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

 $\|T\|=\sup_{\|x\|\leq 1}\|Tx\|.$

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

To say $T\in\mathscr{B}(X)$ is invertible means that there is some $S\in\mathscr{B}(X)$ 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 $T\in\mathscr{B}(X)$, $\ker T=\{0\}$, 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 $T\in\mathscr{B}(X)$ is the set of all $\lambda\in\mathbb{C}$ 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 $\epsilon>0$ there are finitely many points $x_{1},\ldots,x_{r}\in S$ such that $S\subseteq\bigcup_{k=1}^{r}B_{\epsilon}(x_{i})$, where $B_{\epsilon}(x)$ is the open ball of radius $\epsilon$ 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 $\mathscr{B}_{0}(X,Y)$ denote the set of compact linear maps $X\to Y$. It does not take long to prove that $\mathscr{B}_{0}(X)$ is an ideal in the algebra $\mathscr{B}(X)$.

## 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 $Tx_{a(n)}$ converges in $Y$.

###### Proof.

Suppose that $T$ is compact and let $x_{n}\in X$ be bounded, with

 $M=\sup_{n}\|x_{n}\|<\infty.$

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 $Tx_{n}\in N$, there is some convergent subsequence $Tx_{a(n)}$ 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 $Tx_{a(n)}$ 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 $T\in\mathscr{B}_{0}(X,Y)$, 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

 $T(X)=\bigcup_{n=1}^{\infty}T(U_{n}),$

and one checks that $\bigcup_{n=1}^{\infty}L_{n}$ 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.22 2 Walter Rudin, Functional Analysis, second ed., p. 104, Theorem 4.18.

###### Theorem 3.

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

• If $T\in\mathscr{B}(X,Y)$, $T$ is compact, and $T(X)$ is a closed subset of $Y$, then $\dim T(X)<\infty$.

• $\mathscr{B}_{0}(X,Y)$ is a closed subspace of $\mathscr{B}(X,Y)$.

• If $T\in\mathscr{B}(X)$, $T$ is compact, and $\lambda\neq 0$, then $\dim\ker(T-\lambda\mathrm{id}_{X})<\infty$.

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

###### Proof.

If $T\in\mathscr{B}(X,Y)$ 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 $\dim T(X)<\infty$.

It is straightforward to check that $\mathscr{B}_{0}(X,Y)$ is linear subspace of $\mathscr{B}(X,Y)$. Let $T$ be in the closure of $\mathscr{B}_{0}(X,Y)$ and let $U$ be the open unit ball in $X$. We wish to show that $T(U)$ is totally bounded. Let $\epsilon>0$. As $T$ is in the closure of $\mathscr{B}_{0}(X,Y)$, there is some $S\in\mathscr{B}_{0}(X,Y)$ with $\|S-T\|<\epsilon$. As $S$ is compact, its image $S(U)$ is totally bounded, so there are finitely many $Sx_{1},\ldots,Sx_{r}\in S(U)$, with $x_{1},\ldots,x_{r}\in U$, such that $S(U)\subseteq\bigcup_{k=1}^{r}B_{\epsilon}(Sx_{k})$. If $x\in U$, then

 $\|Sx-Tx\|=\|(S-T)x\|\leq\|S-T\|\|x\|<\|S-T\|<\epsilon.$

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

 $\|Tx-Tx_{k}\|\leq\|Tx-Sx\|+\|Sx-Sx_{k}\|+\|Sx_{k}-Tx_{k}\|<3\epsilon,$

so $T(U)\subseteq\bigcup_{k=1}^{r}B_{3\epsilon}(Tx_{k})$, showing that $T(U)$ is totally bounded and hence that $T$ is a compact operator.

If $T\in\mathscr{B}(X)$ is compact and $\lambda\neq 0$, 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 $\lambda\neq 0$: 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 $\dim S(Y)<\infty$, i.e. $\dim Y<\infty$.

