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. Trace Class Operators

We again assume $$$H$$$ is a separable, infinite-dimensional Hilbert space.

Proposition III.19

For $$$A\geq 0$$$ in $$$\mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ and orthonormal bases $$$\mathopen{}\left({e}_{n}\right)\mathclose{}_{n=1}^{\infty}$$$ and $$$\mathopen{}\left({f}_{n}\right)\mathclose{}_{n=1}^{\infty}$$$, we have $$$\sum_{n=1}^{\infty}{} \mathopen{}\left\langle{}A\mathopen{}\left( {e}_{n}\right)\mathclose{}, {e}_{n}\right\rangle\mathclose{} = \sum_{n=1}^{\infty}{} \mathopen{}\left\langle{}A\mathopen{}\left( {f}_{n}\right)\mathclose{}, {f}_{n}\right\rangle\mathclose{}$$$.

Proof. Since we may interchange the order of summands in a double series with nonnegative terms, $$\sum_{n}{} \mathopen{}\left\langle{}A\mathopen{}\left( {e}_{n}\right)\mathclose{}, {e}_{n}\right\rangle\mathclose{} = \sum_{n}{} \mathopen{}\left\langle{}{A}^{\frac{1}{2}}\mathopen{}\left( {e}_{n}\right)\mathclose{}, {A}^{\frac{1}{2}}\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rangle\mathclose{} = \sum_{n}{} {\mathopen{}\left\lVert{}{A}^{\frac{1}{2}}\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rVert\mathclose{}}^{2} = \sum_{n}{} \sum_{m}{} {\mathopen{}\left\lvert{}\mathopen{}\left\langle{}{A}^{\frac{1}{2}}\mathopen{}\left( {e}_{n}\right)\mathclose{}, {f}_{m}\right\rangle\mathclose{}\right\rvert\mathclose{}}^{2} = \sum_{n}{} \sum_{m}{} {\mathopen{}\left\lvert{}\mathopen{}\left\langle{}{A}^{\frac{1}{2}}\mathopen{}\left( {f}_{m}\right)\mathclose{}, {e}_{n}\right\rangle\mathclose{}\right\rvert\mathclose{}}^{2} = \sum_{m}{} \sum_{n}{} {\mathopen{}\left\lvert{}\mathopen{}\left\langle{}{A}^{\frac{1}{2}}\mathopen{}\left( {f}_{m}\right)\mathclose{}, {e}_{n}\right\rangle\mathclose{}\right\rvert\mathclose{}}^{2} = \sum_{m}{} \mathopen{}\left\langle{}A\mathopen{}\left( {f}_{m}\right)\mathclose{}, {f}_{m}\right\rangle\mathclose{} \text{.}$$

Notation III.20

Let $$${\mathcal{L}\mathopen{}\left( H\right)\mathclose{}}^{+}= \mathopen{}\left\{\, A\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}\,\middle\vert\, , A\geq 0, \,\right\}\mathclose{}$$$. Define $$$\operatorname{Tr} : {\mathcal{L}\mathopen{}\left( H\right)\mathclose{}}^{+} \to \mathopen{}\left[0, {+}\infty\right]\mathclose{}$$$ by $$$\operatorname{Tr}\mathopen{}\left( A\right)\mathclose{}= \sum_{n=1}^{\infty}{} \mathopen{}\left\langle{}A\mathopen{}\left( {e}_{n}\right)\mathclose{}, {e}_{n}\right\rangle\mathclose{}$$$.

