Lecture Notes in Functional Analysis

by William L. Paschke

edition 0.9

Contents

Frontmatter

I. Hilbert Space

II. Bounded Operators

III. Compact Operators

IV. The Spectral Theorem

Index and References

## III. Compact Operators

Definition III.1

A linear map $$$T : X \to Y$$$ between normed linear spaces is called compact if $$$\overline{T\mathopen{}\left( B\right)\mathclose{}}$$$ is compact for every norm-bounded subset $$$B$$$ of $$$X$$$.

Remark III.2

It is enough to check for $$$\mathrm{B}_{X}\mathopen{}\left( 1\right)\mathclose{}= \mathopen{}\left\{\, x\in X\,\middle\vert\, , \mathopen{}\left\lVert{}x\right\rVert\mathclose{}\leq 1, \,\right\}\mathclose{}$$$ since each norm-bounded subset $$$B\subseteq r\mathrm{B}_{X}\mathopen{}\left( 1\right)\mathclose{}$$$ for some $$$r\gt 0$$$.

Remark III.3

Compact operators are bounded operators because compact subsets of a metric space are bounded in the metric.

Remark III.4

For $$$T\in \mathcal{L}\mathopen{}\left( X, Y\right)\mathclose{}$$$, $$$\operatorname{dim}\mathopen{}\left( T\mathopen{}\left( X\right)\mathclose{}\right)\mathclose{}\lt \infty$$$ implies that $$$T$$$ is compact.

Remark III.5

If $$$X$$$ is infinite dimensional, then $$$\mathrm{B}_{X}\mathopen{}\left( 1\right)\mathclose{}$$$ (which equals $$$\overline{ \mathrm{B}_{X}\mathopen{}\left( 1\right)\mathclose{} }$$$) is non-compact, so $$$\mathrm{I}_{X}$$$ is a bounded operator but not a compact operator on $$$X$$$.

Recall a metric space is compact if and only if every sequence has a convergent subsequence.

Proposition III.6

An operator $$$T\in \mathcal{L}\mathopen{}\left( X, Y\right)\mathclose{}$$$ is compact if and only if for every bounded sequence $$$\mathopen{}\left({x}_{n}\right)\mathclose{}$$$ in $$$X$$$, the sequence $$$\mathopen{}\left(T\mathopen{}\left( {x}_{n}\right)\mathclose{}\right)\mathclose{}$$$ has a convergent subsequence.

Proof. (⇒) $$$T$$$ compact, $$$\mathopen{}\left({x}_{n}\right)\mathclose{}$$$ bounded implies $$$\overline{ \mathopen{}\left(T\mathopen{}\left( {x}_{n}\right)\mathclose{}\right)\mathclose{} }$$$ compact.

(⇐) $$$B$$$ bounded, $$$\mathopen{}\left({y}_{n}\right)\mathclose{}$$$ a sequence in $$$\overline{T\mathopen{}\left( B\right)\mathclose{}}$$$ then get $$$\mathopen{}\left({x}_{n}\right)\mathclose{}$$$ in $$$B$$$ such that $$$\mathopen{}\left\lVert{}T\mathopen{}\left( {x}_{n}\right)\mathclose{}-{y}_{n}\right\rVert\mathclose{}\lt \frac{1}{n}$$$. Since then $$$\mathopen{}\left(T\mathopen{}\left( {x}_{{n}_{k}}\right)\mathclose{}\right)\mathclose{}$$$ converges for subsequence $$$\mathopen{}\left({x}_{{n}_{k}}\right)\mathclose{}$$$, get $$$\mathopen{}\left({y}_{{n}_{k}}\right)\mathclose{}$$$ to converge (to the same limit).

