Compact operators on Banach spaces

Jordan Bell
November 12, 2017

1 Introduction

In this note I prove several things about compact linear operators from one Banach space to another, especially from a Banach space to itself. Some of these may things be simpler to prove for compact operators on a Hilbert space, but since often in analysis we deal with compact operators from one Banach space to another, such as from a Sobolev space to an Lp space, and since the proofs here are not absurdly long, I think it’s worth the extra time to prove all of this for Banach spaces. The proofs that I give are completely detailed, and one should be able to read them without using a pencil and paper. When I want to use a fact that is not obvious but that I do not wish to prove, I give a precise statement of it, and I verify that its hypotheses are satisfied.

2 Preliminaries

If X and Y are normed spaces, let (X,Y) be the set of bounded linear maps XY. It is straightforward to check that (X,Y) is a normed space with the operator norm

T=supx1Tx.

If X is a normed space and Y is a Banach space, one proves that (X,Y) is a Banach space.11 1 Walter Rudin, Functional Analysis, second ed., p. 92, Theorem 4.1. Let (X)=(X,X). If X is a Banach space then so is (X), and it is straightforward to verify that (X) is a Banach algebra.

To say T(X) is invertible means that there is some S(X) such that ST=idX and TS=idX, and we write T-1=S. It follows from the open mapping theorem that if T(X), kerT={0}, and T(X)=X, then T is invertible (i.e. if a bounded linear map is bijective then its inverse is also a bounded linear map, where we use the open mapping theorem to show that the inverse is continuous).

The spectrum σ(T) of T(X) is the set of all λ such that T-λidX is not invertible. If T-λidX is not injective, we say that λ is an eigenvalue of T, and then there is some nonzero xker(T-λidX), which thus satisfies Tx=λx; we call any nonzero element of ker(T-λidX) an eigenvector of T. The point spectrum of T is the set of eigenvalues of T.

We say that a subset of a topological space is precompact if its closure is compact. The Heine-Borel theorem states that a subset S of a complete metric space M is precompact if and only if it is totally bounded: to be totally bounded means that for every ϵ>0 there are finitely many points x1,,xrS such that Sk=1rBϵ(xi), where Bϵ(x) is the open ball of radius ϵ and center x.

If X and Y are Banach spaces and B1(0) is the open unit ball in X, a linear map T:XY is said to be compact if T(B1(0)) is precompact; equivalently, if T(B1(0)) is totally bounded. Check that a linear map T:XY is compact if and only if the image of every bounded set is precompact. Thus, if we want to prove that a linear map is compact we can show that the image of the open unit ball is precompact, while if we know that a linear map is compact we can use that the image of every bounded set is precompact. It is straightforward to prove that a compact linear map is bounded. Let 0(X,Y) denote the set of compact linear maps XY. It does not take long to prove that 0(X) is an ideal in the algebra (X).

3 Basic facts about compact operators

Theorem 1.

Let X and Y be Banach spaces. If T:XY is linear, then T is compact if and only if xnX being a bounded sequence implies that there is a subsequence xa(n) such that Txa(n) converges in Y.

Proof.

Suppose that T is compact and let xnX be bounded, with

M=supnxn<.

Let V be the closed ball in X of radius M and center 0. V is bounded in X, so N=T(V)¯ is compact in Y. As TxnN, there is some convergent subsequence Txa(n) that converges to some yN.

Suppose that if xn is a bounded sequence in X then there is a subsequence such that Txa(n) is convergent, let U be the open unit ball in X, and let ynT(U) be a sequence. It is a fact that a subset of a metric space is precompact if and only if every sequence has a subsequence that converges to some element in the space; this is not obvious, but at least we are only taking as given a fact about metric spaces. (What we have asserted is that a set in a metric space is precompact if and only if it is sequentially precompact.) As yn are in the image of T, there is a subsequence such that ya(n) is convergent and this implies that T(U) is precompact, and so T is a compact operator. ∎

Theorem 2.

Let X and Y be Banach spaces. If T0(X,Y), then T(X) is separable.

Proof.

Let Un be the closed ball of radius n in X. As T(Un)¯ is a compact metric space it is separable, and hence T(Un), a subset of it, is separable too, say with dense subset Ln. We have

T(X)=n=1T(Un),

and one checks that n=1Ln is a dense subset of the right-hand side, showing that T(X) is separable. ∎

The following theorem gathers some important results about compact operators.22 2 Walter Rudin, Functional Analysis, second ed., p. 104, Theorem 4.18.