Proposition III.21

For $$$A\in {\mathcal{L}\mathopen{}\left( H\right)\mathclose{}}^{+}$$$, $$$\operatorname{Tr}\mathopen{}\left( A\right)\mathclose{}\lt \infty$$$ if and only if $$$A\in \mathcal{K}\mathopen{}\left( H\right)\mathclose{}$$$ and $$$\sum{}{λ}_{n}\lt \infty$$$ where $$$\mathopen{}\left({λ}_{n}\right)\mathclose{}$$$ are the eigenvalues of $$$A$$$. In this case, $$$\operatorname{Tr}\mathopen{}\left( A\right)\mathclose{}= \sum{}{λ}_{n}$$$ where the eigenvalues $$${λ}_{n}$$$ are counted according to multiplicity (that is, if the dimension of $$$\operatorname{Ker}\mathopen{}\left( λ-A\right)\mathclose{}$$$ is $$$d$$$, then $$$λ$$$ appears $$$d$$$ times in $$$\sum{}{λ}_{n}$$$).

Proof. (⇒) Pick an orthonormal basis $$$\mathopen{}\left({e}_{n}\right)\mathclose{}_{n=1}^{\infty}$$$. Then $$$\operatorname{Tr}\mathopen{}\left( A\right)\mathclose{}= \sum{} {\mathopen{}\left\lVert{}{A}^{\frac{1}{2}}\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rVert\mathclose{}}^{2} \lt \infty$$$. For $$$x\in H$$$, $$A\mathopen{}\left( x\right)\mathclose{}= {A}^{\frac{1}{2}}\mathopen{}\left( {A}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}= {A}^{\frac{1}{2}}\sum_{n}{} \mathopen{}\left\langle{}{A}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}, {e}_{n}\right\rangle\mathclose{}{e}_{n} = \sum_{n}{} \mathopen{}\left\langle{}{A}^{\frac{1}{2}}\mathopen{}\left( x\right)\mathclose{}, {e}_{n}\right\rangle\mathclose{}{A}^{\frac{1}{2}}\mathopen{}\left( {e}_{n}\right)\mathclose{} = \sum_{n}{} \mathopen{}\left({A}^{\frac{1}{2}}\mathopen{}\left( {e}_{n}\right)\mathclose{}\otimes {A}^{\frac{1}{2}}\mathopen{}\left( {e}_{n}\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{} \text{.}$$ Further, $$\sum_{n}{} \mathopen{}\left\lVert{}{A}^{\frac{1}{2}}\mathopen{}\left( {e}_{n}\right)\mathclose{}\otimes {A}^{\frac{1}{2}}\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rVert\mathclose{} = \sum_{n}{} {\mathopen{}\left\lVert{}{A}^{\frac{1}{2}}\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rVert\mathclose{}}^{2} \lt \infty \text{.}$$ The series $$$\sum_{n}{} {A}^{\frac{1}{2}}\mathopen{}\left( {e}_{n}\right)\mathclose{}\otimes {A}^{\frac{1}{2}}\mathopen{}\left( {e}_{n}\right)\mathclose{}$$$ converges in norm with sum $$$A$$$. This puts $$$A\in \overline{\mathcal{F}\mathopen{}\left( H\right)\mathclose{}}= \mathcal{K}\mathopen{}\left( H\right)\mathclose{}$$$. So there exists an orthonormal basis $$$\mathopen{}\left({f}_{m}\right)\mathclose{}_{m=1}^{\infty}$$$ and a sequence of nonnegative real numbers $$$\mathopen{}\left({λ}_{i}\right)\mathclose{}_{i=1}^{\infty}$$$ such that $$${λ}_{m} \to 0$$$, $$$A= \sum_{m}{} {λ}_{m}{f}_{m}\otimes {f}_{m}$$$. Furthermore, $$\sum_{m}{} \mathopen{}\left\langle{}A\mathopen{}\left( {f}_{m}\right)\mathclose{}, {f}_{m}\right\rangle\mathclose{} = \sum_{m}{} {λ}_{m}\mathopen{}\left\langle{}{f}_{m}, {f}_{m}\right\rangle\mathclose{} = \sum_{m}{}{λ}_{m} \text{.}$$

(⇐) If you know $$$A\in {\mathcal{K}\mathopen{}\left( H\right)\mathclose{}}^{+}$$$, then $$$A= \sum{} {λ}_{m}{f}_{m}\otimes {f}_{m}$$$ as above, and $$\operatorname{Tr}\mathopen{}\left( A\right)\mathclose{}= \sum{} \mathopen{}\left\langle{}A\mathopen{}\left( {f}_{m}\right)\mathclose{}, {f}_{m}\right\rangle\mathclose{} = \sum{} {λ}_{m} \lt \infty \text{.}$$

