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

## F. Self-Adjoint Operators

Call an operator $$$A\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ self-adjoint if $$$A= A^{*}$$$. There are lots of self-adjoint operators, such as $$$T^{*}T$$$, $$$T+ T^{*}$$$, and $$$\mathrm{i}\mathopen{}\left(T- T^{*}\right)\mathclose{}$$$ for any $$$T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$. The main results in this section are that every self-adjoint operator has real spectrum, and that the polynomial calculus for a self-adjoint operator extends to a functional calculus for arbitrary complex continuous functions on the spectrum of the operator. For the real spectrum result, we will need the following lemma.

Lemma II.53

Let $$$T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$. If there exists $$$r\gt 0$$$ such that $$$\mathopen{}\left\lvert{}\mathopen{}\left\langle{}T\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}\right\rvert\mathclose{}\geq r{\mathopen{}\left\lVert{}x\right\rVert\mathclose{}}^{2}$$$ for all $$$x\in H$$$, then $$$T$$$ is invertible.

Proof. By Cauchy-Schwartz, $$$\mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}\mathopen{}\left\lVert{}x\right\rVert\mathclose{}\geq r{\mathopen{}\left\lVert{}x\right\rVert\mathclose{}}^{2}$$$ so $$$\mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}\geq r\mathopen{}\left\lVert{}x\right\rVert\mathclose{}$$$ for all $$$x$$$. It then follows that the range of $$$T$$$ is closed: $$$T\mathopen{}\left( {x}_{n}\right)\mathclose{} \to y$$$ implies that $$$\mathopen{}\left({x}_{n}\right)\mathclose{}$$$ is Cauchy, as $$$\mathopen{}\left\lVert{}{x}_{n}-{x}_{m}\right\rVert\mathclose{}\leq \frac{1}{r}\mathopen{}\left\lVert{}T\mathopen{}\left( {x}_{n}\right)\mathclose{}-T\mathopen{}\left( {x}_{m}\right)\mathclose{}\right\rVert\mathclose{}$$$, which implies $$${x}_{n} \to x$$$, which implies $$$y= T\mathopen{}\left( x\right)\mathclose{}$$$. The range of $$$T$$$ is dense because if $$$w\in {T\mathopen{}\left( H\right)\mathclose{}}^{\perp}$$$, then $$$\mathopen{}\left\langle{}w, T\mathopen{}\left( x\right)\mathclose{}\right\rangle\mathclose{}= 0$$$ for all $$$x$$$ implies $$$\mathopen{}\left\langle{} T^{*}\mathopen{}\left( w\right)\mathclose{}, x\right\rangle\mathclose{}= 0$$$ so $$$T^{*}\mathopen{}\left( w\right)\mathclose{}= 0$$$, and $$$r{\mathopen{}\left\lVert{}w\right\rVert\mathclose{}}^{2}\leq \mathopen{}\left\lvert{}\mathopen{}\left\langle{}T\mathopen{}\left( w\right)\mathclose{}, w\right\rangle\mathclose{}\right\rvert\mathclose{}= \mathopen{}\left\lvert{}\mathopen{}\left\langle{}w, T^{*}\mathopen{}\left( w\right)\mathclose{}\right\rangle\mathclose{}\right\rvert\mathclose{}= 0$$$. It follows that $$$T\mathopen{}\left( H\right)\mathclose{}= H$$$. Further, $$$T\mathopen{}\left( x\right)\mathclose{}= 0$$$ implies $$$r{\mathopen{}\left\lVert{}x\right\rVert\mathclose{}}^{2}= 0$$$ implies $$$x= 0$$$. $$$T$$$ is one-to-one, so it has an inverse map. Get the linear inverse $$${T}^{-1} : H \to H$$$, which is bounded because $$$r\mathopen{}\left\lVert{}{T}^{-1}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}T\mathopen{}\left( {T}^{-1}\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}x\right\rVert\mathclose{}$$$. So $$${T}^{-1}\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ and $$$\mathopen{}\left\lVert{}{T}^{-1}\right\rVert\mathclose{}\leq \frac{1}{r}$$$.

Proposition II.54

An operator $$$A\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ is self-adjoint if and only if $$$\mathopen{}\left\langle{}A\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}$$$ is real for all $$$x\in H$$$.

Proof.

