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. Weak*-Topology

Throughout, $$$X$$$ is a normed linear space. $$$X^{*}$$$ is the dual space (a Banach space). We have a norm, $$$\mathopen{}\left\lVert{}f\right\rVert\mathclose{}= \sup_{ \mathopen{}\left\lVert{}x\right\rVert\mathclose{}= 1 }{} \mathopen{}\left\lvert{}f\mathopen{}\left( x\right)\mathclose{}\right\rvert\mathclose{}$$$. $$$τ\mathopen{}\left( X^{*}\right)\mathclose{}$$$ is the norm topology on $$$X^{*}$$$.

Skirmish III.29

1. For $$$y\in X$$$, get $$$ŷ\in \mathcal{L}\mathopen{}\left( X^{*}, \mathbb{C}\right)\mathclose{}$$$ such that $$$ŷ\mathopen{}\left( f\right)\mathclose{}= f\mathopen{}\left( y\right)\mathclose{}$$$. Note that $$$\mathopen{}\left\lVert{}ŷ\right\rVert\mathclose{}= \mathopen{}\left\lVert{}y\right\rVert\mathclose{}$$$ from Hahn-Banach.
2. $$$ŷ$$$ is continuous on $$$τ\mathopen{}\left( X^{*}\right)\mathclose{}$$$.

Definition III.30

The weak*-topology $$$\mathop{\mathrm{w}^*}$$$ on $$$X^{*}$$$ is the one given by the neighborhoods $$$\mathop{\mathrm{N}}\mathopen{}\left( f\right)\mathclose{}$$$ of $$$f\in X^{*}$$$ defined as follows: $$\mathop{\mathrm{N}}\mathopen{}\left( f\right)\mathclose{}= \mathop{\mathrm{N}}\mathopen{}\left( f, {x}_{1}, \dotsc, {x}_{n}, ε\right)\mathclose{}= \mathopen{}\left\{\, g\in X^{*}\,\middle\vert\, , \mathopen{}\left\lvert{}f\mathopen{}\left( {x}_{i}\right)\mathclose{}-g\mathopen{}\left( {x}_{i}\right)\mathclose{}\right\rvert\mathclose{}\lt ε, , \forall{}\mathopen{}\left( i\right)\mathclose{}, \,\right\}\mathclose{}$$.

Remark III.31

1. These neighborhoods form a neighborhood base and determine a Hausdorff topology on $$$X^{*}$$$.
2. Each $$$\mathop{\mathrm{N}}\mathopen{}\left( f\right)\mathclose{}$$$ is norm open: $$$\mathop{\mathrm{w}^*}\subseteq τ\mathopen{}\left( X^{*}\right)\mathclose{}$$$.
3. $$$\mathop{\mathrm{w}^*}$$$ is the weakest topology on $$$X^{*}$$$ making each $$$ŷ$$$ continuous.
4. The addition map $$$X^{*}\times X^{*} \to X^{*}$$$ and the scalar multiplication map $$$\mathbb{C}\times X^{*} \to X^{*}$$$ are both $$$\mathop{\mathrm{w}^*}$$$ continuous; that is, $$$\mathopen{}\left( X^{*}, \mathop{\mathrm{w}^*}\right)\mathclose{}$$$ is a topological vector space.

Lemma III.32 (Workhorse Lemma)

Let $$$\mathopen{}\left({f}_{α}\right)\mathclose{}$$$ be a net in $$${X}^{*}$$$. Then $$${f}_{α} \to f$$$ in $$$\mathop{\mathrm{w}^*}$$$ if and only if $$${f}_{α}\mathopen{}\left( x\right)\mathclose{} \to f\mathopen{}\left( x\right)\mathclose{}$$$ for all $$$x\in X$$$.