If $\dim X=\infty$ and $T\in\mathscr{B}(X)$ is compact, suppose by contradiction that $0\not\in\sigma(T)$. So $T$ is invertible, with $TT^{-1}=\mathrm{id}_{X}$. As $\mathscr{B}_{0}(X)$ is an ideal in the algebra $\mathscr{B}(X)$, $\mathrm{id}_{X}$ 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 $\dim X=\infty$. ∎

## 4 Dual spaces

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

 $\langle x,\lambda\rangle=\lambda(x),\qquad x\in X,\lambda\in X^{*}.$

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.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

 $\|x\|=\sup_{\lambda\in V}|\langle x,\lambda\rangle|,\qquad x\in X.$
###### 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 $\lambda_{0}(x_{0})=\|x_{0}\|$ and such that if $x\in X$ then $|\lambda_{0}(x)|\leq\|x\|$.44 4 Walter Rudin, Functional Analysis, second ed., p. 58, Theorem 3.3. That is, that there is some $\lambda_{0}\in V$ such that $\lambda_{0}(x_{0})=\|x_{0}\|$. Hence

 $\sup_{\lambda\in V}|\langle x_{0},\lambda\rangle|\geq|\langle x_{0},\lambda_{0% }\rangle|=|\lambda_{0}(x_{0})|=\|x_{0}\|.$

If $\lambda\in V$, then

 $|\langle x_{0},\lambda\rangle|=|\lambda(x_{0})|\leq\|\lambda\|\|x_{0}\|\leq\|x% _{0}\|,$

so

 $\sup_{\lambda\in V}|\langle x_{0},\lambda\rangle|\leq\|x_{0}\|.$

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

 $(\phi x)(\lambda)=\langle x,\lambda\rangle,\qquad x\in X,\lambda\in X^{*}.$

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

If $X$ and $Y$ are normed spaces and $T\in\mathscr{B}(X,Y)$, 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 $X\to\mathbb{C}$. $T^{*}$ is called the adjoint of $T$. It is straightforward to check that $T^{*}$ is linear and that it satisfies, for $S=T^{*}$,

 $\langle Tx,\lambda\rangle=\langle x,S\lambda\rangle,\qquad x\in X,\lambda\in Y% ^{*}.$ (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,

 $\displaystyle\|T\|$ $\displaystyle=$ $\displaystyle\sup_{\|x\|\leq 1}\|Tx\|$ $\displaystyle=$ $\displaystyle\sup_{\|x\|\leq 1}\sup_{\|\lambda\|\leq 1}|\langle Tx,\lambda\rangle|$ $\displaystyle=$ $\displaystyle\sup_{\|x\|\leq 1}\sup_{\|\lambda\|\leq 1}|\langle x,T^{*}\lambda\rangle|$ $\displaystyle=$ $\displaystyle\sup_{\|\lambda\|\leq 1}\sup_{\|x\|\leq 1}|T^{*}\lambda(x)|$ $\displaystyle=$ $\displaystyle\sup_{\|\lambda\|\leq 1}\|T^{*}\lambda\|$ $\displaystyle=$ $\displaystyle\|T^{*}\|.$

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

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.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 $\lambda_{n}$ 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 $T\in\mathscr{B}(X,Y)$, then $T$ is compact if and only if $T^{*}$ is compact.

###### Proof.

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

If $M$ is a metric space with metric $\rho$ and $\mathcal{F}$ is a set of functions $M\to\mathbb{C}$, we say that $\mathcal{F}$ is equicontinuous if for every $\epsilon>0$ there exists a $\delta>0$ such that if $f\in\mathcal{F}$ and $\rho(x,y)<\delta$ then $|f(x)-f(y)|<\epsilon$. We say that $\mathcal{F}$ is pointwise bounded if for every $x\in M$ there is some $m(x)<\infty$ such that if $f\in\mathcal{F}$ and $x\in M$ then $|f(x)|\leq m(x)$. The Arzelà-Ascoli theorem66 6 Walter Rudin, Real and Complex Analysis, third ed., p. 245, Theorem 11.28. states that if $(M,\rho)$ is a separable metric space and $\mathcal{F}$ is a set of functions $M\to\mathbb{C}$ that is equicontinuous and pointwise bounded, then for every sequence $f_{n}\in\mathcal{F}$ 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 $f_{n}:\overline{T(V)}\to\mathbb{C}$ by

 $f_{n}(y)=\langle y,\lambda_{n}\rangle=\lambda_{n}(y).$

For $y_{1},y_{2}\in\overline{T(V)}$ we have

 $|f_{n}(y_{1})-f_{n}(y_{2})|=|\lambda_{n}(y_{1}-y_{2})|\leq\|\lambda_{n}\|\|y_{% 1}-y_{2}\|\leq\|y_{1}-y_{2}\|.$