Proposition III.22

For $$$T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$, the following are equivalent:

(a)
There exist $$$\mathopen{}\left\{\, {x}_{1}, {x}_{2}, \dotsc\,\right\}\mathclose{}$$$ and $$$\mathopen{}\left\{\, {y}_{1}, {y}_{2}, \dotsc\,\right\}\mathclose{}\in H$$$ such that $$\sum_{j}{} \mathopen{}\left\lVert{}{x}_{j}\right\rVert\mathclose{}\mathopen{}\left\lVert{}{y}_{j}\right\rVert\mathclose{} \lt \infty$$ and $$$T= \sum_{j}{} {x}_{j}\otimes {y}_{j}$$$.
(b)
For all orthonormal sets $$$\mathopen{}\left\{\, {e}_{n}\,\right\}\mathclose{}$$$ and $$$\mathopen{}\left\{\, {f}_{n}\,\right\}\mathclose{}$$$ in $$$H$$$, $$$\sum_{n}{} \mathopen{}\left\lvert{}\mathopen{}\left\langle{}T\mathopen{}\left( {e}_{n}\right)\mathclose{}, {f}_{n}\right\rangle\mathclose{}\right\rvert\mathclose{} \lt \infty$$$.
(c)
$$$\operatorname{Tr}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{}\lt \infty$$$.

Proof. ((a)⇒(b)) Using the Cauchy-Schwarz inequality in $$$\mathrm{l}^{0}$$$ and Bessel's inequality, we have $$\sum_{n}{} \mathopen{}\left\lvert{}\mathopen{}\left\langle{}T\mathopen{}\left( {e}_{n}\right)\mathclose{}, {f}_{n}\right\rangle\mathclose{}\right\rvert\mathclose{} = \sum_{n}{} \mathopen{}\left\lvert{}\sum_{j}{} \mathopen{}\left\langle{}{e}_{n}, {y}_{j}\right\rangle\mathclose{}\mathopen{}\left\langle{}{x}_{j}, {f}_{n}\right\rangle\mathclose{} \right\rvert\mathclose{} \leq \sum_{n}{} \sum_{j}{} \mathopen{}\left\lvert{}\mathopen{}\left\langle{}{e}_{n}, {y}_{j}\right\rangle\mathclose{}\right\rvert\mathclose{}\mathopen{}\left\lvert{}\mathopen{}\left\langle{}{x}_{j}, {f}_{n}\right\rangle\mathclose{}\right\rvert\mathclose{} = \sum_{j}{} \sum_{n}{} \mathopen{}\left\lvert{}\mathopen{}\left\langle{}{e}_{n}, {y}_{j}\right\rangle\mathclose{}\right\rvert\mathclose{}\mathopen{}\left\lvert{}\mathopen{}\left\langle{}{x}_{j}, {f}_{n}\right\rangle\mathclose{}\right\rvert\mathclose{} \leq \sum_{j}{} {\mathopen{}\left(\sum_{n}{} {\mathopen{}\left\lvert{}\mathopen{}\left\langle{}{e}_{n}, {y}_{j}\right\rangle\mathclose{}\right\rvert\mathclose{}}^{2} \right)\mathclose{}}^{\frac{1}{2}}{\mathopen{}\left(\sum_{n}{} {\mathopen{}\left\lvert{}\mathopen{}\left\langle{}{x}_{j}, {f}_{n}\right\rangle\mathclose{}\right\rvert\mathclose{}}^{2} \right)\mathclose{}}^{\frac{1}{2}} \leq \sum_{j}{} \mathopen{}\left\lVert{}{y}_{j}\right\rVert\mathclose{}\mathopen{}\left\lVert{}{x}_{j}\right\rVert\mathclose{} \leq \infty \text{.}$$

