Compact operators on Banach spaces

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 $L^p$ 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 $ℬ(X,Y)$ be the set of bounded linear maps $X\to Y$. It is straightforward to check that $ℬ(X,Y)$ is a normed space with the operator norm

$$\parallel T\parallel =\underset{\parallel x\parallel \le 1}{sup}\parallel Tx\parallel .$$

If $X$ is a normed space and $Y$ is a Banach space, one proves that $ℬ(X,Y)$ is a Banach space. Let $ℬ(X)=ℬ(X,X)$. If $X$ is a Banach space then so is $ℬ(X)$, and it is straightforward to verify that $ℬ(X)$ is a Banach algebra.

To say $T\in ℬ(X)$ is invertible means that there is some $S\in ℬ(X)$ such that $ST={\mathrm{id}}_X$ and $TS={\mathrm{id}}_X$, and we write $T^{-1}=S$. It follows from the open mapping theorem that if $T\in ℬ(X)$, $\ker T=\{0\}$, 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 $\sigma (T)$ of $T\in ℬ(X)$ is the set of all $\lambda \in ℂ$ such that $T-\lambda {\mathrm{id}}_X$ is not invertible. If $T-\lambda {\mathrm{id}}_X$ is not injective, we say that $\lambda$ is an eigenvalue of $T$, and then there is some nonzero $x\in \ker (T-\lambda {\mathrm{id}}_X)$, which thus satisfies $Tx=\lambda x$; we call any nonzero element of $\ker (T-\lambda {\mathrm{id}}_X)$ 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 $x_1,\mathrm{\dots },x_r\in S$ such that $S\subseteq {\bigcup }_{k=1}^r{B}_{ϵ}(x_i)$, where $B_{ϵ}(x)$ is the open ball of radius $ϵ$ and center $x$.

If $X$ and $Y$ are Banach spaces and $B_1(0)$ is the open unit ball in $X$, a linear map $T:X\to Y$ is said to be compact if $T(B_1(0))$ is precompact; equivalently, if $T(B_1(0))$ is totally bounded. Check that a linear map $T:X\to Y$ 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 ${ℬ}_0(X,Y)$ denote the set of compact linear maps $X\to Y$. It does not take long to prove that ${ℬ}_0(X)$ is an ideal in the algebra $ℬ(X)$.

3 Basic facts about compact operators

Theorem 1.

Let $X$ and $Y$ be Banach spaces. If $T:X\to Y$ is linear, then $T$ is compact if and only if $x_n\in X$ being a bounded sequence implies that there is a subsequence $x_{a(n)}$ such that $T{x}_{a(n)}$ converges in $Y$.

Proof.

Suppose that $T$ is compact and let $x_n\in X$ be bounded, with

Let $V$ be the closed ball in $X$ of radius $M$ and center $0$. $V$ is bounded in $X$, so $N=\overline{T(V)}$ is compact in $Y$. As $T{x}_n\in N$, there is some convergent subsequence $T{x}_{a(n)}$ that converges to some $y\in N$.

Suppose that if $x_n$ is a bounded sequence in $X$ then there is a subsequence such that $T{x}_{a(n)}$ is convergent, let $U$ be the open unit ball in $X$, and let $y_n\in T(U)$ 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 $y_n$ are in the image of $T$, there is a subsequence such that $y_{a(n)}$ 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 $T\in {ℬ}_0(X,Y)$, then $T(X)$ is separable.

Proof.

Let $U_n$ be the closed ball of radius $n$ in $X$. As $\overline{T(U_n)}$ is a compact metric space it is separable, and hence $T(U_n)$, a subset of it, is separable too, say with dense subset $L_n$. We have

$$T(X)=\bigcup _{n=1}^{\mathrm{\infty }}T(U_n),$$

and one checks that ${\bigcup }_{n=1}^{\mathrm{\infty }}{L}_n$ 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.

Theorem 3.

Let $X$ and $Y$ be Banach spaces.

