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

## B. Orthogonal Projection

The key to many computations to follow in this chapter can be found in Exercise I.45.

Exercise I.45

Prove that in an inner product space, $$${\mathopen{}\left\lVert{}x+y\right\rVert\mathclose{}}^{2}= {\mathopen{}\left\lVert{}x\right\rVert\mathclose{}}^{2}+2\Re\mathopen{}\left( \mathopen{}\left\langle{}x, y\right\rangle\mathclose{}\right)\mathclose{}+{\mathopen{}\left\lVert{}y\right\rVert\mathclose{}}^{2}$$$.

Theorem I.46 (Parallelogram Law)

For any $$$x$$$ and $$$y$$$ in a normed linear space $$$X$$$, $$${\mathopen{}\left\lVert{}x+y\right\rVert\mathclose{}}^{2}+{\mathopen{}\left\lVert{}x-y\right\rVert\mathclose{}}^{2}= 2\mathopen{}\left({\mathopen{}\left\lVert{}x\right\rVert\mathclose{}}^{2}+{\mathopen{}\left\lVert{}y\right\rVert\mathclose{}}^{2}\right)\mathclose{}$$$.

Theorem I.47

For any $$$x$$$ and $$$y$$$ in a normed linear space $$$X$$$, if $$$\mathopen{}\left\langle{}x, y\right\rangle\mathclose{}= 0$$$ then $$${\mathopen{}\left\lVert{}x+y\right\rVert\mathclose{}}^{2}= {\mathopen{}\left\lVert{}x\right\rVert\mathclose{}}^{2}+{\mathopen{}\left\lVert{}y\right\rVert\mathclose{}}^{2}$$$.

Definition I.48

For an element $$$x$$$ and a subset $$$S$$$ of a normed linear space $$$X$$$, the distance from $$$x$$$ to $$$S$$$ in $$$X$$$, written $$$\operatorname{d}\mathopen{}\left( x, S\right)\mathclose{}$$$, is $$$\inf_{y\in S}{}\operatorname{d}\mathopen{}\left( x, y\right)\mathclose{}$$$.

Theorem I.49

For a closed convex (non-empty) subset $$$E$$$ of a Hilbert space $$$H$$$ and a vector $$$x$$$ in $$$H$$$, there exists a vector $$$y$$$ in $$$E$$$ such that $$$\mathopen{}\left\lVert{}x-y\right\rVert\mathclose{}\lt \mathopen{}\left\lVert{}x-w\right\rVert\mathclose{}$$$ for all $$$w$$$ in $$$E$$$ distinct from $$$y$$$. That is, there exists a unique point in $$$E$$$ nearest to $$$x$$$.

Proof. Let $$$\mathopen{}\left({y}_{n}\right)\mathclose{}_{n=1}^{\infty}$$$ be a sequence of elements of $$$E$$$ such that $$$\mathopen{}\left(\mathopen{}\left\lVert{}x-{y}_{n}\right\rVert\mathclose{}\right)\mathclose{}_{n=1}^{\infty}$$$ converges to $$$\operatorname{d}\mathopen{}\left( x, E\right)\mathclose{}$$$. We claim that $$$\mathopen{}\left({y}_{n}\right)\mathclose{}_{n=1}^{\infty}$$$ is Cauchy. Indeed, using Theorem I.46, $$2\mathopen{}\left({\mathopen{}\left\lVert{}x-{y}_{n}\right\rVert\mathclose{}}^{2}+{\mathopen{}\left\lVert{}x-{y}_{m}\right\rVert\mathclose{}}^{2}\right)\mathclose{}= {\mathopen{}\left\lVert{}{y}_{m}-{y}_{n}\right\rVert\mathclose{}}^{2}+{\mathopen{}\left\lVert{}2\mathopen{}\left(x-\frac{{y}_{m}+{y}_{n}}{2}\right)\mathclose{}\right\rVert\mathclose{}}^{2}\geq {\mathopen{}\left\lVert{}{y}_{m}-{y}_{n}\right\rVert\mathclose{}}^{2}+4{\operatorname{d}\mathopen{}\left( x, E\right)\mathclose{}}^{2} \text{.}$$ Note that $$$\frac{{y}_{m}+{y}_{n}}{2}$$$ is in $$$E$$$ because $$$E$$$ is convex. So $${\mathopen{}\left\lVert{}{y}_{m}-{y}_{n}\right\rVert\mathclose{}}^{2}\leq 2\mathopen{}\left(\mathopen{}\left({\mathopen{}\left\lVert{}x-{y}_{n}\right\rVert\mathclose{}}^{2}-{\operatorname{d}\mathopen{}\left( x, E\right)\mathclose{}}^{2}\right)\mathclose{}+\mathopen{}\left({\mathopen{}\left\lVert{}x-{y}_{m}\right\rVert\mathclose{}}^{2}-{\operatorname{d}\mathopen{}\left( x, E\right)\mathclose{}}^{2}\right)\mathclose{}\right)\mathclose{}$$ and it follows that $$$\mathopen{}\left({y}_{n}\right)\mathclose{}_{n=1}^{\infty}$$$ is Cauchy.

