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. Completeness

Definition I.28

A function $$$d : S\times S \to \mathbb{R}$$$ is a metric on a set $$$S$$$ if

1. $$$d\mathopen{}\left( x, y\right)\mathclose{}\geq 0$$$;
2. $$$d\mathopen{}\left( x, y\right)\mathclose{}= 0$$$ if and only if $$$x= y$$$;
3. $$$d\mathopen{}\left( x, y\right)\mathclose{}= d\mathopen{}\left( y, x\right)\mathclose{}$$$; and
4. $$$d\mathopen{}\left( x, z\right)\mathclose{}\leq d\mathopen{}\left( x, y\right)\mathclose{}+d\mathopen{}\left( y, z\right)\mathclose{}$$$
for all $$$x$$$, $$$y$$$, and $$$z$$$ in $$$S$$$.

Definition I.29

A set with a metric is a metric space.

Proposition I.30

A normed linear space $$$X$$$ is a metric space with metric given by $$$d\mathopen{}\left( x, y\right)\mathclose{}= \mathopen{}\left\lVert{}x-y\right\rVert\mathclose{}$$$.

Proof. We prove one condition is satisfied and leave the remaining conditions as Exercise I.31. $$d\mathopen{}\left( w, x\right)\mathclose{}+d\mathopen{}\left( x, y\right)\mathclose{}= \mathopen{}\left\lVert{}w-x\right\rVert\mathclose{}+\mathopen{}\left\lVert{}x-y\right\rVert\mathclose{}\geq \mathopen{}\left\lVert{}\mathopen{}\left(w-x\right)\mathclose{}+\mathopen{}\left(x-y\right)\mathclose{}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}w-y\right\rVert\mathclose{}= d\mathopen{}\left( y, w\right)\mathclose{} \text{.}$$

Exercise I.31

Complete the proof of Proposition I.30.

Unless otherwise specified, convergence of sequences (and therefore of series) in a normed linear space is with respect to the metric of Proposition I.30, and thus the topology (open sets, closed sets, etc.) comes from the norm.

Definition I.32

A sequence $$$\mathopen{}\left({α}_{n}\right)\mathclose{}_{n=1}^{\infty}$$$ in a metric space $$$M$$$ with metric $$$d$$$ is Cauchy if $$$\lim_{n\to\infty}{} \sup_{m\geq n}{}d\mathopen{}\left( {α}_{n}, {α}_{m}\right)\mathclose{} = 0$$$.

Definition I.33

A metric space $$$M$$$ is complete if every Cauchy sequence converges to a limit in $$$M$$$.

Proposition I.34

A normed linear space $$$X$$$ with norm $$$\mathopen{}\left\lVert{}\cdot\right\rVert\mathclose{}$$$ is complete if and only if $$$\sum_{n=1}^{\infty}{}{x}_{n}$$$ converges in $$$X$$$ whenever $$$\sum_{n=1}^{\infty}{}\mathopen{}\left\lVert{}{x}_{n}\right\rVert\mathclose{}\lt \infty$$$.

Proof. Assume first that $$$X$$$ is complete and suppose $$${x}_{1}$$$, $$${x}_{2}$$$, …, are elements of $$$X$$$ satisfying $$$\sum_{n=1}^{\infty}{}\mathopen{}\left\lVert{}{x}_{n}\right\rVert\mathclose{}\lt \infty$$$. Define $$${s}_{n}= \sum_{j=1}^{n}{}{x}_{j}$$$. Notice that for $$$m\gt n$$$, $$\mathopen{}\left\lVert{}{s}_{m}-{s}_{n}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}\sum_{j=n+1}^{m}{}{x}_{j}\right\rVert\mathclose{}\leq \sum_{j= n+1 }^{ m }{} \mathopen{}\left\lVert{}{x}_{j}\right\rVert\mathclose{} \leq \sum_{j= n+1 }^{ \infty }{} \mathopen{}\left\lVert{}{x}_{j}\right\rVert\mathclose{} \text{,}$$ which converges to zero as $$$n$$$ goes to $$$\infty$$$. So $$$\mathopen{}\left({s}_{n}\right)\mathclose{}_{n=1}^{\infty}$$$ is Cauchy and has a limit, $$$s$$$ say, in $$$X$$$. Since the sequence of partial sums converges to $$$s$$$, $$$\sum_{n=1}^{\infty}{}{x}_{n}$$$ converges and its sum is $$$s$$$.

