2. The same inequality follows for f ∈ Hk(Rn) by

taking limits. But also, this shows that f is the L∞ limit of a sequence

of continuous functions; it must therefore be equal almost everywhere to a

continuous function, and the result follows.

Corollary 2.34 If f ∈ Hk(Ω) with k > n/2 + s then f ∈ Cs(Ω) with

‖f‖Cs ≤ ‖f‖Hk .

Proof Take f ∈ Hk(Ω): then there is an extension of f to Rn, E[f ], such

that

‖E[f ]‖Hk(Rn) ≤ CEk ‖f‖Hk(Ω).

By the previous theorem, E[f ] ∈ C0(Rn) with

‖E[f ]‖C0(Rn) ≤ Ck,n‖E[f ]‖Hk(Rn) ≤ Ck,nCEk ‖f‖Hk(Ω).


Clearly, therefore, f ∈ C0(Ω) with

‖f‖C0(Ω) ≤ CΩ,k,n‖f‖Hk(Ω).

The results for higher derivatives follow by considering each ∂αf in turn.

Actually we can do better, and show that f must be Holder continuous.

Theorem 2.35 If f ∈ Hk(Rn) with n/2 < k < n/2 + 1 then there is a

constant C such that

|f(x)− f(y)| ≤ Ck‖f‖Hk |x− y|k−(n/2) for all x, y ∈ Rn.

Proof See examples.

2.9.1 Compact embeddings

We say that space X is compactly embedded in Y if a bounded set in X is

a precompact subset of Y , i.e. if any bounded sequence that is bounded in

the norm of X has a subsequence that converges in the norm of Y . The

prototype such theorem for us is the Arzela-Ascoli Theorem – for a proof

see the examples.

Theorem 2.36 (Arzela-Ascoli Theorem) Let K ⊂ Rn be compact and

let fn be a sequence in C0(K) that is

(i) bounded, i.e. ‖fn‖∞ ≤M for some M > 0, and

(ii) equicontinuous, i.e. for every ε > 0, there exists a δ > 0 such that

|x− y| < δ ⇒ |fn(x)− fn(y)| < ε

for every x ∈ K, and for every n ∈ N.

Then fn has a subsequence that converges uniformly on K.

Using this result, the Holder continuity above implies that if Ω is bounded,

the embedding of Hk(Ω) into C0(Ω) is compact, i.e. a bounded sequence in

Hk(Ω) has a subsequence that converges uniformly on Ω.

The following result is similar, but arguably more important.


Theorem 2.37 If Ω is bounded and ∂Ω is Ck then Hk+1(Ω) is compactly

embedded in Hk(Ω).

Proof Choose Ω∗ with Ω ⊂⊂ Ω∗. Take a sequence fn that is bounded

in Hk+1(Ω), ‖fn‖Hk+1(Ω) ≤M , and extend to E[fn], bounded in Hk+1(Ω∗)

with ‖E[fn]‖Hk+1(Rn) ≤M ′. For any g ∈ Hk+1(Ω), let

gR = F−1[gχ|ξ|≤R];

then there exists a constant C such that

‖g − gR‖Hk ≤C

R‖g‖Hk+1 ;

Indeed,

‖g − gR‖2Hk ≤ C∫|ξ|≥R

(1 + |ξ|2)k|g(ξ)|2 dξ

=C

1 +R2

∫|ξ|≥R

(1 + |ξ|2)k+1|g(ξ)|2 dξ

≤ C

1 +R2‖g‖2Hk+1 .

Similarly, gR ∈ H l(Rn) for every l ∈ N;

‖gR‖Hl(Rn) ≤ C∫|ξ|≤R

(1 + |ξ|2)l|g(ξ)|2 dξ

≤ C(1 +R2)l−(k+1)

∫|ξ|≤R

(1 + |ξ|2)k+1|g(ξ)|2 dξ

= CR,l‖g‖2Hk+1 .

So

‖gR‖Crb (Rn) ≤ CR,r‖g‖Hk+1 .

It follows that if U is any bounded open set containing Ω∗, then (E[fn])Rhas a subsequence that converges in Ckb (U), and hence in Hk(U). apply-

ing a standard diagonal argument, we can obtain a subsequence such that

(E[fnj ])m converges in Hk(D) for every m.

From this it follows that E[fnj ] is Cauchy in Hk(U): given ε > 0, first

choose m such that

‖E[fi]− E[fi]m‖Hk ≤√C

m‖E[fi]‖Hk+1 ≤

√CM ′

m< ε/3.


Then choose N such that

‖E[fni ]m − E[fnj ]m‖Hk(U) ≤ ε/3 for all i, j ≥ N.

It follows that

‖E[fni ]− E[fnj ]‖Hk(U) ≤ ‖E[fni ]− E[fni ]m‖Hk(U) + ‖E[fni ]m − E[fnj ]m‖Hk(U)

+ ‖E[fnj ]m − E[fnj ]‖Hk(U)

≤ ε for all i, j ≥ N,

and so E[fni ] is Cauchy in Hk(U). It follows that fni is Cauchy in Hk(Ω),

and since Hk(Ω) is complete, fni converges.

2.9.2 Sobolev spaces and Lp spaces

What happens if f ∈ Hk with k < n/2? To consider this, we need the fact

that the Fourier transform maps Lp into Lq, with (p, q) conjugate, such that

‖f‖Lp ≤ Cp‖f‖Lq . (2.14)

This is certainly not obvious; it follows ‘by interpolation’ from the fact that

it maps L2 to L2 and L1 to L∞.

Theorem 2.38 If f ∈ Hk(Rn) with k ≤ n/2 then f ∈ Lp(Rn) for any

p ∈ [2, 2nn−2k ).

Proof Using (2.14) we have (for f ∈ Hk ∩S (Rn))

‖f‖qLp ≤ Cp‖f‖Lq

= Cqp

∫|f(ξ)|q dξ

= Cqp

∫1

(1 + |ξ|2)kq/2(1 + |ξ|2)kq/2|f(ξ)|q dξ

≤ Cqp(∫

1

(1 + |ξ|2)kq/2

2/(2−q)dξ

)(2−q)/2(∫(1 + |ξ|2)k|f(ξ)|2 dξ

)q/2= Cqp

(∫1

(1 + |ξ|2)kq/(2−q)dξ

)(2−q)/2‖f‖q

Hk(Rn)

≤ Cp,k‖f‖qHk(Rn).


The condition

kq

2− q>n

2⇔ 1

q<

2k + n

2n

ensures that the first integral is finite; this translates to

1

p> 1− 2k + n

2n⇔ p <

2n

n− 2k.

Note that if k = n/2 this shows that f ∈ Lp for any 2 ≤ p < ∞. When

k < n/2 the result is also true for the endpoint p = 2n/(n − 2k), which is

the most useful; this requires the ‘death by Young’s inequality’ approach.

Corollary 2.39 If f ∈ Hk(Ω) with k ≤ n/2 then f ∈ Lp(Ω) for p ∈[2, 2n

n−2k ).

Proof Extend f ∈ Hk(Ω) to E[f ] ∈ Hk(Rn), apply the above result, then

restrict back to Ω; clearly ‖f‖Lp(Ω) ≤ ‖E[f ]‖Lp(Rn).

2.9.3 Scaling and inequalities

We have (almost) shown that if f ∈ Hk(Rn) then f ∈ Lp(Rn) with p =

2n/(n− 2k). Why is this what we should expect?

Suppose that we do have an inequality

‖f‖Lp ≤ C‖f‖Hk .

What happens if we rescale the function f(x) and consider instead fλ(x) =

f(λx). This should satisfy the same inequality

‖fλ‖Lp ≤ C‖fλ‖Hk .

Now,

‖fλ‖pLp =

∫Rn

|f(λx)|p dx = λ−n∫Rn

|f(y)|p dy,

and so

‖fλ‖Lp = λ−n/p‖f‖Lp .


Each derivative also brings out a factor of λ, and so

‖∂αf‖L2 = λ|α|−n/2‖f‖L2 .

Putting |α| = k, we would expect

k − n

2= −n

p,

which gives p = 2n/(n− 2k).

2.9.4 Boundary values and the trace operator

If Ω is a bounded Ck domain then we say that f ∈ Hk(∂Ω) if

(Φ−1j )∗(ψjf) ∈ Hk(Rn−1)

for every j, where ψj is a partition of unity as constructed in the proof of

Theorem 2.27, and define

‖f‖2Hk(∂Ω) =m∑j=1

‖(Φ−1j )∗(ψjf)‖2Hk(Rn−1).

(One can show that any choice of Φj and ψj will give equivalent norms.)

Theorem 2.40 Suppose that Ω is a bounded C1 domain. Then there exists

a bounded linear operator

T : H1(Ω)→ L2(∂Ω)

such that for every f ∈ H1(Ω) ∩ C0(Ω), Tu = u|∂Ω.

Proof We first prove the result for f ∈ H1(Rn+) and show that there is a

bounded linear operator from H1(Rn+)∩C1(Rn+) into L2(∂Rn+) = L2(Rn−1).

For any f ∈ H1(Rn+) ∩ C1(Rn+) such that f(x)→ 0 as xn →∞, we have∫∂Rn

+

|f(x′, 0)|2 dx′ = −∫Rn

+

∂n(|f |2) dx

= −∫Rn

+

2f∂nf dx

≤∫|f |2 + |∂nf |2 dx

≤ ‖f‖2H1(Rn+).


Now for f ∈ H1(Ω) ∩ C1(Ω), we use the partition of unity and the above

result for Rn+ so that ‖f‖L2(∂Ω) ≤ ‖f‖H1(Ω). Define Tf = f |∂Ω.

Given any f ∈ H1(Ω), there is a sequence fn ∈ C∞(Ω) that converges

to f in H1(Ω); in particular fn ∈ H1(Ω) ∩ C1(Ω), so we can define Tf by

taking limits. Since fn → f uniformly on Ω (see examples) the result holds

as stated.

Theorem 2.41 f ∈ H10 (Ω) iff f ∈ H1(Ω) and Tf = 0.

3

Elliptic PDEs

We now prove existence and uniqueness result for the weak formulation of

the elliptic problem

Lu = f u|∂Ω = 0, (3.1)

where

Lu = −n∑

i,j=1

∂

∂xi

(aij(x)

∂u

∂xj

)+

n∑i=1

bi(x)∂u

∂xi+ c(x)u. (3.2)

For simplicity we will suppose that the coefficients (aij , bi, and c) are smooth

functions of x.

First we want to recast this equation in a weak form. If we multiply the

equation by a function ϕ ∈ C∞c (Ω) and integrate by parts we obtain

n∑i,j=1

(aij∂ju, ∂iϕ) +n∑i=1

(bi∂iu, ϕ) + (cu, ϕ) = (f, ϕ).