As a Hilbert space, $$$H$$$ is complete, so the sequence $$$\mathopen{}\left({y}_{n}\right)\mathclose{}_{n=1}^{\infty}$$$ converges to some element $$$y$$$ of $$$H$$$. Because $$$E$$$ is closed, $$$y\in E$$$. Thus $$$\mathopen{}\left\lVert{}x-y\right\rVert\mathclose{}= \operatorname{d}\mathopen{}\left( x, E\right)\mathclose{}$$$, and for any element $$$w$$$ of $$$E$$$ other than $$$y$$$, $$2\mathopen{}\left({\mathopen{}\left\lVert{}x-y\right\rVert\mathclose{}}^{2}+{\mathopen{}\left\lVert{}x-w\right\rVert\mathclose{}}^{2}\right)\mathclose{}= {\mathopen{}\left\lVert{}w-y\right\rVert\mathclose{}}^{2}+4{\mathopen{}\left\lVert{}x-\frac{w+y}{2}\right\rVert\mathclose{}}^{2}\gt 4{\operatorname{d}\mathopen{}\left( x, E\right)\mathclose{}}^{2} \text{.}$$ So $$$2{\mathopen{}\left\lVert{}x-w\right\rVert\mathclose{}}^{2}\gt 2{\operatorname{d}\mathopen{}\left( x, E\right)\mathclose{}}^{2}$$$ and $$$\mathopen{}\left\lVert{}x-w\right\rVert\mathclose{}\gt \operatorname{d}\mathopen{}\left( x, E\right)\mathclose{}$$$.

Definition I.50

Let $$$W$$$ be a closed subspace of a Hilbert space $$$H$$$. The orthogonal projection of an element $$$x\in H$$$ on $$$W$$$ is the point in $$$W$$$ nearest to $$$x$$$. The operator that maps each element of $$$H$$$ to its orthogonal projection on $$$W$$$ is called the projection of $$$H$$$ on $$$W$$$.

Exercise I.51

Show that the projection of a Hilbert space onto a closed subspace is a linear operator.

Definition I.52

For an inner product space $$$V$$$, elements $$$x$$$ and $$$y$$$ are orthogonal, denoted $$$x\perp y$$$, if $$$\mathopen{}\left\langle{}x, y\right\rangle\mathclose{}= 0$$$. .

Definition I.53

Let $$$S$$$ be a subset of an inner product space $$$V$$$. The orthogonal complement of $$$S$$$ in $$$V$$$, written $$${S}^{\perp}$$$, is the set of all vectors $$$x$$$ in $$$V$$$ such that $$$x\perp y$$$ for all vectors $$$y$$$ in $$$S$$$.

Exercise I.54

Prove that for any subset $$$S$$$ of an inner product space $$$V$$$, $$${S}^{\perp}$$$ is a subspace of $$$V$$$.