Example III.7

With $$${φ}_{1}$$$, $$$\dotsc$$$, $$${φ}_{n}$$$ in $$$X^{*}$$$ and $$${y}_{1}$$$, $$$\dotsc$$$, $$${y}_{n}$$$ in $$$X$$$, $$T\mathopen{}\left( x\right)\mathclose{}= \sum_{i=1}^{n}{} {φ}_{n}\mathopen{}\left( x\right)\mathclose{}{y}_{i} \text{.}$$

Example III.8

The Volterra operator $$T : \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \to \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \text{,}$$ defined by $$\mathopen{}\left(T\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= \mathopen{}\left\langle{}f, {χ}_{\mathopen{}\left[0, x\right]\mathclose{}}\right\rangle\mathclose{} \text{,}$$ is compact. To see why, first notice $$$\mathopen{}\left\lvert{}T\mathopen{}\left( f\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}f\right\rVert\mathclose{}_{2}\mathopen{}\left\lVert{}{χ}_{\mathopen{}\left[0, x\right]\mathclose{}}\right\rVert\mathclose{}_{2}= \mathopen{}\left\lVert{}f\right\rVert\mathclose{}_{2}\sqrt{x}\leq \mathopen{}\left\lVert{}f\right\rVert\mathclose{}_{2}$$$ and also $$$\mathopen{}\left\lvert{}T\mathopen{}\left( f\mathopen{}\left( {x}_{1}\right)\mathclose{}\right)\mathclose{}-T\mathopen{}\left( f\mathopen{}\left( {x}_{2}\right)\mathclose{}\right)\mathclose{}\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}f\right\rVert\mathclose{}_{2}\sqrt{\mathopen{}\left\lvert{}{x}_{1}-{x}_{2}\right\rvert\mathclose{}}$$$ because $$$\mathopen{}\left\lVert{}{χ}_{\mathopen{}\left[0, {x}_{1}\right]\mathclose{}}-{χ}_{\mathopen{}\left[0, {x}_{2}\right]\mathclose{}}\right\rVert\mathclose{}= \sqrt{\mathopen{}\left\lvert{}{x}_{1}-{x}_{2}\right\rvert\mathclose{}}$$$ (so in particular, $$$\operatorname{Ran}\mathopen{}\left( T\right)\mathclose{}\subseteq \mathrm{C}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}$$$). Let $$$\mathopen{}\left({f}_{n}\right)\mathclose{}$$$ be a bounded sequence in $$$\mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}$$$, say $$$\mathopen{}\left\lVert{}{f}_{n}\right\rVert\mathclose{}_{2}\leq M$$$ for all $$$n$$$. Then $$$\mathopen{}\left\lVert{}T\mathopen{}\left( {f}_{n}\right)\mathclose{}\right\rVert\mathclose{}_{\infty}\leq M$$$ and $$$\mathopen{}\left\lvert{}T\mathopen{}\left( {f}_{n}\mathopen{}\left( {x}_{1}\right)\mathclose{}\right)\mathclose{}-T\mathopen{}\left( {f}_{n}\mathopen{}\left( {x}_{2}\right)\mathclose{}\right)\mathclose{}\right\rvert\mathclose{}\leq M\sqrt{\mathopen{}\left\lvert{}{x}_{1}-{x}_{2}\right\rvert\mathclose{}}$$$. That is to say, you can make this small for all $$$n$$$ simultaneously by making $$$\mathopen{}\left\lvert{}{x}_{1}-{x}_{2}\right\rvert\mathclose{}$$$ sufficiently small, and $$$\mathopen{}\left(T\mathopen{}\left( {f}_{n}\right)\mathclose{}\right)\mathclose{}$$$ is pointwise (here in fact uniformly) bounded, and equicontinuous. Then Arzela's Theorem says there exists a uniformly convergent subsequence of $$$\mathopen{}\left(T\mathopen{}\left( {f}_{n}\right)\mathclose{}\right)\mathclose{}$$$, say $$$\mathopen{}\left( T\mathopen{}\left( {f}_{{n}_{k}}\right)\mathclose{} \right)\mathclose{}$$$, with $$$\mathopen{}\left\lVert{}T\mathopen{}\left( {f}_{{n}_{k}}\right)\mathclose{}-g\right\rVert\mathclose{}_{\infty} \to 0$$$ for some $$$g\in \mathrm{C}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}$$$. So, $$${\mathopen{}\left(\mathopen{}\left\lVert{}T\mathopen{}\left( {f}_{{n}_{k}}\right)\mathclose{}-g\right\rVert\mathclose{}_{2}\right)\mathclose{}}^{2}= \int _{0}^{1}{} {\mathopen{}\left\lvert{}T\mathopen{}\left( {f}_{{n}_{k}}\right)\mathclose{}-g\right\rvert\mathclose{}}^{2} \leq {\mathopen{}\left(\mathopen{}\left\lVert{}T\mathopen{}\left( {f}_{{n}_{k}}\right)\mathclose{}-g\right\rVert\mathclose{}_{\infty}\right)\mathclose{}}^{2}$$$, which makes $$$T\mathopen{}\left( {f}_{{n}_{k}}\right)\mathclose{} \to g$$$ in $$$\mathrm{L}^{\mathrm{2}}$$$.