Theorem 3.

Let X and Y be Banach spaces.

  • If T(X,Y), T is compact, and T(X) is a closed subset of Y, then dimT(X)<.

  • 0(X,Y) is a closed subspace of (X,Y).

  • If T(X), T is compact, and λ0, then dimker(T-λidX)<.

  • If dimX=, T(X), and T is compact, then 0σ(T).

Proof.

If T(X,Y) is compact and T(X) is closed then as a closed subspace of a Banach space, T(X) is itself a Banach space. Of course T:XT(X) is surjective, and X is a Banach space so by the open mapping theorem T:XT(X) is an open map. Let TxT(X). As T is an open map, T(B1(x)) is open, and hence T(B1(x))¯ is a neighborhood of Tx. But because T is compact and B1(x) is bounded, T(B1(x))¯ is compact. Hence T(B1(x))¯ is a compact neighborhood of Tx. As every element of T(X) has a compact neighborhood, T(X) is locally compact. But a locally compact topological vector space is finite dimensional, so dimT(X)<.

It is straightforward to check that 0(X,Y) is linear subspace of (X,Y). Let T be in the closure of 0(X,Y) and let U be the open unit ball in X. We wish to show that T(U) is totally bounded. Let ϵ>0. As T is in the closure of 0(X,Y), there is some S0(X,Y) with S-T<ϵ. As S is compact, its image S(U) is totally bounded, so there are finitely many Sx1,,SxrS(U), with x1,,xrU, such that S(U)k=1rBϵ(Sxk). If xU, then

Sx-Tx=(S-T)xS-Tx<S-T<ϵ.

Let xU. Then there is some k such that SxBϵ(Sxk), and

Tx-TxkTx-Sx+Sx-Sxk+Sxk-Txk<3ϵ,

so T(U)k=1rB3ϵ(Txk), showing that T(U) is totally bounded and hence that T is a compact operator.

If T(X) is compact and λ0, let Y=ker(T-λidX). (If λ is not an eigenvalue of T, then Y={0}.) Y is a closed subspace of X, and hence is itself a Banach space. If yY then Ty=λyY. Define S:YY by Sy=Ty=λy, and as T is compact so is S. Now we use the hypothesis that λ0: if yY, then S(1λy)=y, so S:YY is surjective. We have shown that S:YY is compact and that S(Y) is a closed subset of Y (as it is equal to Y), and as a closed image of a compact operator is finite dimensional, we obtain dimS(Y)<, i.e. dimY<.

If dimX= and T(X) is compact, suppose by contradiction that 0σ(T). So T is invertible, with TT-1=idX. As 0(X) is an ideal in the algebra (X), idX is compact. Of course idX(X)=X is a closed subset of X. But we proved that if the image of a compact linear operator is closed then that image is finite dimensional, contradicting dimX=. ∎

4 Dual spaces

If X is a normed space, let X*=(X,), the set of bounded linear maps X. X* is called the dual space of X, and is a Banach space since is a Banach space. Define ,:X×X* by

x,λ=λ(x),xX,λX*.

This is called the dual pairing of X and X*.

The following theorem gives an expression for the norm of an element of the dual space.33 3 Walter Rudin, Functional Analysis, second ed., p. 94, Theorem 4.3 (b).

Theorem 4.

If X is a normed space and V is the closed unit ball in X*, then

x=supλV|x,λ|,xX.
Proof.

It follows from the Hahn-Banach extension theorem that if x0X, then there is some λ0X* such that λ0(x0)=x0 and such that if xX then |λ0(x)|x.44 4 Walter Rudin, Functional Analysis, second ed., p. 58, Theorem 3.3. That is, that there is some λ0V such that λ0(x0)=x0. Hence

supλV|x0,λ||x0,λ0|=|λ0(x0)|=x0.

If λV, then

|x0,λ|=|λ(x0)|λx0x0,

so

supλV|x0,λ|x0.

Let X be a Banach space. For xX, it is apparent that λλ(x) is a linear map X*. From Theorem 4, it is bounded, with norm x. Define ϕ:XX** by

(ϕx)(λ)=x,λ,xX,λX*.