Write

B(u, ϕ) =

n∑i,j=1

(aij∂ju, ∂iϕ) +

n∑i=1

(bi∂iu, ϕ) + (cu, ϕ),

and note that for a fixed smooth u, the left-hand side defines a linear map

57

58 3 Elliptic PDEs

from C∞c (Ω) into R, such that

|B(u, ϕ)| ≤n∑

i,j=1

‖aij∂ju‖L2‖∂iϕ‖L2 +

n∑i=1

‖bi∂iu‖L2‖ϕ‖L2 + ‖cu‖L2‖ϕ‖L2

≤n∑

i,j=1

‖aij‖L∞‖∂ju‖L2‖ϕ‖H1 +n∑i=1

‖bi‖L∞‖∂iu‖L2‖ϕ‖H1 + ‖c‖L∞‖u‖L2‖ϕ‖H1

≤ C(aij , bi, c)‖u‖H1‖ϕ‖H1 .

Note also that if f ∈ L2 (say) then

|(f, ϕ)| ≤ ‖f‖L2‖ϕ‖L2 ≤ ‖f‖L2‖ϕ‖H1 .

Thus if B(u, ϕ) = (f, ϕ) for every ϕ ∈ C∞c (Ω) then it also holds for every

ϕ ∈ H10 (Ω), using the density of C∞c (Ω) in H1

0 (Ω).

So we have shown that if u is a smooth solution of (3.1), then in fact we

must have

B(u, ϕ) = (f, ϕ) for all ϕ ∈ H10 (Ω), (3.3)

which is the weak form of (3.1). One can show – see examples for a simpler

case – that if u is smooth and (3.3) holds then u satisfies (3.1) in the classical

sense. We can weaken (3.3) just a little further, by noting that the right-hand

side in fact defines a linear functional on H10 (Ω), and choosing to consider

instead the problem

B(u, ϕ) = F (ϕ) for all ϕ ∈ H10 (Ω)

for some F ∈ H−1(Ω).

The problem is now in the right form to apply the Lax–Milgram Lemma.

We have already shown that our B is bounded from H10 × H1

0 into R; all

that is left is to show that B is coercive, i.e. that there exists a β > 0 such

that

B(u, u) ≥ β‖u‖2H1 for all u ∈ H10 (Ω).

This isn’t quite true; but we will show that there is a constant λ > 0 such

that B(u, v) + λ(u, v) is coercive.

In order to do this we will require the assumption that L is uniformly

elliptic: there exists a θ > 0 such that

n∑i,j=1

aij(x)ξiξj ≥ θ|ξ|2. (3.4)

Elliptic PDEs 59

So now

|B(u, u)| ≥ θn∑j=1

‖∂iu‖2L2 −(

maxi‖bi‖L∞

)‖∇u‖L2‖u‖L2 − ‖c‖L∞‖u‖2L2 .

We use “Young’s inequality with ε”,

ab ≤ ε

2a2 +

1

2εb2

on the second term,

b‖∇u‖L2‖u‖L2 ≤θ

2‖∇u‖2 +

1

2θb2‖u‖2L2 ,

so that

|B(u, u)| ≥ θ

2‖∇u‖2L2 − λ′‖u‖2L2 , (3.5)

where

λ′ =1

2θ

(maxi‖bi‖L∞

)2

+ ‖c‖L∞ .

So finally

|B(u, u)| ≥ θ

2‖u‖2H1 − λ0‖u‖2L2 ,

where λ0 = λ′ + (θ/2). (We could also use Poincare’s inequality to get a

better constant in front of the ‖u‖2L2 term.)

We therefore obtain the following theorem:

Theorem 3.1 Let L be a uniformly elliptic operator, and B the associated

bilinear form. Then there exists a constant λ0 > 0 such that for any λ ≥ λ0

the equation

B(u, ϕ) + λ(u, ϕ) = (f, ϕ) for all ϕ ∈ H10 (Ω)

has a unique solution u ∈ H10 (Ω), and

‖u‖H1 ≤ c‖f‖H−1 .

Proof The Lax–Milgram Lemma can be applied to the bilinear formB′(u, v) =

B(u, v) + λ(u, v).

60 3 Elliptic PDEs

Note that if bi = c = 0 then we can take λ0 = 0 in the previous theorem

(and we would only need to use the Riesz representation theorem rather

than the full Lax–Milgram Lemma).

But we should be able to better if f is nicer...

3.1 Elliptic regularity

Suppose that f ∈ L2(Ω) and consider again Poisson’s equation −∆u = f .

Suppose that u is smooth; then we can take the inner product of the equation

with ∂2ku and write∫

Ω(∆u)(∂2

ku) dx =

n∑i=1

∫Ω

(∂2i u)(∂2

ku) dx

= −n∑i=1

∫Ω

(∂iu)(∂i∂2ku) dx+

n∑i=1

∫∂Ω

(∂iu)(∂2ku)ni dS

=n∑i=1

∫Ω

(∂i∂ku)2 dx

+n∑i=1

∫∂Ω

(∂iu)(∂2ku)ni − (∂iu)(∂i∂ku)nk dS,

Summing over k this yields

n∑i,k=1

‖∂i∂ku‖2L2 = ‖∆u‖2L2 + boundary terms,

i.e. the second derivatives are controlled by the Laplacian. Since −∆u = f

this yields ∑|α|=2

‖∂αu‖2L2 = ‖f‖2L2 + boundary terms.

3.1.1 Interior regularity

It is relatively easy to show that if Ω′ ⊂⊂ Ω then the H2(Ω′) of u is bounded

by the L2(Ω) norm of f (‘interior regularity’). We give a less-than rigorous

proof, which nevertheless contains the main idea.


Theorem 3.2 If f ∈ L2(Ω), and u ∈ H10 (Ω) satisfies Lu = f , where L is a

uniformly elliptic operator as in (3.2) (plus (3.4)), then for every Ω′ ⊂⊂ Ω

there exists a constant cΩ′

‖u‖H2(Ω′) ≤ cΩ′(‖u‖H1(Ω) + ‖f‖L2(Ω)). (3.6)

Of course, if we know where u comes from (for example, Theorem 3.1),

then we already have ‖u‖H1(Ω) ≤ c‖f‖L2(Ω), and (3.6) becomes

‖u‖H2(Ω′) ≤ CΩ′‖f‖L2(Ω). (3.7)

Another remark is in order; if one looks carefully at the proof, we do not

in fact need the L2 norm of f on the whole of Ω, but only on a compact

subset of Ω, say Ω that contains Ω′. This means that if f ∈ C0(Ω), then

f ∈ L2(Ω), and hence u ∈ H2(Ω′). This observation will be useful below in

Corollary 3.4.

Proof Here is a non-rigorous proof for the simple case of Poisson’s equation.

Take Ω′ ⊂⊂ Ω, and choose a cut-off function ζ ∈ C∞c (Ω) such that 0 ≤ ζ ≤ 1

and ζ ≡ 1 on Ω′. Now, assuming that u is smooth,∫Ωζ2|∆u|2 dx =

∑i,j

∫Ω

(∂2i u)(∂2

j u)ζ2 dx

= −∑i,j

∫Ω

(∂iu)(∂i∂2j u)ζ2 dx− 2

∑i,j

∫Ω

(∂iu)(∂2j u)ζ(∂iζ) dx

=∑i,j

∫Ω

(∂i∂ju)2ζ2 dx

+ 2∑i,j

∫Ω

(∂iu)(∂i∂ju)ζ(∂jζ)− (∂iu)(∂2j u)ζ(∂iζ) dx.

(Note that this is the same as multiplying by the n ‘test functions’ ζ2∂2j u

and then summing over j.) The left-hand side is bounded by∫Ωζ2|∆u|2 dx ≤

∫Ω|∆u|2 dx =

∫Ω|f |2 dx,

62 3 Elliptic PDEs

so we just have to estimate the messy looking terms (with derivatives of ζ).

Since ζ is fixed and smooth, we have ‖∂jζ‖2L∞ ≤ c, so∣∣∣∣∫Ω

(∂iu)(∂i∂ju)ζ(∂jζ) dx

∣∣∣∣ ≤ ‖∂jζ‖L∞ ∫Ω|∂iu||∂i∂ju||ζ| dx

≤ 2n‖∂jζ‖2L∞∫

Ω|∂iu|2 dx+

1

8n

∫Ω

(∂i∂ju)2ζ2 dx;

we’ve used ab ≤ 2na2 + b2/8n on the integrand here.

So we have∑i,j

∫Ω

(∂i∂ju)2ζ2 dx ≤ ‖f‖2L2(Ω) +∑i,j

4cn‖∂iu‖2L2(Ω) +1

4n

∫Ω

(∂i∂ju)2ζ2

+∑i,j

4cn‖∂iu‖2L2(Ω) +1

4n

∫Ω

(∂2j u)ζ2 dx

≤ ‖f‖2L2(Ω) + 8cn2‖u‖2H1(Ω) +1

2

∫Ω

(∂i∂ju)2ζ2 dx,

from which it follows that

∑|α|=2

‖∂αu‖2L2(Ω′) =n∑

i,j=1

‖∂i∂ju‖2L2(Ω′)

≤∑i,j

∫Ω

(∂i∂ju)2ζ2 dx

≤ 2‖f‖2L2(Ω) + 16cn2‖u‖2H1(Ω),

and the result follows since

‖u‖2H2(Ω′) =∑|α|=2

‖∂αu‖2L2(Ω′) + ‖u‖2H1(Ω′).

The lack of rigour is due to the fact that our solution u does not satisfy

the equation −∆u = f , but the weak form (∇u,∇v) = (f, v) for every

v ∈ H10 (Ω), and that as far as we know u is only in H1

0 (Ω) and isn’t smooth.

How do we get round this? We want to do something like choosing v = ζ2∆u

as a test function, but since we only have u ∈ H10 , this isn’t smooth enough

(we’d have v ∈ H−1, in fact).


In order to deal with this we use difference quotients; given a function u

we define

Dhi u(x) =

u(x+ hei)− u(x)

h,

where ei is a unit vector in the ith direction. Then one can show that if

u ∈ H1(Ω), for any Ω′ ⊂⊂ Ω

‖Dhi u‖L2(Ω′) ≤ ‖∂iu‖L2(Ω),

i.e. ‘differences look like derivatives’, and more importantly that if Ω′ ⊂⊂ Ω

and for every h < dist(Ω′, ∂Ω) we have

‖Dhi u‖L2(Ω′) ≤ C,

then ∂iu ∈ L2(Ω′) and

‖∂iu‖L2(Ω′) ≤ C.

In other words, we can recover weak derivatives by the same sort of limiting

process that we can define classical derivatives.

In the rigorous version of the proof one takes the n test functions v =

D−hk (ζ2Dhkv) with k = 1, . . . , n; note that these are elements of H1

0 (Ω) if

h is small enough. Essentially one then repeats the estimates in the above

theorem in this setting, showing that the estimates are uniform when h is

sufficiently small.

