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. Tensor Notation

For $$$x$$$ and $$$y$$$ in $$$H$$$, define $$$x\otimes y : H \to H$$$ by $$$\mathopen{}\left(x\otimes y\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}= \mathopen{}\left\langle{}ξ, y\right\rangle\mathclose{}x$$$. Then $$$\mathopen{}\left\lVert{}\mathopen{}\left(x\otimes y\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}ξ\right\rVert\mathclose{}\mathopen{}\left\lVert{}y\right\rVert\mathclose{}\mathopen{}\left\lVert{}x\right\rVert\mathclose{}$$$ and so $$$\mathopen{}\left\lVert{}\mathopen{}\left(x\otimes y\right)\mathclose{}\mathopen{}\left( y\right)\mathclose{}\right\rVert\mathclose{}= {\mathopen{}\left\lVert{}y\right\rVert\mathclose{}}^{2}\mathopen{}\left\lVert{}x\right\rVert\mathclose{}$$$. Thus $$$x\otimes y\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$, and $$$\mathopen{}\left\lVert{}x\otimes y\right\rVert\mathclose{}= \mathopen{}\left\lVert{}x\right\rVert\mathclose{}\mathopen{}\left\lVert{}y\right\rVert\mathclose{}$$$. Finally, $$\mathopen{}\left\langle{}\mathopen{}\left(x\otimes y\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}, ν\right\rangle\mathclose{}= \mathopen{}\left\langle{}\mathopen{}\left\langle{}y, ξ\right\rangle\mathclose{}x, ν\right\rangle\mathclose{}= \mathopen{}\left\langle{}y, ξ\right\rangle\mathclose{}\mathopen{}\left\langle{}x, y\right\rangle\mathclose{} \text{,}$$ and $$\mathopen{}\left\langle{}ξ, \mathopen{}\left(y\otimes x\right)\mathclose{}\mathopen{}\left( ν\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{}ξ, \mathopen{}\left\langle{}ν, x\right\rangle\mathclose{}y\right\rangle\mathclose{}= \mathopen{}\left\langle{}x, y\right\rangle\mathclose{}\mathopen{}\left\langle{}y, ξ\right\rangle\mathclose{} \text{.}$$Thus $$$\mathopen{}\left(x\otimes y\right)\mathclose{}^{*}= y\otimes x$$$.

For $$$T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$, we have $$$T\mathopen{}\left(x\otimes y\right)\mathclose{}= T\mathopen{}\left( x\right)\mathclose{}\otimes y$$$ and $$$\mathopen{}\left(\mathopen{}\left(x\otimes y\right)\mathclose{}T\right)\mathclose{}^{*}= T^{*}\mathopen{}\left( y\otimes x\right)\mathclose{}= T^{*}\mathopen{}\left( y\right)\mathclose{}\otimes x= \mathopen{}\left(x\otimes T^{*}\mathopen{}\left( y\right)\mathclose{}\right)\mathclose{}^{*}$$$. So, $$$\mathopen{}\left(x\otimes y\right)\mathclose{}\mathopen{}\left( T\right)\mathclose{}= x\otimes T^{*}\mathopen{}\left( y\right)\mathclose{}$$$. In particular, $$$\mathopen{}\left({x}_{1}\otimes {y}_{1}\right)\mathclose{}\mathopen{}\left( {x}_{2}\otimes {y}_{2}\right)\mathclose{}= \mathopen{}\left\langle{}{x}_{2}, {y}_{1}\right\rangle\mathclose{}\mathopen{}\left({x}_{1}\otimes {y}_{2}\right)\mathclose{}$$$. Notice for $$$y\neq 0$$$, $$$\operatorname{Ran}\mathopen{}\left( x\otimes y\right)\mathclose{}= \mathbb{C}x$$$.