Hence for $\epsilon>0$, if $n\geq 1$ and $\|y_{1}-y_{2}\|<\epsilon$ then $|f_{n}(y_{1})-f_{n}(y_{2})|<\epsilon$. This shows that $\{f_{n}\}$ is equicontinuous. If $y\in\overline{T(V)}$, then, for any $n\geq 1$,

 $|f_{n}(y)|=|\lambda_{n}(y)|\leq\|\lambda_{n}\|\|y\|\leq\|y\|,$

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

 $\displaystyle\|T^{*}\lambda_{a(n)}-T^{*}\lambda_{a(m)}\|$ $\displaystyle=$ $\displaystyle\|\lambda_{a(n)}\circ T-\lambda_{a(m)}\circ T\|$ $\displaystyle=$ $\displaystyle\sup_{x\in V}|\lambda_{a(n)}(Tx)-\lambda_{a(m)}(Tx)|$ $\displaystyle=$ $\displaystyle\sup_{x\in V}|\lambda_{a(n)}(Tx)-\lambda_{a(m)}(Tx)|$ $\displaystyle=$ $\displaystyle\sup_{x\in V}|f_{a(n)}(Tx)-f_{a(m)}(Tx)|$ $\displaystyle<$ $\displaystyle\epsilon.$

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 $T^{*}\in\mathscr{B}(Y^{*},X^{*})$ 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 $\phi:X\to X^{**}$ defined by $(\phi x)\lambda=\lambda(x)$, $x\in X$, $\lambda\in X^{*}$, is an isometric isomorphism $X\to\phi(X)$. 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

 $\langle\lambda,\psi Tx\rangle=\langle Tx,\lambda\rangle=\langle x,T^{*}\lambda% \rangle=\langle T^{*}\lambda,\phi x\rangle=\langle\lambda,T^{**}\phi x\rangle.$

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

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

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.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 $\dim X<\infty$, 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 $T\in\mathscr{B}(X)$ is compact and $\lambda\neq 0$ then $T-\lambda\mathrm{id}_{X}$ has closed image.88 8 Walter Rudin, Functional Analysis, second ed., p. 107, Theorem 4.23.

###### Theorem 7.

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

###### Proof.

According to Theorem 3, $\dim\ker(T-\lambda\mathrm{id}_{X})<\infty$, 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 $S\in\mathscr{B}(N,X)$. 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 $\|Sx\|\geq r\|x\|$.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 $x\in N$ such that $\|Sx\|. So for each $n\geq 1$, let $x_{n}\in N$ with $\|Sx_{n}\|<\frac{1}{n}\|x_{n}\|$, and put $v_{n}=\frac{x_{n}}{\|x_{n}\|}$, so that $\|v_{n}\|=1$ and $\|Sv_{n}\|<\frac{1}{n}$. As $T$ is compact, there is some subsequence such that $Tv_{a(n)}$ converges, say to $v$. Combining this with $Sv_{n}\to 0$ we get $\lambda v_{a(n)}\to v$. On the one hand, $\|\lambda v_{a(n)}\|=|\lambda|\|v_{a(n)}\|=|\lambda|$, so $\|v\|=|\lambda|$. 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

 $Sv=\lim_{n\to\infty}S(\lambda v_{a(n)})=\lambda\lim_{n\to\infty}Sv_{a(n)}=0.$

Because $S$ is injective and $Sv=0$, we get $v=0$, contradicting $\|v\|=|\lambda|>0$. 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 $0$.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, $T\in\mathscr{B}(X)$ 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 $T\in\mathscr{B}(X)$ is compact and $\lambda\neq 0$, then the operator $T-\lambda\mathrm{id}_{X}$ 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 $\lambda\neq 0$ 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 $\lambda\neq 0$ is not an eigenvalue of $T$ then $\lambda\not\in\sigma(T)$. This is an instance of the Fredholm alternative.