If f is smoother, then u is smoother:

Theorem 3.3 Let u ∈ H10 (Ω) satisfy Lu = f , as in the above theorem. If

f ∈ Hs(Ω) then u ∈ Hs+2loc (Ω); for any Ω′ ⊂⊂ Ω there exists a constant cΩ′,s

such that

‖u‖Hs+2(Ω′)| ≤ c‖f‖Hs(Ω).

Proof Again, we dispense with rigour: suppose that −∆u = f . Then the

derivative ∂αu satisfies

−∆(∂αu) = ∂αf.

If f ∈ Hs(Ω), ∂αf ∈ L2(Ω) for any α with |α| = s; so ∂αu ∈ H2(Ω′) by

the preceding interior regularity result. In the rigorous version, one chooses

test functions v = (−1)|α|∂αw with w ∈ C∞c (Ω) and with |α| = s + 1, see

examples.

64 3 Elliptic PDEs

Corollary 3.4 If f ∈ C∞(Ω) and u ∈ H10 (Ω) is a weak solution of Lu = f

then u ∈ C∞(Ω).

Proof Take Ω′ ⊂⊂ Ω′′ ⊂⊂ Ω. If f ∈ C∞(Ω) then f ∈ C∞(Ω′′). It follows

that f ∈ Hk(Ω′′) for every k ∈ N. It then follows that u ∈ Hk+2(Ω′) for

every k ∈ N, and hence in Cr(Ω′) for every r. Since this holds for every

Ω′ ⊂⊂ Ω, u ∈ C∞(Ω).

Corollary 3.5 If u ∈ H10 (Ω) satisfies the weak form of

Lu = λu

then u ∈ C∞(Ω), and hence is a classical solution of Lu = λu.

Proof Suppose that u ∈ Hr(Ω′) for any Ω′ ⊂⊂ Ω. Now pick a particular

Ω′ ⊂⊂ Ω, and find Ω′′ such that Ω′ ⊂⊂ Ω′′ ⊂⊂ Ω. We know that u ∈Hr(Ω′′), and then u ∈ Hr+2(Ω′). Induction finishes the proof.

3.1.2 Boundary regularity

To show that if f ∈ L2(Ω) then in fact u ∈ H2(Ω) requires more work; but

the idea is clear if one considers the simple case of −∆u = f in the upper

half plane, which we treat now.

Suppose that u is smooth and has compact support in Rn+, contained in

D+ = B(0, R) ∩ Rn+, and that u satisfies

(∇u,∇v) = (g, v) for all v ∈ H10 (D+),

where D+ is B(0, R) ∩ Rn+. Then

‖u‖H2(Rn+) ≤ c‖g‖Rn

+.

Take v = ∂2ku as a test function. If k 6= n then note that v is still zero on

xn = 0, so is still an allowable test function. Then∫Rn

+

g∂2ku =

∑i

∫Rn

+

(∂iu)(∂i∂2ku) dx

= −∑i

∫Rn

+

(∂i∂ku)(∂i∂ku) dx,


where we have integrated by parts with respect to xk. Since k 6= n, we do

not include any terms from the boundary xn = 0; the boundary terms that

we would pick up vanish.

Now use Cauchy-Schwarz:∑i

‖∂k∂iu‖2L2(Rn+) ≤ ‖g‖L2(Rn

+)‖∂2ku‖L2(Rn

+)

≤ ‖g‖L2(Rn+)

(∑i

‖∂k∂iu‖L2(Rn+)

),

whence ∑i

‖∂k∂iu‖2L2(Rn+) ≤ ‖g‖L2(Rn

+).

This shows that every second partial derivative apart from ∂2nu is in L2(Rn+).

But the equation −∆u = f shows that

∂2nu = f −

n=1∑j=1

∂2j u.

Since each term on the right-hand side is in L2(Rn+), so is the left-hand side.

To make this work rigorously, one uses a cutoff function, a partition of

unity, and a straightening of the boundary. Note that when one uses the

coordinate transformation to straighten the boundary, one necessarily intro-

duces other terms in the equation (even if we start with just the Laplacian,

we end up having to consider a more general elliptic equation). We state

the theorem, including the higher regularity result.

Theorem 3.6 Suppose that Ω is a bounded domain with a Cr+2 boundary.

Then if u ∈ H10 (Ω) is a weak solution of Lu = f and f ∈ Hr(Ω) then in

fact u ∈ Hr+2(Ω) and

‖u‖H2(Ω) ≤ c(‖u‖H1(Ω) + ‖f‖L2(Ω)).

Corollary 3.7 Suppose that for some λ > 0, there exists a solution u ∈H1

0 (Ω) of the eigenvalue problem

Lu = λu.

Then u ∈ C∞(Ω).

4

Eigenvalues, eigenfunctions, and bases in Hilbertspaces

4.1 Eigenvalues and the spectrum

Let H be a Hilbert space and T ∈ B(H,H), then the point spectrum of T

consists of the set of all eigenvalues,

σp(T ) = λ ∈ C : Tx = λx for some non-zero x ∈ H.

If A is a linear operator on a finite-dimensional space V then λ ∈ C is an

eigenvalue of A if

Ax = λx for some non-zero x ∈ V.

In this case λ is an eigenvalue if and only if A− λI is not invertible (recall

that you can find the eigenvalues of an n×n matrix by solving det(A−λI) =

0). However, this is no longer true in infinite-dimensional spaces, and the

spectrum of A is potentially larger than the set of eigenvalues.

Definition 4.1 The resolvent set of T , R(T ), is

R(T ) = λ ∈ C : T − λI has a bounded inverse defined on all of H.

The spectrum σ(T ) of T ∈ B(H,H) is the complement of R(T ),

σ(T ) = C \R(T ),

i.e. the spectrum of T is the set of all complex λ for which T − λI does not

have a bounded inverse defined on all of H.

66


A bounded linear operator T ∈ L (H,H) is symmetric iff

(Tu, v) = (u, Tv) for all u, v ∈ H.

For compact symmetric operators, the spectrum of T consists only of the

eigenvalues of T (and perhaps zero). We will not prove this here, and content

ourselves with investigating the eigenvalue problem.

4.2 The spectrum of a compact symmetric operator

Theorem 4.2 Let T ∈ L (H,H) be symmetric. Then all the eigenvalues

of T are real and the eigenvectors corresponding to distinct eigenvalues are

orthogonal.

Proof Suppose that Tx = λx with x 6= 0. Then

λ‖x‖2 = (λx, x) = (Tx, x) = (x, T ∗x) = (x, Tx) = (x, λx) = λ‖x‖2,

i.e. λ = λ.

Now if λ and µ are distinct eigenvalues with Tx = λx and Ty = µy then

0 = (Tx, y)− (x, Ty) = (λx, y)− (x, µy) = (λ− µ)(x, y),

and so (x, y) = 0.

It is not immediately obvious that the following result has anything to do

with eigenvalues.

Theorem 4.3 Let H be a Hilbert space and T ∈ L (H,H) a symmetric

operator. Then

(a) (Tx, x) is real for all x ∈ H and

(b) ‖T‖ = sup|(Tx, x)| : x ∈ H, ‖x‖ = 1.

Proof For (a) we have

(Tx, x) = (x, Tx) = (Tx, x),

and so (Tx, x) is real. Now let M = sup|(Tx, x) : x ∈ H, ‖x‖ = 1.Clearly

|(Tx, x)| ≤ ‖Tx‖‖x‖ ≤ ‖T‖‖x‖2 = ‖T‖

68 4 Eigenvalues, eigenfunctions, and bases in Hilbert spaces

when ‖x‖ = 1, and so M ≤ ‖T‖.For any u, v ∈ H we have

4(Tu, v) = (T (u+ v), u+ v)− (T (u− v), u− v)

≤ M(‖u+ v‖2 + ‖u− v‖2)

≤ 2M(‖u‖2 + ‖v‖2)

using the parallelogram law. If Tu 6= 0 choose

v =‖u‖‖Tu‖

Tu

to obtain, since ‖v‖ = ‖u‖, that

4‖u‖‖Tu‖ ≤ 4M‖u‖2,

i.e. ‖Tu‖ ≤M‖u‖. This also holds if Tu = 0. It follows that ‖T‖ ≤M and

therefore that ‖T‖ = M .

Clearly |λ| ≤ ‖T‖op for any λ ∈ σp(T ), since if Tx = λx then

|λ|‖x‖ = |λx| = ‖Tx‖ ≤ ‖T‖op‖x‖.

We now consider eigenvalues of compact self-adjoint linear operators on a

Hilbert space. It is convenient to restrict attention to Hilbert spaces over C,

but this is no restriction, since we can always consider the ‘complexification’

of a real Hilbert space1.

1 Let H be a Hilbert space over R, and define its complexification HC as the vector space

HC = x+ iy : x, y ∈ H,

equipped with operations + and ∗ defined via

(x+ iy) + (w + iz) = (x+ w) + i(y + z), x, y, w, z ∈ V

and

(a+ ib) ∗ (x+ iy) = (ax− by) + i(bx+ ay) a, b ∈ R, x, y ∈ V.

Then equipped with the inner product

(x+ iy, w + iz)HC = (x,w) + i(y, w)− i(x, z) + (y, z)

HC is a Hilbert space. Just as we can complexify a Hilbert space H to give HC, we cancomplexify a linear operator T that acts on H to a linear operator TC that acts on HC: givenT ∈ B(H,H), extend T to a linear operator T : HC → HC via the definition

T (x+ iy) = Tx+ iTy x, y ∈ H.

Then T ∈ B(HC, HC), any eigenvalue of T is an eigenvalue of T , and any real eigenvalue of T

is an eigenvalue of T . If T is symmetric then T is symmetric.


Let’s think again about the weak formulation of the elliptic problem; given

f ∈ L2, find u ∈ H10 (Ω) such that

B(u, v) = (f, v) for all v ∈ H10 (Ω).

We could define a ‘solution mapping’ for this problem: if f ∈ L2(Ω) then

Tf = u, where u is the solution.

We can observe that

(i) T is linear; and

(ii) T maps L2(Ω) into H10 (Ω) continuously, i.e.

‖Tf‖H10 (Ω) ≤ c‖f‖L2(Ω).

(In fact our elliptic regularity tells us more, than T maps L2 into H2

continuously.)

Since H10 (Ω) ⊂ L2(Ω), T is clearly a linear map from L2 into itself, with

the additional property that a bounded set in L2 is mapped into a bounded

set inH10 . SinceH1 is compactly embedded in L2, this means that a bounded

set in L2 becomes a (pre)compact set in L2 (precompact means that its

closure is compact). T is therefore an example of a compact map:

Definition 4.4 A map T : X → Y is compact if whenever U is a bounded

subset of X, the closure of T (U) is a compact subset of Y . Equivalently,

if xn is a bounded sequence in X, then Txn has a subsequence that

converges in Y .

