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 $\mathcal{B}(X,Y)$ be the set of bounded linear maps $X\to Y$. It is straightforward to check that $\mathcal{B}(X,Y)$ 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 $\mathcal{B}(X,Y)$ is a Banach space.^{1}^{1} 1 Walter Rudin, Functional Analysis, second ed., p. 92, Theorem 4.1. Let $\mathcal{B}(X)=\mathcal{B}(X,X)$. If $X$ is a Banach space then so is $\mathcal{B}(X)$, and it is straightforward to verify that $\mathcal{B}(X)$ is a Banach algebra.
To say $T\in \mathcal{B}(X)$ is invertible means that there is some $S\in \mathcal{B}(X)$ 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 $T\in \mathcal{B}(X)$, $\mathrm{ker}T=\{0\}$, 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 $T\in \mathcal{B}(X)$ is the set of all $\lambda \in \u2102$ 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 $x\in \mathrm{ker}(T\lambda {\mathrm{id}}_{X})$, which thus satisfies $Tx=\lambda x$; we call any nonzero element of $\mathrm{ker}(T\lambda {\mathrm{id}}_{X})$ 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 HeineBorel 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 $\u03f5>0$ there are finitely many points ${x}_{1},\mathrm{\dots},{x}_{r}\in S$ such that $S\subseteq {\bigcup}_{k=1}^{r}{B}_{\u03f5}({x}_{i})$, where ${B}_{\u03f5}(x)$ is the open ball of radius $\u03f5$ 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 ${\mathcal{B}}_{0}(X,Y)$ denote the set of compact linear maps $X\to Y$. It does not take long to prove that ${\mathcal{B}}_{0}(X)$ is an ideal in the algebra $\mathcal{B}(X)$.
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
$$ 
Let $V$ be the closed ball in $X$ of radius $M$ and center $0$. $V$ is bounded in $X$, so $N=\overline{T(V)}$ is compact in $Y$. As $T{x}_{n}\in N$, there is some convergent subsequence $T{x}_{a(n)}$ that converges to some $y\in N$.
Suppose that if ${x}_{n}$ is a bounded sequence in $X$ then there is a subsequence such that $T{x}_{a(n)}$ is convergent, let $U$ be the open unit ball in $X$, and let ${y}_{n}\in T(U)$ be a sequence. It is a fact that a subset of a metric space is precompact if and only if every sequence has a subsequence that converges to some element in the space; this is not obvious, but at least we are only taking as given a fact about metric spaces. (What we have asserted is that a set in a metric space is precompact if and only if it is sequentially precompact.) As ${y}_{n}$ are in the image of $T$, there is a subsequence such that ${y}_{a(n)}$ is convergent and this implies that $T(U)$ is precompact, and so $T$ is a compact operator. ∎
Theorem 2.
Let $X$ and $Y$ be Banach spaces. If $T\in {\mathcal{B}}_{0}(X,Y)$, then $T(X)$ is separable.
Proof.
Let ${U}_{n}$ be the closed ball of radius $n$ in $X$. As $\overline{T({U}_{n})}$ is a compact metric space it is separable, and hence $T({U}_{n})$, a subset of it, is separable too, say with dense subset ${L}_{n}$. We have
$$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 righthand side, showing that $T(X)$ is separable. ∎
The following theorem gathers some important results about compact operators.^{2}^{2} 2 Walter Rudin, Functional Analysis, second ed., p. 104, Theorem 4.18.
Theorem 3.
Let $X$ and $Y$ be Banach spaces.

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

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

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

•
If $dimX=\mathrm{\infty}$, $T\in \mathcal{B}(X)$, and $T$ is compact, then $0\in \sigma (T)$.
Proof.
If $T\in \mathcal{B}(X,Y)$ 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 $$.