• If $T\in ℬ(X,Y)$, $T$ is compact, and $T(X)$ is a closed subset of $Y$, then .

• ${ℬ}_0(X,Y)$ is a closed subspace of $ℬ(X,Y)$.

• If $T\in ℬ(X)$, $T$ is compact, and $\lambda \ne 0$, then .

• If $dimX=\mathrm{\infty }$, $T\in ℬ(X)$, and $T$ is compact, then $0\in \sigma (T)$.

Proof.

If $T\in ℬ(X,Y)$ 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 $T:X\to T(X)$ is surjective, and $X$ is a Banach space so by the open mapping theorem $T:X\to T(X)$ is an open map. Let $Tx\in T(X)$. As $T$ is an open map, $T(B_1(x))$ is open, and hence $\overline{T(B_1(x))}$ is a neighborhood of $Tx$. But because $T$ is compact and $B_1(x)$ is bounded, $\overline{T(B_1(x))}$ is compact. Hence $\overline{T(B_1(x))}$ 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 .

It is straightforward to check that ${ℬ}_0(X,Y)$ is linear subspace of $ℬ(X,Y)$. Let $T$ be in the closure of ${ℬ}_0(X,Y)$ 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 ${ℬ}_0(X,Y)$, there is some $S\in {ℬ}_0(X,Y)$ with . As $S$ is compact, its image $S(U)$ is totally bounded, so there are finitely many $S{x}_1,\mathrm{\dots },S{x}_r\in S(U)$, with $x_1,\mathrm{\dots },x_r\in U$, such that $S(U)\subseteq {\bigcup }_{k=1}^r{B}_{ϵ}(S{x}_k)$. If $x\in U$, then

Let $x\in U$. Then there is some $k$ such that $Sx\in B_{ϵ}(S{x}_k)$, and

so $T(U)\subseteq {\bigcup }_{k=1}^r{B}_{3ϵ}(T{x}_k)$, showing that $T(U)$ is totally bounded and hence that $T$ is a compact operator.

If $T\in ℬ(X)$ is compact and $\lambda \ne 0$, let $Y=\ker (T-\lambda {\mathrm{id}}_X)$. (If $\lambda$ is not an eigenvalue of $T$, then $Y=\{0\}$.) $Y$ is a closed subspace of $X$, and hence is itself a Banach space. If $y\in Y$ then $Ty=\lambda y\in Y$. Define $S:Y\to Y$ by $Sy=Ty=\lambda y$, and as $T$ is compact so is $S$. Now we use the hypothesis that $\lambda \ne 0$: if $y\in Y$, then $S(\frac{1}{\lambda }y)=y$, so $S:Y\to Y$ is surjective. We have shown that $S:Y\to Y$ 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 , i.e. .

If $dimX=\mathrm{\infty }$ and $T\in ℬ(X)$ is compact, suppose by contradiction that $0\notin \sigma (T)$. So $T$ is invertible, with $T{T}^{-1}={\mathrm{id}}_X$. As ${ℬ}_0(X)$ is an ideal in the algebra $ℬ(X)$, ${\mathrm{id}}_X$ is compact. Of course ${\mathrm{id}}_X(X)=X$ 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 $dimX=\mathrm{\infty }$. ∎

4 Dual spaces

If $X$ is a normed space, let $X^{*}=ℬ(X,ℂ)$, the set of bounded linear maps $X\to ℂ$. $X^{*}$ is called the dual space of $X$, and is a Banach space since $ℂ$ is a Banach space. Define $⟨\cdot ,\cdot ⟩:X\times X^{*}\to ℂ$ by

$$⟨x,\lambda ⟩=\lambda (x),x\in X,\lambda \in X^{*}.$$

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.

Theorem 4.

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

$$\parallel x\parallel =\underset{\lambda \in V}{sup}|⟨x,\lambda ⟩|,x\in X.$$

Proof.