Note that a compact operator must be bounded, since otherwise there

exists a sequence in H with ‖xn‖ = 1 but ‖Txn‖ → ∞, and clearly Txncannot have a convergent subsequence.

If T ∈ L (H,K) has finite-dimensional range then T is compact, since

any bounded sequence in a finite-dimensional space has a convergent subse-

quence.

We now show that any compact symmetric operator has at least one

eigenvalue.

Theorem 4.5 Let H be a Hilbert space and T ∈ L (H,H) a compact sym-

metric operator. Then at least one of ±‖T‖op is an eigenvalue of T .


Proof We assume that T 6= 0, otherwise the result is trivial. From Theorem

4.3,

‖T‖op = sup‖x‖=1

|(Tx, x)|.

Thus there exists a sequence xn, of unit vectors, such that

(Txn, xn)→ ±‖T‖op = α.

Since T is compact there is a subsequence xnj such that Txnj is convergent

to some y. Relabel xnj as xn again.

Now consider

‖Txn − αxn‖2 = ‖Txn‖2 + α2 − 2α(Txn, xn)

≤ 2α2 − 2α(Txn, xn);

by the choice of xn, the right-hand side tends to zero as n→∞. It follows,

since Txn → y, that

αxn → y,

and since α 6= 0 is fixed we must have xn → x for some x ∈ H. Therefore

Txn → Tx = αx. It follows that

Tx = αx

and clearly x 6= 0, since ‖y‖ = |α|‖x‖ = ‖T‖op 6= 0.

Note that since any eigenvalue must satisfy |λ| ≤ ‖T‖op, it follows that

for compact symmetric operators

‖T‖op = sup|λ| : λ ∈ σp(T ).

Proposition 4.6 Let T be a compact self-adjoint operator on a Hilbert space

H. Then σp(T ) is either finite or consists of a countable sequence tending

to zero.

Proof Suppose that T has infinitely many eigenvalues that do not form a

sequence tending to zero. Then for some ε > 0 there exists a sequence of

distinct eigenvalues with |λn| > ε. Let xn be a corresponding sequence of

eigenvectors with ‖xn‖ = 1; then

‖Txn − Txm‖2 = (Txn − Txm, Txn − Txm) = |λn|2 + |λm|2 ≥ 2ε2

since (xn, xm) = 0. It follows that Txn can have no convergent subse-

quence, which contradicts the compactness of T .


4.3 Bases in Hilbert spaces

In an infinite-dimensional separable Hilbert space the best that we can hope

for in terms of a basis is to find a countable set ej∞j=1, in terms of which

to expand any x ∈ H as potentially infinite series1,

x =∞∑j=1

αjej . (4.1)

A set ej∞j=1 is a basis for H if every x ∈ H can be written uniquely in the

form (4.1) for some αj ∈ K. If in addition ej∞j=1 is an orthonormal set (i.e.

(ei, ej) = δij) then we refer to it as an orthonormal basis. We concentrate

on orthonormal bases.

Lemma 4.7 Let H be a Hilbert space and ej∞j=1 an orthonormal sequence

in H. Then for any x ∈ H∞∑j=1

|(x, ej)|2 ≤ ‖x‖2

(‘Bessel’s inequality’); consequently for any x ∈ H the series

∞∑n=1

(x, en)en

converges to some y ∈ H with ‖y‖2 =∑∞

j=1 |(x, en)|2.

For en to be a basis we need (of course) y = x.

Proof Let us denote by xk the partial sum

xk =

k∑j=1

(x, ej)ej with ‖xk‖2 =

k∑j=1

|(x, ej)|2.

Therefore

‖x− xk‖2 = (x− xk, x− xk)= ‖x‖2 − (xk, x)− (x, xk) + ‖xk‖2

= ‖x‖2 −k∑j=1

(x, ej)(ej , x)−k∑j=1

(x, ej)(x, ej) + ‖xk‖2

= ‖x‖2 − ‖xk‖2.1 equality here means that the partial sums converge to x in the norm of H


It follows that for every k

k∑j=1

|(x, ej)|2 = ‖xk‖2 ≤ ‖x‖2 − ‖x− xk‖2 ≤ ‖x‖2.

Thus if m > n

‖xm − xn‖2 =

∥∥∥∥∥∥m∑

j=n+1

(x, ej)ej

∥∥∥∥∥∥2

=

m∑j=n+1

|(x, ej)|2,

which shows that xm is a Cauchy sequence converging to sum y; the

expression for the norm follows since the norm of the limit is the limit of

the norms.

We now show give criteria for en to form a basis for H.

Proposition 4.8 Let E = ej∞j=1 be an orthonormal set in a Hilbert space

H. Then the following are equivalent to the statement that E is an orthonor-

mal basis for H:

(a) x =∑∞

n=1(x, en)en for all x ∈ H;

(b) ‖x‖2 =∑∞

n=1 |(x, en)|2 for all x ∈ H; and

(c) (x, en) = 0 for all n implies that x = 0.

(d) the linear span of en∞n=1 (the set of all finite linear combinations)

is dense in H, i.e. for any ε > 0 there exist N and αj, j = 1, . . . , N ,

such that

‖x−N∑j=1

αjej‖ < ε.

Proof If E is an orthonormal basis for H then we can write

x =∞∑j=1

αjej , i.e. x = limj→∞

n∑j=1

αjej .

Clearly if k ≤ n we have

(

n∑j=1

αjej , ek) = αk,

and using the properties of the inner product of limits we obtain αk = (x, ek)

and hence (a) holds. The same argument shows that if we assume (a) then

this expansion is unique and so E is a basis.


(a) ⇒ (b) is immediate from Lemma 4.7.

(b) ⇒ (c) is immediate since ‖x‖ = 0 implies that x = 0.

(c) ⇒ (a) Take x ∈ H and let

y = x−∞∑j=1

(x, ej)ej .

For each m ∈ N we have, using the continuity of the inner product,

(y, em) = (x, em)− limn→∞

n∑j=1

(x, ej)ej , em

= 0

since eventually n ≥ m. It follows from (c) that y = 0.

Finally, it is clear that (a) ⇒ (d). We show that (d) ⇒ (c): suppose that

(x, ej) = 0 for every j, and choose xn contained in the linear span of the

ej such that xn → x. Then

‖x‖2 = (x, x) = (x, limn→∞

xn) = limn→∞

(x, xn) = 0,

since xn is a finite linear combination of the ej.

Lemma 4.9 (Gram–Schmidt orthonormalisation) If ej∞j=1 is a lin-

early independent set, then there exists an orthonormal set ej∞j=1 such

that

span(e1, . . . , ek) = span(e1, . . . , ek)

for every k ∈ N.

Proof The proof is inductive. Suppose that we have already found e1, . . . , en.Then set

en+1 = en+1 =

n∑j=1

(en+1, ej)ej then en+1 =en+1

‖en+1‖.

It is easy to check that the new set e1, . . . , en+1 has the desired properties;

and the induction starts with e1 = e1/‖e1‖.

Proposition 4.10 A Hilbert space H is separable iff it has a countable

orthonormal basis.


Proof Given a countable orthonormal basis, consider finite linear combina-

tions with rational coefficients. This is countable and dense.

Given a countable dense subset, remove elements so that wj∞j=1 is a

linearly independent set whose closed linear span (elements that can be

approximated by finite linear combinations) is all of H. Then apply the

Gram–Schmidt process to produce an orthonormal set whose closed linear

span is still all of H.

Theorem 4.11 Any infinite-dimensional separable Hilbert space is isometric

to `2 (i.e. there exists a linear isomorphism that preserves the norm).

Proof H has a countable orthonormal basis ej∞j=1. Given x ∈ H, x =∑∞j=1(x, ej)ej , and ‖x‖2 =

∑∞j=1 |(x, ej)|2. Set xj = (x, ej), and consider

the sequence i(x) = x = (x1, x2, x3, . . .). Clearly x ∈ `2 and ‖x‖`2 = ‖x‖.The map i is linear and invertible, with i−1(x) =

∑∞j=1 xjej .

4.4 Eigenfunctions as basis elements; the Hilbert–Schmidt

Theorem

We can use the Gram–Schmidt idea to show:

Lemma 4.12 If T ∈ L (H,H) is compact and symmetric, then every non-

zero eigenvalue has only a finite number of linearly independent eigenvectors.

Proof Suppose that for some eigenvalue λ 6= 0 there exists an infinite num-

ber of linearly independent eigenvectors wj∞j=1. Using the Gram–Schmidt

process we can find a countably infinite collection of orthonormal eigenvec-

tors ej∞j=1 (any linear combination of the wj is still an eigenvector with

the same eigenvalue, since T (∑

j αjwj) =∑

j αjTwJ = λ(∑

j αjwj). The

proof now follows that of Proposition 4.6.

In order to prove the main theorem on eigenvalues and eigenfunctions of

compact symmetric operators we will require the following simple lemma.

Lemma 4.13 Let T ∈ L (H,H) be symmetric and let S be a closed linear

subspace of H such that TS ⊆ S. Then TS⊥ ⊆ S⊥.

4.4 Eigenfunctions as basis elements; the Hilbert–Schmidt Theorem 75

Proof Let x ∈ S⊥ and y ∈ S. Then Ty ∈ S and so (Ty, x) = (y, Tx) = 0

for all y ∈ S, i.e. Tx ∈ S⊥.

Theorem 4.14 (Hilbert–Schmidt Theorem). Let H be a Hilbert space and

T ∈ B(H,H) be a compact self-adjoint operator. Then there exists a finite

or countably infinite orthonormal sequence wn of eigenvectors of T with

corresponding non-zero real eigenvalues λn such that for all x ∈ H

Tx =∑j

λj(x,wj)wj . (4.2)

Consequently there exists an orthonormal basis for H consisting of eigen-

vectors of T , i.e. a set ej such that every x ∈ H can be written

x =∑j

(x, ej)ej .

Proof By Theorem 4.5 there exists a w1 such that ‖w1‖ = 1 and Tw1 =

±‖T‖w1. Consider the subspace of H perpendicular to w1,

H2 = w⊥1 .

Then since T is self-adjoint, Lemma 4.13 shows that T leaves H2 invariant.

If we consider T2 = T |H2 then we have T2 ∈ B(H2, H2) with T2 compact;

this operator is still self-adjoint, since for all x, y ∈ H2

(x, T2y) = (x, Ty) = (T ∗x, y) = (Tx, y) = (T2x, y).

Now apply Theorem 4.5 to the operator T2 on the Hilbert space H2 find

an eigenvalue λ2 = ±‖T2‖ and an eigenvector w2 ∈ H2 with ‖w2‖ = 1.

Continue this process as long as Tn 6= 0.

If Tn = 0 for some n then for any x ∈ H we have

y := x−n−1∑j=1

(x,wj)wj ∈ Hn.

Then

0 = Tny = Ty = Tx−n−1∑j=1

(x,wj)Twj = Tx−n−1∑j=1

λj(x,wj)wj

which is (4.2).


