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90

Chapter 5

Banach Spaces

Many linear equations may be formulated in terms of a suitable linear operator

acting on a Banach space. In this chapter, we study Banach spaces and linear oper-

ators acting on Banach spaces in greater detail. We give the definition of a Banach

space and illustrate it with a number of examples. We show that a linear operator

is continuous if and only if it is bounded, define the norm of a bounded linear op-

erator, and study some properties of bounded linear operators. Unbounded linear

operators are also important in applications: for example, differential operators are

typically unbounded. We will study them in later chapters, in the simpler context

of Hilbert spaces.

5.1 Banach spaces

A normed linear space is a metric space with respect to the metric d derived from

its norm, where d(x, y) = x y.

Definition 5.1 A Banach space is a normed linear space that is a complete metric

space with respect to the metric derived from its norm.

The following examples illustrate the definition. We will study many of these

examples in greater detail later on, so we do not present proofs here.

Example 5.2 For 1 p

92 Banach Spaces

Example 5.3 The space C([a, b]) of continuous, real-valued (or complex-valued)

functions on [a, b] with the sup-norm is a Banach space. More generally, the space

C(K) of continuous functions on a compact metric space K equipped with the

sup-norm is a Banach space.

Example 5.4 The space Ck([a, b]) of k-times continuously differentiable functions

on [a, b] is not a Banach space with respect to the sup-norm for k 1, since

the uniform limit of continuously differentiable functions need not be differentiable.

We define the Ck-norm by

fCk = f + f + . . .+ f

(k).

Then Ck([a, b]) is a Banach space with respect to the Ck-norm. Convergence with

respect to the Ck-norm is uniform convergence of functions and their first k deriva-

tives.

Example 5.5 For 1 p < , the sequence space `p(N) consists of all infinite

sequences x = (xn)n=1 such that

n=1

|xn|p

Banach spaces 93

For p = , the space L ([a, b]) consists of the Lebesgue measurable functions

f : [a, b] R (or C) that are essentially bounded on [a, b], meaning that f is

bounded on a subset of [a, b] whose complement has measure zero. The norm on

L ([a, b]) is the essential supremum

f = inf {M | |f(x)| M a.e. in [a, b]} .

More generally, if is a measurable subset of Rn, which could be equal to Rn itself,

then Lp() is the set of Lebesgue measurable functions f : R (or C) whose

pth power is Lebesgue integrable, with the norm

fp =

(

|f(x)|p dx

)1/p.

We identify functions that differ on a set of measure zero. For p = , the space

L() is the space of essentially bounded Lebesgue measurable functions on

with the essential supremum as the norm. The spaces Lp() are Banach spaces for

1 p .

Example 5.7 The Sobolev spaces, W k,p, consist of functions whose derivatives

satisfy an integrability condition. If (a, b) is an open interval in R, then we define

W k,p ((a, b)) to be the space of functions f : (a, b) R (or C) whose derivatives of

order less than or equal to k belong to Lp ((a, b)), with the norm

fW k,p =

kj=0

ba

f (j)(x)p dx1/p .

The derivatives f (j) are defined in a weak, or distributional, sense as we explain

later on. More generally, if is an open subset of Rn, then W k,p() is the set of

functions whose partial derivatives of order less than or equal to k belong to Lp().

Sobolev spaces are Banach spaces. We will give more detailed definitions of these

spaces, and state some of their main properties, in Chapter 12.

A closed linear subspace of a Banach space is a Banach space, since a closed

subset of a complete space is complete. Infinite-dimensional subspaces need not

be closed, however. For example, infinite-dimensional Banach spaces have proper

dense subspaces, something which is difficult to visualize from our intuition of finite-

dimensional spaces.

Example 5.8 The space of polynomial functions is a linear subspace of C ([0, 1]),

since a linear combination of polynomials is a polynomial. It is not closed, and

Theorem 2.9 implies that it is dense in C ([0, 1]). The set {f C ([0, 1]) | f(0) = 0}

is a closed linear subspace of C ([0, 1]), and is a Banach space equipped with the

sup-norm.

94 Banach Spaces

Example 5.9 The set `c(N) of all sequences of the form (x1, x2, . . . , xn, 0, 0, . . .)

whose terms vanish from some point onwards is an infinite-dimensional linear sub-

space of `p(N) for any 1 p . The subspace `c(N) is not closed, so it is not a

Banach space. It is dense in `p(N) for 1 p

Bounded linear maps 95

the coefficients cn in the expansion

f(x) =

n=0

cnfn(x)

by equating the values of f and the series at the points x = 2km for k N

and m = 0, 1, . . . , 2k. The uniform continuity of f implies that the resulting series

converges uniformly to f .

5.2 Bounded linear maps

A linear map or linear operator T between real (or complex) linear spaces X , Y is

a function T : X Y such that

T (x + y) = Tx+ Ty for all , R (or C) and x, y X.

A linear map T : X X is called a linear transformation of X , or a linear operator

on X . If T : X Y is one-to-one and onto, then we say that T is nonsingular or

invertible, and define the inverse map T1 : Y X by T1y = x if and only if

Tx = y, so that TT1 = I , T1T = I . The linearity of T implies the linearity of

T1.

If X , Y are normed spaces, then we can define the notion of a bounded linear

map. As we will see, the boundedness of a linear map is equivalent to its continuity.

Definition 5.12 Let X and Y be two normed linear spaces. We denote both the

X and Y norms by . A linear map T : X Y is bounded if there is a constant

M 0 such that

Tx Mx for all x X. (5.1)

If no such constant exists, then we say that T is unbounded. If T : X Y is a

bounded linear map, then we define the operator norm or uniform norm T of T

by

T = inf{M | Tx Mx for all x X}. (5.2)

We denote the set of all linear maps T : X Y by L(X,Y ), and the set of all

bounded linear maps T : X Y by B(X,Y ). When the domain and range spaces

are the same, we write L(X,X) = L(X) and B(X,X) = B(X).

Equivalent expressions for T are:

T = supx6=0

Tx

x; T = sup

x1

Tx; T = supx=1

Tx. (5.3)

We also use the notation Rmn, or Cmn, to denote the space of linear maps from

Rn to Rm, or Cn to Cm, respectively.

96 Banach Spaces

Example 5.13 The linear map A : R R defined by Ax = ax, where a R, is

bounded, and has norm A = |a|.

