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

## A. Two Basic Theorems

Theorem II.14 (Baire's Theorem)

Let $$$X$$$ be a complete metric space.

1. If $$$\mathopen{}\left({U}_{n}\right)\mathclose{}_{n=1}^{\infty}$$$ is a sequence of open dense subsets of $$$X$$$, then $$$\bigcap_{n=1}^{\infty}{}{U}_{n}$$$ is dense in $$$X$$$.
2. $$$X$$$ is not a countable union of nowhere-dense sets.

More simply, if $$$X$$$ is the union of countably many sets, when we close them, at least one of the closures contains an open set.

Definition II.15

Let $$$X$$$ and $$$Y$$$ be topological spaces. A map $$$f : X \to Y$$$ is open if $$$f\mathopen{}\left( U\right)\mathclose{}$$$ is open in $$$Y$$$ whenever $$$U$$$ is open in $$$X$$$.

Remark II.16

If $$$X$$$ and $$$Y$$$ are normed linear spaces and $$$f$$$ is linear, then $$$f$$$ commutes with translations and dilations, so $$$f$$$ is open if and only if $$$f\mathopen{}\left( \mathrm{B}_{X}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{}$$$ contains a ball centered at $$$0$$$ in $$$Y$$$, where $$$\mathrm{B}_{X}\mathopen{}\left( 1\right)\mathclose{}$$$ is the unit ball in $$$X$$$.

Theorem II.17 (Open Mapping Theorem)

Let $$$X$$$ and $$$Y$$$ be Banach Spaces. If an operator $$$T\in \mathcal{L}\mathopen{}\left( X, Y\right)\mathclose{}$$$ is surjective, then $$$T$$$ is open.

Proof. Let $$$\mathrm{B}_{X}\mathopen{}\left( r, {x}_{0}\right)\mathclose{}$$$ denote the open ball of radius $$$r$$$ around $$${x}_{0}$$$ in $$$X$$$, and likewise for balls in $$$Y$$$. Write $$$\mathrm{B}\mathopen{}\left( r\right)\mathclose{}= \mathrm{B}_{X}\mathopen{}\left( r, 0\right)\mathclose{}$$$. It will suffice to show that $$$T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{}$$$ contains a ball about $$$0$$$ in $$$Y$$$. Since $$$X= \bigcup_{n=1}^{\infty}{} \mathrm{B}\mathopen{}\left( n\right)\mathclose{}$$$ and $$$T$$$ is surjective, we have that $$$Y= \bigcup_{n=1}^{\infty}{} T\mathopen{}\left( \mathrm{B}\mathopen{}\left( n\right)\mathclose{}\right)\mathclose{}$$$. The map $$$y\mapsto ny$$$ is a homeomorphism of $$$Y$$$ that maps $$$T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{}$$$ to $$$T\mathopen{}\left( \mathrm{B}\mathopen{}\left( n\right)\mathclose{}\right)\mathclose{}$$$, so it follows from Baire's Theorem that $$$\overline{ T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} }$$$ contains an open ball about $$$0$$$. Thus there exists $$${y}_{0}\in Y$$$ and an $$$r\gt 0$$$ such that $$$\mathrm{B}_{Y}\mathopen{}\left( 4r, {y}_{0}\right)\mathclose{}\subseteq \overline{ T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} }$$$. Now pick $$${y}_{1}= T\mathopen{}\left( x\right)\mathclose{}$$$ with $$$x\in \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}$$$ such that $$$\mathopen{}\left\lVert{}{y}_{1}-{y}_{0}\right\rVert\mathclose{}\lt 2r$$$. Then $$$\mathrm{B}_{Y}\mathopen{}\left( 2r, {y}_{1}\right)\mathclose{}\subseteq \mathrm{B}_{Y}\mathopen{}\left( 4r, {y}_{0}\right)\mathclose{}\subseteq \overline{ T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} }$$$. If $$$\mathopen{}\left\lVert{}y\right\rVert\mathclose{}\lt 2r$$$, then $$${y}_{1}-y\in \mathrm{B}_{Y}\mathopen{}\left( 2r, {y}_{1}\right)\mathclose{}\subseteq \mathrm{B}_{Y}\mathopen{}\left( 4r, {y}_{0}\right)\mathclose{}\subseteq \overline{ T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} }$$$, and hence $$${y}_{1}-y\in \overline{ T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} }$$$. Then $$y= {y}_{1}+\mathopen{}\left(y-{y}_{1}\right)\mathclose{}= T\mathopen{}\left( x\right)\mathclose{}+\mathopen{}\left(y-{y}_{1}\right)\mathclose{}\in T\mathopen{}\left( x\right)\mathclose{}+\overline{ T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} }= \overline{ T\mathopen{}\left( \mathopen{}\left({x}_{1}+\mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{}\right)\mathclose{} }\subseteq \overline{ T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 2\right)\mathclose{}\right)\mathclose{} }$$. So what we have is that $$$\mathrm{B}_{Y}\mathopen{}\left( 2r, 0\right)\mathclose{}\subseteq \overline{ T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 2\right)\mathclose{}\right)\mathclose{} }$$$, and hence $$$\mathrm{B}_{Y}\mathopen{}\left( r, 0\right)\mathclose{}\subseteq \overline{ T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} }$$$.