If Tn is never zero then consider

yn := x−n−1∑j=1

(x,wj)wj ∈ Hn.

Then we have

‖x‖2 = ‖yn‖2 +

n−1∑j=1

|(x,wj)|2,

and so ‖yn‖ ≤ ‖x‖. It follows that∥∥∥∥∥∥Tx−n−1∑j=1

λj(x,wj)wj

∥∥∥∥∥∥ = ‖Tyn‖ ≤ ‖Tn‖‖yn‖ = |λn|‖x‖,

and since |λn| → 0 as n→∞ we have (4.2).

Finally, let ej be an orthonormal basis for KerT ; each ej is an eigen-

vector of T with eigenvalue zero, and since Tej = 0 but Twj = λjwj with

λj 6= 0, we know that (wj , ek) = 0 for all j, k. So wj ∪ ek is a countable

orthonormal set in H.

Now, (??) implies that

T

x− ∞∑j=1

(x,wj)wj

= 0,

i.e. that x−∑∞

j=1(x,wj)wj ∈ KerT , and therefore

x−∞∑j=1

(x,wj)wj =∞∑k=1

αkek,

since ek is a basis for KerT . It follows that wj∪ek is an orthonormal

basis for H.

Note that it follows that when KerT is trivial, i.e. when T is invertible,

that the eigenvectors corresponding to non-zero eigenvalues of T span H.

4.5 Eigenvalues and eigenfunctions for elliptic PDEs

Consider the operator

Lu = −n∑

i,j=1

∂i(aij(x)∂ju), (4.3)

4.5 Eigenvalues and eigenfunctions for elliptic PDEs 77

where aij = aji, and the weak form of the equation Lu = f ,

B(u, v) =∑i,j

∫Ωaij(x)(∂iu)(∂iv) dx =

∫fv dx = (f, v) for all v ∈ H1

0 (Ω).

We know that B is symmetric.

We have already observe that the solution mapping u = Tf is compact.

To show that it is symmetric, take f, g ∈ L2(Ω) and let u = Tf and v = Tg.

Then u, v ∈ H10 (Ω) are both allowable ‘test functions’, and so

(f, Tg) = (f, v) = B(u, v)︸︷︷︸Lu=f

= B(v, u) = (g, u)︸︷︷︸Lv=g

= (g, Tf).

Corollary 4.15 If L is a second order elliptic operator (i.e. (3.4) is satisfied)

as in (4.3) with aij = aji then there exists a finite or countably infinite

sequence of C∞(Ω) eigenfunctions un satisfying

Lun = λnun,

where λn →∞ as n→∞. These eigenfunctions form an orthonormal basis

for L2(Ω).

Proof We know that T is a compact symmetric operator from L2(Ω) into

itself; let us show in addition that KerT = 0. Suppose that f ∈ KerT ;

this means that Tf = 0, i.e. that

0 = B(0, v) = (f, v) for all v ∈ H10 (Ω),

which implies that f = 01. It follows that T has a countable set of eigen-

functions un with Tun = λnun, where λn → 0 as n → ∞, which form an

orthonormal basis for L2(Ω).

Since Tun is the solution of the weak form of the equation for f = un, we

have

B(λnun, v) = (un, v) for all v ∈ H10 (Ω),

so that

B(un, v) =

(unλn, v

)for all v ∈ H1

0 (Ω);

set µn = λ−1n . Corollary 3.5 shows that u ∈ C∞(Ω).

1 This is not entirely straightforward, since f ∈ L2(Ω) but we have to take v ∈ H10 (Ω); see

examples.

5

Linear parabolic problems

We now want to consider linear parabolic problems, the model of which is

∂u

∂t−∆u = f(x, t) u|∂Ω = 0 (5.1)

subject to the initial condition

u(x, 0) = u0(x).

What sort of solution do we expect? Assuming that u is smooth, if we

multiply by u and integrate over Ω then we obtain∫Ωut(x, t)u(x, t) dx+

∫Ω|∇u(x, t)|2 dx =

∫f(x, t)u(x, t) dx,

so that

1

2

∫Ω|u(x, t)|2 dx+

∫Ω|∇u(x, t)|2 dx ≤ ‖f(·, t)‖L2(Ω)‖u(·, t)‖L2(Ω),

i.e.

1

2

d

dt‖u(·, t)‖2L2(Ω) + ‖∇u(·, t)‖2L2(Ω) ≤ c‖f(·, t)‖L2(Ω)‖∇u(·, t)‖L2(Ω),

using Poincare’s inequality (‖u‖L2 ≤ c‖∇u‖L2). Now we can use Young’s

inequality (2ab ≤ a2 + b2) on the right-hand side to obtain

d

dt‖u(·, t)‖2L2(Ω) + ‖∇u(·, t)‖2L2(Ω) ≤ c

2‖f(·, t)‖2L2(Ω).

Integrating this equation from 0 to t gives

‖u(·, t)‖2L2(Ω) +

∫ t

0‖∇u(·, s)‖2L2(Ω) ds ≤ ‖u0(·)‖2L2(Ω) +c2

∫ t

0‖f(·, s)‖2L2(Ω) ds.

78


This tells us that if u0 ∈ L2(Ω) and∫ T

0‖f(·, s)‖2L2(Ω) ds <∞

(we will weaken this second condition), then we should expect a solution

u(x, t) for which u(x, t) ∈ L2(Ω) for every t ≥ 0,

sup0≤t≤T

‖u(·, t)‖2L2(Ω) ≤M := ‖u0(·)‖2L2(Ω) + c2

∫ T

0‖f(·, s)‖2L2(Ω) ds

and such that the H1 norm of u is square integrable in time,∫ t

0‖∇u(·, s)‖2L2(Ω) ds ≤M.

Writing this more compactly, we expect u ∈ L∞(0, T ;L2) ∩ L2(0, T ;H1).

Moreover, since

ut = ∆u+ f(x, t),

and

|(∆u, v)| = |(∇u,∇v)| ≤ ‖∇u‖L2‖∇v‖L2 ,

implies that ‖∆u‖H−1 ≤ ‖∇u‖L2 , it follows that ut ∈ L2(0, T ;H−1).

A weak solution of (5.1) will be a solution u ∈ L2(0, T ;H10 ) with ut ∈

L2(0, T ;H−1); as we will see, other properties of u (like u ∈ C0([0, T ];L2))

will follow from this.

5.1 Banach-space valued function spaces and ‘Sobolev-type’

results

Notation becomes much easier if we suppress the x dependence, i.e. we

think of a solution u(x, t) as a map u : [0, T ] → X, where X is some space

of functions on Ω. Then the fact that u(t) ∈ X expresses the fact that for

each t, u(t) depends on X.

We now make the Lp(0, T ;X) notation more explicit. Let X be a Banach

space. We say that a function f : (0, T )→ X is in Lp(0, T ;X), 1 ≤ p <∞,

if the map t 7→ ‖f(t)‖X is measurable and∫ T

0‖f(t)‖pX dt <∞;

80 5 Linear parabolic problems

the norm in Lp(0, T ;X) is

‖f‖Lp(0,T ;X) =

(∫ T

0‖f(t)‖X dt

)1/p

.

We have f ∈ L∞(0, T ;X) is the map t 7→ ‖f(t)‖X is measurable and the

‖f‖L∞(0,T ;X) = ess sup0≤t≤T ‖f(t)‖X

is finite.

If X is complete then Lp(0, T ;X) is complete. If X is a Hilbert space

then L2(0, T ;X) is a Hilbert space. (See examples.)

The space C0([0, T ];X) consists of all functions that are continuous into

X, i.e. for which

‖f(t)− f(s)‖X → 0 as s→ t.

Just as we had to deal with weak spatial derivatives, we now also have to

deal with weak time derivatives: if u ∈ L1loc(0, T ;X) then ut ∈ L1

loc(0, T ;X)

is its weak time derivative if∫ T

0ut ϕdt = −

∫ T

0uϕt dt

for every ϕ ∈ C∞c (0, T ).

This is a little bit naughty, since we haven’t discussed integration of

Banach-space valued functions; the definition of L1(0, T ;X) only involved

integrating norms of f(t). If X is reflexive (see later) then provided that f

is integrable (can be approximated by step functions) and f ∈ L1(0, T ;X)

one can define the integral of f . Assuming that∥∥∥∥∫ T

0f(t) dt

∥∥∥∥X

≤∫ T

0‖f(t)‖X dt,

we can readily prove the following useful lemma.

Lemma 5.1 Suppose that un has weak time derivative un, and that

un → u and un → v in L1(0, T ;X).

Then v = u.


Proof Taking ϕ ∈ C∞c (0, T ), we have∫ T

0un ϕdt = −

∫ T

0un ϕt dt. (5.2)

Consider the left-hand side. We have∥∥∥∥∫ T

0(un − v)ϕdt

∥∥∥∥X

≤∫ T

0‖(un − v)ϕ(t)‖X dt

≤∫ T

0|ϕ(t)|‖un − v‖X dt

≤ ‖ϕ‖L∞(0,T )

∫ T

0‖un − v‖X dt,

and so the left-hand side of the equality (5.2) tends to∫v ϕdt.

Similarly, the right-hand side tends to

−∫uϕt dt,

whence v = u as claimed.

We now want to prove some ‘Sobolev’ type results for Banach-space valued

function spaces. The first is a generalisation of H1(0, T ) ⊂ C0([0, T ]).

Lemma 5.2 Suppose that f ∈ H1(0, T ;X), i.e. that f ∈ L2(0, T ;X) and

f ∈ L2(0, T ;X). Then

f(t) = f(s) +

∫ t

sf(τ) dτ for all 0 ≤ s ≤ t ≤ T, (5.3)

we have f ∈ C0([0, T ];X), and

sup0≤t≤T

‖f(t)‖X ≤ C‖f‖H1(0,T ;X). (5.4)

If you look carefully at the proof, f, f ∈ L1(0, T ;X) is sufficient for all

these results.

Proof Extend u to be zero outside [0, T ], and then mollify with respect to

t to give a function uh : [0, T ]→ X that is smooth in t, and is such that

uh, uh → u, u in L2(0, T ;X).


Now, for h > 0 we have

uh(t) = uh(s) +

∫ t

suh(τ) dτ,

and a variant of the argument of Lemma 5.1 gives (5.3). That f(t) is con-

tinuous into X follows since the integral of an L1 function is continuous.

For the norm bound, observe that it follows from (5.3) that

‖f(t)‖X ≤ ‖f(0)‖X +

∫ t

0‖dotf(τ)‖X dτ

≤ ‖f(0)‖X +

(∫ T

0dτ

)1/2(∫ t

0‖f(τ)‖2X dτ

)1/2

= ‖f(0)‖X + T 1/2‖f‖L2(0,T ;X),

and also

‖f(0)‖X ≤ ‖f(t)‖X +

∫ t

0‖f(τ)‖X dτ‖ ≤ ‖f(t)‖X + T 1/2‖f‖L2(0,T ;X).