Example 5.14 The identity map I : X X is bounded on any normed space X ,

and has norm one. If a map has norm zero, then it is the zero map 0x = 0.

Linear maps on infinite-dimensional normed spaces need not be bounded.

Example 5.15 Let X = C([0, 1]) consist of the smooth functions on [0, 1] that

have continuous derivatives of all orders, equipped with the maximum norm. The

space X is a normed space, but it is not a Banach space, since it is incomplete.

The differentiation operator Du = u is an unbounded linear map D : X X . For

example, the function u(x) = ex is an eigenfunction of D for any R, meaning

that Du = u. Thus Du/u = || may be arbitrarily large. The unboundedness

of differential operators is a fundamental difficulty in their study.

Suppose that A : X Y is a linear map between finite-dimensional real linear

spaces X , Y with dimX = n, dimY = m. We choose bases {e1, e2, . . . , en} of X

and {f1, f2, . . . , fm} of Y . Then

A (ej) =

mi=1

aijfi,

for a suitable m n matrix (aij) with real entries. We expand x X as

x =

ni=1

xiei, (5.4)

where xi R is the ith component of x. It follows from the linearity of A that

A

nj=1

xjej

= mi=1

yifi,

where

yi =

nj=1

aijxj .

Thus, given a choice of bases for X , Y we may represent A as a linear map A :

Rn Rm with matrix A = (aij), wherey1y2...

ym

=

a11 a12 a1na21 a22 a2n...

.... . .

...

am1 am2 amn

x1x2...

xn

. (5.5)

Bounded linear maps 97

We will often use the same notation A to denote a linear map on a finite-dimensional

space and its associated matrix, but it is important not to confuse the geometrical

notion of a linear map with the matrix of numbers that represents it.

Each pair of norms on Rn and Rm induces a corresponding operator, or matrix,

norm on A. We first consider the Euclidean norm, or 2-norm, A2 of A. The

Euclidean norm of a vector x is given by x22 = (x, x), where (x, y) = xT y. From

(5.3), we may compute the Euclidean norm of A by maximizing the function Ax22on the unit sphere x22 = 1. The maximizer x is a critical point of the function

f(x, ) = (Ax,Ax) {(x, x) 1} ,

where is a Lagrange multiplier. Computing f and setting it equal to zero, we

find that x satisfies

ATAx = x. (5.6)

Hence, x is an eigenvector of the matrix ATA and is an eigenvalue. The matrix

ATA is an n n symmetric matrix, with real, nonnegative eigenvalues. At an

eigenvector x of ATA that satisfies (5.6), normalized so that x2 = 1, we have

(Ax,Ax) = . Thus, the maximum value of Ax22 on the unit sphere is the

maximum eigenvalue of ATA.

We define the spectral radius r(B) of a matrix B to be the maximum absolute

value of its eigenvalues. It follows that the Euclidean norm of A is given by

A2 =r (ATA). (5.7)

In the case of linear maps A : Cn Cm on finite dimensional complex linear

spaces, equation (5.7) holds with ATA replaced by AA, where A is the Hermitian

conjugate of A. Proposition 9.7 gives a formula for the spectral radius of a bounded

operator in terms of the norms of its powers.

To compute the maximum norm of A, we observe from (5.5) that

|yi| |ai1||x1|+ |ai2||x2|+ . . .+ |ain||xn|

(|ai1|+ |ai2|+ . . .+ |ain|) x.

Taking the maximum of this equation with respect to i and comparing the result

with the definition of the operator norm, we conclude that

A max1im

(|ai1|+ |ai2|+ . . .+ |ain|) .

Conversely, suppose that the maximum on the right-hand side of this equation is

attained at i = i0. Let x be the vector with components xj = sgn ai0j , where sgn

is the sign function,

sgnx =

1 if x > 0,

0 if x = 0,

1 if x < 0.

(5.8)

98 Banach Spaces

Then, if A is nonzero, we have x = 1, and

Ax = |ai01|+ |ai02|+ . . .+ |ai0n|.

Since A Ax, we obtain that

A max1im

(|ai1|+ |ai2|+ . . .+ |ain|) .

Therefore, we have equality, and the maximum norm of A is given by the maximum

row sum,

A = max1im

n

j=1

|aij |

. (5.9)A similar argument shows that the sum norm of A is given by the maximum column

sum

A1 = max1jn

{m

i=1

|aij |

}.

For 1 < p

Bounded linear maps 99

for each i N, and in order for Ax to belong to `(N), we require that

supiN

j=1

|aij |

0

such that Tx 1 whenever x . For any x X , with x 6= 0, we define x by

x = x

x.

100 Banach Spaces

Then x , so T x 1. It follows from the linearity of T that

Tx =x

T x Mx,

where M = 1/. Thus T is bounded.

The proof shows that if a linear map is continuous at zero, then it is continuous

at every point. A nonlinear map may be bounded but discontinuous, or continuous

at zero but discontinuous at other points.

The following theorem, sometimes called the BLT theorem for bounded linear

transformation has many applications in defining and studying linear maps.

Theorem 5.19 (Bounded linear transformation) Let X be a normed linear

space and Y a Banach space. If M is a dense linear subspace of X and

T : M X Y

is a bounded linear map, then there is a unique bounded linear map T : X Y

such that Tx = Tx for all x M . Moreover,T = T.

Proof. For every x X , there is a sequence (xn) in M that converges to x. We

define

Tx = limn

Txn.

This limit exists because (Txn) is Cauchy, since T is bounded and (xn) Cauchy,

and Y is complete. We claim that the value of the limit does not depend on the

sequence in M that is used to approximate x. Suppose that (xn) and (xn) are any

two sequences in M that converge to x. Then

xn xn xn x+ x x

n,

and, taking the limit of this equation as n, we see that

limn

xn xn = 0.

It follows that

Txn Txn T xn x

n 0 as n.

Hence, (Txn) and (Txn) converge to the same limit.

The map T is an extension of T , meaning that Tx = Tx, for all x M , because

if x M , we can use the constant sequence with xn = x for all n to define Tx. The

linearity of T follows from the linearity of T .

The fact that T is bounded follows from the inequalityTx = limn

Txn limn

T xn = T x .

It also follows thatT T. Since Tx = Tx for x M , we have T = T.