It is straightforward to check that ${\mathcal{B}}_{0}(X,Y)$ is linear subspace of $\mathcal{B}(X,Y)$. Let $T$ be in the closure of ${\mathcal{B}}_{0}(X,Y)$ and let $U$ be the open unit ball in $X$. We wish to show that $T(U)$ is totally bounded. Let $\u03f5>0$. As $T$ is in the closure of ${\mathcal{B}}_{0}(X,Y)$, there is some $S\in {\mathcal{B}}_{0}(X,Y)$ with $$. 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 $S(U)\subseteq {\bigcup}_{k=1}^{r}{B}_{\u03f5}(S{x}_{k})$. If $x\in U$, then
$$ 
Let $x\in U$. Then there is some $k$ such that $Sx\in {B}_{\u03f5}(S{x}_{k})$, and
$$ 
so $T(U)\subseteq {\bigcup}_{k=1}^{r}{B}_{3\u03f5}(T{x}_{k})$, showing that $T(U)$ is totally bounded and hence that $T$ is a compact operator.
If $T\in \mathcal{B}(X)$ is compact and $\lambda \ne 0$, let $Y=\mathrm{ker}(T\lambda {\mathrm{id}}_{X})$. (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 $T\in \mathcal{B}(X)$ is compact, suppose by contradiction that $0\notin \sigma (T)$. So $T$ is invertible, with $T{T}^{1}={\mathrm{id}}_{X}$. As ${\mathcal{B}}_{0}(X)$ is an ideal in the algebra $\mathcal{B}(X)$, ${\mathrm{id}}_{X}$ 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 ${X}^{*}=\mathcal{B}(X,\u2102)$, the set of bounded linear maps $X\to \u2102$. ${X}^{*}$ is called the dual space of $X$, and is a Banach space since $\u2102$ is a Banach space. Define $\u27e8\cdot ,\cdot \u27e9:X\times {X}^{*}\to \u2102$ by
$$\u27e8x,\lambda \u27e9=\lambda (x),x\in X,\lambda \in {X}^{*}.$$ 
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.^{3}^{3} 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
$$\parallel x\parallel =\underset{\lambda \in V}{sup}\u27e8x,\lambda \u27e9,x\in X.$$ 
Proof.
It follows from the HahnBanach 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 $.^{4}^{4} 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})=\parallel {x}_{0}\parallel $. Hence
$$\underset{\lambda \in V}{sup}\u27e8{x}_{0},\lambda \u27e9\ge \u27e8{x}_{0},{\lambda}_{0}\u27e9={\lambda}_{0}({x}_{0})=\parallel {x}_{0}\parallel .$$ 
If $\lambda \in V$, then
$$\u27e8{x}_{0},\lambda \u27e9=\lambda ({x}_{0})\le \parallel \lambda \parallel \parallel {x}_{0}\parallel \le \parallel {x}_{0}\parallel ,$$ 
so
$$\underset{\lambda \in V}{sup}\u27e8{x}_{0},\lambda \u27e9\le \parallel {x}_{0}\parallel .$$ 
∎
Let $X$ be a Banach space. For $x\in X$, it is apparent that $\lambda \mapsto \lambda (x)$ is a linear map ${X}^{*}\to \u2102$. From Theorem 4, it is bounded, with norm $\parallel x\parallel $. Define $\varphi :X\to {X}^{**}$ by
$$(\varphi x)(\lambda )=\u27e8x,\lambda \u27e9,x\in X,\lambda \in {X}^{*}.$$ 
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 ${X}^{*}\to \u2102$ is of the form $\varphi (x)$ for some $x\in X$.
5 Adjoints
If $X$ and $Y$ are normed spaces and $T\in \mathcal{B}(X,Y)$, 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 $X\to \u2102$. ${T}^{*}$ is called the adjoint of $T$. It is straightforward to check that ${T}^{*}$ is linear and that it satisfies, for $S={T}^{*}$,
$$\u27e8Tx,\lambda \u27e9=\u27e8x,S\lambda \u27e9,x\in X,\lambda \in {Y}^{*}.$$  (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 $  
$=$  $\underset{\parallel x\parallel \le 1}{sup}\underset{\parallel \lambda \parallel \le 1}{sup}\u27e8Tx,\lambda \u27e9$  
$=$  $\underset{\parallel x\parallel \le 1}{sup}\underset{\parallel \lambda \parallel \le 1}{sup}\u27e8x,{T}^{*}\lambda \u27e9$  
$=$  $\underset{\parallel \lambda \parallel \le 1}{sup}\underset{\parallel x\parallel \le 1}{sup}{T}^{*}\lambda (x)$  
$=$  $\underset{\parallel \lambda \parallel \le 1}{sup}\parallel {T}^{*}\lambda \parallel $  
$=$  $\parallel {T}^{*}\parallel .$ 
In particular, ${T}^{*}\in \mathcal{B}({Y}^{*},{X}^{*})$.