It is apparent that ϕ is a linear map. By Theorem 4, if xX then ϕx=x, so ϕ is an isometry. Let ϕxnϕ(X) be a Cauchy sequence. ϕ-1:ϕ(X)X is an isometry, so ϕ-1ϕxn is a Cauchy sequence, i.e. xn is a Cauchy sequence, and so, as X is a Banach space, xn converges to some x. Then ϕxn converges to ϕx, and thus ϕ(X) is a complete metric space. But a subset of a complete metric space is closed if and only if it is complete, so ϕ(X) is a closed subspace of X**. Hence, ϕ(X) is a Banach space and ϕ:Xϕ(X) is an isometric isomorphism. A Banach space is said to be reflexive if ϕ(X)=X**, i.e. if every bounded linear map X* is of the form ϕ(x) for some xX.

5 Adjoints

If X and Y are normed spaces and T(X,Y), define T*:Y*X* by T*λ=λT; as T*λ is the composition of two bounded linear maps it is indeed a bounded linear map X. T* is called the adjoint of T. It is straightforward to check that T* is linear and that it satisfies, for S=T*,

Tx,λ=x,Sλ,xX,λY*. (1)

On the other hand, suppose that S:Y*X* is a function that satisfies (1). Let λY*, and let xX. Then

(Sλ)(x)=λ(Tx)=(T*λ)(x).

This is true for all x, so Sλ=T*λ, and that is true for all λ, so S=T*. Thus T*λ=λT is the unique function Y*X* that satisfies (1), not just the unique bounded linear map that does. (That is, satisfying (1) completely determines a function.)

Using Theorem 4,

T = supx1Tx
= supx1supλ1|Tx,λ|
= supx1supλ1|x,T*λ|
= supλ1supx1|T*λ(x)|
= supλ1T*λ
= T*.

In particular, T*(Y*,X*).

In the following we prove that the adjoint T* of a compact operator T is itself a compact operator, and that if the adjoint of a bounded linear operator is compact then the original operator is compact.55 5 Walter Rudin, Functional Analysis, second ed., p. 105, Theorem 4.19. In the proof we only show that if we take any sequence λn in the closed unit ball then it has a subsequence such that Tλa(n) converges. Check that it suffices merely to do this rather than showing that this happens for any bounded sequence.

Theorem 5.

If X and Y are Banach spaces and T(X,Y), then T is compact if and only if T* is compact.

Proof.

Suppose that T(X,Y) is compact, and let λnY*, n1, be a sequence in the closed unit ball in Y*.

If M is a metric space with metric ρ and is a set of functions M, we say that is equicontinuous if for every ϵ>0 there exists a δ>0 such that if f and ρ(x,y)<δ then |f(x)-f(y)|<ϵ. We say that is pointwise bounded if for every xM there is some m(x)< such that if f and xM then |f(x)|m(x). The Arzelà-Ascoli theorem66 6 Walter Rudin, Real and Complex Analysis, third ed., p. 245, Theorem 11.28. states that if (M,ρ) is a separable metric space and is a set of functions M that is equicontinuous and pointwise bounded, then for every sequence fn there is a subsequence that converges uniformly on every compact subset of M.

Let V be the closed unit ball in X. As T is a compact operator, T(V)¯ is compact and therefore separable, because any compact metric space is separable. Define fn:T(V)¯ by

fn(y)=y,λn=λn(y).

For y1,y2T(V)¯ we have

|fn(y1)-fn(y2)|=|λn(y1-y2)|λny1-y2y1-y2.

Hence for ϵ>0, if n1 and y1-y2<ϵ then |fn(y1)-fn(y2)|<ϵ. This shows that {fn} is equicontinuous. If yT(V)¯, then, for any n1,

|fn(y)|=|λn(y)|λnyy,

showing that {fn} is pointwise bounded. Therefore we can apply the Arzelà-Ascoli theorem: there is a subsequence fa(n) such that fa(n) converges uniformly on every compact subset of T(V)¯, in particular on T(V)¯ itself and therefore on any subset of it, in particular T(V). We are done using the Arzelà-Ascoli theorem: we used it to prove that there is a subsequence fa(n) that converges uniformly on T(V).