Suppose next that $$$\sum_{n=1}^{\infty}{}\mathopen{}\left\lVert{}{x}_{n}\right\rVert\mathclose{}\lt \infty$$$ implies $$$\sum_{n=1}^{\infty}{}{x}_{n}$$$ converges in $$$X$$$ for any sequence $$$\mathopen{}\left({x}_{n}\right)\mathclose{}_{n=1}^{\infty}$$$ with elements in $$$X$$$. Let $$$\mathopen{}\left({y}_{k}\right)\mathclose{}_{k=1}^{\infty}$$$ be a Cauchy sequence in $$$X$$$. Because $$$\mathopen{}\left({y}_{k}\right)\mathclose{}_{k=1}^{\infty}$$$ is Cauchy, there exists a strictly increasing sequence $$$\mathopen{}\left({k}_{n}\right)\mathclose{}_{n=1}^{\infty}$$$ of positive integers such that $$$\mathopen{}\left\lVert{}{y}_{{k}_{n}}-{y}_{{k}_{n+1}}\right\rVert\mathclose{}\lt \frac{1}{{2}^{n}}$$$ for each $$$n$$$. For each $$$n$$$, let $$${x}_{n}= {y}_{{k}_{n}}-{y}_{{k}_{n+1}}$$$. Thus $$$\sum_{n=1}^{\infty}{} \mathopen{}\left\lVert{}{x}_{n}\right\rVert\mathclose{}\lt \infty$$$ and, by assumption, $$$\sum_{n=1}^{\infty}{}{x}_{n}$$$ converges in $$$X$$$. The partial sums of $$$\sum_{n=1}^{\infty}{}{x}_{n}$$$ telescope: $$\mathopen{}\left({y}_{{k}_{1}}-{y}_{{k}_{2}}\right)\mathclose{}+\dotsb+\mathopen{}\left({y}_{{k}_{n-1}}-{y}_{{k}_{n}}\right)\mathclose{}= {y}_{{k}_{1}}-{y}_{{k}_{n}} \text{.}$$ So $$$\mathopen{}\left({y}_{{k}_{n}}\right)\mathclose{}_{n=1}^{\infty}$$$ converges. Because a Cauchy sequence with a convergent subsequence converges, the original sequence $$$\mathopen{}\left({y}_{k}\right)\mathclose{}_{k=1}^{\infty}$$$ converges.

Definition I.35

A complete normed linear space is called a Banach space.

Exercise I.36

Prove that any finite dimensional normed linear space is a Banach space.

Exercise I.37

Prove that the space of continuous functions on a compact metric space is a Banach space using the supremum norm.

Definition I.38

A complete inner product space is called a Hilbert space.

Recall (an excellent reference for rudiments of measure theory is chapter 11 of Rudin's Principles [10], also known as the blue Rudin) for a measure space $$$\mathopen{}\left(Ω, μ\right)\mathclose{}$$$ and a non-negative measurable function $$$g : Ω \to \mathopen{}\left[0, \infty\right]\mathclose{}$$$, the integral $$$\int _{Ω}{}g\,\mathrm{d}μ$$$ is a well-defined number in $$$\mathopen{}\left[0, \infty\right]\mathclose{}$$$ and integration over $$$Ω$$$ with respect to $$$μ$$$ is well-behaved on non-negative measurable functions. For example, if $$$\mathopen{}\left({g}_{n}\right)\mathclose{}_{n=1}^{\infty}$$$ is a sequence of nonnegative measurable functions then $$\sum_{n=1}^{\infty}{} \int _{Ω}{}{g}_{n}\,\mathrm{d}μ = \int _{Ω}{} \sum_{n=1}^{\infty}{}{g}_{n} \,\mathrm{d}μ\text{.}$$ Also, if $$$h$$$ is a complex-valued measurable function on $$$Ω$$$ and $$$\int _{Ω}{}{\mathopen{}\left\lvert{}h\right\rvert\mathclose{}}_{}\,\mathrm{d}μ\lt \infty$$$, then $$$\int _{Ω}{}{h}_{}\,\mathrm{d}μ$$$ is well-defined in $$$\mathbb{C}$$$.