It follows from the Hahn-Banach extension theorem that if $x_0\in X$, then there is some ${\lambda }_0\in X^{*}$ such that ${\lambda }_0(x_0)=\parallel x_0\parallel$ and such that if $x\in X$ then $|{\lambda }_0(x)|\le \parallel x\parallel$. That is, that there is some ${\lambda }_0\in V$ such that ${\lambda }_0(x_0)=\parallel x_0\parallel$. Hence

$$\underset{\lambda \in V}{sup}|⟨x_0,\lambda ⟩|\ge |⟨x_0,{\lambda }_0⟩|=|{\lambda }_0(x_0)|=\parallel x_0\parallel .$$

If $\lambda \in V$, then

$$|⟨x_0,\lambda ⟩|=|\lambda (x_0)|\le \parallel \lambda \parallel \parallel x_0\parallel \le \parallel x_0\parallel ,$$

so

$$\underset{\lambda \in V}{sup}|⟨x_0,\lambda ⟩|\le \parallel x_0\parallel .$$

∎

Let $X$ be a Banach space. For $x\in X$, it is apparent that $\lambda \mapsto \lambda (x)$ is a linear map $X^{*}\to ℂ$. From Theorem 4, it is bounded, with norm $\parallel x\parallel$. Define $\varphi :X\to X^{**}$ by

$$(\varphi x)(\lambda )=⟨x,\lambda ⟩,x\in X,\lambda \in X^{*}.$$

It is apparent that $\varphi$ is a linear map. By Theorem 4, if $x\in X$ then $\parallel \varphi x\parallel =\parallel x\parallel$, so $\varphi$ is an isometry. Let $\varphi x_n\in \varphi (X)$ be a Cauchy sequence. ${\varphi }^{-1}:\varphi (X)\to X$ is an isometry, so ${\varphi }^{-1}\varphi x_n$ is a Cauchy sequence, i.e. $x_n$ is a Cauchy sequence, and so, as $X$ is a Banach space, $x_n$ converges to some $x$. Then $\varphi x_n$ converges to $\varphi x$, and thus $\varphi (X)$ is a complete metric space. But a subset of a complete metric space is closed if and only if it is complete, so $\varphi (X)$ is a closed subspace of $X^{**}$. Hence, $\varphi (X)$ is a Banach space and $\varphi :X\to \varphi (X)$ is an isometric isomorphism. A Banach space is said to be reflexive if $\varphi (X)=X^{**}$, i.e. if every bounded linear map $X^{*}\to ℂ$ is of the form $\varphi (x)$ for some $x\in X$.

5 Adjoints

If $X$ and $Y$ are normed spaces and $T\in ℬ(X,Y)$, define $T^{*}:Y^{*}\to X^{*}$ by $T^{*}\lambda =\lambda \circ T$; as $T^{*}\lambda$ is the composition of two bounded linear maps it is indeed a bounded linear map $X\to ℂ$. $T^{*}$ is called the adjoint of $T$. It is straightforward to check that $T^{*}$ is linear and that it satisfies, for $S=T^{*}$,

$$⟨Tx,\lambda ⟩=⟨x,S\lambda ⟩,x\in X,\lambda \in Y^{*}.$$ (1)

On the other hand, suppose that $S:Y^{*}\to X^{*}$ is a function that satisfies (1). Let $\lambda \in Y^{*}$, and let $x\in X$. Then

$$(S\lambda )(x)=\lambda (Tx)=(T^{*}\lambda )(x).$$

This is true for all $x$, so $S\lambda =T^{*}\lambda$, and that is true for all $\lambda$, so $S=T^{*}$. Thus $T^{*}\lambda =\lambda \circ T$ is the unique function $Y^{*}\to X^{*}$ that satisfies (1), not just the unique bounded linear map that does. (That is, satisfying (1) completely determines a function.)