((b)⇒(c)) Recall the polar decomposition (Proposition II.68): $$$\mathopen{}\left\lvert{}T\right\rvert\mathclose{}= {\mathopen{}\left( T^{*}T\right)\mathclose{}}^{\frac{1}{2}}$$$ and $$$T= V\mathopen{}\left\lvert{}T\right\rvert\mathclose{}$$$ where $$$V^{*}V$$$ is the projection on $$$\overline{\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( H\right)\mathclose{}}= \overline{ T^{*}\mathopen{}\left( H\right)\mathclose{}}$$$. Let $$$\mathopen{}\left\{\, {ẽ}_{n}, {ẽ}_{n}, \dotsc\,\right\}\mathclose{}$$$ be an orthonormal basis for $$$\operatorname{Ker}\mathopen{}\left( T\right)\mathclose{}= \operatorname{Ker}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{}$$$, and let $$$\mathopen{}\left\{\, {e}_{1}, {e}_{2}, \dotsc\,\right\}\mathclose{}$$$ be an orthonormal basis for $$$\overline{\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( H\right)\mathclose{}}= { \mathopen{}\left(\operatorname{Ker}\mathopen{}\left( T\right)\mathclose{}\right)\mathclose{} }^{\perp}$$$. Then $$\operatorname{Tr}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{}= \sum_{n}{} \mathopen{}\left\langle{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( {e}_{n}\right)\mathclose{}, {e}_{n}\right\rangle\mathclose{} +\sum_{n}{} \mathopen{}\left\langle{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( {ẽ}_{n}\right)\mathclose{}, {ẽ}_{n}\right\rangle\mathclose{} = \sum_{n}{} \mathopen{}\left\langle{}\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( {e}_{n}\right)\mathclose{}, V^{*}V\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rangle\mathclose{} = \sum_{n}{} \mathopen{}\left\langle{}V\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( {e}_{n}\right)\mathclose{}, V\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rangle\mathclose{} = \sum_{n}{} \mathopen{}\left\langle{}T\mathopen{}\left( {e}_{n}\right)\mathclose{}, V\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rangle\mathclose{} \leq \sum_{n}{} \mathopen{}\left\lvert{}\mathopen{}\left\langle{}T\mathopen{}\left( {e}_{n}\right)\mathclose{}, {f}_{n}\right\rangle\mathclose{}\right\rvert\mathclose{} \text{,}$$ where $$${f}_{n}= V\mathopen{}\left( {e}_{n}\right)\mathclose{}$$$. Notice $$$\mathopen{}\left\langle{}{f}_{n}, {f}_{j}\right\rangle\mathclose{}= \mathopen{}\left\langle{}V\mathopen{}\left( {e}_{n}\right)\mathclose{}, V\mathopen{}\left( {e}_{j}\right)\mathclose{}\right\rangle\mathclose{}= \mathopen{}\left\langle{} V^{*}V\mathopen{}\left( {e}_{n}\right)\mathclose{}, {e}_{j}\right\rangle\mathclose{}= \mathopen{}\left\langle{}{e}_{n}, {e}_{j}\right\rangle\mathclose{}$$$, so the $$$\mathopen{}\left({f}_{n}\right)\mathclose{}$$$ are orthonormal. By (b), $$$\operatorname{Tr}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{}\lt \infty$$$.