Let ϵ>0. As fa(n) converges uniformly on T(V), there is some N such that if n,mN and yT(V), then |fa(n)(y)-fa(m)(y)|<ϵ. Thus, if n,mN,

T*λa(n)-T*λa(m) = λa(n)T-λa(m)T
= supxV|λa(n)(Tx)-λa(m)(Tx)|
= supxV|λa(n)(Tx)-λa(m)(Tx)|
= supxV|fa(n)(Tx)-fa(m)(Tx)|
< ϵ.

This means that T*λa(n)X* is a Cauchy sequence. As X* is a Banach space, this sequence converges, and therefore T* is a compact operator.

Suppose that T*(Y*,X*) is compact. Therefore, by what we showed in the first half of the proof we have that T**:X**Y** is compact. If V be the closed unit ball in X**, then T**(V) is totally bounded.

We have seen that ϕ:XX** defined by (ϕx)λ=λ(x), xX, λX*, is an isometric isomorphism Xϕ(X). Let ψ:YY** be the same for Y, and let U be the closed unit ball in X. If xX and λY* then

λ,ψTx=Tx,λ=x,T*λ=T*λ,ϕx=λ,T**ϕx.

Therefore ψT=T**ϕ. If xU then ϕxV, as ϕ is an isometry. Hence if xU then ψTx=T**ϕxT**(V), thus

ψT(U)T**(V).

As ψT(U) is contained in a totally bounded set it is itself totally bounded, and as ψ is an isometry, it follows that T(U) is totally bounded. Hence T is a compact operator. ∎

6 Complemented subspaces

If M is a closed subspace of a topological vector space X and there exists a closed subspace N of X such that

X=M+N,MN={0},

we say that M is complemented in X and that X is the direct sum of M and N, which we write as X=MN.

We are going to use the following lemma to prove the theorem that comes after it.77 7 Walter Rudin, Functional Analysis, second ed., p. 106, Lemma 4.21.

Lemma 6.

If X is a locally convex topological vector space and M is a subspace of X with dimX<, then M is complemented in X.

In particular, a normed space is locally convex so the lemma applies to normed spaces. In the following theorem we prove that if T(X) is compact and λ0 then T-λidX has closed image.88 8 Walter Rudin, Functional Analysis, second ed., p. 107, Theorem 4.23.

Theorem 7.

If X is a Banach space, T(X) is compact, and λ0, then the image of T-λidX is closed.

Proof.

According to Theorem 3, dimker(T-λidX)<, and we can then use Lemma 6: ker(T-λidX) is a finite dimensional subspace of the locally convex space X, so there is a closed subspace N of X such that X=ker(T-λidX)N.

Define S:NX by Sx=Tx-λx, so S(N,X). It is apparent that T(X)=S(N) and that S is injective, and we shall prove that S(N) is closed. To show that S(N) is closed, check that it suffices to prove that S is bounded below: that there is some r>0 such that if xN then Sxrx.99 9 A common way of proving that a linear operator is invertible is by proving that it has dense image and that it is bounded below: bounded below implies injective and bounded below and dense image imply surjective.

Suppose by contradiction that for every r>0 there is some xN such that Sx<rx. So for each n1, let xnN with Sxn<1nxn, and put vn=xnxn, so that vn=1 and Svn<1n. As T is compact, there is some subsequence such that Tva(n) converges, say to v. Combining this with Svn0 we get λva(n)v. On the one hand, λva(n)=|λ|va(n)=|λ|, so v=|λ|. On the other hand, since λva(n)N and N is closed, we get vN. S is continuous and λva(n)0, so

Sv=limnS(λva(n))=λlimnSva(n)=0.

Because S is injective and Sv=0, we get v=0, contradicting v=|λ|>0. Therefore S is bounded below, and hence has closed image, completing the proof. ∎

The following theorem states that the point spectrum of a compact operator is countable and bounded, and that if there is a limit point of the point spectrum it is 0.1010 10 Walter Rudin, Functional Analysis, second ed., p. 107, Theorem 4.24. By countable we mean bijective with a subset of the integers.

Theorem 8.

If X is a Banach space, T(X) is compact, and r>0, then there are only finitely many eigenvalues λ of T such that |λ|>r.