Using Theorem 4,

$\parallel T\parallel$ $=$ $\underset{\parallel x\parallel \le 1}{sup}\parallel Tx\parallel$

$=$ $\underset{\parallel x\parallel \le 1}{sup}\underset{\parallel \lambda \parallel \le 1}{sup}|⟨Tx,\lambda ⟩|$

$=$ $\underset{\parallel x\parallel \le 1}{sup}\underset{\parallel \lambda \parallel \le 1}{sup}|⟨x,T^{*}\lambda ⟩|$

$=$ $\underset{\parallel \lambda \parallel \le 1}{sup}\underset{\parallel x\parallel \le 1}{sup}|T^{*}\lambda (x)|$

$=$ $\underset{\parallel \lambda \parallel \le 1}{sup}\parallel T^{*}\lambda \parallel$

$=$ $\parallel T^{*}\parallel .$

In particular, $T^{*}\in ℬ(Y^{*},X^{*})$.

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. In the proof we only show that if we take any sequence ${\lambda }_n$ in the closed unit ball then it has a subsequence such that $T{\lambda }_{a(n)}$ 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 $T\in ℬ(X,Y)$, then $T$ is compact if and only if $T^{*}$ is compact.

Proof.

Suppose that $T\in ℬ(X,Y)$ is compact, and let ${\lambda }_n\in Y^{*}$, $n\ge 1$, be a sequence in the closed unit ball in $Y^{*}$.

If $M$ is a metric space with metric $\rho$ and $ℱ$ is a set of functions $M\to ℂ$, we say that $ℱ$ is equicontinuous if for every $ϵ>0$ there exists a $\delta >0$ such that if $f\in ℱ$ and then . We say that $ℱ$ is pointwise bounded if for every $x\in M$ there is some such that if $f\in ℱ$ and $x\in M$ then $|f(x)|\le m(x)$. The Arzelà-Ascoli theorem states that if $(M,\rho )$ is a separable metric space and $ℱ$ is a set of functions $M\to ℂ$ that is equicontinuous and pointwise bounded, then for every sequence $f_n\in ℱ$ 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, $\overline{T(V)}$ is compact and therefore separable, because any compact metric space is separable. Define $f_n:\overline{T(V)}\to ℂ$ by

$$f_n(y)=⟨y,{\lambda }_n⟩={\lambda }_n(y).$$

For $y_1,y_2\in \overline{T(V)}$ we have

$$|f_n(y_1)-f_n(y_2)|=|{\lambda }_n(y_1-y_2)|\le \parallel {\lambda }_n\parallel \parallel y_1-y_2\parallel \le \parallel y_1-y_2\parallel .$$

Hence for $ϵ>0$, if $n\ge 1$ and then . This shows that $\{f_n\}$ is equicontinuous. If $y\in \overline{T(V)}$, then, for any $n\ge 1$,

$$|f_n(y)|=|{\lambda }_n(y)|\le \parallel {\lambda }_n\parallel \parallel y\parallel \le \parallel y\parallel ,$$

showing that $\{f_n\}$ is pointwise bounded. Therefore we can apply the Arzelà-Ascoli theorem: there is a subsequence $f_{a(n)}$ such that $f_{a(n)}$ converges uniformly on every compact subset of $\overline{T(V)}$, in particular on $\overline{T(V)}$ 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 $f_{a(n)}$ that converges uniformly on $T(V)$.

Let $ϵ>0$. As $f_{a(n)}$ converges uniformly on $T(V)$, there is some $N$ such that if $n,m\ge N$ and $y\in T(V)$, then . Thus, if $n,m\ge N$,

$\parallel T^{*}{\lambda }_{a(n)}-T^{*}{\lambda }_{a(m)}\parallel$ $=$ $\parallel {\lambda }_{a(n)}\circ T-{\lambda }_{a(m)}\circ T\parallel$

$=$ $\underset{x\in V}{sup}|{\lambda }_{a(n)}(Tx)-{\lambda }_{a(m)}(Tx)|$

$=$ $\underset{x\in V}{sup}|{\lambda }_{a(n)}(Tx)-{\lambda }_{a(m)}(Tx)|$

$=$ $\underset{x\in V}{sup}|f_{a(n)}(Tx)-f_{a(m)}(Tx)|$

$ϵ.$

This means that $T^{*}{\lambda }_{a(n)}\in X^{*}$ is a Cauchy sequence. As $X^{*}$ is a Banach space, this sequence converges, and therefore $T^{*}$ is a compact operator.