((c)⇒(a)) For an orthonormal basis $$$\mathopen{}\left({e}_{n}\right)\mathclose{}_{n=1}^{\infty}$$$, $$$\operatorname{Tr}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{}= \sum{} {\mathopen{}\left\lVert{}{\mathopen{}\left\lvert{}T\right\rvert\mathclose{}}^{\frac{1}{2}}\mathopen{}\left( {e}_{n}\right)\mathclose{}\right\rVert\mathclose{}}^{2} \lt \infty$$$. So, $$\mathopen{}\left\lvert{}T\right\rvert\mathclose{}= \sum{} {\mathopen{}\left\lvert{}T\right\rvert\mathclose{}}^{\frac{1}{2}}\mathopen{}\left( {e}_{n}\right)\mathclose{}\otimes {\mathopen{}\left\lvert{}T\right\rvert\mathclose{}}^{\frac{1}{2}}\mathopen{}\left( {e}_{n}\right)\mathclose{}$$ (as in the proof of Proposition III.21) and $$T= V\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{}= \sum{} V{\mathopen{}\left\lvert{}T\right\rvert\mathclose{}}^{\frac{1}{2}}\mathopen{}\left( {e}_{n}\right)\mathclose{}\otimes {\mathopen{}\left\lvert{}T\right\rvert\mathclose{}}^{\frac{1}{2}}\mathopen{}\left( {e}_{n}\right)\mathclose{} \text{.}$$ Since $$$\mathopen{}\left\lVert{}V\right\rVert\mathclose{}= 1$$$ we can take $$${x}_{n}= V{\mathopen{}\left\lvert{}T\right\rvert\mathclose{}}^{\frac{1}{2}}\mathopen{}\left( {e}_{n}\right)\mathclose{}$$$ and $$${y}_{n}= {\mathopen{}\left\lvert{}T\right\rvert\mathclose{}}^{\frac{1}{2}}\mathopen{}\left( {e}_{n}\right)\mathclose{}$$$ to get (a).

Definition III.23

Call $$$T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ a trace-class operator if $$$\operatorname{Tr}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{}\lt \infty$$$. Write $$$\mathcal{T}\mathopen{}\left( H\right)\mathclose{}$$$ for the set of all trace-class operators.

Proposition III.24

$$$\mathcal{T}\mathopen{}\left( H\right)\mathclose{}$$$ is an ideal of $$$\mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ with $$$\mathcal{F}\mathopen{}\left( H\right)\mathclose{}\subseteq \mathcal{T}\mathopen{}\left( H\right)\mathclose{}\subseteq \mathcal{K}\mathopen{}\left( H\right)\mathclose{} \text{,}$$$ and $$$T\in \mathcal{T}\mathopen{}\left( H\right)\mathclose{}$$$ implies $$$T^{*}\in \mathcal{T}\mathopen{}\left( H\right)\mathclose{}$$$.

Proof. By Proposition III.22, the generic trace-class operator is $$$T= \sum{} {x}_{j}\otimes {y}_{j}$$$, where $$$\sum{} \mathopen{}\left\lVert{}{x}_{j}\right\rVert\mathclose{}\mathopen{}\left\lVert{}{y}_{j}\right\rVert\mathclose{} \lt \infty$$$. The partial sums are in $$$\mathcal{F}\mathopen{}\left( H\right)\mathclose{}$$$ and converge in norm to $$$T$$$, so $$$T\in \mathcal{K}\mathopen{}\left( H\right)\mathclose{}$$$. The sum of two such operators is another operator of the same type, and for $$$S\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ we have $$$S\mathopen{}\left( T\right)\mathclose{}= \sum{} S\mathopen{}\left( {x}_{j}\right)\mathclose{}\otimes {y}_{j}$$$ and $$$T\mathopen{}\left( S\right)\mathclose{}= \sum{} {x}_{j}\otimes S^{*}\mathopen{}\left( {y}_{j}\right)\mathclose{}$$$. Finally, $$$T^{*}= \sum{} {y}_{j}\otimes {x}_{j}$$$.

