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. Hilbert-Schmidt Operators

Definition III.44

An operator $$$T$$$ is said to be Hilbert-Schmidt if there exists an orthonormal basis $$$\mathopen{}\left({e}_{n}\right)\mathclose{}$$$ for $$$H$$$ such that $$\sum_{n}{} {\mathopen{}\left\lVert{}T\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rVert\mathclose{}}^{2} \lt \infty \text{.}$$ Denote by $$$\mathcal{HS}\mathopen{}\left( H\right)\mathclose{}$$$ the set of all Hilbert-Schmidt operators on $$$H$$$.

Remark III.45

1. $$$T$$$ is Hilbert-Schmidt if and only if $$$T^{*}\mathopen{}\left( T\right)\mathclose{}= {\mathopen{}\left\lvert{}T\right\rvert\mathclose{}}^{2}\in \mathcal{T}\mathopen{}\left( H\right)\mathclose{}$$$ since $$\sum_{n}{} {\mathopen{}\left\lVert{}T\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rVert\mathclose{}}^{2} = \sum_{n}{} \mathopen{}\left\langle{}T\mathopen{}\left( {e}_{n}\right)\mathclose{}, T\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rangle\mathclose{} = \sum_{n}{} \mathopen{}\left\langle{} T^{*}\mathopen{}\left( T\mathopen{}\left( {e}_{n}\right)\mathclose{}\right)\mathclose{}, {e}_{n}\right\rangle\mathclose{} \text{.}$$
2. It follows from above and Proposition III.29 that $$$\mathcal{HS}\mathopen{}\left( H\right)\mathclose{}$$$ is an ideal of $$$\mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$.
3. $$$T\in \mathcal{HS}\mathopen{}\left( H\right)\mathclose{}$$$ implies $$$\sum{} {\mathopen{}\left\lVert{}T\mathopen{}\left( {f}_{n}\right)\mathclose{}\right\rVert\mathclose{}}^{2} \lt \infty$$$ for any orthonormal basis $$$\mathopen{}\left({f}_{n}\right)\mathclose{}$$$ of $$$H$$$.

Proposition III.46

$$$\mathcal{T}\mathopen{}\left( H\right)\mathclose{}\subseteq \mathcal{HS}\mathopen{}\left( H\right)\mathclose{}\subseteq \mathcal{K}\mathopen{}\left( H\right)\mathclose{}$$$.

Proof. $$$T\in \mathcal{T}\mathopen{}\left( H\right)\mathclose{}$$$ implies $$$T^{*}T\in \mathcal{T}\mathopen{}\left( H\right)\mathclose{}$$$, which gives $$$T\in \mathcal{HS}\mathopen{}\left( H\right)\mathclose{}$$$. Then $$${\mathopen{}\left\lvert{}T\right\rvert\mathclose{}}^{2}\in \mathcal{K}\mathopen{}\left( H\right)\mathclose{}$$$, which implies $$$\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\in \mathcal{K}\mathopen{}\left( H\right)\mathclose{}$$$ as an easy consequence of the spectral theorem for compact self-adjoint operators. Now polarize to get $$$T= V\mathopen{}\left\lvert{}T\right\rvert\mathclose{}$$$ for some $$$V\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$. Therefore, $$$T\in \mathcal{K}\mathopen{}\left( H\right)\mathclose{}$$$.

As expected, not all compact operators are Hilbert-Schmidt. An instance of this is the diagonal operator $$$D$$$ on $$$H$$$ given in terms of an orthonormal basis $$$\mathopen{}\left({e}_{n}\right)\mathclose{}$$$ by $$$D\mathopen{}\left( {e}_{n}\right)\mathclose{}= \frac{1}{\sqrt{n}}{e}_{n}$$$, since $$\sum{} {\mathopen{}\left\lVert{}D\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rVert\mathclose{}}^{2} = \sum{} {\mathopen{}\left\lvert{}\frac{1}{\sqrt{n}}\right\rvert\mathclose{}}^{2}{\mathopen{}\left\lVert{}{e}_{n}\right\rVert\mathclose{}}^{2} = \sum{} \frac{1}{n} = \infty \text{.}$$ Hence $$$D$$$ is not Hilbert-Schmidt but is compact since $$$\frac{1}{\sqrt{n}} \to 0$$$.