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

We have seen that $\varphi :X\to X^{**}$ defined by $(\varphi x)\lambda =\lambda (x)$, $x\in X$, $\lambda \in X^{*}$, is an isometric isomorphism $X\to \varphi (X)$. Let $\psi :Y\to Y^{**}$ be the same for $Y$, and let $U$ be the closed unit ball in $X$. If $x\in X$ and $\lambda \in Y^{*}$ then

$$⟨\lambda ,\psi Tx⟩=⟨Tx,\lambda ⟩=⟨x,T^{*}\lambda ⟩=⟨T^{*}\lambda ,\varphi x⟩=⟨\lambda ,T^{**}\varphi x⟩.$$

Therefore $\psi T=T^{**}\varphi$. If $x\in U$ then $\varphi x\in V$, as $\varphi$ is an isometry. Hence if $x\in U$ then $\psi Tx=T^{**}\varphi x\in T^{**}(V)$, thus

$$\psi T(U)\subseteq T^{**}(V).$$

As $\psi T(U)$ is contained in a totally bounded set it is itself totally bounded, and as $\psi$ 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

$$X=M+N,M\cap N=\{0\},$$

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

We are going to use the following lemma to prove the theorem that comes after it.

Lemma 6.

If $X$ is a locally convex topological vector space and $M$ is a subspace of $X$ with , 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 $T\in ℬ(X)$ is compact and $\lambda \ne 0$ then $T-\lambda {\mathrm{id}}_X$ has closed image.

Theorem 7.

If $X$ is a Banach space, $T\in ℬ(X)$ is compact, and $\lambda \ne 0$, then the image of $T-\lambda {\mathrm{id}}_X$ is closed.

Proof.

According to Theorem 3, , and we can then use Lemma 6: $\ker (T-\lambda {\mathrm{id}}_X)$ is a finite dimensional subspace of the locally convex space $X$, so there is a closed subspace $N$ of $X$ such that $X=\ker (T-\lambda {\mathrm{id}}_X)\oplus N$.

Define $S:N\to X$ by $Sx=Tx-\lambda x$, so $S\in ℬ(N,X)$. 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 $x\in N$ then $\parallel Sx\parallel \ge r\parallel x\parallel$.

Suppose by contradiction that for every $r>0$ there is some $x\in N$ such that . So for each $n\ge 1$, let $x_n\in N$ with , and put $v_n=\frac{x_n}{\parallel x_n\parallel }$, so that $\parallel v_n\parallel =1$ and . As $T$ is compact, there is some subsequence such that $T{v}_{a(n)}$ converges, say to $v$. Combining this with $S{v}_n\to 0$ we get $\lambda v_{a(n)}\to v$. On the one hand, $\parallel \lambda v_{a(n)}\parallel =|\lambda |\parallel v_{a(n)}\parallel =|\lambda |$, so $\parallel v\parallel =|\lambda |$. On the other hand, since $\lambda v_{a(n)}\in N$ and $N$ is closed, we get $v\in N$. $S$ is continuous and $\lambda v_{a(n)}\to 0$, so

$$Sv=\underset{n\to \mathrm{\infty }}{lim}S(\lambda v_{a(n)})=\lambda \underset{n\to \mathrm{\infty }}{lim}S{v}_{a(n)}=0.$$

Because $S$ is injective and $Sv=0$, we get $v=0$, contradicting $\parallel v\parallel =|\lambda |>0$. 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 $0$. By countable we mean bijective with a subset of the integers.

Theorem 8.