###### Theorem 9 (Fredholm alternative).

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

###### Proof.

Suppose that $T-\lambda\mathrm{id}_{X}$ is injective and let $V_{n}=(T-\lambda\mathrm{id}_{X})^{n}X$, $n\geq 1$. 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

 $V_{1}\supseteq V_{2}\supseteq\cdots$

Certainly $V_{n}$ is a normed vector space. Define $T_{n}\in\mathscr{B}(V_{n})$ 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 $T_{n}\in\mathscr{B}(V_{n})$ 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 $n\geq 1$ then $V_{n}$ is a closed subspace of $X$.

Suppose by contradiction that there is some $x\not\in(T-\lambda\mathrm{id}_{X})X=V_{1}$. 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 $x\not\in(T-\lambda\mathrm{id}_{X})X$, we have $x-(T-\lambda\mathrm{id}_{X})y\neq 0$. 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 $(T-\lambda\mathrm{id}_{X})^{n}x\neq(T-\lambda\mathrm{id}_{X})^{n+1}y$, and as $y\in X$ was arbitrary,

 $(T-\lambda\mathrm{id}_{X})^{n}x\not\in(T-\lambda\mathrm{id}_{X})^{n+1}X.$

However, of course $(T-\lambda\mathrm{id}_{X})^{n}x\in V_{n}$, so if $n\geq 1$ 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 $0, then there is some $x\in M$ with $\|x\|=1$ and $\inf_{y\in X}\|x-y\|\geq r$.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 $n\geq 1$, using Riesz’s lemma there is some $v_{n}\in V_{n}$, $\|v_{n}\|=1$, such that

 $\inf_{y\in V_{n+1}}\|v_{n}-y\|\geq\frac{1}{2};$

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 $n,m\geq 1$, then $(T-\lambda\mathrm{id}_{X})v_{m}\in V_{m+1}$ and check that $Tv_{m+n}\subseteq V_{m+n}$, so

 $Tv_{m}-Tv_{m+n}=\lambda v_{m}+(T-\lambda\mathrm{id}_{X})v_{m}-Tv_{m+n}\in% \lambda v_{m}+V_{m+1}.$

From this and the definition of the sequence $v_{m}$, we get

 $\|Tv_{m}-Tv_{m+n}\|\geq|\lambda|\cdot\frac{1}{2}.$

That is, the distance between any two terms in $Tv_{m}$ is $\geq\frac{|\lambda|}{2}$, which is a fixed positive constant, hence $Tv_{m}$ has no convergent subsequence. But $\|v_{m}\|=1$, so $v_{m}$ is bounded and therefore, as $T$ is compact, the sequence $Tv_{m}$ 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 $\mu\in X^{*}$, $\langle\lambda\mathrm{id}_{X}x,\mu\rangle=\mu(\lambda x)=\lambda\mu(x)=\langle x% ,\lambda\mathrm{id}_{X^{*}}\mu\rangle$, 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 $S\in\mathscr{B}(X)$ and $S^{**}:X^{**}\to X^{**}$ is injective then $S$ is injective; this is proved using the fact that $\phi:X\to X^{**}$ defined by $(\phi x)(\lambda)=\lambda(x)$ is an isometric isomorphism $X\to\phi(X)$. 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 $\mathcal{F}$ is a set of functions $M\to\mathbb{C}$, we say that $\mathcal{F}$ is equicontinuous if for all $\epsilon>0$ there is some $\delta>0$ such that $f\in\mathcal{F}$ and $\rho(x,y)<\delta$ imply that $|f(x)-f(y)|<\epsilon$. We say that $\mathcal{F}$ is pointwise bounded if for all $x\in M$ there is some $m(x)$ such that if $f\in\mathcal{F}$ then $|f(x)|\leq m(x)$. The Arzelà-Ascoli theorem states that if $M$ is a separable metric space and $\mathcal{F}$ is equicontinuous and pointwise bounded, then every sequence in $\mathcal{F}$ 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 $M\to\mathbb{C}$. It does not take long to prove that with the norm $\|f\|=\sup_{x\in M}|f(x)|$, $C(M)$ is a Banach space.