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.^{5}^{5} 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 $T\in \mathcal{B}(X,Y)$, then $T$ is compact if and only if ${T}^{*}$ is compact.
Proof.
Suppose that $T\in \mathcal{B}(X,Y)$ 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 $\mathcal{F}$ is a set of functions $M\to \u2102$, we say that $\mathcal{F}$ is equicontinuous if for every $\u03f5>0$ there exists a $\delta >0$ such that if $f\in \mathcal{F}$ and $$ then $$. We say that $\mathcal{F}$ is pointwise bounded if for every $x\in M$ there is some $$ such that if $f\in \mathcal{F}$ and $x\in M$ then $f(x)\le m(x)$. The ArzelàAscoli theorem^{6}^{6} 6 Walter Rudin, Real and Complex Analysis, third ed., p. 245, Theorem 11.28. states that if $(M,\rho )$ is a separable metric space and $\mathcal{F}$ is a set of functions $M\to \u2102$ that is equicontinuous and pointwise bounded, then for every sequence ${f}_{n}\in \mathcal{F}$ there is a subsequence that converges uniformly on every compact subset of $M$.
Let $V$ be the closed unit ball in $X$. As $T$ is a compact operator, $\overline{T(V)}$ is compact and therefore separable, because any compact metric space is separable. Define ${f}_{n}:\overline{T(V)}\to \u2102$ by
$${f}_{n}(y)=\u27e8y,{\lambda}_{n}\u27e9={\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 .$$ 
Hence for $\u03f5>0$, if $n\ge 1$ and $$ 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 ,$$ 
showing that $\{{f}_{n}\}$ is pointwise bounded. Therefore we can apply the ArzelàAscoli theorem: there is a subsequence ${f}_{a(n)}$ such that ${f}_{a(n)}$ converges uniformly on every compact subset of $\overline{T(V)}$, in particular on $\overline{T(V)}$ itself and therefore on any subset of it, in particular $T(V)$. We are done using the ArzelàAscoli theorem: we used it to prove that there is a subsequence ${f}_{a(n)}$ that converges uniformly on $T(V)$.
Let $\u03f5>0$. As ${f}_{a(n)}$ converges uniformly on $T(V)$, there is some $N$ such that if $n,m\ge N$ and $y\in T(V)$, then $$. 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 $  
$=$  $\underset{x\in V}{sup}{\lambda}_{a(n)}(Tx){\lambda}_{a(m)}(Tx)$  
$=$  $\underset{x\in V}{sup}{\lambda}_{a(n)}(Tx){\lambda}_{a(m)}(Tx)$  
$=$  $\underset{x\in V}{sup}{f}_{a(n)}(Tx){f}_{a(m)}(Tx)$  
$$  $\u03f5.$ 
This means that ${T}^{*}{\lambda}_{a(n)}\in {X}^{*}$ is a Cauchy sequence. As ${X}^{*}$ is a Banach space, this sequence converges, and therefore ${T}^{*}$ is a compact operator.
Suppose that ${T}^{*}\in \mathcal{B}({Y}^{*},{X}^{*})$ is compact. Therefore, by what we showed in the first half of the proof we have that ${T}^{**}:{X}^{**}\to {Y}^{**}$ is compact. If $V$ be the closed unit ball in ${X}^{**}$, then ${T}^{**}(V)$ is totally bounded.
