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

## C. Invertible Operators

In what follows, let $$$E$$$ be a Banach space.

Definition II.29

Denote by $$${\mathcal{L}\mathopen{}\left( E\right)\mathclose{}}^{-1}$$$ the set of invertible operators in $$$\mathcal{L}\mathopen{}\left( E\right)\mathclose{}$$$.

We remark that all of the results (except for those specifically for operators on Hilbert space) in this and the next two sections are valid, with essentially the same proofs, in the more general setting of a Banach algebra (i.e., a complete normed algebra) with multiplicative unit. Here, the unit is $$$\mathrm{I}$$$, the identity operator on $$$E$$$.

Theorem II.30

If $$$A\in \mathcal{L}\mathopen{}\left( E\right)\mathclose{}$$$ and $$$\mathopen{}\left\lVert{}\mathrm{I}-A\right\rVert\mathclose{}\lt 1$$$, then $$$A\in {\mathcal{L}\mathopen{}\left( E\right)\mathclose{}}^{-1}$$$, and $$$\mathopen{}\left\lVert{}\mathrm{I}-{A}^{-1}\right\rVert\mathclose{}\leq \frac{\mathopen{}\left\lVert{}\mathrm{I}-A\right\rVert\mathclose{}}{\mathopen{}\left(1-\mathopen{}\left\lVert{}\mathrm{I}-A\right\rVert\mathclose{}\right)\mathclose{}}$$$.

Proof. Notice that $$$\sum_{n=0}^{\infty}{}\mathopen{}\left\lVert{}{\mathopen{}\left(\mathrm{I}-A\right)\mathclose{}}^{n}\right\rVert\mathclose{}\leq \sum_{n=0}^{\infty}{}{\mathopen{}\left\lVert{}\mathrm{I}-A\right\rVert\mathclose{}}^{n}\lt \infty$$$. Because $$$\mathcal{L}\mathopen{}\left( E\right)\mathclose{}$$$ is complete, this means the series $$$\sum_{n=0}^{\infty}{}{\mathopen{}\left(\mathrm{I}-A\right)\mathclose{}}^{n}$$$ converges in $$$\mathcal{L}\mathopen{}\left( E\right)\mathclose{}$$$. Call the sum $$$T$$$. Then $$$\mathopen{}\left(\mathrm{I}-A\right)\mathclose{}T= T-\mathrm{I}$$$, so $$$AT= \mathrm{I}$$$, and likewise $$$TA= \mathrm{I}$$$. We have $$\mathopen{}\left\lVert{}\mathrm{I}-{A}^{-1}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}\mathrm{I}-T\right\rVert\mathclose{}= \mathopen{}\left\lVert{}\sum_{n=1}^{\infty}{}{\mathopen{}\left(\mathrm{I}-A\right)\mathclose{}}^{n}\right\rVert\mathclose{}\leq \sum_{n=1}^{\infty}{} {\mathopen{}\left\lVert{}\mathopen{}\left(\mathrm{I}-A\right)\mathclose{}\right\rVert\mathclose{}}^{n} = \frac{\mathopen{}\left\lVert{}\mathrm{I}-A\right\rVert\mathclose{}}{\mathopen{}\left(1-\mathopen{}\left\lVert{}\mathrm{I}-A\right\rVert\mathclose{}\right)\mathclose{}} \text{.}$$

Corollary II.31

The set of invertible operators $$${\mathcal{L}\mathopen{}\left( E\right)\mathclose{}}^{-1}$$$ is an open subset of $$$\mathcal{L}\mathopen{}\left( E\right)\mathclose{}$$$.

Proof. If $$$B\in {\mathcal{L}\mathopen{}\left( E\right)\mathclose{}}^{-1}$$$, $$$A\in \mathcal{L}\mathopen{}\left( E\right)\mathclose{}$$$, and $$$\mathopen{}\left\lVert{}A-B\right\rVert\mathclose{}\leq \frac{1}{\mathopen{}\left\lVert{}{B}^{-1}\right\rVert\mathclose{}}$$$, then $$$A\in {\mathcal{L}\mathopen{}\left( E\right)\mathclose{}}^{-1}$$$, since $$$\mathopen{}\left\lVert{}A{B}^{-1}-I\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}A-B\right\rVert\mathclose{}\mathopen{}\left\lVert{}{B}^{-1}\right\rVert\mathclose{}\lt 1$$$, so $$$A{B}^{-1}\in {\mathcal{L}\mathopen{}\left( E\right)\mathclose{}}^{-1}$$$, so $$$A\in {\mathcal{L}\mathopen{}\left( E\right)\mathclose{}}^{-1}$$$.