The following theorem shows that if T(X) is compact and λ0, then the operator T-λidX is injective if and only if it is surjective.1111 11 Paul Garrett, Compact operators on Banach spaces: Fredholm-Riesz, http://www.math.umn.edu/~garrett/m/fun/fredholm-riesz.pdf This tells us that if λ0 is not an eigenvalue of T, then T-λidX is both injective and surjective, and hence is invertible, which means that if λ0 is not an eigenvalue of T then λσ(T). This is an instance of the Fredholm alternative.

Theorem 9 (Fredholm alternative).

Let X be a Banach space, T(X) be compact, and λ0. T-λidX is injective if and only if it is surjective.

Proof.

Suppose that T-λidX is injective and let Vn=(T-λidX)nX, n1. If (T-λidX)nxVn, then, as (T-λidX)xX, we have

(T-λidX)n-1(T-λidX)xVn-1,

so VnVn-1. Thus

V1V2

Certainly Vn is a normed vector space. Define Tn(Vn) by Tnx=Tx, namely, Tn is the restriction of T to Vn.

As T is a compact operator, by Theorem 7 we get that V1=(T-λidX)(X) is closed. Hence V1 is a Banach space, being a closed subspace of a Banach space. Assume as induction hypothesis that Vn is a closed subset of X. Thus Vn is a Banach space, and Tn(Vn) is a compact operator, as it is the restriction of the compact operator T to Vn. Therefore by Theorem 7, the image of Tn-λidX is closed, but this image is precisely Vn+1. Therefore, if n1 then Vn is a closed subspace of X.

Suppose by contradiction that there is some x(T-λidX)X=V1. If yX then

(T-λidX)nx-(T-λidX)n+1y=(T-λidX)n(x-(T-λidX)y).

As x(T-λidX)X, we have x-(T-λidX)y0. As we have supposed that T-λidX is injective, any positive power of it is injective, and hence the right hand side of the above equation is not 0. Thus (T-λidX)nx(T-λidX)n+1y, and as yX was arbitrary,

(T-λidX)nx(T-λidX)n+1X.

However, of course (T-λidX)nxVn, so if n1 then Vn strictly contains Vn+1.

Riesz’s lemma states that if M is a normed space, N is a proper closed subspace of M, and 0<r<1, then there is some xM with x=1 and infyXx-yr.1212 12 Paul Garrett, Riesz’s lemma, http://www.math.umn.edu/~garrett/m/fun/riesz_lemma.pdf In this reference, Riesz’s lemma is stated for Banach spaces, but the proof in fact works for normed spaces with no modifications. For each n1, using Riesz’s lemma there is some vnVn, vn=1, such that

infyVn+1vn-y12;

we proved that each Vn is closed and that Vn is a strictly decreasing sequence to allow us to use Riesz’s lemma.

If n,m1, then (T-λidX)vmVm+1 and check that Tvm+nVm+n, so

Tvm-Tvm+n=λvm+(T-λidX)vm-Tvm+nλvm+Vm+1.

From this and the definition of the sequence vm, we get

Tvm-Tvm+n|λ|12.

That is, the distance between any two terms in Tvm is |λ|2, which is a fixed positive constant, hence Tvm has no convergent subsequence. But vm=1, so vm is bounded and therefore, as T is compact, the sequence Tvm has a convergent subsequence, a contradiction. Therefore T-λidX is surjective.

Suppose that T-λidX is surjective. One checks that if a bounded linear operator is surjective then its adjoint is injective. For xX and μX*, λidXx,μ=μ(λx)=λμ(x)=x,λidX*μ, so (λidX)*=λidX*. Hence (T-λidX)*=T*-λidX*. T is compact so T* is compact. As T*-λidX* is injective and T* is compact, T*-λidX* is surjective, whence its adjoint T**-λidX**:X**X** is injective. One checks that if S(X) and S**:X**X** is injective then S is injective; this is proved using the fact that ϕ:XX** defined by (ϕx)(λ)=λ(x) is an isometric isomorphism Xϕ(X). Using this, T-λidX is injective, completing the proof. ∎

7 Compact metric spaces

In the proof of Theorem 5 we stated the Arzelà-Ascoli theorem. First we state definitions again. If M is a metric space with metric ρ and is a set of functions M, we say that is equicontinuous if for all ϵ>0 there is some δ>0 such that f and ρ(x,y)<δ imply that |f(x)-f(y)|<ϵ. We say that is pointwise bounded if for all xM there is some m(x) such that if f then |f(x)|m(x). The Arzelà-Ascoli theorem states that if M is a separable metric space and is equicontinuous and pointwise bounded, then every sequence in has a sequence that converges uniformly on every compact subset of M.1313 13 Walter Rudin, Real and Complex Analysis, third ed., p. 245, Theorem 11.28.