Notation III.9

For normed linear spaces $$$X$$$ and $$$Y$$$, let $$$\mathcal{K}\mathopen{}\left( X, Y\right)\mathclose{}$$$ be the set of compact operators: $$$T\in \mathcal{L}\mathopen{}\left( X, Y\right)\mathclose{}$$$ such that $$$T$$$ is compact.

Proposition III.10

If $$$Y$$$ is a Banach space, then $$$\mathcal{K}\mathopen{}\left( X, Y\right)\mathclose{}$$$ is a closed subspace of $$$\mathcal{L}\mathopen{}\left( X, Y\right)\mathclose{}$$$.

Proof. If $$$S$$$ and $$$T$$$ are in $$$\mathcal{K}\mathopen{}\left( X, Y\right)\mathclose{}$$$ and $$$\mathopen{}\left({x}_{n}\right)\mathclose{}$$$ is bounded in $$$X$$$, then there exists $$$\mathopen{}\left( {x}_{n}'\right)\mathclose{}$$$ such that $$$\mathopen{}\left(T\mathopen{}\left( {x}_{n}'\right)\mathclose{}\right)\mathclose{}$$$ converges, and there is also a further subsequence $$$\mathopen{}\left( {x}_{n}''\right)\mathclose{}$$$ such that $$$\mathopen{}\left(S\mathopen{}\left( {x}_{n}''\right)\mathclose{}\right)\mathclose{}$$$ converges. Then we have that $$$\mathopen{}\left( \mathopen{}\left(S+T\right)\mathclose{}\mathopen{}\left( \mathopen{}\left( {x}_{n}'' \right)\mathclose{}\right)\mathclose{} \right)\mathclose{}$$$ is convergent, which shows $$$S+T\in \mathcal{K}\mathopen{}\left( X, Y\right)\mathclose{}$$$.