Exercise I.55

Prove that for any subset $$$S$$$ of an inner product space $$$V$$$, $$$S\subseteq {\mathopen{}\left( {S}^{\perp}\right)\mathclose{}}^{\perp}$$$.

Exercise I.56

Prove that for any subspace $$$W$$$ of an inner product space $$$V$$$, $$$W\cap {W}^{\perp}= \mathopen{}\left\{\, 0\,\right\}\mathclose{}$$$.

Proposition I.57

For any subset $$$S$$$ of an inner product space $$$V$$$, $$${S}^{\perp}$$$ is a closed subspace of $$$V$$$.

Proof. The subset $$${S}^{\perp}$$$ is a subspace by Exercise I.54. To show that it is closed, consider a sequence $$$\mathopen{}\left({x}_{n}\right)\mathclose{}_{n=1}^{\infty}$$$ of elements of $$${S}^{\perp}$$$ that converges in $$$V$$$ to a vector $$$x$$$. Then using Proposition I.3, $$\mathopen{}\left\lvert{}\mathopen{}\left\langle{}x, y\right\rangle\mathclose{}\right\rvert\mathclose{}= \mathopen{}\left\lvert{}\mathopen{}\left\langle{}x-{x}_{n}, y\right\rangle\mathclose{}\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}y\right\rVert\mathclose{}\mathopen{}\left\lVert{}x-{x}_{n}\right\rVert\mathclose{}$$ for all $$$y$$$ in $$$S$$$. Taking limits on each side completes the proof.

Definition I.58

A vector space $$$V$$$ is the direct sum of two subspaces $$$U$$$ and $$$W$$$, which we write as $$$V= U\oplus W$$$, if each vector $$$v$$$ in $$$V$$$ can be written uniquely as the sum of a vector $$$x$$$ in $$$U$$$ and a vector $$$y$$$ in $$$W$$$.

Theorem I.59

For any closed subspace $$$W$$$ of a Hilbert space $$$H$$$, we have $$$H= W\oplus {W}^{\perp}$$$.

Proof. Take an element $$$x$$$ in the Hilbert space $$$H$$$. By Theorem I.49 there exists a point $$$y$$$ in $$$W$$$ nearest to $$$x$$$. In particular, $$$\mathopen{}\left\lVert{}x-y\right\rVert\mathclose{}= \min_{w\in W}{}\mathopen{}\left\lVert{}x-w\right\rVert\mathclose{}$$$. We claim the orthogonal decomposition we seek is $$$x= y+\mathopen{}\left(x-y\right)\mathclose{}$$$.

We must first show that $$$x-y$$$ is in $$${W}^{\perp}$$$. Indeed, for $$$w\in W$$$ and $$$ε\gt 0$$$, $${\mathopen{}\left\lVert{}x-y+εw\right\rVert\mathclose{}}^{2}= {\mathopen{}\left\lVert{}x-y\right\rVert\mathclose{}}^{2}+2ε\Re\mathopen{}\left( \mathopen{}\left\langle{}x-y, w\right\rangle\mathclose{}\right)\mathclose{}+{ε}^{2}{\mathopen{}\left\lVert{}w\right\rVert\mathclose{}}^{2}\geq {\mathopen{}\left\lVert{}x-y\right\rVert\mathclose{}}^{2}$$ since $$$y-εw$$$ is in $$$W$$$. So, for $$$ε\gt 0$$$, $$$2ε\Re\mathopen{}\left( \mathopen{}\left\langle{}x-y, w\right\rangle\mathclose{}\right)\mathclose{}+{ε}^{2}{\mathopen{}\left\lVert{}w\right\rVert\mathclose{}}^{2}\geq 0$$$. Dividing by $$$ε$$$ gives $$$2\Re\mathopen{}\left( \mathopen{}\left\langle{}x-y, w\right\rangle\mathclose{}\right)\mathclose{}+ε{\mathopen{}\left\lVert{}w\right\rVert\mathclose{}}^{2}\geq 0$$$. Thus $$$\Re\mathopen{}\left( \mathopen{}\left\langle{}x-y, w\right\rangle\mathclose{}\right)\mathclose{}\geq 0$$$ for all $$$w\in W$$$. Repeating the same argument with $$${-}w$$$ in place of $$$w$$$ leads to the conclusion $$$\Re\mathopen{}\left( \mathopen{}\left\langle{}x-y, w\right\rangle\mathclose{}\right)\mathclose{}\leq 0$$$ for all $$$w\in W$$$, meaning, in fact, $$$\Re\mathopen{}\left( \mathopen{}\left\langle{}x-y, w\right\rangle\mathclose{}\right)\mathclose{}= 0$$$ for all $$$w\in W$$$. Repeating the above with $$$\mathrm{i}w$$$ and $$${-}\mathrm{i}w$$$ shows that $$$\mathopen{}\left\langle{}x-y, w\right\rangle\mathclose{}= 0$$$ for all $$$w\in W$$$. Thus, $$$x-y$$$ is in $$${W}^{\perp}$$$.