Corollary II.32

The inversion map $$$A\mapsto {A}^{-1}$$$ is continuous from $$${\mathcal{L}\mathopen{}\left( E\right)\mathclose{}}^{-1}$$$ to itself.

Proof. The norm inequality in Theorem II.30 shows that inversion is continuous at $$$\mathrm{I}$$$. If $$$\mathopen{}\left(A\right)\mathclose{}_{n}$$$ is a sequence in $$${\mathcal{L}\mathopen{}\left( E\right)\mathclose{}}^{-1}$$$ converging to $$$A$$$ in $$${\mathcal{L}\mathopen{}\left( E\right)\mathclose{}}^{-1}$$$, then $$${A}_{n}{A}^{-1} \to \mathrm{I}$$$, so $$$A{A}_{n}^{-1} \to \mathrm{I}$$$ (by continuity at $$$\mathrm{I}$$$), so $$${A}_{n}^{-1} \to {A}^{-1}$$$.

Definition II.33

The spectrum of $$$A\in \mathcal{L}\mathopen{}\left( E\right)\mathclose{}$$$ is $$\mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}= \mathopen{}\left\{\, λ\in \mathbb{C}\,\middle\vert\, λ\mathrm{I}-A\notin {\mathcal{L}\mathopen{}\left( E\right)\mathclose{}}^{-1}\,\right\}\mathclose{} \text{.}$$

In the finite-dimensional case, $$$λ\mathrm{I}-A$$$ is invertible if and only if $$$\operatorname{Ker}\mathopen{}\left( λ\mathrm{I}-A\right)\mathclose{}= \mathopen{}\left\{\, 0\,\right\}\mathclose{}$$$ (use rank + nullity argument). Then $$$\mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}$$$ is the set of eigenvalues. In general, every eigenvalue is in the spectrum. However, the converse does not hold in infinite dimensions.

Example II.34

Let $$$E= \mathrm{l}^{0}$$$ and consider the right shift operator $$R\mathopen{}\left( {x}_{1}, {x}_{2}, \dotsc\right)\mathclose{}= \mathopen{}\left(0, {x}_{1}, {x}_{2}, \dotsc\right)\mathclose{}\text{.}$$ Suppose $$$R\mathopen{}\left( \mathbf{x}\right)\mathclose{}= λ\mathbf{x}$$$, $$$λ\in \mathbb{C}$$$, $$$\mathbf{x}\in E$$$. Then $$$λ{x}_{1}= 0$$$, $$$λ{x}_{2}= {x}_{1}$$$, $$$λ{x}_{i}= {x}_{i-1}$$$, which means $$$\mathbf{x}= \mathbf{0}$$$. So, there are no eigenvalues. However, $$$0\in \mathop{\sigma}\mathopen{}\left( R\right)\mathclose{}$$$ since $$$R$$$ is not invertible. It turns out that $$$\mathop{\sigma}\mathopen{}\left( R\right)\mathclose{}= \mathopen{}\left\{\, z\in \mathbb{C}\,\middle\vert\, \mathopen{}\left\lvert{}z\right\rvert\mathclose{}\leq 1\,\right\}\mathclose{}$$$.

Theorem II.35

For $$$A\in \mathcal{L}\mathopen{}\left( E\right)\mathclose{}$$$, $$$\mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}$$$ is closed and bounded. Moreover, $$$\mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\subseteq \mathopen{}\left\{\, λ\in \mathbb{C}\,\middle\vert\, \mathopen{}\left\lvert{}λ\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}A\right\rVert\mathclose{}\,\right\}\mathclose{}$$$. In particular, $$$\mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}$$$ is compact.