Definition I.39

For a measure space $$$\mathopen{}\left(Ω, μ\right)\mathclose{}$$$ and a real number $$$p$$$ with $$$1\leq p\lt \infty$$$, let $$$\mathrm{M}^{p}\mathopen{}\left( Ω\right)\mathclose{}$$$ (or just $$$\mathrm{M}^{p}$$$) be the set of all measurable complex-valued functions $$$f$$$ on $$$Ω$$$ such that $$\int _{Ω}{}{\mathopen{}\left\lvert{}f\right\rvert\mathclose{}}^{p}\,\mathrm{d}μ\lt \infty \text{.}$$

Exercise I.40

Prove that $$$\mathrm{M}^{\mathrm{2}}$$$ is a vector space.

Hint

Prove as an intermediate step that for functions $$$f$$$ and $$$g$$$ in $$$\mathrm{M}^{\mathrm{2}}$$$, $${\mathopen{}\left(\int _{Ω}{}{\mathopen{}\left\lvert{}f+g\right\rvert\mathclose{}}^{2}\,\mathrm{d}μ\right)\mathclose{}}^{\frac{1}{2}}\leq {\mathopen{}\left(\int _{Ω}{}{\mathopen{}\left\lvert{}f\right\rvert\mathclose{}}^{2}\,\mathrm{d}μ\right)\mathclose{}}^{\frac{1}{2}}+{\mathopen{}\left(\int _{Ω}{}{\mathopen{}\left\lvert{}g\right\rvert\mathclose{}}^{2}\,\mathrm{d}μ\right)\mathclose{}}^{\frac{1}{2}} \text{.}$$

Exercise I.41

Show that $$\mathopen{}\left\langle{}f, g\right\rangle\mathclose{}_{\mathrm{M}^{\mathrm{2}}}= \int _{Ω}{}f\overline{g}\,\mathrm{d}μ$$ defines a positive sesquilinear form on $$$\mathrm{M}^{\mathrm{2}}$$$.

Hint

As an intermediate step, as in the proof of Proposition I.3, prove that for functions $$$f$$$ and $$$g$$$ in $$$\mathrm{M}^{\mathrm{2}}$$$, $${\mathopen{}\left(\int _{Ω}{}{\mathopen{}\left\lvert{}fg\right\rvert\mathclose{}}^{2}\,\mathrm{d}μ\right)\mathclose{}}^{\frac{1}{2}}\leq {\mathopen{}\left(\int _{Ω}{}{\mathopen{}\left\lvert{}f\right\rvert\mathclose{}}^{2}\,\mathrm{d}μ\right)\mathclose{}}^{\frac{1}{2}}{\mathopen{}\left(\int _{Ω}{}{\mathopen{}\left\lvert{}g\right\rvert\mathclose{}}^{2}\,\mathrm{d}μ\right)\mathclose{}}^{\frac{1}{2}} \text{.}$$

Definition I.42

Let $$$N$$$ be the subspace of null vectors (functions) of $$$\mathrm{M}^{\mathrm{2}}$$$. Define $$$\mathrm{L}^{\mathrm{2}}= \mathrm{M}^{\mathrm{2}}/N$$$.

The space $$$\mathrm{L}^{\mathrm{2}}$$$ thus comprises the $$$\mathrm{M}^{\mathrm{2}}$$$ functions modulo agreement almost everywhere.

Theorem I.43

The space $$$\mathrm{L}^{\mathrm{2}}$$$ is a Hilbert space with respect to the inner product $$\mathopen{}\left\langle{}f, g\right\rangle\mathclose{}_{\mathrm{L}^{\mathrm{2}}}= \int _{Ω}{}f\overline{g}\,\mathrm{d}μ \text{.}$$