We will conclude the proof by showing that $$$\mathrm{B}_{Y}\mathopen{}\left( \frac{r}{2}, 0\right)\mathclose{}\subseteq \overline{ T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{} }$$$. Now, since $$$T$$$ commutes with dilations, we have that $$$\mathopen{}\left\lVert{}y\right\rVert\mathclose{}\lt {2}^{{-}n}r$$$ implies $$$y\in \overline{ T\mathopen{}\left( \mathrm{B}\mathopen{}\left( {2}^{{-}n} \right)\mathclose{}\right)\mathclose{} }$$$. Suppose that $$$\mathopen{}\left\lVert{}y\right\rVert\mathclose{}\lt \frac{r}{2}$$$. We can find $$${x}_{1}\in \mathrm{B}_{X}\mathopen{}\left( \frac{1}{2}, 0\right)\mathclose{}$$$ such that $$$\mathopen{}\left\lVert{}y-T\mathopen{}\left( {x}_{1}\right)\mathclose{}\right\rVert\mathclose{}\lt \frac{r}{4}$$$. By induction we get $$${x}_{n}\in \mathrm{B}\mathopen{}\left( {2}^{{-}n} \right)\mathclose{}$$$ such that $$$\mathopen{}\left\lVert{}y-\sum_{i=1}^{n}{} T\mathopen{}\left( {x}_{i}\right)\mathclose{} \right\rVert\mathclose{}\lt r{2}^{{-}n-1}$$$. Since $$$X$$$ is complete the series $$$\sum_{n=1}^{\infty}{} {x}_{n}$$$ converges in $$$X$$$. Call the sum $$$s$$$. Then $$$\mathopen{}\left\lVert{}s\right\rVert\mathclose{}\lt \sum_{n=1}^{\infty}{} {2}^{{-}n} = 1$$$ and $$$y= T\mathopen{}\left( s\right)\mathclose{}$$$ (because $$$T$$$ is bounded). Thus $$$y\in T\mathopen{}\left( \mathrm{B}\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{}$$$ for all $$$y$$$ such that $$$\mathopen{}\left\lVert{}y\right\rVert\mathclose{}\lt \frac{r}{2}$$$.

Corollary II.18

If $$$X$$$ and $$$Y$$$ are Banach spaces and $$$T$$$ in $$$\mathcal{L}\mathopen{}\left( X, Y\right)\mathclose{}$$$ is one-to-one and onto, then $$${T}^{-1}$$$ is bounded.

Remark II.19

If $$$X$$$ and $$$Y$$$ are normed linear spaces, we'll define the norm on $$$X\oplus Y$$$ by $$$\mathopen{}\left\lVert{}\mathopen{}\left(x, y\right)\mathclose{}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}x\right\rVert\mathclose{}+\mathopen{}\left\lVert{}y\right\rVert\mathclose{}$$$ (for now).

Definition II.20

The graph of a linear map $$$T : X \to Y$$$ between vector spaces is (of course) the subspace $$$\mathop{\mathcal{G}}\mathopen{}\left( T\right)\mathclose{}$$$ of $$$X\oplus Y$$$ defined by $$$\mathop{\mathcal{G}}\mathopen{}\left( T\right)\mathclose{}= \mathopen{}\left\{\, \mathopen{}\left(x, T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}\,\middle\vert\, x\in X\,\right\}\mathclose{}$$$.

When $$$X$$$ and $$$Y$$$ are normed linear spaces and $$$T$$$ is bounded, it is straightforward to show that $$$\mathop{\mathcal{G}}\mathopen{}\left( T\right)\mathclose{}$$$ is closed in $$$X\oplus Y$$$, normed as above (or in any fashion that makes the inclusions from $$$X$$$ and $$$Y$$$ to $$$X\oplus Y$$$ and the projections from $$$X\oplus Y$$$ onto $$$X$$$ and $$$Y$$$ continuous). For Banach spaces, the converse is true.

Theorem II.21 (Closed Graph Theorem)

If $$$T : X \to Y$$$ is a linear map between Banach spaces whose graph is closed, then $$$T$$$ is bounded.

Proof. Define $$${φ}_{1}\in \mathcal{L}\mathopen{}\left( \mathop{\mathcal{G}}\mathopen{}\left( T\right)\mathclose{}, X\right)\mathclose{}$$$ by $$${φ}_{1}\mathopen{}\left( x, T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}= x$$$ and $$${φ}_{2}\in \mathcal{L}\mathopen{}\left( \mathop{\mathcal{G}}\mathopen{}\left( T\right)\mathclose{}, Y\right)\mathclose{}$$$ by $$${φ}_{2}\mathopen{}\left( x, T\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}= T\mathopen{}\left( x\right)\mathclose{}$$$. It is easily seen that $$$X\oplus Y$$$ is complete, so our hypothesis makes $$$\mathop{\mathcal{G}}\mathopen{}\left( T\right)\mathclose{}$$$ complete. Notice that $$${φ}_{1}$$$ is bijective. By Corollary II.18, $$${φ}_{1}^{-1}$$$ is bounded. Thus $$${φ}_{2}\circ {φ}_{1}^{-1}= T$$$ is bounded.