1. (⇒) $$$\mathopen{}\left\langle{}A\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}= \mathopen{}\left\langle{}x, A\mathopen{}\left( x\right)\mathclose{}\right\rangle\mathclose{}$$$ implies $$$\mathopen{}\left\langle{}A\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}= \overline{\mathopen{}\left\langle{}A\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}}$$$.
2. (⇐) $$$\mathopen{}\left\langle{}A\mathopen{}\left( x+y\right)\mathclose{}, x+y\right\rangle\mathclose{}= \mathopen{}\left\langle{}A\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}+\mathopen{}\left\langle{}A\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{}+\mathopen{}\left\langle{}A\mathopen{}\left( y\right)\mathclose{}, x\right\rangle\mathclose{}+\mathopen{}\left\langle{}A\mathopen{}\left( y\right)\mathclose{}, y\right\rangle\mathclose{}\in \mathbb{R}$$$ by hypothesis. Then $$$\mathopen{}\left\langle{}A\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{}+\mathopen{}\left\langle{}A\mathopen{}\left( y\right)\mathclose{}, x\right\rangle\mathclose{}\in \mathbb{R}$$$ so $$$\Im\mathopen{}\left( \mathopen{}\left\langle{}A\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{}\right)\mathclose{}= {-} \Im\mathopen{}\left( \mathopen{}\left\langle{}A\mathopen{}\left( y\right)\mathclose{}, x\right\rangle\mathclose{}\right)\mathclose{}$$$. Replacing $$$y$$$ by $$$\mathrm{i}y$$$ makes the element $$${-}\mathrm{i}\mathopen{}\left\langle{}A\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{}+\mathrm{i}\mathopen{}\left\langle{}A\mathopen{}\left( y\right)\mathclose{}, x\right\rangle\mathclose{}\in \mathbb{R}$$$, which implies $$$\Re\mathopen{}\left( \mathopen{}\left\langle{}A\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{}\right)\mathclose{}= \Re\mathopen{}\left( \mathopen{}\left\langle{}A\mathopen{}\left( y\right)\mathclose{}, x\right\rangle\mathclose{}\right)\mathclose{}$$$. Hence $$$\overline{ \mathopen{}\left\langle{}A\mathopen{}\left( y\right)\mathclose{}, x\right\rangle\mathclose{} }= \mathopen{}\left\langle{}A\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{}$$$ implies $$$\mathopen{}\left\langle{}A\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{}= \mathopen{}\left\langle{}x, A\mathopen{}\left( y\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{} A^{*}\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{}$$$ implies $$$\mathopen{}\left\langle{}\mathopen{}\left(A- A^{*}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{}= 0$$$.

Proposition II.55

$$$A= A^{*}\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ implies $$$\mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\subseteq \mathbb{R}$$$.

Proof. Let $$$λ= α+\mathrm{i}β$$$ for $$$α$$$ and $$$β$$$ in $$$\mathbb{R}$$$, with $$$β\neq 0$$$. $$$\mathopen{}\left\lvert{}\mathopen{}\left\langle{}\mathopen{}\left(λ-A\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}\right\rvert\mathclose{}= \mathopen{}\left\lvert{}\mathopen{}\left\langle{}\mathopen{}\left(α-A\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}, x\right\rangle\mathclose{}+\mathrm{i}β\mathopen{}\left\langle{}x, x\right\rangle\mathclose{}\right\rvert\mathclose{}\geq \mathopen{}\left\lvert{}β\right\rvert\mathclose{}\mathopen{}\left\langle{}x, x\right\rangle\mathclose{}= \mathopen{}\left\lvert{}β\right\rvert\mathclose{}{\mathopen{}\left\lVert{}x\right\rVert\mathclose{}}^{2}$$$ so $$$λ-A\in { \mathcal{L}\mathopen{}\left( X\right)\mathclose{} }^{-1}$$$ by Lemma II.53.

Proposition II.56

Let $$$T$$$ be a normal operator and let $$$p$$$ be a polynomial with complex coefficients. Then $$$p\mathopen{}\left( T\right)\mathclose{}$$$ is normal, and $$$\mathopen{}\left\lVert{}p\mathopen{}\left( T\right)\mathclose{}\right\rVert\mathclose{}= \max_{λ\in \mathop{\sigma}\mathopen{}\left( T\right)\mathclose{}}{}\mathopen{}\left\lvert{}p\mathopen{}\left( λ\right)\mathclose{}\right\rvert\mathclose{}$$$.