Proof. Take a sequence $$$\mathopen{}\left({f}_{n}\right)\mathclose{}_{n=1}^{\infty}$$$ of functions in $$$\mathrm{L}^{\mathrm{2}}$$$ such that $$$\sum_{n=1}^{\infty}{}\mathopen{}\left\lVert{}{f}_{n}\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}}\lt \infty$$$ (note that the norm on $$$\mathrm{L}^{\mathrm{2}}$$$ comes from the inner product on $$$\mathrm{L}^{\mathrm{2}}$$$ as in Proposition I.21). We may assume that the measure space is σ-finite (that is, a countable union of sets of finite measure), for we may fix function representatives for each $$${f}_{n}$$$ and work in $$\bigcup_{n,k=1}^{\infty}{} \mathopen{}\left\{\, x\,\middle\vert\, , \mathopen{}\left\lvert{}{f}_{n}\mathopen{}\left( x\right)\mathclose{}\right\rvert\mathclose{}\geq \frac{1}{k}, \,\right\}\mathclose{} \text{.}$$ For any $$$g\in \mathrm{L}^{\mathrm{2}}$$$ such that $$$g\geq 0$$$, $$\sum_{n=1}^{\infty}{} \int _{Ω}{}g\mathopen{}\left\lvert{}{f}_{n}\right\rvert\mathclose{}\,\mathrm{d}μ \leq \sum_{n=1}^{\infty}{} \mathopen{}\left\lVert{}g\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}}\mathopen{}\left\lVert{}{f}_{n}\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}} \text{.}$$ So, by the monotone convergence theorem, $$\int _{Ω}{} \sum_{n=1}^{\infty}{}g\mathopen{}\left\lvert{}{f}_{n}\right\rvert\mathclose{} \,\mathrm{d}μ\lt \infty\text{.}$$ Thus $$$g\sum_{n=1}^{\infty}{}\mathopen{}\left\lvert{}{f}_{n}\right\rvert\mathclose{}\lt \infty$$$ almost everywhere.

Setting $$$g= \chi_{E}$$$ for measurable sets $$$E$$$ with finite nonzero measure shows that $$$\sum_{n=1}^{\infty}{}\mathopen{}\left\lvert{}{f}_{n}\right\rvert\mathclose{}\lt \infty$$$ almost everywhere, and thus $$$\sum_{n=1}^{\infty}{}{f}_{n}$$$ converges almost everywhere. Let $$$S= \sum_{n=1}^{\infty}{}{f}_{n}$$$ (defined almost everywhere) and $$${S}_{m}= \sum_{n=1}^{m}{}{f}_{n}$$$. Then, using Fatou's Lemma, $${\mathopen{}\left(\int _{Ω}{}{\mathopen{}\left\lvert{}S-{S}_{N}\right\rvert\mathclose{}}^{2}\,\mathrm{d}μ\right)\mathclose{}}^{\frac{1}{2}}= {\mathopen{}\left(\int _{Ω}{}\lim_{m\to\infty}{}{\mathopen{}\left\lvert{}{S}_{m}-{S}_{N}\right\rvert\mathclose{}}^{2}\,\mathrm{d}μ\right)\mathclose{}}^{\frac{1}{2}}\leq \liminf_{m\to\infty}{} {\mathopen{}\left(\int _{Ω}{}{\mathopen{}\left\lvert{}{S}_{m}-{S}_{N}\right\rvert\mathclose{}}^{2}\,\mathrm{d}μ\right)\mathclose{}}^{\frac{1}{2}} = \liminf_{m\to\infty}{} {\mathopen{}\left(\int _{Ω}{}{\mathopen{}\left\lvert{}\sum_{n=N+1}^{m}{}\mathopen{}\left\lvert{}{f}_{n}\right\rvert\mathclose{}\right\rvert\mathclose{}}^{2}\,\mathrm{d}μ\right)\mathclose{}}^{\frac{1}{2}} \leq \liminf_{m\to\infty}{} \sum_{n=N+1}^{m}{}\mathopen{}\left\lVert{}{f}_{n}\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}} = \sum_{n=N+1}^{\infty}{}\mathopen{}\left\lVert{}{f}_{n}\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}} \text{.}$$ So $$\mathopen{}\left\lVert{}S-{S}_{N}\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}}\leq \sum_{n=1}^{\infty}{}\mathopen{}\left\lVert{}{f}_{n}\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}}$$ This makes $$$S\in \mathrm{L}^{\mathrm{2}}$$$ (because $$$S= S-{S}_{N}+{S}_{N}$$$ ), and $$$\lim_{N\to\infty}{}\mathopen{}\left\lVert{}S-{S}_{N}\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}}= 0$$$ because $$$\sum_{n=1}^{\infty}{}\mathopen{}\left\lVert{}{f}_{n}\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}}\lt \infty$$$.