Example III.25

The Volterra operator $$$T$$$ is not trace-class. Look at $$${e}_{n}\mathopen{}\left( t\right)\mathclose{}= \sqrt{2}\cos\mathopen{}\left( nπt\right)\mathclose{}$$$. Then $$T\mathopen{}\left( {e}_{n}\mathopen{}\left( t\right)\mathclose{}\right)\mathclose{}= \frac{\sqrt{2}}{nπ}\sin\mathopen{}\left( nπt\right)\mathclose{}= \frac{{f}_{n}\mathopen{}\left( t\right)\mathclose{}}{nπ} \text{,}$$ where $$${f}_{n}\mathopen{}\left( t\right)\mathclose{}= \sqrt{2}\sin\mathopen{}\left( nπt\right)\mathclose{}$$$. $$$\mathopen{}\left\langle{}T\mathopen{}\left( {e}_{n}\right)\mathclose{}, {f}_{n}\right\rangle\mathclose{}= \frac{1}{nπ}$$$, so $$$T\notin \mathcal{T}\mathopen{}\left( H\right)\mathclose{}$$$ by Proposition III.22. On the other hand, $$${T}^{2}$$$ is trace-class because, as we have seen above, $${T}^{2}= {ξ}_{1}\otimes \mathopen{}\left({ξ}_{0}-{ξ}_{1}\right)\mathclose{}-\frac{1}{{π}^{2}}\sum_{n=1}^{\infty}{} \frac{1}{{n}^{2}}{e}_{n}\otimes {e}_{n} \text{,}$$ where $$${ξ}_{j}\mathopen{}\left( t\right)\mathclose{}= {t}^{j}$$$ and $$${e}_{n}\mathopen{}\left( t\right)\mathclose{}= \sqrt{2}\sin\mathopen{}\left( nπt\right)\mathclose{}$$$.

Proposition III.26

For $$$T\in \mathcal{T}\mathopen{}\left( H\right)\mathclose{}$$$, written $$$T= \sum_{j}{} {x}_{j}\otimes {y}_{j}$$$ with $$$\sum_{j}{} \mathopen{}\left\lVert{}{x}_{j}\right\rVert\mathclose{}\mathopen{}\left\lVert{}{y}_{j}\right\rVert\mathclose{} \lt \infty$$$, we have

1. $$$\sum_{j}{} \mathopen{}\left\langle{}{x}_{j}, {y}_{j}\right\rangle\mathclose{}$$$ converges;
2. $$$\sum_{n}{} \mathopen{}\left\langle{}T\mathopen{}\left( {e}_{n}\right)\mathclose{}, {e}_{n}\right\rangle\mathclose{}$$$ converges for any orthonormal basis $$$\mathopen{}\left({e}_{n}\right)\mathclose{}_{n=1}^{\infty}$$$, and $$\sum_{n}{} \mathopen{}\left\langle{}T\mathopen{}\left( {e}_{n}\right)\mathclose{}, {e}_{n}\right\rangle\mathclose{} = \sum_{j}{} \mathopen{}\left\langle{}{x}_{j}, {y}_{j}\right\rangle\mathclose{} \text{.}$$

Proof.