Scalar multiplication is clearly ok. For closed, suppose $$$\mathopen{}\left({T}_{n}\right)\mathclose{}$$$ is a sequence of compact operators converging in norm to $$$T\in \mathcal{L}\mathopen{}\left( X, Y\right)\mathclose{}$$$ ($$$\mathopen{}\left\lVert{}{T}_{n}-T\right\rVert\mathclose{} \to 0$$$). Take $$$\mathopen{}\left({x}_{n}\right)\mathclose{}$$$ in $$$X$$$ with $$$\mathopen{}\left\lVert{}{x}_{n}\right\rVert\mathclose{}\leq M$$$ for all $$$n$$$. Using compactness of $$${T}_{1}$$$, $$${T}_{2}$$$, …, get successive subsequences $$$\mathopen{}\left( {x}_{n}'\right)\mathclose{}$$$, $$$\mathopen{}\left( {x}_{n}''\right)\mathclose{}$$$, … such that $$$\mathopen{}\left( {T}_{1}\mathopen{}\left( {x}_{n}'\right)\mathclose{} \right)\mathclose{}$$$ converges, $$$\mathopen{}\left( {T}_{j}\mathopen{}\left( {x}_{n}''\right)\mathclose{} \right)\mathclose{}$$$ converges for $$$j$$$ in $$$\mathopen{}\left\{\, 1, 2\,\right\}\mathclose{}$$$, $$$\mathopen{}\left( {T}_{j}\mathopen{}\left( {x}_{n}'''\right)\mathclose{} \right)\mathclose{}$$$ converges for $$$j$$$ in $$$\mathopen{}\left\{\, 1, 2, 3\,\right\}\mathclose{}$$$, etc. Let $$$\mathopen{}\left({x}_{{m}_{k}}\right)\mathclose{}$$$ be the $$$k$$$th item in the $$$m$$$th sequence. This makes $$$\mathopen{}\left({x}_{{m}_{k}}\right)\mathclose{}_{k=1}^{\infty}$$$ eventually a subsequence of $$$\mathopen{}\left( {x}_{n}'\right)\mathclose{}$$$, $$$\mathopen{}\left( {x}_{n}''\right)\mathclose{}$$$, …. It follows that $$$\mathopen{}\left( {T}_{n}\mathopen{}\left( {x}_{{m}_{k}}\right)\mathclose{} \right)\mathclose{}_{k=1}^{\infty}$$$ converges for every $$$n$$$. Now $$\mathopen{}\left\lVert{}T\mathopen{}\left( {x}_{{m}_{i}}\right)\mathclose{}-T\mathopen{}\left( {x}_{{m}_{k}}\right)\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}\mathopen{}\left(T-{T}_{n}\right)\mathclose{}\mathopen{}\left( {x}_{{m}_{i}}\right)\mathclose{}\right\rVert\mathclose{}+\mathopen{}\left\lVert{}{T}_{n}\mathopen{}\left( {x}_{{m}_{i}}-{x}_{{m}_{k}}\right)\mathclose{}\right\rVert\mathclose{}+\mathopen{}\left\lVert{}\mathopen{}\left({T}_{n}-T\right)\mathclose{}\mathopen{}\left( {x}_{{m}_{k}}\right)\mathclose{}\right\rVert\mathclose{}\leq 2M\mathopen{}\left\lVert{}T-{T}_{n}\right\rVert\mathclose{}+\mathopen{}\left\lVert{}{T}_{n}\mathopen{}\left( {x}_{{m}_{i}}-{x}_{{m}_{k}}\right)\mathclose{}\right\rVert\mathclose{}\text{.}$$ Given $$$ε\gt 0$$$, there exists $$$n$$$ such that $$$\mathopen{}\left\lVert{}T\mathopen{}\left( {x}_{{m}_{i}}\right)\mathclose{}-T\mathopen{}\left( {x}_{{m}_{k}}\right)\mathclose{}\right\rVert\mathclose{}\lt \frac{ε}{2}+\mathopen{}\left\lVert{}{T}_{n}\mathopen{}\left( {x}_{{m}_{i}}-{x}_{{m}_{k}}\right)\mathclose{}\right\rVert\mathclose{}$$$ and this is less than $$$\frac{ε}{2}+\frac{ε}{2}= ε$$$ for sufficiently large $$$i$$$ and $$$k$$$. So the sequence is Cauchy, and we're done since $$$Y$$$ is complete.

The spectral theory of compact operators has the following special features.

Theorem III.11

Let $$$T$$$ be a compact operator on a Banach space $$$X$$$. Then

1. $$$\mathopen{}\left(λ-T\right)\mathclose{}\mathopen{}\left( X\right)\mathclose{}$$$ is closed for every $$$λ\in \mathbb{C}\setminus \mathopen{}\left\{\, 0\,\right\}\mathclose{}$$$.
2. $$$\mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}\setminus \mathopen{}\left\{\, 0\,\right\}\mathclose{}$$$ consists of eigenvalues of $$$T$$$; and
3. $$$\mathopen{}\left\{\, λ\in \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}\,\middle\vert\, , \mathopen{}\left\lvert{}λ\right\rvert\mathclose{}\geq δ, \,\right\}\mathclose{}$$$ is finite for every $$$δ\gt 0$$$.