We have seen that $\varphi :X\to {X}^{**}$ defined by $(\varphi x)\lambda =\lambda (x)$, $x\in X$, $\lambda \in {X}^{*}$, is an isometric isomorphism $X\to \varphi (X)$. Let $\psi :Y\to {Y}^{**}$ be the same for $Y$, and let $U$ be the closed unit ball in $X$. If $x\in X$ and $\lambda \in {Y}^{*}$ then
$$\u27e8\lambda ,\psi Tx\u27e9=\u27e8Tx,\lambda \u27e9=\u27e8x,{T}^{*}\lambda \u27e9=\u27e8{T}^{*}\lambda ,\varphi x\u27e9=\u27e8\lambda ,{T}^{**}\varphi x\u27e9.$$ 
Therefore $\psi T={T}^{**}\varphi $. If $x\in U$ then $\varphi x\in V$, as $\varphi $ is an isometry. Hence if $x\in U$ then $\psi Tx={T}^{**}\varphi x\in {T}^{**}(V)$, thus
$$\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\},$$ 
we say that $M$ is complemented in $X$ and that $X$ is the direct sum of $M$ and $N$, which we write as $X=M\oplus N$.
We are going to use the following lemma to prove the theorem that comes after it.^{7}^{7} 7 Walter Rudin, Functional Analysis, second ed., p. 106, Lemma 4.21.
Lemma 6.
If $X$ is a locally convex topological vector space and $M$ is a subspace of $X$ with $$, 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 $T\in \mathcal{B}(X)$ is compact and $\lambda \ne 0$ then $T\lambda {\mathrm{id}}_{X}$ has closed image.^{8}^{8} 8 Walter Rudin, Functional Analysis, second ed., p. 107, Theorem 4.23.
Theorem 7.
If $X$ is a Banach space, $T\in \mathcal{B}(X)$ is compact, and $\lambda \ne 0$, then the image of $T\lambda {\mathrm{id}}_{X}$ is closed.
Proof.
According to Theorem 3, $$, and we can then use Lemma 6: $\mathrm{ker}(T\lambda {\mathrm{id}}_{X})$ is a finite dimensional subspace of the locally convex space $X$, so there is a closed subspace $N$ of $X$ such that $X=\mathrm{ker}(T\lambda {\mathrm{id}}_{X})\oplus N$.
Define $S:N\to X$ by $Sx=Tx\lambda x$, so $S\in \mathcal{B}(N,X)$. It is apparent that $T(X)=S(N)$ and that $S$ is injective, and we shall prove that $S(N)$ is closed. To show that $S(N)$ is closed, check that it suffices to prove that $S$ is bounded below: that there is some $r>0$ such that if $x\in N$ then $\parallel Sx\parallel \ge r\parallel x\parallel $.^{9}^{9} 9 A common way of proving that a linear operator is invertible is by proving that it has dense image and that it is bounded below: bounded below implies injective and bounded below and dense image imply surjective.
Suppose by contradiction that for every $r>0$ there is some $x\in N$ such that $$. 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 $0$.^{10}^{10} 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, $T\in \mathcal{B}(X)$ 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 $T\in \mathcal{B}(X)$ is compact and $\lambda \ne 0$, then the operator $T\lambda {\mathrm{id}}_{X}$ is injective if and only if it is surjective.^{11}^{11} 11 Paul Garrett, Compact operators on Banach spaces: FredholmRiesz, http://www.math.umn.edu/~garrett/m/fun/fredholmriesz.pdf 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, $T\in \mathcal{B}(X)$ 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
$${(T\lambda {\mathrm{id}}_{X})}^{n1}(T\lambda {\mathrm{id}}_{X})x\in {V}_{n1},$$ 
so ${V}_{n}\supseteq {V}_{n1}$. Thus
$${V}_{1}\supseteq {V}_{2}\supseteq \mathrm{\cdots}$$ 
Certainly ${V}_{n}$ is a normed vector space. Define ${T}_{n}\in \mathcal{B}({V}_{n})$ 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 ${T}_{n}\in \mathcal{B}({V}_{n})$ 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 $$, then there is some $x\in M$ with $\parallel x\parallel =1$ and ${inf}_{y\in X}\parallel xy\parallel \ge r$.^{12}^{12} 12 Paul Garrett, Riesz’s lemma, http://www.math.umn.edu/~garrett/m/fun/riesz_lemma.pdf In this reference, Riesz’s lemma is stated for Banach spaces, but the proof in fact works for normed spaces with no modifications. For each $n\ge 1$, using Riesz’s lemma there is some ${v}_{n}\in {V}_{n}$, $\parallel {v}_{n}\parallel =1$, such that
$$\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}.$$ 
That is, the distance between any two terms in $T{v}_{m}$ is $\ge \frac{\lambda }{2}$, which is a fixed positive constant, hence $T{v}_{m}$ has no convergent subsequence. But $\parallel {v}_{m}\parallel =1$, so ${v}_{m}$ is bounded and therefore, as $T$ is compact, the sequence $T{v}_{m}$ has a convergent subsequence, a contradiction. Therefore $T\lambda {\mathrm{id}}_{X}$ is surjective.