Bounded linear maps 101

Finally, we show that T is the unique bounded linear map from X to Y that

coincides with T on M . Suppose that T is another such map, and let x be any

point in X , We choose a sequence (xn) in M that converges to x. Then, using the

continuity of T , the fact that T is an extension of T , and the definition of T , we see

that

T x = limn

T xn = limn

Txn = Tx.

We can use linear maps to define various notions of equivalence between normed

linear spaces.

Definition 5.20 Two linear spaces X , Y are linearly isomorphic if there is a one-

to-one, onto linear map T : X Y . If X and Y are normed linear spaces and

T , T1 are bounded linear maps, then X and Y are topologically isomorphic. If

T also preserves norms, meaning that Tx = x for all x X , then X , Y are

isometrically isomorphic.

When we say that two normed linear spaces are isomorphic we will usually

mean that they are topologically isomorphic. We are often interested in the case

when we have two different norms defined on the same space, and we would like to

know if the norms define the same topologies.

Definition 5.21 Let X be a linear space. Two norms 1 and 2 on X are

equivalent if there are constants c > 0 and C > 0 such that

cx1 x2 Cx1 for all x X. (5.10)

Theorem 5.22 Two norms on a linear space generate the same topology if and

only if they are equivalent.

Proof. Let 1 and 2 be two norms on a linear space X . We consider the

identity map

I : (X, 1) (X, 2).

From Corollary 4.20, the topologies generated by the two norms are the same if and

only if I and I1 are continuous. Since I is linear, it is continuous if and only if it

is bounded. The boundedness of the identity map and its inverse is equivalent to

the existence of constants c and C such that (5.10) holds.

Geometrically, two norms are equivalent if the unit ball of either one of the

norms is contained in a ball of finite radius of the other norm.

We end this section by stating, without proof, a fundamental fact concerning

linear operators on Banach spaces.

Theorem 5.23 (Open mapping) Suppose that T : X Y is a one-to-one, onto

bounded linear map between Banach spaces X , Y . Then T1 : Y X is bounded.

102 Banach Spaces

This theorem states that the existence of the inverse of a continuous linear map

between Banach spaces implies its continuity. Contrast this result with Example 4.9.

5.3 The kernel and range of a linear map

The kernel and range are two important linear subspaces associated with a linear

map.

Definition 5.24 Let T : X Y be a linear map between linear spaces X , Y . The

null space or kernel of T , denoted by kerT , is the subset of X defined by

kerT = {x X | Tx = 0} .

The range of T , denoted by ranT , is the subset of Y defined by

ranT = {y Y | there exists x X such that Tx = y} .

The word kernel is also used in a completely different sense to refer to the

kernel of an integral operator. A map T : X Y is one-to-one if and only if

kerT = {0}, and it is onto if and only if ranT = Y .

Theorem 5.25 Suppose that T : X Y is a linear map between linear spaces X ,

Y . The kernel of T is a linear subspace of X , and the range of T is a linear subspace

of Y . If X and Y are normed linear spaces and T is bounded, then the kernel of T

is a closed linear subspace.

Proof. If x1, x2 kerT and 1, 2 R (or C), then the linearity of T implies

that

T (1x1 + 2x2) = 1Tx1 + 2Tx2 = 0,

so 1x1 +2x2 kerT . Therefore, kerT is a linear subspace. If y1, y2 ranT , then

there are x1, x2 X such that Tx1 = y1 and Tx2 = y2. Hence

T (1x1 + 2x2) = 1Tx1 + 2Tx2 = 1y1 + 2y2,

so 1y1 + 2y2 ranT . Therefore, ranT is a linear subspace.

Now suppose that X and Y are normed spaces and T is bounded. If (xn) is a

sequence of elements in kerT with xn x in X , then the continuity of T implies

that

Tx = T(

limn

xn

)= lim

nTxn = 0,

so x kerT , and kerT is closed.

The nullity of T is the dimension of the kernel of T , and the rank of T is the

dimension of the range of T . We now consider some examples.

The kernel and range of a linear map 103

Example 5.26 The right shift operator S on `(N) is defined by

S(x1, x2, x3, . . .) = (0, x1, x2, . . .),

and the left shift operator T by

T (x1, x2, x3, . . .) = (x2, x3, x4, . . .).

These maps have norm one. Their matrices are the infinite-dimensional Jordan

blocks,

[S] =

0 0 0 . . .

1 0 0 . . .

0 1 0 . . ....

......

. . .

, [T ] =

0 1 0 . . .

0 0 1 . . .

0 0 0 . . ....

......

. . .

.The kernel of S is {0} and the range of S is the subspace

ranS = {(0, x2, x3, . . .) `(N)} .

The range of T is the whole space `(N), and the kernel of T is the one-dimensional

subspace

kerT = {(x1, 0, 0, . . .) | x1 R} .

The operator S is one-to-one but not onto, and T is onto but not one-to-one. This

cannot happen for linear maps T : X X on a finite-dimensional space X , such

as X = Rn. In that case, kerT = {0} if and only if ranT = X .

Example 5.27 An integral operator K : C([0, 1]) C([0, 1])

Kf(x) =

10

k(x, y)f(y) dy

is said to be degenerate if k(x, y) is a finite sum of separated terms of the form

k(x, y) =

ni=1

i(x)i(y),

where i, i : [0, 1] R are continuous functions. We may assume without loss of

generality that {1, . . . , n} and {1, . . . , n} are linearly independent. The range

of K is the finite-dimensional subspace spanned by {1, 2, . . . , n}, and the kernel

of K is the subspace of functions f C([0, 1]) such that 10

f(y)i(y) dy = 0 for i = 1, . . . , n.

Both the range and kernel are closed linear subspaces of C([0, 1]).

104 Banach Spaces

Example 5.28 Let X = C([0, 1]) with the maximum norm. We define the integral

operator K : X X by

Kf(x) =

x0

f(y) dy. (5.11)

An integral operator like this one, with a variable range of integration, is called a

Volterra integral operator. Then K is bounded, with K 1, since

Kf sup0x1

x0

|f(y)| dy

10

|f(y)| dy f.

In fact, K = 1, since K(1) = x and x = 1. The range of K is the set

of continuously differentiable functions on [0, 1] that vanish at x = 0. This is a

linear subspace of C([0, 1]) but it is not closed. The lack of closure of the range

of K is due to the smoothing effect of K, which maps continuous functions to

differentiable functions. The problem of inverting integral operators with similar

properties arises in a number of inverse problems, where one wants to reconstruct

a source distribution from remotely sensed data. Such problems are ill-posed and

require special treatment.