Proof. (⇒) Suppose $$${f}_{α} \to f$$$ in $$$\mathop{\mathrm{w}^*}$$$. For $$$x\in X$$$, $$$ε\gt 0$$$, define $$$V\mathopen{}\left( f, x, ε\right)\mathclose{}= \mathopen{}\left\{\, g\in X^{*}\,\middle\vert\, , \mathopen{}\left\lvert{}f\mathopen{}\left( x\right)\mathclose{}-g\mathopen{}\left( x\right)\mathclose{}\right\rvert\mathclose{}\lt ε, \,\right\}\mathclose{}$$$. Then $$${f}_{α} \mathbin{\overset{\mathop{\mathrm{w}^*}}{\to}}f$$$ implies $$${f}_{α}\in V\mathopen{}\left( f, x, ε\right)\mathclose{}$$$ eventually for all $$$x\in X$$$ and $$$ε\gt 0$$$, which implies $$$\mathopen{}\left\lvert{}{f}_{α}\mathopen{}\left( x\right)\mathclose{}-f\mathopen{}\left( x\right)\mathclose{}\right\rvert\mathclose{}\lt ε$$$ eventually for all $$$x\in X$$$ and $$$ε\gt 0$$$, which implies $$${f}_{α}\mathopen{}\left( x\right)\mathclose{} \to f\mathopen{}\left( x\right)\mathclose{}$$$ for all $$$x\in X$$$.

(⇐) Suppose $$${f}_{α}\mathopen{}\left( x\right)\mathclose{} \to f\mathopen{}\left( x\right)\mathclose{}$$$ for all $$$x\in X$$$. Let $$${x}_{i}\in X$$$ for $$$i\in \mathopen{}\left\{\, 1, 2, \dotsc, n\,\right\}\mathclose{}$$$ and let $$$ε\gt 0$$$ be given. $$${f}_{α}\mathopen{}\left( x\right)\mathclose{} \to f\mathopen{}\left( x\right)\mathclose{}$$$ for all $$$x$$$ implies $$${f}_{α}\mathopen{}\left( {x}_{i}\right)\mathclose{} \to f\mathopen{}\left( {x}_{i}\right)\mathclose{}$$$ for all $$$i$$$. This implies $$${f}_{α}\in \mathop{\mathrm{N}}\mathopen{}\left( f, {x}_{i}, ε\right)\mathclose{}$$$ for all $$$i$$$ eventually. Then for all $$$i$$$, there exists $$${α}_{i}$$$ such that for all $$$α\gt {α}_{i}$$$, $$${f}_{α}\in \mathop{\mathrm{N}}\mathopen{}\left( f, {x}_{i}, ε\right)\mathclose{}$$$. Pick $$$β\gt {α}_{i}$$$ for all $$$i$$$. Then $$${f}_{β}\in \bigcap_{i=1}^{n}{} \mathop{\mathrm{N}}\mathopen{}\left( f, {x}_{i}, ε\right)\mathclose{} = \mathop{\mathrm{N}}\mathopen{}\left( f, {x}_{1}, \dotsc, {x}_{n}, ε\right)\mathclose{}$$$ which gives $$${f}_{α} \mathbin{\overset{\mathop{\mathrm{w}^*}}{\to}}f$$$.

Example III.33

Let $$$X= \mathrm{L}^{\mathrm{1}}\mathopen{}\left( \mathbb{R}\right)\mathclose{}$$$. Then $$$X^{*}= \mathrm{L}^{\mathrm{∞}}\mathopen{}\left( \mathbb{R}\right)\mathclose{}$$$. For $$$F\in X$$$, set $$${f}_{n}\mathopen{}\left( F\right)\mathclose{}= \int _{n}^{\infty}{}F$$$.