We are going to use a converse of the Arzelà-Ascoli theorem in the case of a compact metric space.1414 14 John B. Conway, A Course in Functional Analysis, second ed., p. 175, Theorem 3.8. Let M be a compact metric space and let C(M) be the set of continuous functions M. It does not take long to prove that with the norm f=supxM|f(x)|, C(M) is a Banach space.

Theorem 10.

Let (M,ρ) be a compact metric space and let C(M). is precompact in C(M) if and only if is bounded and equicontinuous.

Proof.

Suppose that is bounded and equicontinuous. To say that is bounded is to say that there is some C such that if f then fC, and this implies that is pointwise bounded. As M is compact it is separable, so the Arzelà-Ascoli theorem tells us that every sequence in has a subsequence that converges on every compact subset of M. To say that a sequence of functions M converges uniformly on the compact subsets of M is to say that the sequence converges in the norm of the Banach space C(M), and thus if is a subset of C(M), then to say that every sequence in has a subsequence that converges uniformly on every compact subset of M is to say that is precompact in C(M).

In the other direction, suppose that is precompact. Hence it is totally bounded in C(M). It is straightforward to verify that is bounded. We have to show that is equicontinuous. Let ϵ>0. As is totally bounded, there are f1,,fn such that k=1nBϵ/3(fk). As each fk:M is continuous and M is compact, there is some δk>0 such that if ρ(x,y)<δk then |fk(x)-fk(y)|<ϵ3. Let δ=min1knδk. If f and ρ(x,y)<δ, then, taking k such that f-fk<ϵ3,

|f(x)-f(y)| |f(x)-fk(x)|+|fk(x)-fk(y)|+|fk(y)-f(y)
< f-fk+ϵ3+fk-f
< ϵ3+ϵ3+ϵ3
= ϵ,

showing that is equicontinuous. ∎

We now show that if M is a compact metric space then the Banach space C(M) has the approximation property: every compact linear operator C(M)C(M) is the limit of a sequence of bounded finite rank operators.1515 15 John B. Conway, A Course in Functional Analysis, second ed., p. 176, Theorem 3.11.

Theorem 11.

If (M,ρ) is a compact metric space, then 00(C(M)) is a dense subset of 0(C(M)).

Proof.

Let T0(C(M)), let V be the closed unit ball in C(M), and let ϵ>0. Because T(V) is precompact in C(M), by Theorem 10 it is bounded and equicontinuous. Then there is some δ>0 such that if TfT(V) and ρ(x,y)<δ then |(Tf)(x)-(Tf)(y)|<ϵ. M is compact, so there are x1,,xnM such that M=j=1nBδ(xj). It is a fact that there is a partition of unity that is subordinate to this open covering of M: there are continuous functions ϕ1,,ϕn:M[0,1] such that if xM then j=1nϕj(x)=1, and ϕj(x)=0 if xBδ(xj).1616 16 John B. Conway, Functional Analysis, second ed., p. 139, Theorem 6.5. Define Tϵ:C(M)C(M) by

Tϵf=j=1n(Tf)(xj)ϕj.

It is apparent that Tϵ is linear. Tϵfj=1nTf=nTf, so TenT. And the image of Tϵ is contained in the span of {ϕ1,,ϕn}. Therefore Tϵ00(C(M)).

If fV and xM, then for each j either xBδ(xj), in which case |(Tf)(x)-(Tf)(xj)|<ϵ, or xBδ(xj), in which case ϕj(x)=0. This gives us

|((Tf)(x)-(Tϵf)(x)| = |(Tf)(x)j=1nϕj(x)-j=1n(Tf)(xj)ϕj(x)|
= |j=1n((Tf)(x)-(Tf)(xj))ϕj(x)|
j=1n|(Tf)(x)-(Tf)(xj)|ϕj(x)
< j=1nϵϕj(x)
= ϵ,

showing that Tf-Tϵf<ϵ, and as this is true for all fV we get T-Tϵ<ϵ. ∎