Example 5.29 Consider the operator T = I +K on C([0, 1]), where K is defined

in (5.11), which is a perturbation of the identity operator by K. The range of T

is the whole space C([0, 1]), and is therefore closed. To prove this statement, we

observe that g = Tf if and only if

f(x) +

x0

f(y) dy = g(x).

Writing F (x) = x0 f(y) dy, we have F

= f and

F + F = g, F (0) = 0.

The solution of this initial value problem is

F (x) =

x0

e(xy)g(y) dy.

Differentiating this expression with respect to x, we find that f is given by

f(x) = g(x)

x0

e(xy)g(y) dy.

Thus, the operator T = I +K is invertible on C([0, 1]) and

(I +K)1

= I L,

where L is the Volterra integral operator

Lg(x) =

x0

e(xy)g(y) dy.

The kernel and range of a linear map 105

The following result provides a useful way to show that an operator T has closed

range. It states that T has closed range if one can estimate the norm of the solution

x of the equation Tx = y in terms of the norm of the right-hand side y. In that

case, it is often possible to deduce the existence of solutions (see Theorem 8.18).

Proposition 5.30 Let T : X Y be a bounded linear map between Banach

spaces X , Y . The following statements are equivalent:

(a) there is a constant c > 0 such that

cx Tx for all x X;

(b) T has closed range, and the only solution of the equation Tx = 0 is x = 0.

Proof. First, suppose that T satisfies (a). Then Tx = 0 implies that x = 0, so

x = 0. To show that ranT is closed, suppose that (yn) is a convergent sequence in

ranT , with yn y Y . Then there is a sequence (xn) in X such that Txn = yn.

The sequence (xn) is Cauchy, since (yn) is Cauchy and

xn xm 1

cT (xn xm) =

1

cyn ym.

Hence, since X is complete, we have xn x for some x X . Since T is bounded,

we have

Tx = limn

Txn = limn

yn = y,

so y ranT , and ranT is closed.

Conversely, suppose that T satisfies (b). Since ranT is closed, it is a Banach

space. Since T : X Y is one-to-one, the operator T : X ranT is a one-to-

one, onto map between Banach spaces. The open mapping theorem, Theorem 5.23,

implies that T1 : ranT X is bounded, and hence that there is a constant C > 0

such that T1y Cy for all y ranT .Setting y = Tx, we see that cx Tx for all x X , where c = 1/C.

Example 5.31 Consider the Volterra integral operator K : C([0, 1]) C([0, 1])

defined in (5.11). Then

K [cosnpix] =

x0

cosnpiy dy =sinnpix

npi.

We have cosnpix = 1 for every n N, but K [cosnpix] 0 as n . Thus,

it is not possible to estimate f in terms of Kf, consistent with the fact that

the range of K is not closed.

106 Banach Spaces

5.4 Finite-dimensional Banach spaces

In this section, we prove that every finite-dimensional (real or complex) normed lin-

ear space is a Banach space, that every linear operator on a finite-dimensional space

is continuous, and that all norms on a finite-dimensional space are equivalent. None

of these statements is true for infinite-dimensional linear spaces. As a result, topo-

logical considerations can often be neglected when dealing with finite-dimensional

spaces but are of crucial importance when dealing with infinite-dimensional spaces.

We begin by proving that the components of a vector with respect to any basis

of a finite-dimensional space can be bounded by the norm of the vector.

Lemma 5.32 Let X be a finite-dimensional normed linear space with norm ,

and {e1, e2, . . . , en} any basis of X . There are constants m > 0 and M > 0 such

that if x =n

i=1 xiei, then

mn

i=1

|xi| x Mn

i=1

|xi| . (5.12)

Proof. By the homogeneity of the norm, it suffices to prove (5.12) for x X such

thatn

i=1 |xi| = 1. The cube

C =

{(x1, . . . , xn) R

n n

i=1

|xi| = 1

}is a closed, bounded subset of Rn, and is therefore compact by the Heine-Borel

theorem. We define a function f : C X by

f ((x1, . . . , xn)) =n

i=1

xiei.

For (x1, . . . , xn) Rn and (y1, . . . , yn) R

n, we have

f ((x1, . . . , xn)) f ((y1, . . . , yn))

ni=1

|xi yi|ei,

so f is continuous. Therefore, since : X R is continuous, the map

(x1, . . . , xn) 7 f ((x1, . . . , xn))

is continuous. Theorem 1.68 implies that f is bounded on C and attains its

infimum and supremum. Denoting the minimum by m 0 and the maximum by

M m, we obtain (5.12). Let (x1, . . . , xn) be a point in C where f attains its

minimum, meaning that

x1e1 + . . .+ xnen = m.

The linear independence of the basis vectors {e1, . . . , en} implies that m 6= 0, so

m > 0.

Finite-dimensional Banach spaces 107

This result is not true in an infinite-dimensional space because, if a basis consists

of vectors that become almost parallel, then the cancellation in linear combina-

tions of basis vectors may lead to a vector having large components but small norm.

Theorem 5.33 Every finite-dimensional normed linear space is a Banach space.

Proof. Suppose that (xk)k=1 is a Cauchy sequence in a finite-dimensional normed

linear space X . Let {e1, . . . , en} be a basis of X . We expand xk as

xk =

ni=1

xi,kei,

where xi,k R. For 1 i n, we consider the real sequence of ith components,

(xi,k)k=1. Equation (5.12) implies that

|xi,j xi,k| 1

mxj xk,

so (xi,k)k=1 is Cauchy. Since R is complete, there is a yi R, such that

limk

xi,k = yi.

We define y X by

y =

ki=1

yiei.

Then, from (5.12),

xk y M

ni=1

|xi,k yi| ei,

and hence xk y as k . Thus, every Cauchy sequence in X converges, and X

is complete.

Since a complete space is closed, we have the following corollary.

Corollary 5.34 Every finite-dimensional linear subspace of a normed linear space

is closed.

In Section 5.2, we proved explicitly the boundedness of linear maps on finite-

dimensional linear spaces with respect to certain norms. In fact, linear maps on

finite-dimensional spaces are always bounded.

Theorem 5.35 Every linear operator on a finite-dimensional linear space is bounded.