1. We may assume $$$λ= 1$$$. Suppose $$${x}_{n}-T\mathopen{}\left( {x}_{n}\right)\mathclose{} \to w$$$. Let $$${d}_{n}= \operatorname{d}\mathopen{}\left( {x}_{n}, \operatorname{Ker}\mathopen{}\left( 1-T\right)\mathclose{}\right)\mathclose{}= \inf_{T\mathopen{}\left( y\right)\mathclose{}= y}{}\mathopen{}\left\lVert{}{x}_{n}-y\right\rVert\mathclose{}$$$. Harmlessly perturb each $$${x}_{n}$$$ by a vector in $$$\operatorname{Ker}\mathopen{}\left( 1-T\right)\mathclose{}$$$ to make $$$\mathopen{}\left\lVert{}{x}_{n}\right\rVert\mathclose{}\lt {d}_{n}+1$$$. We claim that $$$\sup_{n}{}{d}_{n}\lt \infty$$$. Otherwise, pass to a subsequence and assume $$${d}_{n} \to \infty$$$. Let $$${z}_{n}= \frac{{x}_{n}}{{d}_{n}+1}$$$, so $$$\mathopen{}\left\lVert{}{z}_{n}\right\rVert\mathclose{}\lt 1$$$. Again passing to a subsequence, we may assume $$$T\mathopen{}\left( {z}_{n}\right)\mathclose{} \to y$$$, because $$$T$$$ is compact. Also $$${z}_{n}-T\mathopen{}\left( {z}_{n}\right)\mathclose{}= \frac{{x}_{n}-T\mathopen{}\left( {x}_{n}\right)\mathclose{}}{{d}_{n}+1} \to 0$$$. It follows that $$${z}_{n} \to y$$$ and that $$$T\mathopen{}\left( y\right)\mathclose{}= y$$$. Thus $$\frac{{d}_{n}}{{d}_{n}+1}\leq \frac{\mathopen{}\left\lVert{}{x}_{n}-\mathopen{}\left({d}_{n}+1\right)\mathclose{}y\right\rVert\mathclose{}}{{d}_{n}+1}= \mathopen{}\left\lVert{}{z}_{n}-y\right\rVert\mathclose{} \to 0 \text{,}$$ a contradiction. With the claim established, we know that $$$\mathopen{}\left({x}_{n}\right)\mathclose{}$$$ is bounded, so after passing to a subsequence we get that $$$T\mathopen{}\left( {x}_{n}\right)\mathclose{} \to y$$$, so $$${x}_{n} \to w+y$$$, and thus $$$w= \lim_{n}{} \mathopen{}\left(1-T\right)\mathclose{}{x}_{n} = \mathopen{}\left(1-T\right)\mathclose{}\mathopen{}\left(w+y\right)\mathclose{}$$$.
2. Suppose contrariwise that $$$λ\in \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}\setminus \mathopen{}\left\{\, 0\,\right\}\mathclose{}$$$ but $$$\operatorname{Ker}\mathopen{}\left( λ-T\right)\mathclose{}= \mathopen{}\left\{\, 0\,\right\}\mathclose{}$$$. As usual, we may assume $$$λ= 1$$$. The range of $$$1-T$$$ cannot be all of $$$X$$$, since if it were, the injective operator $$$1-T$$$ would be invertible by the open mapping theorem. Thus $$$\mathopen{}\left(1-T\right)\mathclose{}\mathopen{}\left( X\right)\mathclose{}$$$ is a closed proper subspace of $$$X$$$, and $$$1-T$$$ is a linear homeomorphism of $$$X$$$ with $$$\mathopen{}\left(1-T\right)\mathclose{}\mathopen{}\left( X\right)\mathclose{}$$$. It follows that the subspaces $$${\mathopen{}\left(1-T\right)\mathclose{}}^{n}\mathopen{}\left( X\right)\mathclose{}$$$ ($$$n\in \mathopen{}\left\{\, 0, 1, 2, \dotsc\,\right\}\mathclose{}$$$) are all closed, with $$${\mathopen{}\left(1-T\right)\mathclose{}}^{n+1}\mathopen{}\left( X\right)\mathclose{}\subsetneq {\mathopen{}\left(1-T\right)\mathclose{}}^{n}\mathopen{}\left( X\right)\mathclose{}$$$ for each $$$n$$$, Lemma I.65 yields a sequence $$$\mathopen{}\left({x}_{n}\right)\mathclose{}$$$ with $$\mathopen{}\left\lVert{}{x}_{n}\right\rVert\mathclose{}= 1 \text{,}$$ $${x}_{n}\in {\mathopen{}\left(1-T\right)\mathclose{}}^{n}\mathopen{}\left( X\right)\mathclose{} \text{,}$$ $$\operatorname{d}\mathopen{}\left( {x}_{n}, {\mathopen{}\left(1-T\right)\mathclose{}}^{n+1}\mathopen{}\left( X\right)\mathclose{}\right)\mathclose{}\gt \frac{1}{2}$$ for every $$$n$$$. Take $$$m\gt n$$$ and write $$$y= \mathopen{}\left(T-1\right)\mathclose{}\mathopen{}\left( {x}_{n}\right)\mathclose{}+\mathopen{}\left(1-T\right)\mathclose{}\mathopen{}\left( {x}_{m}\right)\mathclose{}-{x}_{m}$$$. Then $$$y\in {\mathopen{}\left(T-1\right)\mathclose{}}^{n+1}\mathopen{}\left( X\right)\mathclose{}$$$. Since $$$T\mathopen{}\left( {x}_{n}\right)\mathclose{}-T\mathopen{}\left( {x}_{m}\right)\mathclose{}= {x}_{n}+y$$$, we conclude that $$$\mathopen{}\left\lVert{}T\mathopen{}\left( {x}_{n}\right)\mathclose{}-T\mathopen{}\left( {x}_{m}\right)\mathclose{}\right\rVert\mathclose{}\gt \frac{1}{2}$$$ for all $$$m\neq n$$$, which contradicts $$$T\in \mathcal{K}\mathopen{}\left( X\right)\mathclose{}$$$.