Proof. Consider the continuous function $$$F : \mathbb{C} \to \mathcal{L}\mathopen{}\left( E\right)\mathclose{}$$$ defined by $$$F\mathopen{}\left( λ\right)\mathclose{}= λI-A$$$. By definition, $$$\mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}= {F}^{-1}\mathopen{}\left( \mathcal{L}\mathopen{}\left( E\right)\mathclose{}\setminus {\mathcal{L}\mathopen{}\left( E\right)\mathclose{}}^{-1}\right)\mathclose{}$$$. However, $$${\mathcal{L}\mathopen{}\left( E\right)\mathclose{}}^{-1}$$$ is open, so $$$\mathcal{L}\mathopen{}\left( E\right)\mathclose{}\setminus {\mathcal{L}\mathopen{}\left( E\right)\mathclose{}}^{-1}$$$ is closed. Thus, because $$$F$$$ is continuous, $$$\mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}$$$ is closed. To show it is bounded, note $$$\mathopen{}\left\lvert{}μ\right\rvert\mathclose{}\gt \mathopen{}\left\lVert{}A\right\rVert\mathclose{}$$$ if and only if $$$\mathopen{}\left\lVert{}\frac{1}{μ}A\right\rVert\mathclose{}\lt 1$$$. Thus $$$I-\frac{1}{μ}A\in {\mathcal{L}\mathopen{}\left( E\right)\mathclose{}}^{-1}$$$ if and only if $$$μI-A\in {\mathcal{L}\mathopen{}\left( E\right)\mathclose{}}^{-1}$$$ if and only if $$$μ\notin \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}$$$. That is, $$$\mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\subseteq \mathopen{}\left\{\, λ\,\middle\vert\, , \mathopen{}\left\lvert{}λ\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}A\right\rVert\mathclose{}, \,\right\}\mathclose{}$$$.

Example II.36

Consider $$$\mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[a, b\right]\mathclose{}\right)\mathclose{}$$$ and $$$f\in \mathrm{C}\mathopen{}\left( \mathopen{}\left[a, b\right]\mathclose{}\right)\mathclose{}$$$. Define $$$\mathrm{M}_{f}\mathopen{}\left( x\mathopen{}\left( t\right)\mathclose{}\right)\mathclose{}$$$ to be $$$f\mathopen{}\left( t\right)\mathclose{}x\mathopen{}\left( t\right)\mathclose{}$$$. We claim $$$\mathop{\sigma}\mathopen{}\left( \mathrm{M}_{f}\right)\mathclose{}= f\mathopen{}\left( \mathopen{}\left[a, b\right]\mathclose{}\right)\mathclose{}$$$. If $$$λ\notin f\mathopen{}\left( \mathopen{}\left[a, b\right]\mathclose{}\right)\mathclose{}$$$, then the continuous function $$$λ-f$$$ has a reciprocal, so $$$λI-\mathrm{M}_{f}= \mathrm{M}_{\mathopen{}\left(λ-f\right)\mathclose{}}$$$ is invertible, that is $$$λ\notin \mathop{\sigma}\mathopen{}\left( \mathrm{M}_{f}\right)\mathclose{}$$$. Conversely, if $$$λ= f\mathopen{}\left( {t}_{0}\right)\mathclose{}$$$ for $$${t}_{0}\in \mathopen{}\left[a, b\right]\mathclose{}$$$, define $$${J}_{n}= \mathopen{}\left\{\, t\in \mathopen{}\left[a, b\right]\mathclose{}\,\middle\vert\, , \mathopen{}\left\lvert{}f\mathopen{}\left( t\right)\mathclose{}-λ\right\rvert\mathclose{}\lt \frac{1}{n}, \,\right\}\mathclose{}$$$. Let $$${δ}_{n}$$$ be the measure of this set. Consider $$${g}_{n}\mathopen{}\left( t\right)\mathclose{}\in \mathrm{L}^{\mathrm{2}}$$$ given by $${g}_{n}\mathopen{}\left( t\right)\mathclose{}= \begin{cases}\frac{1}{\sqrt{{δ}_{n}}}, & t\in {J}_{n} ; \\ 0, & t\notin {J}_{n}\text{.} \end{cases}$$ This makes $$$\mathopen{}\left\lVert{}{g}_{n}\right\rVert\mathclose{}= 1$$$. Suppose $$$T= {\mathopen{}\left(λI-\mathrm{M}_{f}\right)\mathclose{}}^{-1}$$$. Then $$\mathopen{}\left(λI-\mathrm{M}_{f}\right)\mathclose{}{g}_{n}= \begin{cases}\frac{λ-f\mathopen{}\left( t\right)\mathclose{}}{\sqrt{{δ}_{n}}}, & t\in {J}_{n} ; \\ 0, & t\notin {J}_{n}\text{.} \end{cases}$$ Thus $$${\mathopen{}\left\lVert{}\mathopen{}\left(λI-\mathrm{M}_{f}\right)\mathclose{}{g}_{n}\right\rVert\mathclose{}}^{2}\leq \frac{μ\mathopen{}\left( {J}_{n}\right)\mathclose{}}{{n}^{2}{δ}_{n}}= \frac{1}{{n}^{2}}$$$. An operator that takes a sequence of unit vectors to 0 cannot have a bounded inverse, so $$$λ\in \mathop{\sigma}\mathopen{}\left( \mathrm{M}_{f}\right)\mathclose{}$$$.