108 Banach Spaces

Proof. Suppose that A : X Y is a linear map and X is finite dimensional. Let

{e1, . . . , en} be a basis of X . If x =n

i=1 xiei X , then (5.12) implies that

Ax n

i=1

|xi| Aei max1in

{Aei}n

i=1

|xi| 1

mmax

1in{Aei} x,

so A is bounded.

Finally, we show that although there are many different norms on a finite-

dimensional linear space they all lead to the same topology and the same notion of

convergence. This fact follows from Theorem 5.22 and the next result.

Theorem 5.36 Any two norms on a finite-dimensional space are equivalent.

Proof. Let 1 and 2 be two norms on a finite-dimensional space X . We

choose a basis {e1, e2, . . . , en} of X . Then Lemma 5.32 implies that there are strictly

positive constants m1, m2, M1, M2 such that if x =n

i=1 xiei, then

m1

ni=1

|xi| x1 M1

ni=1

|xi| ,

m2

ni=1

|xi| x2 M2

ni=1

|xi| .

Equation (5.10) then follows with c = m2/M1 and C = M2/m1.

5.5 Convergence of bounded operators

The set B(X,Y ) of bounded linear maps from a normed linear space X to a normed

linear space Y is a linear space with respect to the natural pointwise definitions of

vector addition and scalar multiplication:

(S + T )x = Sx+ Tx, (T )x = (Tx).

It is straightforward to check that the operator norm in Definition 5.12,

T = supx6=0

Tx

x,

defines a norm on B(X,Y ), so that B(X,Y ) is a normed linear space.

The composition of two linear maps is linear, and the following theorem states

that the composition of two bounded linear maps is bounded.

Theorem 5.37 Let X , Y , and Z be normed linear spaces. If T B(X,Y ) and

S B(Y, Z), then ST B(X,Z), and

ST S T. (5.13)

Convergence of bounded operators 109

Proof. For all x X we have

STx S Tx S T x.

For example, if T B(X), then T n B(X) and T n Tn. It may well

happen that we have strict inequality in (5.13).

Example 5.38 Consider the linear maps A, B on R2 with matrices

A =

( 0

0 0

), B =

(0 0

0

).

These matrices have the Euclidean (or sum, or maximum) norms A = || and

B = ||, but AB = 0.

A linear space with a product defined on it is called an algebra. The composition

of maps defines a product on the space B(X) of bounded linear maps onX into itself,

so B(X) is an algebra. The algebra is associative, meaning that (RS)T = R(ST ),

but is not commutative, since in general ST is not equal to TS. If S, T B(X), we

define the commutator [S, T ] B(X) of S and T by

[S, T ] = ST TS.

If ST = TS, or equivalently if [S, T ] = 0, then we say that S and T commute.

The convergence of operators in B(X,Y ) with respect to the operator norm is

called uniform convergence.

Definition 5.39 If (Tn) is a sequence of operators in B(X,Y ) and

limn

Tn T = 0

for some T B(X,Y ), then we say that Tn converges uniformly to T , or that Tnconverges to T in the uniform, or operator norm, topology on B(X,Y ).

Example 5.40 Let X = C([0, 1]) equipped with the supremum norm. For kn(x, y)

is a real-valued continuous function on [0, 1] [0, 1], we define Kn B(X) by

Knf(x) =

10

kn(x, y)f(y) dy. (5.14)

Then Kn 0 uniformly as n if

Kn = maxx[0,1]

{ 10

|kn(x, y)| dy

} 0 as n. (5.15)

An example of functions kn satisfying (5.15) is kn(x, y) = xyn.

A basic fact about a space of bounded linear operators that take values in a

Banach space is that it is itself a Banach space.

110 Banach Spaces

Theorem 5.41 If X is a normed linear space and Y is a Banach space, then

B(X,Y ) is a Banach space with respect to the operator norm.

Proof. We have to prove that B(X,Y ) is complete. Let (Tn) be a Cauchy se-

quence in B(X,Y ). For each x X , we have

Tnx Tmx Tn Tm x,

which shows that (Tnx) is a Cauchy sequence in Y . Since Y is complete, there is

a y Y such that Tnx y. It is straightforward to check that Tx = y defines a

linear map T : X Y . We show that T is bounded. For any > 0, let N be such

that Tn Tm < /2 for all n,m N. Take n N. Then for each x X , there

is an m(x) N such that Tm(x)x Tx /2. If x = 1, we have

Tnx Tx Tnx Tm(x)x+ Tm(x)x Tx . (5.16)

It follows that if n N, then

Tx Tnx+ Tx Tnx Tn+

for all x with x = 1, so T is bounded. Finally, from (5.16) it follows that

limn Tn T = 0. Hence, Tn T in the uniform norm.

A particularly important class of bounded operators is the class of compact

operators.

Definition 5.42 A linear operator T : X Y is compact if T (B) is a precompact

subset of Y for every bounded subset B of X .

An equivalent formulation is that T is compact if and only if every bounded

sequence (xn) in X has a subsequence (xnk ) such that (Txnk ) converges in Y . We

do not require that the range of T be closed, so T (B) need not be compact even if B

is a closed bounded set. We leave the proof of the following properties of compact

operators as an exercise.

Proposition 5.43 Let X , Y , Z be Banach spaces. (a) If S, T B(X,Y ) are

compact, then any linear combination of S and T is compact. (b) If (Tn) is a

sequence of compact operators in B(X,Y ) converging uniformly to T , then T is

compact. (c) If T B(X,Y ) has finite-dimensional range, then T is compact. (d)

Let S B(X,Y ), T B(Y, Z). If S is bounded and T is compact, or S is compact

and T is bounded, then TS B(X,Z) is compact.

It follows from parts (a)(b) of this proposition that the space K(X,Y ) of com-

pact linear operators from X to Y is a closed linear subspace of B(X,Y ). Part (d)

implies that K(X) is a two-sided ideal of B(X), meaning that if K K(X), then

AK K(X) and KA K(X) for all A B(X).

Convergence of bounded operators 111

From parts (b)(c), an operator that is the uniform limit of operators with finite

rank, that is with finite-dimensional range, is compact. The converse is also true for

compact operators on many Banach spaces, including all Hilbert spaces, although

there exist separable Banach spaces on which some compact operators cannot be

approximated by finite-rank operators. As a result, compact operators on infinite-

dimensional spaces behave in many respects like operators on finite-dimensional

spaces. We will discuss compact operators on a Hilbert space in greater detail in

Chapter 9.