Proof. Notice that $$$p\mathopen{}\left( T\right)\mathclose{}^{*}= \overline{p}\mathopen{}\left( T^{*}\right)\mathclose{}$$$, where $$$\overline{p}$$$ is the polynomial obtained from $$$p$$$ by conjugating its coefficients. Any polynomial in $$$T$$$ commutes with any polynomial in $$$T^{*}$$$ because $$${T}^{j}$$$ commutes with $$${\mathopen{}\left( T^{*}\right)\mathclose{}}^{k}$$$ for all $$$j$$$ and $$$k$$$ greater than or equal to $$$0$$$. Thus $$$p\mathopen{}\left( T\right)\mathclose{}$$$ is normal. The formula for $$$\mathopen{}\left\lVert{}p\mathopen{}\left( T\right)\mathclose{}\right\rVert\mathclose{}$$$ now follows from Proposition II.44 and Proposition II.52.

Now for the functional calculus for self-adjoint operators $$$A= A^{*}\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$. Let $$$P$$$ be the subalgebra of $$$\mathrm{C}\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}$$$ consisting of polynomials. Then $$$P$$$ is point-separating and nowhere-vanishing on $$$\mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}$$$. It is furthermore self-adjoint as a subalgebra because $$$\mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}$$$ is real. By the Stone-Weierstrass Theorem, then, $$$P$$$ is uniformly dense in $$$\mathrm{C}\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}$$$. By Proposition II.56, the map $$$p\mapsto p\mathopen{}\left( A\right)\mathclose{}$$$ is a linear isometry from $$$P$$$ (with the uniform norm) into $$$\mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$. In fact, it's an isometric *-algebra homomorphism, i.e. $$$p\mathopen{}\left( A\right)\mathclose{}^{*}= \overline{p}\mathopen{}\left( A\right)\mathclose{}$$$ and $$$\mathopen{}\left(pq\right)\mathclose{}\mathopen{}\left( A\right)\mathclose{}= p\mathopen{}\left( A\right)\mathclose{}q\mathopen{}\left( A\right)\mathclose{}$$$. Thus we have the following:

Theorem II.57

Let $$$A= A^{*}$$$. There is an isometric *-isomorphism $$$Φ : \mathrm{C}\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{} \to \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ such that $$$Φ\mathopen{}\left( p\right)\mathclose{}= p\mathopen{}\left( A\right)\mathclose{}$$$ for all $$$p\in P$$$.

Proof. As with any isometry mapping a dense subset of a metric space into a complete metric space, we obtain $$$Φ$$$ unambiguously by defining $$$Φ\mathopen{}\left( f\right)\mathclose{}= \lim_{n}{}{p}_{n}\mathopen{}\left( A\right)\mathclose{}$$$ for any sequence $$$\mathopen{}\left({p}_{n}\right)\mathclose{}$$$ converging uniformly to $$$f$$$ on $$$\mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}$$$. That $$$Φ$$$ is linear and multiplicative follows from taking limits along sequences of polynomials. Likewise $$$Φ\mathopen{}\left( f\right)\mathclose{}^{*}= Φ\mathopen{}\left( \overline{f}\right)\mathclose{}$$$ for $$$f$$$ in $$$\mathrm{C}\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}$$$ because $$$A= A^{*}$$$ and therefore $$$p\mathopen{}\left( A\right)\mathclose{}^{*}= \overline{p}\mathopen{}\left( A\right)\mathclose{}$$$ for every polynomial $$$p$$$.

Better notation: Write $$$f\mathopen{}\left( A\right)\mathclose{}= Φ\mathopen{}\left( f\right)\mathclose{}$$$ for $$$f\in \mathrm{C}\mathopen{}\left( \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\right)\mathclose{}$$$.

Lemma II.58

For $$$A= A^{*}\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$, $$${λ}_{1}\neq {λ}_{2}\in \mathbb{R}$$$, $$$\operatorname{Ker}\mathopen{}\left( {λ}_{1}-A\right)\mathclose{}\perp \operatorname{Ker}\mathopen{}\left( {λ}_{2}-A\right)\mathclose{}$$$.

