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

E. Integral Operators

Let $$$S$$$ be the Volterra operator on $$$\mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}= H$$$. Then $$$\mathopen{}\left(S\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= \int _{0}^{x}{}f$$$. We showed $$${S}^{2}$$$ is trace-class. $$\mathopen{}\left({S}^{2}\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= \int _{0}^{x}{} \mathopen{}\left(S\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( s\right)\mathclose{} \,\mathrm{d}s= \int _{0}^{x}{} \int _{0}^{s}{} f\mathopen{}\left( t\right)\mathclose{} \,\mathrm{d}t \,\mathrm{d}s \text{.}$$ Fubinize to get $$\mathopen{}\left({S}^{2}\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= \int _{0}^{x}{} \int _{t}^{x}{}f\mathopen{}\left( t\right)\mathclose{}\,\mathrm{d}s \,\mathrm{d}t= \int _{0}^{x}{} \mathopen{}\left(x-t\right)\mathclose{}f\mathopen{}\left( t\right)\mathclose{} \,\mathrm{d}t= \int _{0}^{1}{} τ\mathopen{}\left( x, t\right)\mathclose{}f\mathopen{}\left( t\right)\mathclose{} \,\mathrm{d}t$$ where $$τ\mathopen{}\left( x, t\right)\mathclose{}= \begin{cases}x-t, & 0\leq t\leq x; \\ 0, & \text{else,}\end{cases}$$ which is continuous on $$${\mathopen{}\left[0, 1\right]\mathclose{}}^{2}$$$.

We say that $$${S}^{2}$$$ is an integral operator with kernel $$$τ$$$. Our goal in what follows is a formula for the trace of a trace-class operator that can be realized as an integral operator with a continuous kernel. This will need some Fourier series lore, both to furnish a suitable basis for $$$H= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}$$$, and to use Fourier expansions to approximate continuous functions uniformly on $$$\mathopen{}\left[0, 1\right]\mathclose{}$$$. For the basis, let $$${e}_{0}= 1$$$, and then $$${e}_{n}\mathopen{}\left( x\right)\mathclose{}= \sqrt{2}\cos\mathopen{}\left( nπx\right)\mathclose{}$$$ for $$$n\in \mathopen{}\left\{\, 1, 2, \dotsc\,\right\}\mathclose{}$$$. The argument at the end of Chapter 1 for the completeness of the sine basis, crucially modified by extending functions on $$$\mathopen{}\left[0, 1\right]\mathclose{}$$$ evenly to $$$\mathopen{}\left[{-}1, 1\right]\mathclose{}$$$ rather than oddly, shows that the sequence $$$\mathopen{}\left\{\, {e}_{0}, {e}_{1}, {e}_{2}, \dotsc\,\right\}\mathclose{}$$$ is an orthonormal basis for $$$H$$$. (The main reason to use the cosine basis is that the even extension of any continuous function on $$$\mathopen{}\left[0, 1\right]\mathclose{}$$$ is continuous on $$$\mathopen{}\left[{-}1, 1\right]\mathclose{}$$$ with the same value at $$$1$$$ as at $$${-}1$$$.)

Let $$${P}_{n}$$$ be the projection on $$$\operatorname{span}\mathopen{}\left( {e}_{0}, {e}_{1}, \dotsc, {e}_{n}\right)\mathclose{}$$$. Then $$${P}_{n}\mathopen{}\left( f\right)\mathclose{}$$$ is the $$$n$$$th partial sum of the Fourier cosine series of $$$f$$$, $$\sum_{n=0}^{\infty}{} \mathopen{}\left\langle{}f, {e}_{n}\right\rangle\mathclose{}{e}_{n} \text{.}$$ Although $$$\mathopen{}\left\lVert{}{P}_{n}\mathopen{}\left( f\right)\mathclose{}-f\right\rVert\mathclose{}_{\infty}$$$ need not go to zero for continuous $$$f$$$, we get uniform convergence if we consider the Cesàro means $$${C}_{n}= \frac{1}{n+1}\mathopen{}\left({P}_{0}+{P}_{1}+\dotsb+{P}_{n}\right)\mathclose{}$$$ of the series.

Theorem III.41 (Fejer's Theorem)

Let $$$F$$$ be a uniformly bounded equicontinuous family of functions on $$$\mathopen{}\left[0, 1\right]\mathclose{}$$$. Then for all $$$ε\gt 0$$$ there exists $$$N$$$ such that $$$\mathopen{}\left\lVert{}{C}_{n}\mathopen{}\left( f\right)\mathclose{}-f\right\rVert\mathclose{}_{\infty}\lt ε$$$ for all $$$n\gt N$$$ and all $$$f\in F$$$; in particular, $$${C}_{n}\mathopen{}\left( f\right)\mathclose{} \to f$$$ uniformly for all $$$f\in \mathrm{C}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}$$$.

Proof. See for instance Chapter 5 of Young [12] for the proof in the standard situation of a single $$$2π$$$-periodic function $$$g$$$ on $$$\mathbb{R}$$$ and its usual Fourier series, then rescale and specialize to even $$$g$$$. The extension to bounded, equicontinuous families follows effortlessly from the observation that the upper estimates in the proof depend only on $$$\mathopen{}\left\lVert{}g\right\rVert\mathclose{}_{\infty}$$$ and $$$\mathopen{}\left\lvert{}g\mathopen{}\left( x\right)\mathclose{}-g\mathopen{}\left( y\right)\mathclose{}\right\rvert\mathclose{}$$$ for small $$$\mathopen{}\left\lvert{}x-y\right\rvert\mathclose{}$$$.