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

The following theorem shows that if $T\in ℬ(X)$ is compact and $\lambda \ne 0$, then the operator $T-\lambda {\mathrm{id}}_X$ is injective if and only if it is surjective. This tells us that if $\lambda \ne 0$ is not an eigenvalue of $T$, then $T-\lambda {\mathrm{id}}_X$ is both injective and surjective, and hence is invertible, which means that if $\lambda \ne 0$ is not an eigenvalue of $T$ then $\lambda \notin \sigma (T)$. This is an instance of the Fredholm alternative.

Theorem 9 (Fredholm alternative).

Let $X$ be a Banach space, $T\in ℬ(X)$ be compact, and $\lambda \ne 0$. $T-\lambda {\mathrm{id}}_X$ is injective if and only if it is surjective.

Proof.

Suppose that $T-\lambda {\mathrm{id}}_X$ is injective and let $V_n={(T-\lambda {\mathrm{id}}_X)}^{n}X$, $n\ge 1$. If ${(T-\lambda {\mathrm{id}}_X)}^{n}x\in V_n$, then, as $(T-\lambda {\mathrm{id}}_X)x\in X$, we have

$${(T-\lambda {\mathrm{id}}_X)}^{n-1}(T-\lambda {\mathrm{id}}_X)x\in V_{n-1},$$

so $V_n\supseteq V_{n-1}$. Thus

$$V_1\supseteq V_2\supseteq \mathrm{\cdots }$$

Certainly $V_n$ is a normed vector space. Define $T_n\in ℬ(V_n)$ by $T_{n}x=Tx$, namely, $T_n$ is the restriction of $T$ to $V_n$.

As $T$ is a compact operator, by Theorem 7 we get that $V_1=(T-\lambda {\mathrm{id}}_X)(X)$ is closed. Hence $V_1$ is a Banach space, being a closed subspace of a Banach space. Assume as induction hypothesis that $V_n$ is a closed subset of $X$. Thus $V_n$ is a Banach space, and $T_n\in ℬ(V_n)$ is a compact operator, as it is the restriction of the compact operator $T$ to $V_n$. Therefore by Theorem 7, the image of $T_n-\lambda {\mathrm{id}}_X$ is closed, but this image is precisely $V_{n+1}$. Therefore, if $n\ge 1$ then $V_n$ is a closed subspace of $X$.

Suppose by contradiction that there is some $x\notin (T-\lambda {\mathrm{id}}_X)X=V_1$. If $y\in X$ then

$${(T-\lambda {\mathrm{id}}_X)}^{n}x-{(T-\lambda {\mathrm{id}}_X)}^{n+1}y={(T-\lambda {\mathrm{id}}_X)}^n(x-(T-\lambda {\mathrm{id}}_X)y).$$

As $x\notin (T-\lambda {\mathrm{id}}_X)X$, we have $x-(T-\lambda {\mathrm{id}}_X)y\ne 0$. As we have supposed that $T-\lambda {\mathrm{id}}_X$ is injective, any positive power of it is injective, and hence the right hand side of the above equation is not $0$. Thus ${(T-\lambda {\mathrm{id}}_X)}^{n}x\ne {(T-\lambda {\mathrm{id}}_X)}^{n+1}y$, and as $y\in X$ was arbitrary,

$${(T-\lambda {\mathrm{id}}_X)}^{n}x\notin {(T-\lambda {\mathrm{id}}_X)}^{n+1}X.$$

However, of course ${(T-\lambda {\mathrm{id}}_X)}^{n}x\in V_n$, so if $n\ge 1$ then $V_n$ strictly contains $V_{n+1}$.