1. $$${f}_{n} \mathbin{\overset{\mathop{\mathrm{w}^*}}{\to}}0$$$. As $$$n \to \infty$$$, we have $$$\int _{n}^{\infty}{}F \to 0$$$ for all $$$F\in \mathrm{L}^{\mathrm{1}}\mathopen{}\left( \mathbb{R}\right)\mathclose{}$$$ by the dominated convergence theorem. This gives $$${f}_{n}\mathopen{}\left( F\right)\mathclose{} \to 0$$$ and $$${f}_{n} \mathbin{\overset{\mathop{\mathrm{w}^*}}{\to}}0$$$ by Lemma III.32.
2. On the other hand, $$$\mathopen{}\left({f}_{n}\right)\mathclose{}$$$ does not converge to $$$0$$$ in $$$τ\mathopen{}\left( X^{*}\right)\mathclose{}$$$. Indeed, let $$${F}_{n}$$$ be the indicator function of $$$\mathopen{}\left[n, n+1\right]\mathclose{}$$$. Then $$${f}_{n}\mathopen{}\left( {F}_{n}\right)\mathclose{}= 1= \mathopen{}\left\lVert{}{F}_{n}\right\rVert\mathclose{}$$$, so $$$\mathopen{}\left\lVert{}{f}_{n}\right\rVert\mathclose{}\geq 1$$$ for every $$$n$$$.
3. An even easier example of the same phenomenon is furnished by an orthonormal sequence $$$\mathopen{}\left({e}_{n}\right)\mathclose{}$$$ in an infinite-dimensional Hilbert space $$$H$$$. Define $$${e}_{n}\in {H}^{*}$$$ by $$${e}_{n}\mathopen{}\left( x\right)\mathclose{}= \mathopen{}\left\langle{}x, {e}_{n}\right\rangle\mathclose{}$$$. Then $$${e}_{n} \mathbin{\overset{\mathop{\mathrm{w}^*}}{\to}}0$$$ by Bessel's inequality, while $$$\mathopen{}\left\lVert{}{e}_{n}\right\rVert\mathclose{}= 1$$$.

Theorem III.34 (Alaoglu's Theorem)

Let $$$X$$$ be a normed linear space. Then the closed unit ball in $$$X^{*}$$$ is compact in $$$\mathop{\mathrm{w}^*}$$$.

Proof. For $$$x\in X$$$, define $$${I}_{x}= \mathopen{}\left\{\, z\in \mathbb{C}\,\middle\vert\, , \mathopen{}\left\lvert{}z\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}x\right\rVert\mathclose{}, \,\right\}\mathclose{}$$$. Then $$${I}_{x}$$$ is compact in $$$\mathbb{C}$$$. Let $$$I= \prod_{x\in X}{}{I}_{x}$$$. $$$I$$$ is compact by the Tychonoff product theorem (with the product topology). Let $$$B= \mathopen{}\left\{\, f\in X^{*}\,\middle\vert\, , \mathopen{}\left\lVert{}f\right\rVert\mathclose{}\leq 1, \,\right\}\mathclose{}$$$. Define $$$J : B \hookrightarrow I$$$, $$$f\mapsto \mathopen{}\left\{\, f\mathopen{}\left( x\right)\mathclose{}\,\middle\vert\, , x\in X, \,\right\}\mathclose{}$$$.

Claim 1: $$$B\simeq J\mathopen{}\left( B\right)\mathclose{}$$$. $$$J$$$ is injective, as $$$J\mathopen{}\left( f\right)\mathclose{}= J\mathopen{}\left( g\right)\mathclose{}$$$ implies $$$\mathopen{}\left\{\, f\mathopen{}\left( x\right)\mathclose{}\,\middle\vert\, x\in X\,\right\}\mathclose{}= \mathopen{}\left\{\, g\mathopen{}\left( x\right)\mathclose{}\,\middle\vert\, x\in X\,\right\}\mathclose{}$$$ implies $$$f= g$$$. To show $$$J$$$ is bicontinuous (both it and its inverse are continuous) onto $$$J\mathopen{}\left( B\right)\mathclose{}$$$, let $$$\mathopen{}\left({f}_{α}\right)\mathclose{}$$$ be a net in $$$B$$$ converging in $$$\mathop{\mathrm{w}^*}$$$ to $$$f\in B$$$. By Lemma III.32, this is if and only if $$${f}_{α}\mathopen{}\left( x\right)\mathclose{} \to f\mathopen{}\left( x\right)\mathclose{}$$$ for all $$$x\in X$$$ if and only if $$$\mathopen{}\left\{\, {f}_{α}\mathopen{}\left( x\right)\mathclose{}\,\middle\vert\, x\in X\,\right\}\mathclose{} \to \mathopen{}\left\{\, f\mathopen{}\left( x\right)\mathclose{}\,\middle\vert\, x\in X\,\right\}\mathclose{}$$$ if and only if $$$J\mathopen{}\left( {f}_{α}\right)\mathclose{} \to J\mathopen{}\left( f\right)\mathclose{}$$$. This shows $$$J$$$ and $$${J}^{-1}$$$ are continuous, and the claim is proven.