Another type of convergence of linear maps is called strong convergence.

Definition 5.44 A sequence (Tn) in B(X,Y ) converges strongly if

limn

Tnx = Tx for every x X.

Thus, strong convergence of linear maps is convergence of their pointwise values

with respect to the norm on Y . The terminology here is a little inconsistent: strong

and norm convergence mean the same thing for vectors in a Banach space, but

different things for operators on a Banach space. The associated strong topology

on B(X,Y ) is distinct from the uniform norm topology whenever X is infinite-

dimensional, and is not derived from a norm. We leave the proof of the following

theorem as an exercise.

Theorem 5.45 If Tn T uniformly, then Tn T strongly.

The following examples show that strong convergence does not imply uniform

convergence.

Example 5.46 Let X = `2(N), and define the projection Pn : X X by

Pn(x1, x2, . . . , xn, xn+1, xn+2, . . .) = (x1, x2, . . . , xn, 0, 0, . . .).

Then PnPm = 1 for n 6= m, so (Pn) does not converge uniformly. Nevertheless,

if x `2(N) is any fixed vector, we have Pnx x as n . Thus, Pn I

strongly.

Example 5.47 Let X = C([0, 1]), and consider the sequence of continuous linear

functionals Kn : X R, given by

Knf =

10

sin(npix) f(x) dx.

If p is a polynomial, then an integration by parts implies that

Knp =p(0) cos(npi)p(1)

npi+

1

npi

10

cos(npix) p(x) dx.

112 Banach Spaces

Hence, Knp 0 as n . If f C([0, 1]), then by Theorem 2.9 for any > 0

there is a polynomial p such that f p < /2, and there is an N such that

|Knp| < /2 for n N . Since Kn 1 for all n, it follows that

|Knf | Kn f p+ |Knp| <

when n N . Thus, Knf 0 as n for every f C([0, 1]). This result is

a special case of the Riemann-Lebesgue lemma, which we prove in Theorem 11.34.

On the other hand, if fn(x) = sin(npix), then fn = 1 and Knfn = 1/2, which

implies that Kn 1/2. (In fact, Kn = 2/pi for each n.) Hence, Kn 0

strongly, but not uniformly.

A third type of convergence of operators, weak convergence, may be defined

using the notion of weak convergence in a Banach space, given in Definition 5.59

below. We say that Tn converges weakly to T in B(X,Y ) if the pointwise values

Tnx converge weakly to Tx in Y . We will not consider the weak convergence of

operators in this book.

We end this section with two applications of operator convergence. First we

define the exponential of an operator, and use it to solve a linear evolution equation.

If A : X X is a bounded linear operator on a Banach space X , then, by analogy

with the power series expansion of ea, we define

eA = I +A+1

2!A2 +

1

3!A3 + . . .+

1

n!An + . . . . (5.17)

A comparison with the convergent real series

eA = 1 + A+1

2!A2 +

1

3!A3 + . . .+

1

n!An + . . . ,

implies that the series on the right hand side of (5.17) is absolutely convergent in

B(X), and hence norm convergent. It also follows thateA eA.If A and B commute, then multiplication and rearrangement of the series for the

exponentials implies that

eAeB = eA+B.

The solution of the initial value problem for the linear, scalar ODE xt = ax with

x(0) = x0 is given by x(t) = x0eat. This result generalizes to a linear system,

xt = Ax, x(0) = x0, (5.18)

where x : R X , with X a Banach space, and A : X X is a bounded linear

operator on X . The solution of (5.18) is given by

x(t) = etAx0.

Convergence of bounded operators 113

This is a solution because

d

dtetA = AetA,

where the derivative is given by the uniformly convergent limit,

d

dtetA = lim

h0

(eA(t+h) etA

h

)= etA lim

h0

(eAh I

h

)= AetA lim

h0

n=0

1

(n+ 1)!Anhn

= AetA.

An important application of this result is to linear systems of ODEs when x(t)

Rn and A is an n n matrix, but it also applies to linear equations on infinite-

dimensional spaces.

Example 5.48 Suppose that k : [0, 1] [0, 1] R is a continuous function, and

K : C([0, 1]) C([0, 1]) is the integral operator

Ku(x) =

10

k(x, y)u(y) dy.

The solution of the initial value problem

ut(x, t) + u(x, t) =

10

k(x, y)u(y, t) dy, u(x, 0) = u0(x),

with u(, t) C([0, 1]), is u = e(KI)tu0.

The one-parameter family of operators T (t) = etA is called the flow of the

evolution equation (5.18). The operator T (t) maps the solution at time 0 to the

solution at time t. We leave the proof of the following properties of the flow as an

exercise.

Theorem 5.49 If A : X X is a bounded linear operator and T (t) = etA for

t R, then:

(a) T (0) = I ;

(b) T (s)T (t) = T (s+ t) for s, t R;

(c) T (t) I uniformly as t 0.

A family of bounded linear operators {T (t) | t R} that satisfies the proper-

ties (a)(c) in this theorem is called a one-parameter uniformly continuous group.

Properties (a)(b) imply that the operators form a commutative group under com-

position, while (c) states that T : R B(X) is continuous with respect to the

114 Banach Spaces

uniform, or norm, topology on B(X) at t = 0. The group property implies that T

is uniformly continuous on R, meaning that T (t) T (t0) 0 as t t0 for any

t0 R.

Any one-parameter uniformly continuous group of operators can be written as

T (t) = etA for a suitable operator A, called the generator of the group. The

generator A may be recovered from the operators T (t) by

A = limt0

(T (t) I

t

). (5.19)

Many linear partial differential equations can be written as evolution equations of

the form (5.18) in which A is an unbounded operator. Under suitable conditions

on A, there exist solution operators T (t), which may be defined only for t 0, and

which are strongly continuous functions of t, rather than uniformly continuous. The

solution operators are then said to form a C0-semigroup. For an example, see the

discussion of the heat equation in Section 7.3.