3. Suppose contrariwise that for some $$$δ\gt 0$$$ there are distinct $$${λ}_{1}$$$, $$${λ}_{2}$$$, … and non-zero vectors $$${x}_{1}$$$, $$${x}_{2}$$$, … with $$$T\mathopen{}\left( {x}_{j}\right)\mathclose{}= {λ}_{j}{x}_{j}$$$ and $$$\mathopen{}\left\lvert{}{λ}_{j}\right\rvert\mathclose{}\geq δ$$$ for all $$$j$$$. Let $$${Y}_{n}$$$ be the linear span of $$${x}_{1}$$$, $$${x}_{2}$$$, …, $$${x}_{n}$$$. As in elementary linear algebra, the $$${x}_{j}$$$'s are linearly independent, so each $$${Y}_{n}$$$ is properly contained in $$${Y}_{n+1}$$$. By Lemma I.65, we can find unit vectors $$${y}_{n}$$$ with $$${y}_{n}\in {Y}_{n}$$$ and $$$\operatorname{d}\mathopen{}\left( {y}_{n+1}, {Y}_{n}\right)\mathclose{}\gt \frac{1}{2}$$$. Notice that $$$T\mathopen{}\left( {Y}_{n}\right)\mathclose{}\subseteq {Y}_{n}$$$. Take $$$m\gt n$$$. Write $$${y}_{m}= α{x}_{m}+ y'$$$, where $$$y'\in {Y}_{m-1}$$$. We have $$T\mathopen{}\left( {y}_{m}\right)\mathclose{}-T\mathopen{}\left( {y}_{n}\right)\mathclose{}= {λ}_{m}α{x}_{m}+T\mathopen{}\left( y'\right)\mathclose{}-T\mathopen{}\left( {y}_{n}\right)\mathclose{}-{λ}_{m}{y}_{m}+{λ}_{m}{y}_{m}= {λ}_{m}{y}_{m}-{λ}_{m} y'+T\mathopen{}\left( y'\right)\mathclose{}-T\mathopen{}\left( {y}_{n}\right)\mathclose{}\in {λ}_{m}{y}_{m}+{Y}_{m-1} \text{.}$$ We conclude that $$$\mathopen{}\left\lVert{}T\mathopen{}\left( {y}_{n}\right)\mathclose{}, T\mathopen{}\left( {y}_{m}\right)\mathclose{}\right\rVert\mathclose{}\gt \frac{δ}{2}$$$ for all $$$m\neq n$$$, which contradicts $$$T\in \mathcal{K}\mathopen{}\left( X\right)\mathclose{}$$$.
Remark III.12