Integrating this inequality with respect to t from 0 to T , to obtain

‖f(0)‖X ≤1

T

∫ T

0‖f(t)‖X dt+ T 1/2‖f‖L2(0,T ;X)

≤ 1

T 1/2

(∫ T

0‖f(t)‖X dt

)1/2

+ T 1/2‖f‖L2(0,T ;X),

whence (5.4) follows.

In the light of our heuristic discussion of weak solutions, the following

lemma will be more useful. We use 〈f, g〉 to denote the pairing of and

f ∈ H−1 with a g ∈ H10 (we have previously written this f(g)).

Lemma 5.3 Suppose that f ∈ L2(0, T ;H10 ) and f ∈ L2(0, T ;H−1). Then

(i) f ∈ C0([0, T ];L2),

(ii)

d

dt‖f(t)‖2L2 = 2

⟨f , f

⟩,

and

(iii)

sup0≤t≤T

‖f(t)| ≤ C(‖f‖L2(0,T ;H1

0 ) + ‖f‖L2(0,T ;H−1)

).

5.2 Weak solutions of parabolic problems 83

Note that (iii) shows that if fn → f in L2(0, T ;H10 ), fn → f in L2(0, T ;H−1),

then in fact fn → f in C0([0, T ];L2) (i.e. fn(t) → f(t) in L2, uniformly for

all t ∈ [0, T ].)

Proof Extend f by zero outside [0, T ] and mollify to give fh such that

fh → f in L2(0, T ;H10 ) and fh → f in L2(0, T ;H−1).

Then for any h > 0 and any t∗ ∈ [0, T ], since fh is smooth in time

‖fh(t)‖2L2 = ‖fh(t∗)‖2L2 + 2

∫ t

t∗

⟨fh(s), fh(s)

⟩ds.

Now choose t∗ ∈ [0, T ] such that

‖fh(t∗)‖2L2 =1

T

∫ T

0‖fh(t)‖2 dt.

Then

‖fh(t)‖2L2 ≤1

T

∫ T

0‖fh(t)‖2 dt+

∫ T

0‖fh(s)‖H−1‖fh(s)‖H1 ds,

i.e.

sup0≤t≤T

‖uh(t)‖2L2 ≤ C(‖fh‖L2(0,T ;H1)2 + ‖fh‖2L2(0,T ;H−1)

).

Letting h→ 0 gives (iii), (ii), and (i).

5.2 Weak solutions of parabolic problems

Assume that u is smooth, and take the inner product of (5.1) with some

fixed v ∈ C∞c (Ω); then for each fixed t we obtain

(ut, v)− (∆u, v) = (f(t), v),

and integrating by parts we obtain

(ut, v) + (∇u,∇v) = (f(t), v). (5.5)

Arguing as before, using the density of C∞c (Ω) in H10 (Ω), we expect (5.5) to

hold for every v ∈ H10 (Ω). We have made the equation ‘weak in space’. We

make it ‘weak in time’ by allowing ut to be the weak time derivative u.

We have therefore reformulated the problem as: given u0 ∈ L2(Ω) and


f(t) ∈ L2(0, T ;H−1(Ω)), find u ∈ L2(0, T ;H10 (Ω)) with u ∈ L2(0, T ;H−1(Ω))

such that u(0) = u0, and

(u, v) + (∇u,∇v) = 〈f(t), v〉 for all v ∈ H10 (Ω)

and for almost every t ∈ (0, T ) (since changing a function on a set of measure

zero will not affect its weak derivative).

This formulation is equivalent to the following, which will prove useful in

our existence/uniqueness proof: find u with the regularity above such that

u−∆u = f(t)

as an equality in L2(0, T ;H−1).

5.3 Galerkin approximations

We show the existence and uniqueness of weak solutions using the Galerkin

method – we approximate the original infinite-dimensional problem by a se-

quence of finite-dimensional problems, and then take the limit. In order to

do this we will use the orthonormal basis of L2(Ω) given by the eigenfunc-

tions wj∞j=1 of the Laplacian on Ω with Dirichlet boundary conditions (as

in Corollary 4.15). In fact these eigenfunctions are also an orthogonal basis

for H10 (Ω); they are a basis since L2(Ω) ⊃ H1

0 (Ω), and they are orthogonal

since

(wk, wj)H1 = (wk, wj)L2 + (∇wk,∇wj)L2 = δjk + (wk, (−∆)wj)

= δj + λj(wk, wk) = (1 + λj)δjk.

Now, if we take u ∈ L2(Ω) then we can approximate u in the space spanned

by the first n eigenfunctions,

Pnu =

n∑j=1

(u,wj)wj .

The same definition works for u ∈ H10 (Ω) (we still take the inner product in

L2), and for f ∈ H−1(Ω) we define Pnf by

〈Pnf, u〉 = 〈f, Pnu〉 for all u ∈ H10 (Ω).

Note that for u, v ∈ L2,

(u, Pnv) = (Pnu, v) = (Pnu, Pnv)


and similarly in H10 (see examples).

We now have the following.

Lemma 5.4 Let X = H10 (Ω), L2(Ω), or H−1(Ω), and take f ∈ X. Then

‖Pnf‖X ≤ ‖f‖X and Pnf → f in X.

Proof If X = L2 then since wj∞j=1 is an orthonormal basis,

f =∞∑j=1

(f, wj)wj and ‖f‖2L2 =∞∑j=1

|(f, wj)|2,

which implies the result. For X = H10 (Ω), we have

(∇f,∇f) = (f,−∆f) =∞∑j=1

λj |(f, wj)|2,

and again the result follows. For X = H−1(Ω), for the norm bound we have

| 〈Pnf, v〉 | = | 〈f, Pnv〉 | ≤ ‖f‖H−1‖Pnv‖H10≤ ‖f‖H−1‖v‖H1

0,

so that ‖Pnf‖H−1 ≤ ‖f‖H−1 . For the convergence, use the Reisz Represen-

tation Theorem to find a ϕ ∈ H10 such that

〈f, v〉 = (ϕ, v)H10

for all v ∈ H10 .

Then

‖Pnf − f‖H−1 = supv∈H1

0 : ‖v‖H1=1

| 〈Pnf − f, v〉 |,

and for ‖v‖H1 = 1,

| 〈Pnf − f, v〉 | = | 〈f, Pnv − v〉 |= |(ϕ, Pnv − v)H1

0|

= |(Pnϕ− ϕ, v)H10|

≤ ‖Pnϕ− ϕ‖H1‖v‖H1

= ‖Pnϕ− ϕ‖H1 ,

i.e. ‖Pnf − f‖H−1 → 0 as required.


We now find approximate solutions un(t) contained in the linear span of

w1, . . . , wn,

un(t) =n∑j=1

unj(t)wj .

Note that since each wj is a smooth function of x, each of the un is a smooth

function of x (for each fixed t). So we can integrate by parts, etc., rigorously.

We have two ways of viewing the equations for un, which are

(un, wj) +B(un, wj) = 〈f(t), wj〉 j = 1, . . . , n

First, as a set of n ODEs for the coefficients unj , which form a vector

un ∈ Rn,

unj + λjunj = fj(t) := 〈f(t), wj〉 , (5.6)

or as a PDE for the (smooth) function un,

∂un∂t−∆un = Pnf(t). (5.7)

The initial condition is un(0) = Pnu0.

Using standard theory for ODES, the n ODEs in (5.6) have - at least for a

short time - a unique solution un(t). Problems with this solution only arise

if the solutions ‘blows up’, i.e. if its norm tends to infinity. Since

‖un‖2Rn =n∑j=1

|unj |2 = ‖un‖2L2(Ω),

we could also show that the L2 norm of the function un remains bounded.

This is probably easier; in fact, we have already done these calculations in

our ‘heuristic’ analysis: take the inner product of (5.7) with un and integrate

by parts (we and do this because un is smooth), so that

1

2

d

dt‖un‖2L2 + ‖∇un‖2 = 〈Pnf(t), un〉

= 〈f(t), Pnun〉 = 〈f(t), un〉≤ ‖f(t)‖H−1‖un‖H1

≤ ‖f(t)‖H−1‖∇un‖L2 .

Using 2ab ≤ a2 + b2 we obtain

d

dt‖un‖2L2 + ‖∇un‖2 ≤ c2‖f(t)‖2H−1 ,


and hence, integrating from 0 to t,

‖un(t)‖2L2 +

∫ t

0‖∇un(s)‖2 ds ≤ ‖un(0)‖2L2 + c2

∫ t

0‖f(s)‖2H−1 ds

≤ ‖u0‖2L2 + c2

∫ T

0‖f(s)‖2H−1 ds,

for all t ∈ [0, T ].

So we have an approximate solution un that is bounded in L∞(0, T ;L2)

and L2(0, T : H10 ) with the bound independent of n. Arguing as we did

before, we also have a bound on un in L2(0, T ;H−1) that does not depend

on n.

We now need to show that these approximate solutions in fact converge to

some u as n→∞, and that this u satisfies the equation. These are separate

issues; it is possible to have un converges to u but not in a strong enough

sense that u actually satisfies the limiting equation (this will, fortunately,

not be the case here).

In this simple linear case we can show the convergence directly: consider

dundt−∆un = Pnf(t) un(0) = Pnu0

anddumdt−∆um = Pmf(t) um(0) = Pmu0.

We want to show that un (and also un) is Cauchy. To do this, consider

wnm = un − um, with n > m. Then wnm satisfies

dwnmdt

−∆wnm = (Pn − Pm)f(t) wnm(0) = (Pn − Pm)u0.

If we repeat the calculations above we obtain

1

2

d

dt‖wnm‖2L2 + ‖∇wnm‖2L2 = 〈(Pn − Pm)f(t), wnm〉

≤ c‖(Pn − Pm)f(t)‖H−1‖∇wnm‖L2 ,

and sod

dt‖wnm‖2L2 + ‖∇wnm‖2L2 ≤ ‖(Pn − Pm)f(t)‖2H−1 .

Integrating from 0 to T we obtain

‖wnm(T )‖2+

∫ T

0‖∇wnm(s)‖2 ds ≤ ‖wnm(0)‖2L2+

∫ T

0‖(Pn−Pm)f(t)‖2H−1 dt.


Choose ε > 0. Then we can make ‖wnm(0)‖2L2 as small as we wish, since

Pnu0 → u0 in L2. We show that Pnf → f in L2(0, T ;H−1), from which

it follows that Pnf is Cauchy in the same space, and hence that wnm is

Cauchy in L2(0, T ;H1). Consider∫ T

0‖Pnf(t)− f(t)‖2H−1 dt. (5.8)

We know that Pnf(t) → f(t) in H−1 for each fixed t, i.e. that ‖Pnf(t) −f(t)‖2H−1 → 0 for each fixed t; and we also know that

‖Pnf(t)− f(t)‖2H−1 ≤ (‖Pnf(t)‖H−1 + ‖f(t)‖H1 )2 ≤ 4‖f(t)‖2H−1 ∈ L1(0, T ).