Definition I.44

If $$$Ω$$$ is just a set $$$S$$$ and the measure $$$μ$$$ on $$$S$$$ is counting measure, then we write $$$\mathrm{l}^{0}\mathopen{}\left( S\right)\mathclose{}= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( S\right)\mathclose{}$$$. In this case, $$\mathrm{l}^{0}\mathopen{}\left( S\right)\mathclose{}= \mathopen{}\left\{\, ξ : S \to \mathbb{C} \,\middle\vert\, , \sum_{s\in S}{}{\mathopen{}\left\lvert{}ξ\mathopen{}\left( s\right)\mathclose{}\right\rvert\mathclose{}}^{2}\lt \infty, \,\right\}\mathclose{}\text{.}$$ If $$$S$$$ is countably infinite, write $$$\mathrm{l}^{0}= \mathrm{l}^{0}\mathopen{}\left( S\right)\mathclose{}$$$.

We'll see later that all Hilbert spaces are (isomorphic to) $$$\mathrm{l}^{0}\mathopen{}\left( S\right)\mathclose{}$$$ for some $$$S$$$. The only thing that matters is the cardinality of $$$S$$$. In fact, we will commonly think of $$$\mathrm{l}^{0}$$$ as $$$\mathrm{l}^{0}\mathopen{}\left( \mathbb{N}\right)\mathclose{}$$$.

The well-known procedure for completing a metric space adapts readily to normed linear spaces, yielding Banach spaces. In the case of an inner product space, we can extend the inner product to the metric completion to obtain a Hilbert space as follows: Start with an inner product space $$$V$$$. Write $$$S$$$ for the set of all Cauchy sequences in $$$V$$$, which is also a vector space. Define a positive sesquilinear form on $$$S$$$ by $$\mathopen{}\left\langle{}\mathopen{}\left({x}_{n}\right)\mathclose{}_{n=1}^{\infty}, \mathopen{}\left({y}_{n}\right)\mathclose{}_{n=1}^{\infty}\right\rangle\mathclose{}= \lim_{n\to\infty}{} \mathopen{}\left\langle{}{x}_{n}, {y}_{n}\right\rangle\mathclose{} \text{.}$$ This works because $$\mathopen{}\left\lvert{}\mathopen{}\left\langle{}{x}_{n}, {y}_{n}\right\rangle\mathclose{}-\mathopen{}\left\langle{}{x}_{m}, {y}_{m}\right\rangle\mathclose{}\right\rvert\mathclose{}\leq \mathopen{}\left\lvert{}\mathopen{}\left\langle{}{x}_{n}, {y}_{n}\right\rangle\mathclose{}-\mathopen{}\left\langle{}{x}_{m}, {y}_{n}\right\rangle\mathclose{}+\mathopen{}\left\langle{}{x}_{m}, {y}_{n}\right\rangle\mathclose{}-\mathopen{}\left\langle{}{x}_{m}, {y}_{m}\right\rangle\mathclose{}\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}{x}_{n}-{x}_{m}\right\rVert\mathclose{}\mathopen{}\left\lVert{}{y}_{n}\right\rVert\mathclose{}+\mathopen{}\left\lVert{}{x}_{m}\right\rVert\mathclose{}\mathopen{}\left\lVert{}{y}_{n}-{y}_{m}\right\rVert\mathclose{}\text{.}$$ Since the sequences $$$\mathopen{}\left({x}_{n}\right)\mathclose{}_{n=1}^{\infty}$$$ and $$$\mathopen{}\left({y}_{n}\right)\mathclose{}_{n=1}^{\infty}$$$ are Cauchy, they are bounded, and so $$\mathopen{}\left\lVert{}{x}_{n}-{x}_{m}\right\rVert\mathclose{}\mathopen{}\left\lVert{}{y}_{n}\right\rVert\mathclose{}+\mathopen{}\left\lVert{}{x}_{m}\right\rVert\mathclose{}\mathopen{}\left\lVert{}{y}_{n}-{y}_{m}\right\rVert\mathclose{}\leq A\mathopen{}\left\lVert{}{x}_{n}-{x}_{m}\right\rVert\mathclose{}+B\mathopen{}\left\lVert{}{y}_{n}-{y}_{m}\right\rVert\mathclose{}$$ for some real numbers $$$A$$$ and $$$B$$$. Thus the sequence $$$\mathopen{}\left(\mathopen{}\left\langle{}{x}_{n}, {y}_{n}\right\rangle\mathclose{}\right)\mathclose{}_{n=1}^{\infty}$$$ is Cauchy. Letting $$$N$$$ be the set of null vectors of $$$S$$$ (in this case those Cauchy sequences that converge to $$$0$$$), $$$\overline{V}= S/N$$$ is the completion of $$$V$$$.