By Exercise I.56, the decomposition $$$x= y+\mathopen{}\left(x-y\right)\mathclose{}$$$ is unique (since two different decompositions would give you something non-zero in $$$W\cap {W}^{\perp}$$$). Thus $$$H= W\oplus {W}^{\perp}$$$.

Exercise I.60

For $$$S$$$ a subset of an inner product space $$$V$$$, $$${S}^{\perp}= {\operatorname{span}\mathopen{}\left( S\right)\mathclose{}}^{\perp}= {\overline{\operatorname{span}\mathopen{}\left( S\right)\mathclose{}}}^{\perp}$$$.

Proposition I.61

For a subset $$$S$$$ of a Hilbert space $$$H$$$, $$${\mathopen{}\left( {S}^{\perp}\right)\mathclose{}}^{\perp}= \overline{\operatorname{span}\mathopen{}\left( S\right)\mathclose{}}$$$.

Proof. Let $$$W= \overline{\operatorname{span}\mathopen{}\left( S\right)\mathclose{}}$$$ so that $$$W$$$ is a closed subspace of $$$H$$$. We claim that $$$W= {\mathopen{}\left( {W}^{\perp}\right)\mathclose{}}^{\perp}$$$. Indeed, $$$W\subseteq {\mathopen{}\left( {W}^{\perp}\right)\mathclose{}}^{\perp}$$$ by Exercise I.55. Using Theorem I.59, for $$$x\in {\mathopen{}\left( {W}^{\perp}\right)\mathclose{}}^{\perp}$$$, write $$$x= y+w$$$ where $$$w\in W$$$ and $$$y\in {W}^{\perp}$$$. Then $$$\mathopen{}\left\langle{}x, y\right\rangle\mathclose{}= 0$$$, and so $$\mathopen{}\left\langle{}x, y\right\rangle\mathclose{}= \mathopen{}\left\langle{}w+y, y\right\rangle\mathclose{}= \mathopen{}\left\langle{}w, y\right\rangle\mathclose{}+\mathopen{}\left\langle{}y, y\right\rangle\mathclose{}= 0+\mathopen{}\left\langle{}y, y\right\rangle\mathclose{} \text{.}$$ So $$$\mathopen{}\left\langle{}y, y\right\rangle\mathclose{}= 0$$$ and thus $$$y= 0$$$. Hence $$$x= w$$$, so $$$x\in W$$$, and thus $$$W= {\mathopen{}\left( {W}^{\perp}\right)\mathclose{}}^{\perp}$$$. Finally, since $$${W}^{\perp}= {S}^{\perp}$$$ by Exercise I.60, we have $$${\mathopen{}\left( {S}^{\perp}\right)\mathclose{}}^{\perp}= \overline{\operatorname{span}\mathopen{}\left( S\right)\mathclose{}}$$$.