1. Follows from $$$\mathopen{}\left\lvert{}\mathopen{}\left\langle{}{x}_{j}, {y}_{j}\right\rangle\mathclose{}\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}{x}_{j}\right\rVert\mathclose{}\mathopen{}\left\lVert{}{y}_{j}\right\rVert\mathclose{}$$$ and absolute convergence implies convergence.
2. $$\sum_{j}{} \sum_{n}{} \mathopen{}\left\lvert{}\mathopen{}\left\langle{}{e}_{n}, {y}_{j}\right\rangle\mathclose{}\right\rvert\mathclose{}\mathopen{}\left\lvert{}\mathopen{}\left\langle{}{x}_{j}, {e}_{n}\right\rangle\mathclose{}\right\rvert\mathclose{} \leq \sum_{j}{} {\mathopen{}\left(\sum_{n}{} {\mathopen{}\left\lvert{}\mathopen{}\left\langle{}{e}_{n}, {y}_{j}\right\rangle\mathclose{}\right\rvert\mathclose{}}^{2} \right)\mathclose{}}^{\frac{1}{2}}{\mathopen{}\left(\sum_{n}{} {\mathopen{}\left\lvert{}\mathopen{}\left\langle{}{x}_{j}, {e}_{n}\right\rangle\mathclose{}\right\rvert\mathclose{}}^{2} \right)\mathclose{}}^{\frac{1}{2}} = \sum_{j}{} {\mathopen{}\left({\mathopen{}\left\lVert{}{y}_{i}\right\rVert\mathclose{}}^{2}\right)\mathclose{}}^{\frac{1}{2}}{\mathopen{}\left({\mathopen{}\left\lVert{}{x}_{j}\right\rVert\mathclose{}}^{2}\right)\mathclose{}}^{\frac{1}{2}} \leq \infty \text{.}$$ So, both double series $$\sum_{j}{} \sum_{n}{} \mathopen{}\left\langle{}{e}_{n}, {y}_{j}\right\rangle\mathclose{}\mathopen{}\left\langle{}{x}_{j}, {e}_{n}\right\rangle\mathclose{} = \sum_{n}{} \sum_{j}{} \mathopen{}\left\langle{}{e}_{n}, {y}_{j}\right\rangle\mathclose{}\mathopen{}\left\langle{}{x}_{j}, {e}_{n}\right\rangle\mathclose{}$$ converge. Then $$\sum_{j}{} \sum_{n}{} \mathopen{}\left\langle{}{e}_{n}, {y}_{j}\right\rangle\mathclose{}\mathopen{}\left\langle{}{x}_{j}, {e}_{n}\right\rangle\mathclose{} = \sum_{j}{} \mathopen{}\left\langle{}{x}_{j}, {y}_{j}\right\rangle\mathclose{}$$ and $$\sum_{n}{} \sum_{j}{} \mathopen{}\left\langle{}{e}_{n}, {y}_{j}\right\rangle\mathclose{}\mathopen{}\left\langle{}{x}_{j}, {e}_{n}\right\rangle\mathclose{} = \sum_{n}{} \sum_{j}{} \mathopen{}\left\langle{}\mathopen{}\left\langle{}{e}_{n}, {y}_{j}\right\rangle\mathclose{}{x}_{j}, {e}_{n}\right\rangle\mathclose{} = \sum_{n}{} \mathopen{}\left\langle{}\mathopen{}\left(\sum_{j}{} {x}_{j}\otimes {y}_{j} \right)\mathclose{}{e}_{n}, {e}_{n}\right\rangle\mathclose{} = \sum_{n}{} \mathopen{}\left\langle{}T\mathopen{}\left( {e}_{n}\right)\mathclose{}, {e}_{n}\right\rangle\mathclose{} \text{.}$$

Definition III.27

Let $$$\mathopen{}\left\{\, {e}_{n}\,\right\}\mathclose{}$$$ be an orthonormal basis of $$$H$$$. For $$$T\in \mathcal{T}\mathopen{}\left( H\right)\mathclose{}$$$, we define $$$\operatorname{Tr}\mathopen{}\left( T\right)\mathclose{}= \sum{} \mathopen{}\left\langle{}T\mathopen{}\left( {e}_{n}\right)\mathclose{}, {e}_{n}\right\rangle\mathclose{}$$$.

Remark III.28

1. It doesn't matter which orthonormal basis you use.
2. If $$$\sum{} \mathopen{}\left\lVert{}{x}_{j}\right\rVert\mathclose{}\mathopen{}\left\lVert{}{y}_{j}\right\rVert\mathclose{} \lt \infty$$$, $$$\operatorname{Tr}\mathopen{}\left( \sum{} {x}_{j}\otimes {y}_{j} \right)\mathclose{}= \sum{} \mathopen{}\left\langle{}{x}_{j}, {y}_{j}\right\rangle\mathclose{}$$$.
3. $$$\operatorname{Tr} : \mathcal{T}\mathopen{}\left( H\right)\mathclose{} \to \mathbb{C}$$$ is linear.

Proposition III.29

For $$$S\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{}$$$ and $$$T\in \mathcal{T}\mathopen{}\left( H\right)\mathclose{}$$$,

1. $$$\operatorname{Tr}\mathopen{}\left( ST\right)\mathclose{}= \operatorname{Tr}\mathopen{}\left( TS\right)\mathclose{}$$$.
2. $$$\mathopen{}\left\lvert{}\operatorname{Tr}\mathopen{}\left( ST\right)\mathclose{}\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}S\right\rVert\mathclose{}\operatorname{Tr}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{}$$$.