Example III.47

Here is a rich source of examples of Hilbert-Schmidt operators. Let $$$X$$$ be a measure space with measure $$$μ$$$. (Let us assume, for convenience, that $$$H= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( X\right)\mathclose{}$$$ is separable.) Fix $$$k\in \mathrm{L}^{\mathrm{2}}\mathopen{}\left( X\times X, μ\times μ\right)\mathclose{}$$$. Think $$$k= k\mathopen{}\left( t, s\right)\mathclose{}$$$. For $$$t\in X$$$, define the slice $$${k}_{t}$$$ of $$$k$$$ by $$${k}_{t}\mathopen{}\left( s\right)\mathclose{}= k\mathopen{}\left( t, s\right)\mathclose{}$$$. Then by Fubini's theorem each $$${k}_{t}\in H$$$ for almost every $$$t$$$ and $$\int _{X}{} {\mathopen{}\left\lVert{}{k}_{t}\right\rVert\mathclose{}}^{2} \,\mathrm{d}μ\mathopen{}\left( t\right)\mathclose{}= \int _{X\times X}{} {\mathopen{}\left\lvert{}k\right\rvert\mathclose{}}^{2} \,\mathrm{d} \mathopen{}\left(μ\times μ\right)\mathclose{} \lt \infty \text{.}$$ For $$$f\in H$$$, define the function $$$K\mathopen{}\left( f\right)\mathclose{}$$$ almost everywhere by $$\mathopen{}\left(K\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( t\right)\mathclose{}= \int _{X}{} k\mathopen{}\left( t, s\right)\mathclose{}f\mathopen{}\left( s\right)\mathclose{} \,\mathrm{d}μ\mathopen{}\left( s\right)\mathclose{}= \mathopen{}\left\langle{}f, \overline{{k}_{t}}\right\rangle\mathclose{} \text{.}$$ Then $$$K\mathopen{}\left( f\right)\mathclose{}$$$ is measurable (by Fubini). Further, $$\int _{X}{} {\mathopen{}\left\lvert{}\mathopen{}\left(K\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( t\right)\mathclose{}\right\rvert\mathclose{}}^{2} \leq \int _{X}{} {\mathopen{}\left\lVert{}f\right\rVert\mathclose{}}^{2}{\mathopen{}\left\lvert{}\overline{{k}_{t}}\right\rvert\mathclose{}}^{2} \,\mathrm{d}μ\mathopen{}\left( t\right)\mathclose{}= {\mathopen{}\left\lVert{}f\right\rVert\mathclose{}}^{2}\int _{X\times X}{} {\mathopen{}\left\lvert{}k\right\rvert\mathclose{}}^{2} \,\mathrm{d} \mathopen{}\left(μ\times μ\right)\mathclose{} \text{,}$$ so $$$K$$$ maps $$$H$$$ boundedly into itself, with $$$\mathopen{}\left\lVert{}K\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}k\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}\mathopen{}\left( X\times X\right)\mathclose{}}$$$. Let $$$\mathopen{}\left\{\, {e}_{1}, {e}_{2}, \dotsc\,\right\}\mathclose{}$$$ be an orthonormal basis for $$$H$$$. Then by the calculations above $$\sum_{n}{} {\mathopen{}\left\lVert{}K\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rVert\mathclose{}}^{2} = \sum_{n}{} \int _{X}{} {\mathopen{}\left\lvert{}\mathopen{}\left\langle{}{e}_{n}, \overline{{k}_{t}}\right\rangle\mathclose{}\right\rvert\mathclose{}}^{2} \,\mathrm{d}μ\mathopen{}\left( t\right)\mathclose{} = \int _{X}{} \sum_{n}{} {\mathopen{}\left\lvert{}\mathopen{}\left\langle{}{e}_{n}, \overline{{k}_{t}}\right\rangle\mathclose{}\right\rvert\mathclose{}}^{2} \,\mathrm{d}μ\mathopen{}\left( t\right)\mathclose{}= \int _{X}{} {\mathopen{}\left\lVert{}{k}_{t}\right\rVert\mathclose{}}^{2} \,\mathrm{d}μ\mathopen{}\left( t\right)\mathclose{}= {\mathopen{}\left(\mathopen{}\left\lVert{}k\right\rVert\mathclose{}_{\mathrm{L}^{\mathrm{2}}\mathopen{}\left( X\times X\right)\mathclose{}}\right)\mathclose{}}^{2} \text{.}$$ Thus $$$K$$$ is a Hilbert-Schmidt operator on $$$H$$$, and its Hilbert-Schmidt norm (to be defined shortly) coincides with the $$$\mathrm{L}^{\mathrm{2}}$$$ norm of the kernel function $$$k$$$.

We return to the general discussion of Hilbert-Schmidt operators. Let $$$\mathopen{}\left({e}_{n}\right)\mathclose{}$$$ be an orthonormal basis for $$$H$$$.

