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

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 $$$M= H$$$.