Proof. $$$A\mathopen{}\left( {x}_{i}\right)\mathclose{}= {λ}_{i}{x}_{i}$$$ means $${λ}_{1}\mathopen{}\left\langle{}{x}_{1}, {x}_{2}\right\rangle\mathclose{}= \mathopen{}\left\langle{}A\mathopen{}\left( {x}_{1}\right)\mathclose{}, {x}_{2}\right\rangle\mathclose{}= \mathopen{}\left\langle{}{x}_{1}, A\mathopen{}\left( {x}_{2}\right)\mathclose{}\right\rangle\mathclose{}= {λ}_{2}\mathopen{}\left\langle{}{x}_{1}, {x}_{2}\right\rangle\mathclose{}$$ so that $$$\mathopen{}\left\langle{}{x}_{1}, {x}_{2}\right\rangle\mathclose{}= 0$$$.

Lemma II.59

Let $$$M$$$ be a closed invariant subspace for $$$A= A^{*}\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$. Then

1. $$$\mathopen{}\left(A|M\right)\mathclose{}^{*}= A|M$$$.
2. $$$A\mathopen{}\left( {M}^{\perp}\right)\mathclose{}\subseteq {M}^{\perp}$$$.
3. $$$\mathopen{}\left(A|{M}^{\perp}\right)\mathclose{} ^{*}= A|{M}^{\perp}$$$.

Proof.

1. $$$\mathopen{}\left\langle{}A\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{}= \mathopen{}\left\langle{}x, A\mathopen{}\left( y\right)\mathclose{}\right\rangle\mathclose{}$$$ for all $$$x$$$ and $$$y$$$ in $$$M$$$.
2. $$$x\in {M}^{\perp}$$$, $$$y\in M$$$ implies $$$\mathopen{}\left\langle{}A\mathopen{}\left( x\right)\mathclose{}, y\right\rangle\mathclose{}= \mathopen{}\left\langle{}x, A\mathopen{}\left( y\right)\mathclose{}\right\rangle\mathclose{}= 0$$$ (since $$$M$$$ invariant means that $$$y\in M$$$, $$$A\mathopen{}\left( y\right)\mathclose{}\in M$$$).
3. Follows from the two items above.

Theorem II.60

Let $$$A= A^{*}\in \mathcal{K}\mathopen{}\left( H\right)\mathclose{}$$$. Then $$$H$$$ has an orthonormal basis consisting of eigenvectors for $$$A$$$.

Proof. Let $$$M$$$ be the closed span of $$$\operatorname{Ker}\mathopen{}\left( A\right)\mathclose{}$$$ and $$$\bigcup_{ λ\in \mathop{\sigma}\mathopen{}\left( A\right)\mathclose{}\setminus \mathopen{}\left\{\, 0\,\right\}\mathclose{} }{} \operatorname{Ker}\mathopen{}\left( λ-A\right)\mathclose{}$$$, which by Lemma II.58 plainly does have an orthonormal basis of $$$A$$$-eigenvectors. Notice $$$A\mathopen{}\left( M\right)\mathclose{}\subseteq M$$$. It follows from Lemma II.59 that $$$A|{M}^{\perp}$$$ is a self-adjoint compact operator on $$${M}^{\perp}$$$. Suppose that $$${M}^{\perp}\neq 0$$$. Then $$$A|{M}^{\perp}\neq 0$$$ (else $$${M}^{\perp}\subseteq \operatorname{Ker}\mathopen{}\left( A\right)\mathclose{}\subseteq M$$$). Now $$$\mathopen{}\left\lVert{}A|{M}^{\perp}\right\rVert\mathclose{}= ρ\mathopen{}\left( A|{M}^{\perp}\right)\mathclose{}$$$, so $$$A|{M}^{\perp}\neq 0$$$ implies $$$A|{M}^{\perp}$$$ has a nonzero eigenvalue $$$μ$$$. But $$$\operatorname{Ker}\mathopen{}\left( μ-A\right)\mathclose{}\subseteq M$$$, so $$$\operatorname{Ker}\mathopen{}\left( μ-A\right)\mathclose{}\cap {M}^{\perp}= \mathopen{}\left\{\, 0\,\right\}\mathclose{}$$$, which gives a contradiction. We conclude that $$$M= H$$$.

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