So we can use the Dominated Convergence Theorem to show that (5.8)

converges to zero.

So un converges to some u in L2(0, T ;H1) (and also in L∞(0, T ;L2,

but we will soon do better than this). It follows that −∆un → −∆u in

L2(0, T ;H−1); since we have already seen that Pnf → f in L2(0, T : H−1),

it follows from the equation

un = −∆un + Pnf(t)

that un → v in L2(0, T ;H−1). Since un → u in L2(0, T ;H1) implies that

un → u in L2(0, T ;H−1), we can use Lemma 5.1 to deduce that v = u.

We have therefore shown that every term in the equation converges to its

appropriate limit in L2(0, T ;H−1), so u ∈ L2(0, T ;H10 ), u ∈ L2(0, T ;H−1),

and u satisfies

u−∆u = f(t)

as an equality in L2(0, T ;H−1). We remarked already that this was equiva-

lent to the weak form (5.6).

All that is left is to check that u satisfies the initial condition and is unique.

Since un is Cauchy in L2(0, T ;H1) and un is Cauchy in L2(0, T ;H−1), it

follows from Lemma 5.3 that in fact un is Cauchy in C0([0, T ];L2). In

particular, u(0) = limn→∞ un(0) = limn→∞ Pnu0 = u0, as required.

Finally, for uniqueness we again use Lemma 5.3, which ensure that if

du

dt−∆u = f(t) and

dv

dt−∆v = f(t)


with u(0) = v(0) then w = ∆w and

1

2

d

dt‖w‖2L2 = 〈w, w〉 = 〈∆w,w〉 = −‖∇w‖2L2 ≤ 0,

whence ‖w(t)‖2L2 = 0 for all t ≥ 0, i.e. the solutions are unique.

Combining this, we have proved the following:

Theorem 5.5 Given u0 ∈ L2 and f ∈ L2(0, T ;H−1), the equation

ut −∆u = f(t) u|∂Ω = 0 u(0) = u0

has a unique weak solution u, with u ∈ L2(0, T ;H10 ) ∩ C0([0, T ];L2) and

u ∈ L2(0, T ;H−1), such that u(0) = u0 and

u−∆u = f(t)

as an equality in L2(0, T ;H−1).

One can also obtain higher regularity of solutions assuming higher regu-

larity of u0 and f , for example: if u0 ∈ H10 (Ω) and f ∈ L2(0, T ;L2) then in

fact u ∈ L2(0, T ;H2) ∩ C0([0, T ];H10 ) and u ∈ L2(0, T ;L2). (The equation

then holds as an equality in L2(0, T ;L2), and in particular almost every-

where in (0, T ) × Ω). To prove this rigorously one performs the following

heuristic computations using the Galerkin approximations and then takes

limits; but essentially the result follows from the following ‘formal estimates’

(‘formal’ here means ‘assume that everything is smooth enough to carry out

any manipulations you like’). Take the inner product with −∆u to give∫Ωut · (−∆u) dx+ ‖∆u‖2L2 = (f(t),−∆u) ≤ ‖f(t)‖L2‖∆u‖L2 .

The first term is

−n∑

i,j=1

∫Ω

(∂tuj)(∂2i uj) =

n∑i,j=1

∫Ω∂t(∂iuj)(∂iuj) =

1

2

d

dt

n∑i,j=1

∫Ω|∂iuj |2

=1

2

d

dt‖∇u‖2,

so1

2

d

dt‖∇u‖2 + ‖∆u‖2 ≤ ‖f(t)‖L2‖∆u‖L2 ,

which using Young’s inequality gives

d

dt‖∇u‖2 + ‖∆u‖2 ≤ ‖f(t)‖2L2 .


Integrating from 0 to t we obtain

‖∇u(t)‖2L2 +

∫ t

0‖∆u(s)‖2L2 ds ≤ ‖∇u(0)‖2L2 +

∫ t

0‖f(s)‖2L2 ds.

Since u0 ∈ H10 and f ∈ L2(0, T ;L2) this shows that we should expect

u ∈ L∞(0, T ;H10 ) and ∫ t

0‖∆u(s)‖2L2 ds <∞.

Since the theory of elliptic regularity shows that when −∆u = f then

‖u‖H2 ≤ c‖f‖L2 , i.e. that

‖u‖H2 ≤ c‖∆u‖L2

it follows also that u ∈ L2(0, T ;H2). Now, if u ∈ H2, then ∆u ∈ L2. So

from u = ∆u+ f(t) we expect u ∈ L2(0, T ;L2).

We know that if f ∈ L2(0, T ;H10 ) and f ∈ L2(0, T ;H1) then f ∈ C0([0, T ];L2).

One can in fact show - with just a little more work - that the same result hold

if f ∈ L2(0, T ;H1) (H1 now, not H10 ). Using this (which we haven’t proved),

one can show that the increased regularity of u and u implies that u is contin-

uous into H1: indeed, if u ∈ L2(0, T ;H2) then for any j = 1, . . . , n, we have

∂ju ∈ L2(0, T ;H1) and ∂j u ∈ L2(0, T ;H−1). So then ∂ju ∈ C0([0, T ];L2),

which shows that u ∈ C0([0, T ];H1).

There are of course even higher regularity results, e.g. if u0 ∈ Hk and

f ∈ L2(0, T ;Hk−1) then u ∈ L2(0, T ;Hk+1), u ∈ L2(0, T ;Hk−1), and u ∈C0([0, T ];Hk). See Evans (****) for details. One can also prove regularity

in time with additional assumptions on f ; see the examples.

6

Nonlinear equations, weak convergence

In our previous analysis we first obtained uniform estimates on the Galerkin

approximations: we showed that un was bounded in L∞(0, T ;L2) and L2(0, T ;H10 ),

and the un was bounded in L2(0, T ;H−1), with the bounds independent of

n.

Because the equations were linear, we could then in fact show directly that

un was Cauchy in these spaces. But more generally we cannot proceed di-

rectly: we require some compactness results to guarantee that boundedness

of un allows us to find subsequences that converge in some sense. We now

introduce such a notion of convergence, and prove the required compactness

theorem.

6.1 Dual spaces, the Hahn–Banach Theorem, and weak

convergence

We have briefly introduced dual spaces earlier, and we now consider them

in a little more detail.

Recall that if X is a Banach space, its dual space X∗ consists of all

bounded linear functionals from X into K (although we restrict to the case

K = R here), i.e. X∗ = L (X,K). The norm on X∗ is the L (X,K) norm

(‘operator norm’),

‖f‖X∗ = supx∈X: ‖x‖X=1

|f(x)|.

The following theorem is somehow the first and central result in the study

91

92 6 Nonlinear equations, weak convergence

of dual spaces. It shows that a linear functional defined on a subspace of X

can be extended to a linear functional defined on the whole of X without

increasing its norm.

Theorem 6.1 (Hahn–Banach Theorem) Let X be a Banach space, and

U a subspace of X. Suppose that f : U → R is a linear functional on U

such that

|f(x)| ≤M‖x‖ for all x ∈ U.

Then there exists an F ∈ X∗ that extends f (i.e. F (x) = f(x) for all x ∈ U)

and does not increase its norm,

|F (x)| ≤M‖x‖ for all x ∈ X.

For a simple proof in a Hilbert space see the examples. The result in full

generality relies on Zorn’s Lemma, i.e. on the Axiom of Choice. Although it

looks fairly innocent, it has some interesting consequences (see examples).

As an immediate application, we prove the existence of a particularly

useful class of linear functionals, and show therefore that understanding

linear functionals is in some way enough to understand elements of X.

Lemma 6.2 Let x ∈ X. Then there exists an f ∈ X∗ such that ‖f‖X∗ = 1

and f(x) = ‖x‖.

Proof Define f on the linear space U spanned by x as

f(αx) = α‖x‖.

Then f(x) = ‖x‖ and |f(z)| ≤ ‖z‖ for all z ∈ U . Extend f to an f ∈ X∗;then ‖f‖X∗ = 1 and f(x) = f(x) = ‖x‖.

Corollary 6.3 Let x, y ∈ X. If f(x) = f(y) for every f ∈ X∗ then x = y.

Proof If x 6= y then by the previous lemma there exists an f with ‖f‖X∗ = 1

such that f(x)− f(y) = f(x− y) = ‖x− y‖ 6= 0.

We say that xn ∈ X converges weakly to x ∈ X, and write xn x, if

f(xn)→ f(x) for all f ∈ X∗.


If xn → x then xn x; for any f ∈ X∗

|f(xn)− f(x)| ≤ ‖f‖X∗‖xn − x‖X → 0 as n→∞.

Note that in a Hilbert space, where every linear functional is of the form

(y, ·) for some y ∈ H, xn x if

(xn, y)→ (x, y) for all y ∈ H.

This provides an easy example of a sequence that converges weakly but does

not converge; pick a countable orthonormal set ej∞j=1. Then for any y ∈ HBessel’s inequality

∞∑j=1

|(y, ej)|2 ≤ ‖y‖2

shows that the sum converges; it follows that (y, ej) → 0 as j → ∞, and

hence that ej 0. But the sequence ej does not converge (any two

elements are a distance√

2 apart).

Lemma 6.4 If xn x then

‖x‖ ≤ lim infn→∞

‖x‖n.

Proof Choose f ∈ X∗ with ‖f‖X∗ = 1 such that f(x) = ‖x‖. Then

‖x‖ = f(x) = limn→∞

f(xn),

so

‖x‖ ≤ lim infn→∞

|f(xn)| ≤ lim infn→∞

‖f‖X∗‖xn‖X ;

the result follows since ‖f‖X∗ = 1.

In fact any weakly convergent sequence is bounded; this is not straight-

forward (see examples).

Lemma 6.5 Weak limits are unique.

Proof Suppose that xn x and xn y. Then for any f ∈ X∗, f(x) =

limn→∞ f(xn) = f(y). So by Lemma 6.3, x = y.

The following result is often useful.


Lemma 6.6 Suppose that X,Y are Banach spaces, with Y compactly em-

bedded in X. Then if yn y in Y , yn → y in X.

Proof First, observe that yn y in Y . Take f ∈ Y ∗, then Y is a linear

subspace of X so there is an F ∈ X∗ such that F = f on Y , and so

f(yn) = F (yn)→ F (y) = f(y).

Now, if yn 6→ y in X, then there is an ε > 0 and a subsequence (which we

relabel) yn such that ‖yn − y‖X > ε for every j. Since yn is a bounded

sequence in Y , and Y is compactly embedded in X, it has a subsequence

(which we relabel) that converges to some z ∈ X. If yn → z in X then

yn z in X; but weak limits are unique, so z = y, a contradiction.