We now have a complete picture of the spectrum of $$$T\in \mathcal{K}\mathopen{}\left( X\right)\mathclose{}$$$. If $$$X$$$ is finite-dimensional, then of course $$$\mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}$$$ can be any subset of $$$\mathbb{C}$$$ of cardinality not exceeding $$$\operatorname{dim}\mathopen{}\left( X\right)\mathclose{}$$$. If $$$X$$$ is infinite-dimensional, then $$$0$$$ must belong to $$$\mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}$$$ because the unit ball of $$$X$$$ is not compact. The rest of $$$\mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}$$$ consists of either a finite set of non-zero eigenvalues, or a sequence of non-zero eigenvalues converging to $$$0$$$. For every non-zero $$$λ$$$, (eigenvalue or not) the subspace $$$\operatorname{Ker}\mathopen{}\left( λ-T\right)\mathclose{}$$$ is finite-dimensional.

Remark III.13

Composition of a compact operator with a bounded operator (in either order) gives a compact operator. Thus $$$\mathcal{K}\mathopen{}\left( X\right)\mathclose{}$$$ is a two-sided ideal in $$$\mathcal{L}\mathopen{}\left( X\right)\mathclose{}$$$ which is closed if $$$X$$$ is a Banach space. $$$\mathcal{L}\mathopen{}\left( X\right)\mathclose{}/\mathcal{K}\mathopen{}\left( X\right)\mathclose{}$$$ is called the Calkin algebra of $$$X$$$.

Notation III.14

Call $$$T\in \mathcal{L}\mathopen{}\left( X\right)\mathclose{}$$$ finite rank if $$$\operatorname{dim}\mathopen{}\left( T\mathopen{}\left( X\right)\mathclose{}\right)\mathclose{}\lt \infty$$$. Let $$$\mathcal{F}\mathopen{}\left( X\right)\mathclose{}$$$ be the set of finite rank operators in $$$\mathcal{L}\mathopen{}\left( X\right)\mathclose{}$$$.

Remark III.15

$$$\mathcal{F}\mathopen{}\left( X\right)\mathclose{}$$$ is also an ideal in $$$\mathcal{L}\mathopen{}\left( X\right)\mathclose{}$$$. We have $$$\overline{\mathcal{F}\mathopen{}\left( X\right)\mathclose{}}\subseteq \mathcal{K}\mathopen{}\left( X\right)\mathclose{}$$$ if $$$X$$$ is complete. Often, perhaps universally, $$$\overline{\mathcal{F}\mathopen{}\left( X\right)\mathclose{}}= \mathcal{K}\mathopen{}\left( X\right)\mathclose{}$$$ when $$$X$$$ is a Banach space.

Now, specialize to a Hilbert space $$$H$$$.

Proposition III.16

Let $$$T\in \mathcal{K}\mathopen{}\left( H\right)\mathclose{}$$$. Then

1. $$$\overline{T\mathopen{}\left( H\right)\mathclose{}}$$$ is separable.
2. Assuming $$$T\notin \mathcal{F}\mathopen{}\left( H\right)\mathclose{}$$$, let $$$\mathopen{}\left\{\, {e}_{1}, {e}_{2}, \dotsc\,\right\}\mathclose{}$$$ be an orthonormal basis for $$$\overline{T\mathopen{}\left( H\right)\mathclose{}}$$$ and let $$${P}_{n}$$$ be the orthogonal projection of $$$H$$$ on $$$\operatorname{span}\mathopen{}\left( {e}_{1}, \dotsc, {e}_{n}\right)\mathclose{}$$$. Then $$$\mathopen{}\left\lVert{}{P}_{n}\mathopen{}\left( T\right)\mathclose{}-T\right\rVert\mathclose{} \to 0$$$.

Proof.