Thus $$$B$$$ is compact if and only if $$$J\mathopen{}\left( B\right)\mathclose{}$$$ is compact if and only if $$$J\mathopen{}\left( B\right)\mathclose{}$$$ is closed.

Claim 2: $$$J\mathopen{}\left( B\right)\mathclose{}$$$ is closed. Let $$$\mathopen{}\left\{\, J\mathopen{}\left( {f}_{α}\right)\mathclose{}\,\right\}\mathclose{}$$$ be a net in $$$J\mathopen{}\left( B\right)\mathclose{}$$$ and suppose $$$\mathopen{}\left\{\, J\mathopen{}\left( {f}_{α}\right)\mathclose{}\,\right\}\mathclose{} \to \mathopen{}\left\{\, f\mathopen{}\left( x\right)\mathclose{}\,\middle\vert\, x\in X\,\right\}\mathclose{}$$$. Then $$$\mathopen{}\left\{\, {f}_{α}\mathopen{}\left( x\right)\mathclose{}\,\middle\vert\, x\in X\,\right\}\mathclose{} \to \mathopen{}\left\{\, f\mathopen{}\left( x\right)\mathclose{}\,\middle\vert\, x\in X\,\right\}\mathclose{}$$$. We want to show $$$f\in B$$$.

First, we have $$$f\mathopen{}\left( x+y\right)\mathclose{}= f\mathopen{}\left( x\right)\mathclose{}+f\mathopen{}\left( y\right)\mathclose{}$$$ and $$$f\mathopen{}\left( λx\right)\mathclose{}= λf\mathopen{}\left( x\right)\mathclose{}$$$ because these are limits of nets and thus we can use properties of nets in topological vector spaces. That is $$f\mathopen{}\left( x+y\right)\mathclose{}= \lim_{α}{} {f}_{α}\mathopen{}\left( x+y\right)\mathclose{} \text{,}$$ $$f\mathopen{}\left( x\right)\mathclose{}+f\mathopen{}\left( y\right)\mathclose{}= \lim_{α}{} {f}_{α}\mathopen{}\left( x\right)\mathclose{} +\lim_{α}{} {f}_{α}\mathopen{}\left( y\right)\mathclose{} \text{,}$$ etc.

For $$$x\in X$$$, $$$f\mathopen{}\left( x\right)\mathclose{}\in {I}_{x}$$$ implies $$$\mathopen{}\left\lvert{}f\mathopen{}\left( x\right)\mathclose{}\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}x\right\rVert\mathclose{}$$$ for all $$$x$$$. This gives $$$\mathopen{}\left\lVert{}f\right\rVert\mathclose{}= \sup_{ \mathopen{}\left\lVert{}x\right\rVert\mathclose{}\leq 1 }{}\mathopen{}\left\lvert{}f\mathopen{}\left( x\right)\mathclose{}\right\rvert\mathclose{}\leq 1$$$. Thus we have $$$f\in B$$$. From this, we get $$$\mathopen{}\left\{\, f\mathopen{}\left( x\right)\mathclose{}\,\middle\vert\, , x\in X, \,\right\}\mathclose{}= J\mathopen{}\left( f\right)\mathclose{}\in J\mathopen{}\left( B\right)\mathclose{}$$$. Finally, then, $$$J\mathopen{}\left( B\right)\mathclose{}$$$ is closed.