There is another notion of weak convergence, weak-* convergence, which

deals with sequences of elements of X∗. If fn ∈ X∗ then fn converges

weakly-* to f , fn∗ f if

fn(x)→ f(x) for all x ∈ X.

Theorem 6.7 Suppose that X is separable. Then a bounded sequence in X∗

has a weakly-* convergent subsequence.

Proof Let xk be a countable dense subset of X, and fj a sequence in X∗

such that ‖fj‖X∗ ≤M . A standard diagonal argument yields a subsequence

of the fj (which we relabel) such that fj(xk) converges for every k. To

show that in fact fj(x) converges for every x ∈ X, given ε > 0 find xk such

that ‖x− xk‖ < ε/3M , and write

|fi(x)− fj(x)| ≤ |fi(x)− fi(xk)|+ |fi(xk)− fj(xk)|+ |fj(xk)− fj(x)|≤ ‖fi‖X∗‖x− xk‖+ |fi(xk)− fj(xk)|+ ‖fj‖X∗‖xk − x‖

≤ 2ε

3+ |fi(xk)− fj(xk)|.

Since fj(xk) is Cauchy, one can find an N such that the second term is

< ε/3 for all i, j ≥ N . One then shows as in the proof of Theorem 1.3 that

f defined for each x ∈ X by f(x) = limn→∞ fn(x) is an element of X∗ with

‖f‖X∗ ≤M .

We can convert this into a compactness result for bounded subsets of X

under a condition that relates the second dual of X, (X∗)∗, to X itself...


Since X∗ is a Banach space (Theorem 1.3), we can consider its dual, (X∗)∗

(usually written just X∗∗). We already have a large collection of elements

of X∗∗ to hand; given any x ∈ X, define

Lx(f) = f(x) for all f ∈ X∗.

Then Lx is linear in f , and since

|Lx(f)| = |f(x)| ≤ ‖f‖X∗‖x‖X

and if f ∈ X∗ has f(x) = ‖x‖ and ‖f‖X∗ = 1 then

|Lx(f)| = |f(x)| = ‖x‖ = ‖f‖X∗‖x‖X

it follows that ‖Lx‖X∗∗ = ‖x‖X . The map x 7→ Lx maps X isometrically

onto a subspace of X∗∗. If this map is onto, i.e. if for every G ∈ X∗∗

there exists an x such that G = Lx, then X is called reflexive. Loosely, X is

reflexive if X = X∗∗; but the equality here has to be understood in the sense

that the two spaces are isometrically isomorphic under the map x 7→ Lx.

Examples: any Hilbert space is reflexive (since H∗ ' H). Every Lp space

with 1 < p <∞ is reflexive, since (Lp)∗ ' Lq with p, q conjugate: that every

f ∈ Lq(Ω) gives rise to a linear functional on Lp(Ω) follows from Holder’s

inequality: define

Lf (g) :=

∫Ωf(x)g(x) dx,

and then

|Lf (g)| ≤ ‖f‖Lq‖g‖Lp .

To show that ‖Lf‖(Lp)∗ = ‖f‖Lq , observe that if f ∈ Lq then the function

g = f |f |q−2 ∈ Lp, since for (p, q) conjugate p = q/(q − 1), so that

‖g‖Lp =

(∫(|f |q−1)p

)1/p

=

(∫|f |q)

= ‖f‖q/pLq .

With this choice for g, we have

|Lf (g)| =∥∥∥∥∫

Ω|f |q dx

∣∣∣∣ = ‖f‖qLq = ‖f‖Lq‖f‖q−1Lq = ‖f‖Lq‖g‖p(q−1)/q

Lp

= ‖f‖Lq‖g‖Lp ,

which shows that ‖Lf‖(Lp)∗ = ‖f‖Lq as required. Some more work is re-

quired to show that every such linear functional can be written as Lf for

some f .


The dual of L1 is L∞; the dual of L∞ is not L1. So L1 is not reflexive.

Neither is L∞, since X is reflexive iff X∗ is reflexive (a simple but head-

spinning proof). (However, L1 is separable, so we can apply Theorem 6.7 to

deduce that a sequence bounded in L∞ = (L1)∗ has a weakly-* convergent

subsequence.)

We can now prove the following more comfortable compactness result.

Theorem 6.8 Let X be a reflexive Banach space with X∗ separable. Then

any bounded sequence in X has a weakly convergent subsequence.

The result is still true (with a more complicated proof) dropping the

requirement that X is separable.

Proof Take a bounded sequence xn ∈ X. Consider the sequence Gn ∈ X∗∗with

Gn(f) = f(xn) for all f ∈ X∗.

Then ‖Gn‖X∗∗ = ‖xn‖X , so Gn is a bounded sequence in (X∗)∗. Theorem

6.7 implies that there is a subsequence that converges weakly-* in (X∗)∗ to

some G ∈ X∗∗.Since X is reflexive, there exists an x ∈ X such that G = Lx, i.e. such

that

G(f) = f(x) for all f ∈ X∗.

That Gn∗ G in (X∗)∗ says that

f(xn) = Gn(f)→ G(f) = f(x) for all f ∈ X∗.

This is precisely xn x.

6.2 Weak compactness and our linear parabolic PDE

If we return to our uniform bounds on the Galerkin solutions of our PDE

ut −∆u = f(t)

we have un bounded in L2(0, T ;H10 ) (also L∞(0, T ;L2)) and un bounded in

L2(0, T ;H−1).


Using the above compactness theorems, we can find subsequence such that

unj u in L2(0, T ;H10 )

unj

∗ u in L∞(0, T ;L2)

unj

∗ v in L2(0, T ;H−1).

Some comments are in order here: L2(0, T ;H10 ) is a Hilbert space, hence

reflexive, so we can use Theorem 6.8 to find a subsequence for which unj u

in L2(0, T ;H10 ).

L∞(0, T ;L2) is not reflexive. But it is the dual space of L1(0, T ;L2),

which is separable. So we can use Theorem 6.7 to find a sub-subsequence

(a subsequence of the first subsequence) for which (relabelling) we also have

unj

∗ w in L∞(0, T ;L2). L2(0, T ;L2) is a Hilbert subspace of both these

two spaces, so unj u and unj

∗ w in L2(0, T ;L2); in a Hilbert space (in

fact in any reflexive space) weak and weak-* convergence are equivalent, so

by uniqueness of weak limits u = w.

The space L2(0, T ;H−1) is the dual of L2(0, T ;H10 ), so we can use Theo-

rem 6.7 again to find a weakly-* convergent subsequence (we take a further

sub-sub-sub-sequence so as not to lose the two ‘convergences’ that we had

already).

What is, for example, weak-* convergence in L2(0, T ;H10 )? This is the

dual space of L2(0, T ;H−1); so given any ϕ ∈ L2(0, T ;H10 ),∫ T

0

⟨unj (t), ϕ(t)

⟩dt→

∫ T

0〈v(t), ϕ(t)〉 dt.

Similarly, weak convergence of unj to u in L2(0, T ;H10 ) means that for any

ϕ ∈ L2(0, T ;H−1),∫ T

0

⟨ϕ(t), unj (t)

⟩dt→

∫ T

0〈ϕ(t), u(t)〉 dt.

This is already enough to sort out one of our problems, whether v = u.

Take v ∈ H10 and ψ(t) ∈ C∞c (0, T ), then, since vψ(t) ∈ L2(0, T ;H1

0 ),∫ T

0

⟨unj (t), vψ(t)

⟩dt→

∫ T

0〈v(t), vψ(t)〉 dt.


But∫ T

0

⟨unj (t), vψ(t)

⟩dt =

⟨∫ T

0unj (t)ψ(t) dt, v

⟩= −

⟨∫ T

0unj (t)ψ(t) dt, v

⟩= −

∫ T

0

⟨unj (t), ψ(t)v

⟩dt.

Now certainly ψ(t)v ∈ L2(0, T ;H−1), so using the weak convergence of unj

to u in L2(0, T ;H10 ), this expression converges to

−∫ T

0〈u(t), ψ(t)v〉 dt.

We therefore have, for every v ∈ H10 ,∫ T

0〈v(t), vψ(t)〉 dt = −

∫ T

0〈u(t), ψ(t)v〉 dt.

This implies that⟨∫ T

0v(t)ψ(t) dt, v

⟩=

⟨−∫ T

0u(t)ψ(t) dt, v

⟩for every v ∈ H1

0 , i.e. that∫ T

0v(t)ψ(t) dt = −

∫ T

0u(t)ψ(t) dt

as an equality in H−1, so v = u.

We want to show that every term in

un −∆un = Pnf(t)

converges weakly-* to its limit in L2(0, T ;H−1). We have, at least, that

un v in this sense; and we showed before the Pnf(t) → f(t) strongly

in L2(0, T ;H−1), and hence weakly. For the Laplacian term, take ϕ ∈L2(0, T ;H1

0 ); then∫ T

0〈−∆un(t), ϕ(t)〉 dt =

∫ T

0〈un(t),−∆ϕ(t)〉 dt

→∫ T

0〈u(t),−∆ϕ(t)〉 dt

=

∫ T

0〈−∆u(t), ϕ(t)〉 dt,

since ∆φ(t) ∈ L2(0, T ;H−1) and un u in L2(0, T ;H10 ).


It follows that

u−∆u = f(t)

as an equality in L2(0, T ;H−1), with u ∈ L2(0, T ;H10 ) and u ∈ L2(0, T ;H−1).

We therefore have u ∈ C0([0, T ];L2), and it only remains to check that

u(0) = u0.

Last time we used the uniform convergence of un to u (i.e. convergence in

C0([0, T ];L2)), but we need a more roundabout argument now.

First, since we have

u−∆u = f(t)

in L2(0, T ;H−1), choose v ∈ H10 , and then z ∈ C1([0, T ];H1

0 ) ⊂ L2(0, T ;H10 )

with z(T ) = 0 and z(0) = v. Then∫ T

0〈u(t), z(t)〉+ (∇u(t),∇z(t)− 〈f(t), z(t)〉 dt = 0.

Integrating the first term by parts (since u is the weak time derivative) we

have ∫ T

0〈u(t),−z(t)〉+ (∇u(t),∇z(t)− 〈f(t), z(t)〉 dt = (u(0), v).

(We are not just using the definition of the weak time derivative, but also

the result of Lemma 5.2; see the examples.)

For the Galerkin approximations we have

un −∆un = Pnf(t);

taking the inner product with z and integrating in space and time yields∫ T

0〈un(t),−z(t)〉+ (∇un(t),∇z(t))− 〈Pnf(t), z(t)〉 dt = (Pnu0, v).

Using the fact that un u in L2(0, T ;H10 ) and Pnf → f in L2(0, T ;H−1),

we let n→∞ to obtain∫ T

0〈u(t),−z(t)〉+ (∇u(t),∇z(t)− 〈f(t), z(t)〉 dt = (u0, v).

This shows that (u(0), v) = (u0, v) for every v ∈ H10 . So u(0) = u0 as

required.

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