As a second application of operator convergence, we consider the convergence of

approximation schemes. Suppose we want to solve an equation of the form

Au = f, (5.20)

where A : X Y is a nonsingular linear operator between Banach spaces and

f Y is given. Suppose we can approximate (5.20) by an equation

Au = f, (5.21)

whose solution u can be computed more easily. We assume that A : X Y

is a nonsingular linear operator with a bounded inverse. We call the family of

equations (5.21) an approximation scheme for (5.20). For instance, if (5.20) is a

differential equation, then (5.21) may be obtained by a finite difference or finite

element approximation, where is a grid spacing. One complication is that a

numerical approximation A may act on a different space X than the space X .

For simplicity, we suppose that the approximations A may be defined on the same

space as A. The primary requirement of an approximation scheme is that it is

convergent.

Definition 5.50 The approximation scheme (5.21) is convergent to (5.20) if u u

as 0 whenever f f .

We make precise the idea that A approximates A in the following definition of

consistency.

Definition 5.51 The approximation scheme (5.21) is consistent with (5.20) if Av

Av as 0 for each v X .

Convergence of bounded operators 115

In other words, the approximation scheme is consistent if A converges strongly

to A as 0. Consistency on its own is not sufficient to guarantee convergence.

We also need a second property called stability.

Definition 5.52 The approximation scheme (5.21) is stable if there is a constant

M , independent of , such that

A1 M.

Consistency and convergence relate the operators A to A, while stability is

a property of the approximate operators A alone. Stability plays a crucial role

in convergence, because it prevents the amplification of errors in the approximate

solutions as 0.

Theorem 5.53 (Lax equivalence) An consistent approximation scheme is con-

vergent if and only if it is stable.

Proof. First, we prove that a stable scheme is convergent. If Au = f and Au =

f, then

u u = A1 (AuAu+ f f) .

Taking the norm of this equation, using the definition of the operator norm, and

the triangle inequality, we find that

u u A1 (AuAu+ f f) . (5.22)

If the scheme is stable, then

u u M (AuAu+ f f) ,

and if the scheme is consistent, then Au Au as 0. It follows that u u if

f f , and the scheme is convergent.

Conversely, we prove that a convergent scheme is stable. For any f Y , let

u = A1 f . Then, since the scheme is convergent, we have u u as 0, where

u = A1f , so that u is bounded. Thus, there exists a constant Mf , independent

of , such that A1 f Mf . The uniform boundedness theorem, which we do not

prove here, then implies that there exists a constant M such that A1 M , so

the scheme is stable.

An analogous result holds for linear evolution equations of the form (5.18) (see

Strikwerder [53], for example). There is, however, no general criterion for the

convergence of approximation schemes for nonlinear equations.

116 Banach Spaces

5.6 Dual spaces

The dual space of a linear space consists of the scalar-valued linear maps on the

space. Duality methods play a crucial role in many parts of analysis. In this section,

we consider real linear spaces for definiteness, but all the results hold for complex

linear spaces.

Definition 5.54 A scalar-valued linear map from a linear space X to R is called

a linear functional or linear form on X . The space of linear functionals on X is

called the algebraic dual space of X , and the space of continuous linear functionals

on X is called the topological dual space of X .

In terms of the notation in Definition 5.12, the algebraic dual space of X is

L(X,R), and the topological dual space is B(X,R). A linear functional : X R

is bounded if there is a constant M such that

|(x)| Mx for all x X,

and then we define by

= supx6=0

|(x)|

x. (5.23)

If X is infinite dimensional, then L(X,R) is much larger than B(X,R), as we illus-

trate in Example 5.57 below. Somewhat confusingly, both dual spaces are commonly

denoted by X. We will use X to denote the topological dual space of X . Either

dual space is itself a linear space under the operations of pointwise addition and

scalar multiplication of maps, and the topological dual is a Banach space, since R

is complete.

If X is finite dimensional, then L(X,R) = B(X,R), so there is no need to

distinguish between the algebraic and topological dual spaces. Moreover, the dual

space X of a finite-dimensional space X is linearly isomorphic to X . To show this,

we pick a basis {e1, e2, . . . , en} of X . The map i : X R defined by

i

nj=1

xjej

= xi (5.24)is an element of the algebraic dual space X. The linearity of i is obvious.

For example, if X = Rn and

e1 = (1, 0, . . . , 0), e2 = (0, 1, . . . , 0), . . . , en = (0, 0, . . . , 1),

are the coordinate basis vectors, then

i : (x1, x2, . . . , xn) 7 xi

is the map that takes a vector to its ith coordinate.

Dual spaces 117

The action of a general element of the dual space : X R on a vector

x X is given by a linear combination of the components of x, since

(n

i=1

xiei

)=

ni=1

ixi,

where i = (ei) R. It follows that, as a map,

=

ni=1

ii.

Thus, {1, 2, . . . , n} is a basis of X, called the dual basis of {e1, e2, . . . , en}, and

both X and X are linearly isomorphic to Rn. The dual basis has the property that

i(ej) = ij ,

where ij is the Kronecker delta function, defined by

ij =

{1 if i = j,

0 if i 6= j.(5.25)

Although a finite-dimensional space is linearly isomorphic with its dual space,

there is no canonical way to identify the space with its dual; there are many iso-

morphisms, depending on an arbitrary choice of a basis. In the following chapters,

we will study Hilbert spaces, and show that the topological dual space of a Hilbert

space can be identified with the original space in a natural way through the inner

product (see the Riesz representation theorem, Theorem 8.12). The dual of an

infinite-dimensional Banach space is, in general, different from the original space.

Example 5.55 In Section 12.8, we will see that for 1 p < the dual of Lp()

is Lp

(), where 1/p+ 1/p = 1. The Hilbert space L2() is self-dual.

Example 5.56 Consider X = C([a, b]). For any L1([a, b]), the following for-

mula defines a continuous linear functional on X :

(f) =

ba

f(x)(x) dx. (5.26)

Not all continuous linear functionals are of the form (5.26). For example, if x0

[a, b], then the evaluation of f at x0 is a continuous linear functional. That is, if we

define x0 : C([a, b]) R by

x0(f) = f(x0),

then x0 is a continuous linear functional on C([a, b]). A full description of the dual

space of C([a, b]) is not so simple: it may be identified with the space of Radon

measures on [a, b] (see [12], for example).

118 Banach Spaces

One way to obtain a linear functional on a linear space is to start with a linear

functional defined on a subspace, extend a Hamel basis of the subspace to a Hamel

basis of the whole space and extend the functional to the whole space, by use

of linearity and an arbitrary definition of the functional on the additional basis

elements. The next example uses this procedure to obtain a discontinuous linear

functional on C([0, 1]).