1. Since $$$\overline{ T\mathopen{}\left( \mathrm{B}_{H}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} }$$$ is compact, it is separable and there exists a countable subset $$$S\subseteq \mathrm{B}_{H}\mathopen{}\left( 1\right)\mathclose{}$$$ such that $$$\overline{T\mathopen{}\left( S\right)\mathclose{}}= \overline{ T\mathopen{}\left( \mathrm{B}_{H}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} }$$$. So, $$$T\mathopen{}\left( \bigcup_{n=1}^{\infty}{}nS\right)\mathclose{}$$$ is a countable dense subset of $$$\overline{T\mathopen{}\left( H\right)\mathclose{}}$$$.
2. Notice $$$\mathopen{}\left\lVert{}{P}_{n}\mathopen{}\left( y\right)\mathclose{}-y\right\rVert\mathclose{} \to 0$$$ for all $$$y\in \overline{T\mathopen{}\left( H\right)\mathclose{}}$$$ (since we have $$${\mathopen{}\left\lVert{}{P}_{n}\mathopen{}\left( y\right)\mathclose{}-y\right\rVert\mathclose{}}^{2}= \sum_{j=n+1}^{\infty}{}{\mathopen{}\left\lvert{}\mathopen{}\left\langle{}y, {e}_{j}\right\rangle\mathclose{}\right\rvert\mathclose{}}^{2}$$$). $$$\mathopen{}\left\lVert{}{P}_{n+1}\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}-T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}{P}_{n}\mathopen{}\left( T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}-T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}$$$ for all $$$x$$$. So, $$$\mathopen{}\left\lVert{}{P}_{n+1}\mathopen{}\left( T\right)\mathclose{}-T\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}{P}_{n}\mathopen{}\left( T\right)\mathclose{}-T\right\rVert\mathclose{}$$$. Let $$$δ= \inf_{n}{}\mathopen{}\left\lVert{}{P}_{n}\mathopen{}\left( T\right)\mathclose{}-T\right\rVert\mathclose{}$$$. If $$$δ\gt 0$$$, then we get $$${x}_{n}\in H$$$ with $$$\mathopen{}\left\lVert{}{x}_{n}\right\rVert\mathclose{}= 1$$$ such that $$$\mathopen{}\left\lVert{}{P}_{n}\mathopen{}\left( T\mathopen{}\left( {x}_{n}\right)\mathclose{}\right)\mathclose{}-T\mathopen{}\left( {x}_{n}\right)\mathclose{}\right\rVert\mathclose{}\geq \frac{δ}{2}$$$. Compactness of $$$T$$$ gives $$$T\mathopen{}\left( {x}_{{n}_{k}}\right)\mathclose{} \to y$$$ (in $$$\overline{T\mathopen{}\left( H\right)\mathclose{}}$$$). So $$\frac{δ}{2}\leq \mathopen{}\left\lVert{}{P}_{{n}_{k}}T\mathopen{}\left( {x}_{{n}_{k}}\right)\mathclose{}-T\mathopen{}\left( {x}_{{n}_{k}}\right)\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}{P}_{{n}_{k}}T\mathopen{}\left( {x}_{{n}_{k}}\right)\mathclose{}-y\right\rVert\mathclose{}+\mathopen{}\left\lVert{}{P}_{{n}_{k}}\mathopen{}\left( y\right)\mathclose{}-y\right\rVert\mathclose{}+\mathopen{}\left\lVert{}y-T\mathopen{}\left( {x}_{{n}_{k}}\right)\mathclose{}\right\rVert\mathclose{}\leq 2\mathopen{}\left\lVert{}T\mathopen{}\left( {x}_{{n}_{k}}\right)\mathclose{}-y\right\rVert\mathclose{}+\mathopen{}\left\lVert{}{P}_{{n}_{k}}\mathopen{}\left( y\right)\mathclose{}-y\right\rVert\mathclose{} \to 0 \text{,}$$ and $$$\frac{δ}{2}= 0$$$.

Theorem III.17

$$$\overline{\mathcal{F}\mathopen{}\left( H\right)\mathclose{}}= \mathcal{K}\mathopen{}\left( H\right)\mathclose{}= \mathcal{K}\mathopen{}\left( H\right)\mathclose{}^{*}$$$.

Proof. $$${P}_{n}T$$$ and $$$T^{*}{P}_{n}$$$ are in $$$\mathcal{F}\mathopen{}\left( H\right)\mathclose{}$$$.