Every $$$T\in \mathcal{F}\mathopen{}\left( H\right)\mathclose{}$$$ can be written $$$T= \sum_{i=1}^{n}{} {x}_{i}\otimes {y}_{i}$$$. Indeed, let $$$\mathopen{}\left\{\, {e}_{1}, \dotsc, {e}_{n}\,\right\}\mathclose{}$$$ be an orthonormal basis for $$$T\mathopen{}\left( H\right)\mathclose{}$$$. Then $$T\mathopen{}\left( ξ\right)\mathclose{}= \sum_{i=1}^{n}{} \mathopen{}\left\langle{}T\mathopen{}\left( ξ\right)\mathclose{}, {e}_{i}\right\rangle\mathclose{}{e}_{i} = \sum_{i=1}^{n}{} \mathopen{}\left\langle{}ξ, T^{*}\mathopen{}\left( {e}_{i}\right)\mathclose{}\right\rangle\mathclose{}{e}_{i} = \sum_{i=1}^{n}{} \mathopen{}\left({e}_{i}\otimes T^{*}\mathopen{}\left( {e}_{i}\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( ξ\right)\mathclose{}$$ Easily, then $$$\mathcal{F}\mathopen{}\left( H\right)\mathclose{}= H\odot \overline{H}$$$ where $$$\overline{H}= H$$$ with conjugate scalar multiplication.

Until further notice, $$$H$$$ is separable and infinite dimensional.

Notice that if $$$\mathopen{}\left\{\, {e}_{1}, \dotsc, {e}_{n}\,\right\}\mathclose{}$$$ is a finite orthonormal set, then $$$P= \sum_{i=1}^{n}{} {e}_{i}\otimes {e}_{i}$$$ is the projection of $$$H$$$ on $$$\operatorname{span}\mathopen{}\left( {e}_{1}, \dotsc, {e}_{n}\right)\mathclose{}$$$. In particular, $$$P\mathopen{}\left( ξ\right)\mathclose{}= \sum_{i=1}^{n}{} \mathopen{}\left\langle{}ξ, {e}_{i}\right\rangle\mathclose{}{e}_{i}$$$. The spectral theorem for self-adjoint compact operators says if $$$A= A^{*}\in \mathcal{K}\mathopen{}\left( H\right)\mathclose{}$$$, then there exist real $$${λ}_{n}$$$ with $$${λ}_{n} \to 0$$$ and an orthonormal basis $$$\mathopen{}\left\{\, {e}_{1}, {e}_{2}, \dotsc\,\right\}\mathclose{}$$$ for $$$H$$$ such that $$$A= \sum_{n=1}^{\infty}{} λ{e}_{n}\otimes {e}_{n}$$$ (norm-convergent series). In this case $$$A\mathopen{}\left( {e}_{j}\right)\mathclose{}= {λ}_{j}{e}_{j}$$$.