Riesz’s lemma states that if $M$ is a normed space, $N$ is a proper closed subspace of $M$, and , then there is some $x\in M$ with $\parallel x\parallel =1$ and ${inf}_{y\in X}\parallel x-y\parallel \ge r$. For each $n\ge 1$, using Riesz’s lemma there is some $v_n\in V_n$, $\parallel v_n\parallel =1$, such that

$$\underset{y\in V_{n+1}}{inf}\parallel v_n-y\parallel \ge \frac{1}{2};$$

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

If $n,m\ge 1$, then $(T-\lambda {\mathrm{id}}_X)v_m\in V_{m+1}$ and check that $T{v}_{m+n}\subseteq V_{m+n}$, so

$$T{v}_m-T{v}_{m+n}=\lambda v_m+(T-\lambda {\mathrm{id}}_X)v_m-T{v}_{m+n}\in \lambda v_m+V_{m+1}.$$

From this and the definition of the sequence $v_m$, we get

$$\parallel T{v}_m-T{v}_{m+n}\parallel \ge |\lambda |\cdot \frac{1}{2}.$$

That is, the distance between any two terms in $T{v}_m$ is $\ge \frac{|\lambda |}{2}$, which is a fixed positive constant, hence $T{v}_m$ has no convergent subsequence. But $\parallel v_m\parallel =1$, so $v_m$ is bounded and therefore, as $T$ is compact, the sequence $T{v}_m$ has a convergent subsequence, a contradiction. Therefore $T-\lambda {\mathrm{id}}_X$ is surjective.

Suppose that $T-\lambda {\mathrm{id}}_X$ is surjective. One checks that if a bounded linear operator is surjective then its adjoint is injective. For $x\in X$ and $\mu \in X^{*}$, $⟨\lambda {\mathrm{id}}_{X}x,\mu ⟩=\mu (\lambda x)=\lambda \mu (x)=⟨x,\lambda {\mathrm{id}}_{X^{*}}\mu ⟩$, so ${(\lambda {\mathrm{id}}_X)}^{*}=\lambda {\mathrm{id}}_{X^{*}}$. Hence ${(T-\lambda {\mathrm{id}}_X)}^{*}=T^{*}-\lambda {\mathrm{id}}_{X^{*}}$. $T$ is compact so $T^{*}$ is compact. As $T^{*}-\lambda {\mathrm{id}}_{X^{*}}$ is injective and $T^{*}$ is compact, $T^{*}-\lambda {\mathrm{id}}_{X^{*}}$ is surjective, whence its adjoint $T^{**}-\lambda {\mathrm{id}}_{X^{**}}:X^{**}\to X^{**}$ is injective. One checks that if $S\in ℬ(X)$ and $S^{**}:X^{**}\to X^{**}$ is injective then $S$ is injective; this is proved using the fact that $\varphi :X\to X^{**}$ defined by $(\varphi x)(\lambda )=\lambda (x)$ is an isometric isomorphism $X\to \varphi (X)$. Using this, $T-\lambda {\mathrm{id}}_X$ 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 $\rho$ and $ℱ$ is a set of functions $M\to ℂ$, we say that $ℱ$ is equicontinuous if for all $ϵ>0$ there is some $\delta >0$ such that $f\in ℱ$ and imply that . We say that $ℱ$ is pointwise bounded if for all $x\in M$ there is some $m(x)$ such that if $f\in ℱ$ then $|f(x)|\le m(x)$. The Arzelà-Ascoli theorem states that if $M$ is a separable metric space and $ℱ$ is equicontinuous and pointwise bounded, then every sequence in $ℱ$ has a sequence that converges uniformly on every compact subset of $M$.

We are going to use a converse of the Arzelà-Ascoli theorem in the case of a compact metric space. Let $M$ be a compact metric space and let $C(M)$ be the set of continuous functions $M\to ℂ$. It does not take long to prove that with the norm $\parallel f\parallel ={sup}_{x\in M}|f(x)|$, $C(M)$ is a Banach space.

Theorem 10.

Let $(M,\rho )$ be a compact metric space and let $ℱ\subseteq C(M)$. $ℱ$ is precompact in $C(M)$ if and only if $ℱ$ is bounded and equicontinuous.

Proof.