###### Theorem 10.

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

###### Proof.

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

In the other direction, suppose that $\mathcal{F}$ is precompact. Hence it is totally bounded in $C(M)$. It is straightforward to verify that $\mathcal{F}$ is bounded. We have to show that $\mathcal{F}$ is equicontinuous. Let $\epsilon>0$. As $\mathcal{F}$ is totally bounded, there are $f_{1},\ldots,f_{n}\in\mathcal{F}$ such that $\mathcal{F}\subseteq\bigcup_{k=1}^{n}B_{\epsilon/3}(f_{k})$. As each $f_{k}:M\to\mathbb{C}$ is continuous and $M$ is compact, there is some $\delta_{k}>0$ such that if $\rho(x,y)<\delta_{k}$ then $|f_{k}(x)-f_{k}(y)|<\frac{\epsilon}{3}$. Let $\delta=\min_{1\leq k\leq n}\delta_{k}$. If $f\in\mathcal{F}$ and $\rho(x,y)<\delta$, then, taking $k$ such that $\|f-f_{k}\|<\frac{\epsilon}{3}$,

 $\displaystyle|f(x)-f(y)|$ $\displaystyle\leq$ $\displaystyle|f(x)-f_{k}(x)|+|f_{k}(x)-f_{k}(y)|+|f_{k}(y)-f(y)$ $\displaystyle<$ $\displaystyle\|f-f_{k}\|+\frac{\epsilon}{3}+\|f_{k}-f\|$ $\displaystyle<$ $\displaystyle\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}$ $\displaystyle=$ $\displaystyle\epsilon,$

showing that $\mathcal{F}$ 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.1515 15 John B. Conway, A Course in Functional Analysis, second ed., p. 176, Theorem 3.11.

###### Theorem 11.

If $(M,\rho)$ is a compact metric space, then $\mathscr{B}_{00}(C(M))$ is a dense subset of $\mathscr{B}_{0}(C(M))$.

###### Proof.

Let $T\in\mathscr{B}_{0}(C(M))$, let $V$ be the closed unit ball in $C(M)$, and let $\epsilon>0$. 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 $\rho(x,y)<\delta$ then $|(Tf)(x)-(Tf)(y)|<\epsilon$. $M$ is compact, so there are $x_{1},\ldots,x_{n}\in M$ 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 $\phi_{1},\ldots,\phi_{n}:M\to[0,1]$ such that if $x\in M$ then $\sum_{j=1}^{n}\phi_{j}(x)=1$, and $\phi_{j}(x)=0$ if $x\not\in B_{\delta}(x_{j})$.1616 16 John B. Conway, Functional Analysis, second ed., p. 139, Theorem 6.5. Define $T_{\epsilon}:C(M)\to C(M)$ by

 $T_{\epsilon}f=\sum_{j=1}^{n}(Tf)(x_{j})\phi_{j}.$

It is apparent that $T_{\epsilon}$ is linear. $\|T_{\epsilon}f\|\leq\sum_{j=1}^{n}\|T\|\|f\|=n\|T\|\|f\|$, so $\|T_{e}\|\leq n\|T\|$. And the image of $T_{\epsilon}$ is contained in the span of $\{\phi_{1},\ldots,\phi_{n}\}$. Therefore $T_{\epsilon}\in\mathscr{B}_{00}(C(M))$.

If $f\in V$ and $x\in M$, then for each $j$ either $x\in B_{\delta}(x_{j})$, in which case $|(Tf)(x)-(Tf)(x_{j})|<\epsilon$, or $x\not\in B_{\delta}(x_{j})$, in which case $\phi_{j}(x)=0$. This gives us

 $\displaystyle|((Tf)(x)-(T_{\epsilon}f)(x)|$ $\displaystyle=$ $\displaystyle\left|(Tf)(x)\cdot\sum_{j=1}^{n}\phi_{j}(x)-\sum_{j=1}^{n}(Tf)(x_% {j})\phi_{j}(x)\right|$ $\displaystyle=$ $\displaystyle\left|\sum_{j=1}^{n}\left((Tf)(x)-(Tf)(x_{j})\right)\phi_{j}(x)\right|$ $\displaystyle\leq$ $\displaystyle\sum_{j=1}^{n}\left|(Tf)(x)-(Tf)(x_{j})\right|\phi_{j}(x)$ $\displaystyle<$ $\displaystyle\sum_{j=1}^{n}\epsilon\phi_{j}(x)$ $\displaystyle=$ $\displaystyle\epsilon,$

showing that $\|Tf-T_{\epsilon}f\|<\epsilon$, and as this is true for all $f\in V$ we get $\|T-T_{\epsilon}\|<\epsilon$. ∎