Example III.18

Let $$$H$$$ be the space $$$\mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}$$$. Let $$${ξ}_{0}\mathopen{}\left( x\right)\mathclose{}= 1$$$ and $$${ξ}_{1}\mathopen{}\left( x\right)\mathclose{}= x$$$. Let $$$T$$$ be the Volterra operator, defined by $$$\mathopen{}\left(T\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( t\right)\mathclose{}= \int _{0}^{t}{}f\mathopen{}\left( x\right)\mathclose{}\,\mathrm{d}x$$$. Recall $$$\mathopen{}\left( T^{*}\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( t\right)\mathclose{}= \int _{t}^{1}{}f\mathopen{}\left( x\right)\mathclose{}\,\mathrm{d}x$$$. So $$$\mathopen{}\left(\mathopen{}\left(T+ T^{*}\right)\mathclose{}\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( t\right)\mathclose{}= \int _{0}^{1}{}f\mathopen{}\left( x\right)\mathclose{}\,\mathrm{d}x= \mathopen{}\left\langle{}f, {ξ}_{0}\right\rangle\mathclose{}$$$. Then $$$\mathopen{}\left(T+ T^{*}\right)\mathclose{}\mathopen{}\left( f\right)\mathclose{}= \mathopen{}\left\langle{}f, {ξ}_{0}\right\rangle\mathclose{}{ξ}_{0}$$$ and $$$T+ T^{*}= {ξ}_{0}\otimes {ξ}_{0}$$$. Define $$$A= {-}{T}^{2}+{ξ}_{1}\otimes \mathopen{}\left({ξ}_{0}-{ξ}_{1}\right)\mathclose{}$$$, so $$$A\in \mathcal{K}\mathopen{}\left( H\right)\mathclose{}$$$. We claim that $$$A= A^{*}$$$. $${\mathopen{}\left( T^{*}\right)\mathclose{}}^{2}= {\mathopen{}\left({ξ}_{0}\otimes {ξ}_{0}-T\right)\mathclose{}}^{2}= {\mathopen{}\left({ξ}_{0}\otimes {ξ}_{0}\right)\mathclose{}}^{2}-\mathopen{}\left({ξ}_{0}\otimes {ξ}_{0}\right)\mathclose{}T-T\mathopen{}\left({ξ}_{0}\otimes {ξ}_{0}\right)\mathclose{}+{T}^{2}= {ξ}_{0}\otimes {ξ}_{0}-{ξ}_{0}\otimes T^{*}\mathopen{}\left( {ξ}_{0}\right)\mathclose{}-T\mathopen{}\left( {ξ}_{0}\right)\mathclose{}\otimes {ξ}_{0}+{T}^{2}= {ξ}_{0}\otimes {ξ}_{0}-{ξ}_{0}\otimes \mathopen{}\left({ξ}_{0}-{ξ}_{1}\right)\mathclose{}-{ξ}_{1}\otimes {ξ}_{0}+{T}^{2}= {T}^{2}+{ξ}_{0}\otimes {ξ}_{1}-{ξ}_{1}\otimes {ξ}_{0} \text{.}$$ Thus $$${-} A^{*}= {\mathopen{}\left( T^{*}\right)\mathclose{}}^{2}-\mathopen{}\left({ξ}_{0}-{ξ}_{1}\right)\mathclose{}\otimes {ξ}_{1}= {ξ}_{1}\otimes {ξ}_{1}-{ξ}_{1}\otimes {ξ}_{0}+{T}^{2}= {ξ}_{1}\otimes \mathopen{}\left({ξ}_{1}-{ξ}_{0}\right)\mathclose{}+{T}^{2}= {-}A$$$. Consider eigenvalues $$$λ\in \mathbb{R}$$$, $$$A\mathopen{}\left( f\right)\mathclose{}= λ\mathopen{}\left( f\right)\mathclose{}$$$, $$$f\neq 0$$$. There are three cases.