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

In the other direction, suppose that $ℱ$ is precompact. Hence it is totally bounded in $C(M)$. It is straightforward to verify that $ℱ$ is bounded. We have to show that $ℱ$ is equicontinuous. Let $ϵ>0$. As $ℱ$ is totally bounded, there are $f_1,\mathrm{\dots },f_n\in ℱ$ such that $ℱ\subseteq {\bigcup }_{k=1}^n{B}_{ϵ/3}(f_k)$. As each $f_k:M\to ℂ$ is continuous and $M$ is compact, there is some ${\delta }_k>0$ such that if then . Let $\delta ={\min }_{1\le k\le n}{\delta }_k$. If $f\in ℱ$ and , then, taking $k$ such that ,

$|f(x)-f(y)|$ $\le$ $|f(x)-f_k(x)|+|f_k(x)-f_k(y)|+|f_k(y)-f(y)$

$\parallel f-f_k\parallel +{\displaystyle \frac{ϵ}{3}}+\parallel f_k-f\parallel$

${\displaystyle \frac{ϵ}{3}}+{\displaystyle \frac{ϵ}{3}}+{\displaystyle \frac{ϵ}{3}}$

$=$ $ϵ,$

showing that $ℱ$ 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 $C(M)\to C(M)$ is the limit of a sequence of bounded finite rank operators.

Theorem 11.

If $(M,\rho )$ is a compact metric space, then ${ℬ}_{00}(C(M))$ is a dense subset of ${ℬ}_0(C(M))$.

Proof.

Let $T\in {ℬ}_0(C(M))$, 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 $\delta >0$ such that if $Tf\in T(V)$ and then . $M$ is compact, so there are $x_1,\mathrm{\dots },x_n\in M$ such that $M={\bigcup }_{j=1}^n{B}_{\delta }(x_j)$. It is a fact that there is a partition of unity that is subordinate to this open covering of $M$: there are continuous functions ${\varphi }_1,\mathrm{\dots },{\varphi }_n:M\to [0,1]$ such that if $x\in M$ then ${\sum }_{j=1}^n{\varphi }_j(x)=1$, and ${\varphi }_j(x)=0$ if $x\notin B_{\delta }(x_j)$. Define $T_{ϵ}:C(M)\to C(M)$ by

$$T_{ϵ}f=\sum _{j=1}^n(Tf)(x_j){\varphi }_j.$$

It is apparent that $T_{ϵ}$ is linear. $\parallel T_{ϵ}f\parallel \le {\sum }_{j=1}^n\parallel T\parallel \parallel f\parallel =n\parallel T\parallel \parallel f\parallel$, so $\parallel T_e\parallel \le n\parallel T\parallel$. And the image of $T_{ϵ}$ is contained in the span of $\{{\varphi }_1,\mathrm{\dots },{\varphi }_n\}$. Therefore $T_{ϵ}\in {ℬ}_{00}(C(M))$.

If $f\in V$ and $x\in M$, then for each $j$ either $x\in B_{\delta }(x_j)$, in which case , or $x\notin B_{\delta }(x_j)$, in which case ${\varphi }_j(x)=0$. This gives us

$|((Tf)(x)-(T_{ϵ}f)(x)|$ $=$ $\left|(Tf)(x)\cdot {\displaystyle \sum _{j=1}^n}{\varphi }_j(x)-{\displaystyle \sum _{j=1}^n}(Tf)(x_j){\varphi }_j(x)\right|$

$=$ $\left|{\displaystyle \sum _{j=1}^n}\left((Tf)(x)-(Tf)(x_j)\right){\varphi }_j(x)\right|$

$\le$ ${\displaystyle \sum _{j=1}^n}\left|(Tf)(x)-(Tf)(x_j)\right|{\varphi }_j(x)$

${\displaystyle \sum _{j=1}^n}ϵ{\varphi }_j(x)$

$=$ $ϵ,$

showing that , and as this is true for all $f\in V$ we get . ∎