Notation II.37

Unless otherwise specified, in what follows, $$$H$$$ is a Hilbert space. Write $$$λ= λI$$$ for $$$λ\in \mathbb{C}$$$.

Proposition II.38

For $$$T$$$ and $$$S$$$ in $$$\mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$, we have

1. $$$\mathopen{}\left(TS\right)\mathclose{}^{*}= S^{*} T^{*}$$$.
2. $$$S\in { \mathcal{L}\mathopen{}\left( H\right)\mathclose{} }^{-1}$$$ implies $$$S^{*}\in { \mathcal{L}\mathopen{}\left( H\right)\mathclose{} }^{-1}$$$ with $$${\mathopen{}\left( S^{*}\right)\mathclose{}}^{-1}= \mathopen{}\left({S}^{-1}\right)\mathclose{}^{*}$$$.
3. $$$\mathop{\sigma}\mathopen{}\left( T^{*}\right)\mathclose{}= \overline{\mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}}$$$.

Proof.

1. $$$\mathopen{}\left\langle{} \mathopen{}\left(TS\right)\mathclose{}^{*}\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{}= \mathopen{}\left\langle{}x, T\mathopen{}\left( S\mathopen{}\left( y\right)\mathclose{}\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{} T^{*}\mathopen{}\left( x\right)\mathclose{}, S\mathopen{}\left( y\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{} S^{*}\mathopen{}\left( T^{*}\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}, y\right\rangle\mathclose{}$$$.
2. $$$S{S}^{-1}= I= {S}^{-1}S$$$ implies $$$\mathopen{}\left({S}^{-1}\right)\mathclose{}^{*} S^{*}= I^{*}= I= S^{*} \mathopen{}\left({S}^{-1}\right)\mathclose{}^{*}$$$.
3. For $$$λ\notin \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}$$$, $$${\mathopen{}\left(\overline{λ}- T^{*}\right)\mathclose{}}^{-1}= \mathopen{}\left({\mathopen{}\left(λ-T\right)\mathclose{}}^{-1}\right)\mathclose{}^{*}$$$, so $$$\overline{λ}\notin \mathop{\sigma}\mathopen{}\left( T^{*}\right)\mathclose{}$$$ and vice versa.

Example II.39

Let $$$S$$$ be the unilateral shift on $$$\mathrm{l}^{2}$$$. Then $$$\mathop{\sigma}\mathopen{}\left( S\right)\mathclose{}$$$ is the closed unit disc with a proof sketch as follows: For $$$\mathopen{}\left\lvert{}λ\right\rvert\mathclose{}\lt 1$$$, notice $$${\mathbf{ξ}}_{λ}= \mathopen{}\left(1, λ, {λ}^{2}, \dotsc\right)\mathclose{}\in \mathrm{l}^{2}$$$ and $$$S^{*}\mathopen{}\left( {\mathbf{ξ}}_{λ}\right)\mathclose{}= \mathopen{}\left(λ, {λ}^{2}, \dotsc\right)\mathclose{}= λ{\mathbf{ξ}}_{λ}$$$. So, the open unit disc is contained in $$$\mathop{\sigma}\mathopen{}\left( S^{*}\right)\mathclose{}$$$. Then $$$\mathop{\sigma}\mathopen{}\left( S^{*}\right)\mathclose{}$$$ is the closed unit disc and so $$$\mathop{\sigma}\mathopen{}\left( S\right)\mathclose{}$$$ is also the closed unit disk.