Definition III.48

For $$${S}_{1}$$$ and $$${S}_{2}$$$ in $$$\mathcal{HS}\mathopen{}\left( H\right)\mathclose{}$$$, define $$\mathopen{}\left\langle{}{S}_{1}, {S}_{2}\right\rangle\mathclose{}_{\mathcal{HS}}= \sum{} \mathopen{}\left\langle{}{S}_{1}\mathopen{}\left( {e}_{n}\right)\mathclose{}, {S}_{2}\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rangle\mathclose{} = \operatorname{Tr}\mathopen{}\left( {S}_{2}^{*}{S}_{1}\right)\mathclose{} \text{.}$$

This is an inner product. The algebraic properties are obvious, and $$$\mathopen{}\left\langle{}S, S\right\rangle\mathclose{}_{\mathcal{HS}}= 0$$$ implies $$$S\mathopen{}\left( {e}_{n}\right)\mathclose{}= 0$$$ for all $$$n$$$ implies $$$S= 0$$$. We define the Hilbert-Schmidt norm on $$$\mathcal{HS}\mathopen{}\left( H\right)\mathclose{}$$$ by $$$\mathopen{}\left\lVert{}S\right\rVert\mathclose{}_{\mathcal{HS}}= {\mathopen{}\left(\mathopen{}\left\langle{}S, S\right\rangle\mathclose{}_{\mathcal{HS}}\right)\mathclose{}}^{\frac{1}{2}}$$$.

It is routine to show that the series defining the inner product converges absolutely. How does this norm relate to the regular old operator norm? $${\mathopen{}\left\lVert{}S\right\rVert\mathclose{}_{\mathcal{HS}}}^{2}= \sum_{n}{} {\mathopen{}\left\lVert{}S\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rVert\mathclose{}}^{2} \geq {\mathopen{}\left\lVert{}S\mathopen{}\left( {e}_{i}\right)\mathclose{}\right\rVert\mathclose{}}^{2}$$ implies $$\mathopen{}\left\lVert{}S\right\rVert\mathclose{}_{\mathcal{HS}}\geq \mathopen{}\left\lVert{}S\right\rVert\mathclose{} \text{.}$$

Lastly, we want to show that this normed linear space of Hilbert-Schmidt operators on $$$H$$$ is complete.

Proposition III.49

$$$\mathcal{HS}\mathopen{}\left( H\right)\mathclose{}$$$ with $$$\mathopen{}\left\lVert{}\cdot\right\rVert\mathclose{}_{\mathcal{HS}}$$$ is complete (that is, a Hilbert space).

Proof. Take a Cauchy sequence of Hilbert-Schmidt operators $$$\mathopen{}\left({T}_{n}\right)\mathclose{}\subseteq \mathcal{HS}\mathopen{}\left( H\right)\mathclose{}$$$. Then since $$$\mathopen{}\left\lVert{}{T}_{n}-{T}_{m}\right\rVert\mathclose{}_{\mathcal{HS}}\geq \mathopen{}\left\lVert{}{T}_{n}-{T}_{m}\right\rVert\mathclose{}$$$, we have a Cauchy sequence in the regular operator norm which converges to some $$$T\in \mathcal{K}\mathopen{}\left( H\right)\mathclose{}$$$ ($$$\mathopen{}\left\lVert{}{T}_{m}-T\right\rVert\mathclose{} \to 0$$$). Then using Fatou and the fact that a Cauchy sequence is bounded, $$\sum{} {\mathopen{}\left\lVert{}T\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rVert\mathclose{}}^{2} = \sum_{n}{} \lim_{m}{} {\mathopen{}\left\lVert{}{T}_{m}\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rVert\mathclose{}}^{2} \leq \liminf_{m}{} \sum{} {\mathopen{}\left\lVert{}{T}_{m}\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rVert\mathclose{}}^{2} = \liminf_{m}{} {\mathopen{}\left(\mathopen{}\left\lVert{}{T}_{m}\right\rVert\mathclose{}_{\mathcal{HS}}\right)\mathclose{}}^{2} \lt \infty \text{.}$$ Thus we have $$$T\in \mathcal{HS}\mathopen{}\left( H\right)\mathclose{}$$$. Finally, $${\mathopen{}\left\lVert{}{T}_{m}-T\right\rVert\mathclose{}_{\mathcal{HS}}}^{2}= \sum_{n}{} {\mathopen{}\left\lVert{}\mathopen{}\left({T}_{m}-T\right)\mathclose{}\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rVert\mathclose{}}^{2} = \sum_{n}{} \lim_{k}{} {\mathopen{}\left\lVert{}\mathopen{}\left({T}_{m}-{T}_{k}\right)\mathclose{}\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rVert\mathclose{}}^{2} \leq \liminf_{k}{} \sum_{n}{} {\mathopen{}\left\lVert{}\mathopen{}\left({T}_{m}-{T}_{k}\right)\mathclose{}\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rVert\mathclose{}}^{2} = \liminf_{k}{} {\mathopen{}\left(\mathopen{}\left\lVert{}{T}_{m}-{T}_{k}\right\rVert\mathclose{}_{\mathcal{HS}}\right)\mathclose{}}^{2} \to 0$$ as $$$m \to \infty$$$.

Remark III.50

It turns out that $$$\mathcal{HS}\mathopen{}\left( H\right)\mathclose{}\simeq H\otimes \overline{H}$$$, where $$$\overline{H}$$$ is $$$H$$$ with conjugate scalar multiplication.