Theorem III.42

Suppose $$$T\in \mathcal{T}\mathopen{}\left( H\right)\mathclose{}$$$ has the form $$T\mathopen{}\left( f\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}= \int _{0}^{1}{} τ\mathopen{}\left( x, t\right)\mathclose{}f\mathopen{}\left( t\right)\mathclose{} \,\mathrm{d}t$$ for some continuous $$$τ$$$ on $$${\mathopen{}\left[0, 1\right]\mathclose{}}^{2}$$$ (i.e. $$$T$$$ is a trace-class integral operator with continuous kernel). Then $$$\operatorname{Tr}\mathopen{}\left( T\right)\mathclose{}= \int _{0}^{1}{}τ\mathopen{}\left( x, x\right)\mathclose{}\,\mathrm{d}x$$$.

Proof. This proof is taken from Gohberg [3]. With $$$\mathopen{}\left({e}_{n}\right)\mathclose{}_{n=0}^{\infty}$$$ the cosine basis as above, $$$\operatorname{Tr}\mathopen{}\left( T\right)\mathclose{}= \sum_{n}{} \mathopen{}\left\langle{}T\mathopen{}\left( {e}_{n}\right)\mathclose{}, {e}_{n}\right\rangle\mathclose{}$$$, which is the limit of the sequence of Cesàro means. Then, with $$${τ}_{x}\in \mathrm{C}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}$$$ defined by $$${τ}_{x}\mathopen{}\left( t\right)\mathclose{}= τ\mathopen{}\left( x, t\right)\mathclose{}$$$, $$\mathopen{}\left\langle{}T\mathopen{}\left( {e}_{n}\right)\mathclose{}, {e}_{n}\right\rangle\mathclose{}= \int _{0}^{1}{} \mathopen{}\left(T\mathopen{}\left( {e}_{n}\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}{e}_{n}\mathopen{}\left( x\right)\mathclose{} \,\mathrm{d}x= \int _{0}^{1}{} \int _{0}^{1}{} τ\mathopen{}\left( x, t\right)\mathclose{}{e}_{n}\mathopen{}\left( t\right)\mathclose{}{e}_{n}\mathopen{}\left( x\right)\mathclose{} \,\mathrm{d}t \,\mathrm{d}x= \int _{0}^{1}{} \mathopen{}\left(\mathopen{}\left\langle{}{τ}_{x}, {e}_{n}\right\rangle\mathclose{}{e}_{n}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{} \,\mathrm{d}x \text{.}$$ Let $$${A}_{n}$$$ be the $$$n$$$th Cesàro mean of $$$\sum{} \mathopen{}\left\langle{}T\mathopen{}\left( {e}_{n}\right)\mathclose{}, {e}_{n}\right\rangle\mathclose{}$$$ so that $$${A}_{n} \to \operatorname{Tr}\mathopen{}\left( T\right)\mathclose{}$$$. Then $$${A}_{n}= \int _{0}^{1}{} \mathopen{}\left({C}_{n}\mathopen{}\left( {τ}_{x}\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{} \,\mathrm{d}x$$$ and $$\mathopen{}\left\lvert{}{A}_{n}-\int _{0}^{1}{}τ\mathopen{}\left( x, x\right)\mathclose{}\,\mathrm{d}x\right\rvert\mathclose{}= \mathopen{}\left\lvert{}\int _{0}^{1}{} \mathopen{}\left(\mathopen{}\left({C}_{n}\mathopen{}\left( {τ}_{x}\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}-{τ}_{x}\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{} \,\mathrm{d}x\right\rvert\mathclose{}\leq \int _{0}^{1}{} \mathopen{}\left\lVert{}{C}_{n}\mathopen{}\left( {τ}_{x}\right)\mathclose{}-{τ}_{x}\right\rVert\mathclose{}_{\infty} \,\mathrm{d}x\leq \sup_{x}{} \mathopen{}\left\lVert{}{C}_{n}\mathopen{}\left( {τ}_{x}\right)\mathclose{}-{τ}_{x}\right\rVert\mathclose{} \to 0$$ by Fejer for Families because $$$F= \mathopen{}\left\{\, {τ}_{x}\,\middle\vert\, , x\in \mathopen{}\left[0, 1\right]\mathclose{}, \,\right\}\mathclose{}$$$ is uniformly bounded, and also equicontinuous because of the uniform continuity of $$$τ$$$.

In particular, if $$$S$$$ is the Volterra operator, then $$$\operatorname{Tr}\mathopen{}\left( {S}^{2}\right)\mathclose{}= 0$$$.

Corollary III.43

For any $$$r\in \mathrm{C}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}$$$, the operator $$$T$$$ defined by $$\mathopen{}\left(T\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= \int _{0}^{x}{} \mathopen{}\left(x-t\right)\mathclose{}f\mathopen{}\left( t\right)\mathclose{}r\mathopen{}\left( t\right)\mathclose{} \,\mathrm{d}t$$ is trace-class and $$$\operatorname{Tr}\mathopen{}\left( T\right)\mathclose{}= 0$$$.

Proof. If $$$S$$$ is the Volterra operator, and $$$\mathrm{M}_{r}$$$ denotes multiplication by $$$r$$$, then $$$T= {S}^{2}\circ \mathrm{M}_{r}\in \mathcal{T}\mathopen{}\left( H\right)\mathclose{}$$$. The kernel of $$$T$$$ is just $$$τ\mathopen{}\left( x, t\right)\mathclose{}r\mathopen{}\left( t\right)\mathclose{}$$$, which is $$$0$$$ on the diagonal.