Suppose that $T\lambda {\mathrm{id}}_{X}$ is surjective. One checks that if a bounded linear operator is surjective then its adjoint is injective. For $x\in X$ and $\mu \in {X}^{*}$, $\u27e8\lambda {\mathrm{id}}_{X}x,\mu \u27e9=\mu (\lambda x)=\lambda \mu (x)=\u27e8x,\lambda {\mathrm{id}}_{{X}^{*}}\mu \u27e9$, so ${(\lambda {\mathrm{id}}_{X})}^{*}=\lambda {\mathrm{id}}_{{X}^{*}}$. Hence ${(T\lambda {\mathrm{id}}_{X})}^{*}={T}^{*}\lambda {\mathrm{id}}_{{X}^{*}}$. $T$ is compact so ${T}^{*}$ is compact. As ${T}^{*}\lambda {\mathrm{id}}_{{X}^{*}}$ is injective and ${T}^{*}$ is compact, ${T}^{*}\lambda {\mathrm{id}}_{{X}^{*}}$ is surjective, whence its adjoint ${T}^{**}\lambda {\mathrm{id}}_{{X}^{**}}:{X}^{**}\to {X}^{**}$ is injective. One checks that if $S\in \mathcal{B}(X)$ and ${S}^{**}:{X}^{**}\to {X}^{**}$ is injective then $S$ is injective; this is proved using the fact that $\varphi :X\to {X}^{**}$ defined by $(\varphi x)(\lambda )=\lambda (x)$ is an isometric isomorphism $X\to \varphi (X)$. Using this, $T\lambda {\mathrm{id}}_{X}$ is injective, completing the proof. ∎
7 Compact metric spaces
In the proof of Theorem 5 we stated the ArzelàAscoli theorem. First we state definitions again. If $M$ is a metric space with metric $\rho $ and $\mathcal{F}$ is a set of functions $M\to \u2102$, we say that $\mathcal{F}$ is equicontinuous if for all $\u03f5>0$ there is some $\delta >0$ such that $f\in \mathcal{F}$ and $$ imply that $$. 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)\le m(x)$. The ArzelàAscoli 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$.^{13}^{13} 13 Walter Rudin, Real and Complex Analysis, third ed., p. 245, Theorem 11.28.