Example 5.57 Let M = {xn | n = 0, 1, 2, . . .} be the set of monomials in C([0, 1]).

The set M is linearly independent, so it may be extended to a Hamel basis H . Each

f C([0, 1]) can be written uniquely as

f = c1h1 + + cNhN , (5.27)

for suitable basis functions hi H and nonzero scalar coefficients ci. For each

n = 0, 1, 2, . . ., we define n(f) by

n(f) =

{ci if hi = x

n,

0 otherwise.

Due to the uniqueness of the decomposition in (5.27), the functional n is well-

defined. We define a linear functional on C([0, 1]) by

(f) =

n=1

nn(f).

For each f , only a finite number of terms in this sum are nonzero, so is a well-

defined linear functional on C([0, 1]). The functional is unbounded, since for each

n = 0, 1, 2, . . . we have xn = 1 and |(xn)| = n.

A similar construction shows that every infinite-dimensional linear space has

discontinuous linear functionals defined on it. On the other hand, Theorem 5.35

implies that all linear functionals on a finite-dimensional linear space are bounded.

It is not obvious that this extension procedure can be used to obtain bounded

linear functionals on an infinite-dimensional linear space, or even that there are

any nonzero bounded linear functionals at all, because the extension need not be

bounded. In fact, it is possible to maintain boundedness of an extension by a

suitable choice of its values off the original subspace, as stated in the following

version of the Hahn-Banach theorem.

Theorem 5.58 (Hahn-Banach) If Y is a linear subspace of a normed linear space

X and : Y R is a bounded linear functional on Y with = M , then there is

a bounded linear functional : X R on X such that restricted to Y is equal

to and = M .

We omit the proof here. One consequence of this theorem is that there are

enough bounded linear functionals to separate X , meaning that if (x) = (y) for

all X, then x = y (see Exercise 5.6).

Dual spaces 119

Since X is a Banach space, we can form its dual space X, called the bidual of

X . There is no natural way to identify an element of X with an element of the dual

X, but we can naturally identify an element of X with an element of the bidual

X. If x X , then we define Fx X by evaluation at x:

Fx() = (x) for every X. (5.28)

Thus, we may regard X as a subspace of X. If all continuous linear functionals

on X are of the form (5.28), then X = X under the identification x 7 Fx, and

we say that X is reflexive.

Linear functionals may be used to define a notion of convergence that is weaker

than norm, or strong, convergence on an infinite-dimensional Banach space.

Definition 5.59 A sequence (xn) in a Banach space X converges weakly to x,

denoted by xn x as n, if

(xn) (x) as n,

for every bounded linear functional in X.

If we think of a linear functional : X R as defining a coordinate (x)

of x, then weak convergence corresponds to coordinate-wise convergence. Strong

convergence implies weak convergence: if xn x in norm and is a bounded linear

functional, then

|(xn) (x)| = |(xn x)| xn x 0.

Weak convergence does not imply strong convergence on an infinite-dimensional

space, as we will see in Section 8.6.

If X is the dual of a Banach space X , then we can define another type of weak

convergence on X, called weak- convergence, pronounced weak star.

Definition 5.60 Let X be the dual of a Banach space X . We say X is the

weak- limit of a sequence (n) in X if

n(x) (x) as n,

for every x X . We denote weak- convergence by

n .

By contrast, weak convergence of (n) in X means that

F (n) F () as n,

for every F X. If X is reflexive, then weak and weak- convergence in X are

equivalent because every bounded linear functional on X is of the form (5.28). If

X is the dual space of a nonreflexive space X , then weak and weak- convergence

120 Banach Spaces

are different, and it is preferable to use weak- convergence in X instead of weak

convergence.

One reason for the importance of weak- convergence is the following compact-

ness result, called the Banach-Alaoglu theorem.

Theorem 5.61 (Banach-Alaoglu) Let X be the dual space of a Banach space

X . The closed unit ball in X is weak- compact.

We will not prove this result here, but we prove a special case of it in Theo-

rem 8.45.

5.7 References

For more on linear operators in Banach spaces, see Kato [26]. For proofs of the

Hahn-Banach, open mapping, and Banach-Alaoglu theorems, see Folland [12], Reed

and Simon [45], or Rudin [48]. The use of linear and Banach spaces in optimization

theory is discussed in [34]. Applied functional analysis is discussed in Lusternik and

Sobolev [33]. For an introduction to semigroups associated with evolution equations,

see [4]. For more on matrices, see [24]. An introduction to the numerical aspects

of matrices and linear algebra is in [54]. For more on the stability, consistency,

and convergence of finite difference schemes for partial differential equations, see

Strikwerder [53].

5.8 Exercises

Exercise 5.1 Prove that the expressions in (5.2) and (5.3) for the norm of a

bounded linear operator are equivalent.

Exercise 5.2 Suppose that {e1, e2, . . . , en} and {e1, e2, . . . , en} are two bases of

the n-dimensional linear space X , with

ei =n

j=1

Lijej , ei =n

j=1

Lij ej ,

where (Lij) is an invertible matrix with inverse(Lij

), meaning that

nj=1

LijLjk = ik.

Let {1, 2, . . . , n} and {1, 2, . . . , n} be the associated dual bases of X.

Exercises 121

(a) If x =xiei =

xiei X , then prove that the components of x transform

under a change of basis according to

xi = Lijxj . (5.29)

(b) If =ii =

ii X

, then prove that the components of

transform under a change of basis according to

i = Ljij . (5.30)

Exercise 5.3 Let : C([0, 1]) R be the linear functional that evaluates a func-

tion at the origin: (f) = f(0). If C([0, 1]) is equipped with the sup-norm,

f = sup0x1

|f(x)|,

show that is bounded and compute its norm. If C([0, 1]) is equipped with the

one-norm,

f1 =

10

|f(x)|dx,

show that is unbounded.

Exercise 5.4 Consider the 2 2 matrix

A =

(0 a2

b2 0

),

where a > b > 0. Compute the spectral radius r(A) of A. Show that the Euclidean

norms of powers of the matrix are given byA2n = a2nb2n, A2n+1 = a2n+2b2n.Verify that r(A) = limn A

n1/n

.

Exercise 5.5 Define K : C([0, 1])