Proof.

1. Write $$$T= \sum{} {x}_{j}\otimes {y}_{j}$$$ with $$$\sum{} \mathopen{}\left\lVert{}{x}_{j}\right\rVert\mathclose{}\mathopen{}\left\lVert{}{y}_{j}\right\rVert\mathclose{} \lt \infty$$$. Then $$\operatorname{Tr}\mathopen{}\left( ST\right)\mathclose{}= \operatorname{Tr}\mathopen{}\left( \sum{} S\mathopen{}\left( {x}_{j}\right)\mathclose{}\otimes {y}_{j} \right)\mathclose{}= \sum_{j}{} \mathopen{}\left\langle{}S\mathopen{}\left( {x}_{j}\right)\mathclose{}, {y}_{j}\right\rangle\mathclose{} = \sum_{j}{} \mathopen{}\left\langle{}{x}_{j}, S^{*}\mathopen{}\left( {y}_{j}\right)\mathclose{}\right\rangle\mathclose{} = \operatorname{Tr}\mathopen{}\left( \sum{} {x}_{j}\otimes S^{*}\mathopen{}\left( {y}_{j}\right)\mathclose{} \right)\mathclose{}= \operatorname{Tr}\mathopen{}\left( TS\right)\mathclose{} \text{.}$$
2. By the spectral theorem, get an orthonormal basis $$$\mathopen{}\left({e}_{n}\right)\mathclose{}$$$ for $$$H$$$ and (non-negative) $$${λ}_{1}$$$, $$${λ}_{2}$$$, … such that $$$\mathopen{}\left\lvert{}T\right\rvert\mathclose{}= \sum{} {λ}_{n}\mathopen{}\left({e}_{n}\otimes {e}_{n}\right)\mathclose{}$$$. Use the polar decomposition $$$T= V\mathopen{}\left\lvert{}T\right\rvert\mathclose{}$$$. Then $$\mathopen{}\left\lvert{}\operatorname{Tr}\mathopen{}\left( S\mathopen{}\left( T\right)\mathclose{}\right)\mathclose{}\right\rvert\mathclose{}= \mathopen{}\left\lvert{}\operatorname{Tr}\mathopen{}\left( SV\mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{}\right\rvert\mathclose{}= \mathopen{}\left\lvert{}\sum{} \mathopen{}\left\langle{}S\mathopen{}\left( V\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\mathopen{}\left( {e}_{n}\right)\mathclose{}\right)\mathclose{}\right)\mathclose{}, {e}_{n}\right\rangle\mathclose{} \right\rvert\mathclose{}= \mathopen{}\left\lvert{}\sum{} {λ}_{n}\mathopen{}\left\langle{}S\mathopen{}\left( V\mathopen{}\left( {e}_{n}\right)\mathclose{}\right)\mathclose{}, {e}_{n}\right\rangle\mathclose{} \right\rvert\mathclose{}\leq \sum{} {λ}_{n}\mathopen{}\left\lvert{}\mathopen{}\left\langle{}S\mathopen{}\left( V\mathopen{}\left( {e}_{n}\right)\mathclose{}\right)\mathclose{}, {e}_{n}\right\rangle\mathclose{}\right\rvert\mathclose{} \leq \mathopen{}\left\lVert{}SV\right\rVert\mathclose{}\sum{} {λ}_{n} \leq \mathopen{}\left\lVert{}S\right\rVert\mathclose{}\operatorname{Tr}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{}$$ since $$$\mathopen{}\left\lVert{}V\right\rVert\mathclose{}\leq 1$$$ and $$$\sum{} {λ}_{n} = \operatorname{Tr}\mathopen{}\left( \mathopen{}\left\lvert{}T\right\rvert\mathclose{}\right)\mathclose{}$$$.