We are going to use a converse of the ArzelàAscoli theorem in the case of a compact metric space.^{14}^{14} 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 \u2102$. 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 $\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 $\parallel f\parallel \le C$, and this implies that $\mathcal{F}$ is pointwise bounded. As $M$ is compact it is separable, so the ArzelàAscoli 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 \u2102$ 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 $\u03f5>0$. As $\mathcal{F}$ is totally bounded, there are ${f}_{1},\mathrm{\dots},{f}_{n}\in \mathcal{F}$ such that $\mathcal{F}\subseteq {\bigcup}_{k=1}^{n}{B}_{\u03f5/3}({f}_{k})$. As each ${f}_{k}:M\to \u2102$ is continuous and $M$ is compact, there is some ${\delta}_{k}>0$ such that if $$ then $$. Let $\delta ={\mathrm{min}}_{1\le k\le n}{\delta}_{k}$. If $f\in \mathcal{F}$ and $$, 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{\u03f5}{3}}+\parallel {f}_{k}f\parallel $  
$$  $\frac{\u03f5}{3}}+{\displaystyle \frac{\u03f5}{3}}+{\displaystyle \frac{\u03f5}{3}$  
$=$  $\u03f5,$ 
showing that $\mathcal{F}$ is equicontinuous. ∎
We now show that if $M$ is a compact metric space then the Banach space $C(M)$ has the approximation property: every compact linear operator $C(M)\to C(M)$ is the limit of a sequence of bounded finite rank operators.^{15}^{15} 15 John B. Conway, A Course in Functional Analysis, second ed., p. 176, Theorem 3.11.
Theorem 11.
If $(M,\rho )$ is a compact metric space, then ${\mathcal{B}}_{00}(C(M))$ is a dense subset of ${\mathcal{B}}_{0}(C(M))$.
Proof.
Let $T\in {\mathcal{B}}_{0}(C(M))$, let $V$ be the closed unit ball in $C(M)$, and let $\u03f5>0$. Because $T(V)$ is precompact in $C(M)$, by Theorem 10 it is bounded and equicontinuous. Then there is some $\delta >0$ such that if $Tf\in T(V)$ and $$ then $$. $M$ is compact, so there are ${x}_{1},\mathrm{\dots},{x}_{n}\in M$ such that $M={\bigcup}_{j=1}^{n}{B}_{\delta}({x}_{j})$. It is a fact that there is a partition of unity that is subordinate to this open covering of $M$: there are continuous functions ${\varphi}_{1},\mathrm{\dots},{\varphi}_{n}:M\to [0,1]$ such that if $x\in M$ then ${\sum}_{j=1}^{n}{\varphi}_{j}(x)=1$, and ${\varphi}_{j}(x)=0$ if $x\notin {B}_{\delta}({x}_{j})$.^{16}^{16} 16 John B. Conway, Functional Analysis, second ed., p. 139, Theorem 6.5. Define ${T}_{\u03f5}:C(M)\to C(M)$ by
$${T}_{\u03f5}f=\sum _{j=1}^{n}(Tf)({x}_{j}){\varphi}_{j}.$$ 
It is apparent that ${T}_{\u03f5}$ is linear. $\parallel {T}_{\u03f5}f\parallel \le {\sum}_{j=1}^{n}\parallel T\parallel \parallel f\parallel =n\parallel T\parallel \parallel f\parallel $, so $\parallel {T}_{e}\parallel \le n\parallel T\parallel $. And the image of ${T}_{\u03f5}$ is contained in the span of $\{{\varphi}_{1},\mathrm{\dots},{\varphi}_{n}\}$. Therefore ${T}_{\u03f5}\in {\mathcal{B}}_{00}(C(M))$.
If $f\in V$ and $x\in M$, then for each $j$ either $x\in {B}_{\delta}({x}_{j})$, in which case $$, or $x\notin {B}_{\delta}({x}_{j})$, in which case ${\varphi}_{j}(x)=0$. This gives us
$((Tf)(x)({T}_{\u03f5}f)(x)$  $=$  $\left(Tf)(x)\cdot {\displaystyle \sum _{j=1}^{n}}{\varphi}_{j}(x){\displaystyle \sum _{j=1}^{n}}(Tf)({x}_{j}){\varphi}_{j}(x)\right$  
$=$  $\left{\displaystyle \sum _{j=1}^{n}}\left((Tf)(x)(Tf)({x}_{j})\right){\varphi}_{j}(x)\right$  
$\le $  $\sum _{j=1}^{n}}\left(Tf)(x)(Tf)({x}_{j})\right{\varphi}_{j}(x)$  
$$  $\sum _{j=1}^{n}}\u03f5{\varphi}_{j}(x)$  
$=$  $\u03f5,$ 
showing that $$, and as this is true for all $f\in V$ we get $$. ∎