1. ($$$λ\gt 0$$$) Since $$$T\mathopen{}\left( H\right)\mathclose{}\subseteq \mathrm{C}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}$$$, the fundamental theorem of calculus makes $$${T}^{2}\mathopen{}\left( f\right)\mathclose{}\in \mathrm{C}^{1}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}$$$ with $$$\mathopen{}\left({T}^{2}\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}' = T\mathopen{}\left( f\right)\mathclose{}$$$. Writing $$$λ= \frac{1}{{μ}^{2}}$$$, we have $$$f= {μ}^{2}A\mathopen{}\left( f\right)\mathclose{}$$$, and thus $$$f\in \mathrm{C}^{1}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}$$$. It follows that $$$T\mathopen{}\left( f\right)\mathclose{}\in \mathrm{C}^{1}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}$$$, with $$$\mathopen{}\left(T\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}' = f$$$. Further $$f' = {μ}^{2}\mathopen{}\left(A\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}' = {μ}^{2}\mathopen{}\left({-}T\mathopen{}\left( f\right)\mathclose{}+\mathopen{}\left\langle{}f, {ξ}_{0}-{ξ}_{1}\right\rangle\mathclose{}\right)\mathclose{} \text{,}$$ so $$$f' \in \mathrm{C}^{1}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}$$$ and $$f'' = {μ}^{2}\mathopen{}\left({-}f+0\right)\mathclose{}= {-}{μ}^{2}f \text{.}$$ We claim that $$$f\mathopen{}\left( 0\right)\mathclose{}= 0= f\mathopen{}\left( 1\right)\mathclose{}$$$, which amounts to showing $$$\mathopen{}\left(A\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( 0\right)\mathclose{}= 0= \mathopen{}\left(A\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( 1\right)\mathclose{}$$$. Indeed, from $$$\mathopen{}\left(A\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( t\right)\mathclose{}= {-} \mathopen{}\left({T}^{2}\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( t\right)\mathclose{} +\mathopen{}\left\langle{}f, {ξ}_{0}-{ξ}_{1}\right\rangle\mathclose{}t$$$, we immediately get $$$\mathopen{}\left(A\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( 0\right)\mathclose{}= 0$$$, and at the other end $$A\mathopen{}\left( f\mathopen{}\left( 1\right)\mathclose{}\right)\mathclose{}= {-} \mathopen{}\left({T}^{2}\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( 1\right)\mathclose{} +\mathopen{}\left\langle{}f, {ξ}_{0}-{ξ}_{1}\right\rangle\mathclose{}= {-} \int _{0}^{1}{} \mathopen{}\left(T\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{} \,\mathrm{d}x +\mathopen{}\left\langle{}f, {ξ}_{0}-{ξ}_{1}\right\rangle\mathclose{}= {-} \mathopen{}\left\langle{}T\mathopen{}\left( f\right)\mathclose{}, {ξ}_{0}\right\rangle\mathclose{} +\mathopen{}\left\langle{}f, {ξ}_{0}-{ξ}_{1}\right\rangle\mathclose{}= {-} \mathopen{}\left\langle{}f, T^{*}\mathopen{}\left( {ξ}_{0}\right)\mathclose{}\right\rangle\mathclose{} +\mathopen{}\left\langle{}f, {ξ}_{0}-{ξ}_{1}\right\rangle\mathclose{}= 0 \text{.}$$ The general solution to $$$f'' +{μ}^{2}f= 0$$$ is $$$f\mathopen{}\left( x\right)\mathclose{}= a\cos\mathopen{}\left( μx\right)\mathclose{}+b\sin\mathopen{}\left( μx\right)\mathclose{}$$$. The endpoint condition $$$f\mathopen{}\left( 0\right)\mathclose{}= 0$$$ makes $$$a= 0$$$ (and thus $$$b\neq 0$$$). From $$$f\mathopen{}\left( 1\right)\mathclose{}= 0$$$, we deduce that $$$μ$$$ is a nonzero integer multiple of $$$π$$$. We conclude that the positive eigenvalues of $$$A$$$ are $$${λ}_{n}= \frac{1}{{n}^{2}{π}^{2}}$$$ with normalized eigenfunctions $$${e}_{n}= {e}_{n}\mathopen{}\left( x\right)\mathclose{}= \sqrt{2}\sin\mathopen{}\left( nπx\right)\mathclose{}$$$, for $$$n$$$ in $$$\mathopen{}\left\{\, 1, 2, \dotsc\,\right\}\mathclose{}$$$.
2. ($$$λ= 0$$$) $$$\operatorname{Ker}\mathopen{}\left( A\right)\mathclose{}= 0$$$ because $$$A\mathopen{}\left( f\right)\mathclose{}= 0$$$ implies $$${T}^{2}\mathopen{}\left( f\right)\mathclose{}= c{ξ}_{1}$$$, which gives $$$T\mathopen{}\left( f\right)\mathclose{}= c{ξ}_{0}$$$. But then $$$T\mathopen{}\left( f\mathopen{}\left( 0\right)\mathclose{}\right)\mathclose{}= 0$$$, so $$$c= 0$$$.
3. ($$$λ\lt 0$$$) Write $$$λ= {-} \frac{1}{{μ}^{2}}$$$. $$$A\mathopen{}\left( f\right)\mathclose{}= {-} \frac{1}{{μ}^{2}} f$$$ gives $$$f\mathopen{}\left( x\right)\mathclose{}= α{\mathrm{e}}^{μx}+β{\mathrm{e}}^{{-}μx}$$$ (as before) and $$$f\mathopen{}\left( 0\right)\mathclose{}= 0= f\mathopen{}\left( 1\right)\mathclose{}$$$ forces $$$α= 0= β$$$.
We have diagonalized the operator $$$A$$$. That is, the $$${e}_{n}$$$'s form an orthonormal basis for $$$\mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}$$$ (a fact we have already demonstrated less gracefully in an earlier example), and $$A= \frac{1}{{π}^{2}}\sum_{n=1}^{\infty}{} \frac{1}{{n}^{2}}{e}_{n}\otimes {e}_{n} \text{.}$$ Notice that this makes $$$\mathopen{}\left\lVert{}A\right\rVert\mathclose{}= ρ\mathopen{}\left( A\right)\mathclose{}= \frac{1}{{π}^{2}}$$$, which is not obvious from the definition of $$$A$$$.