Let
- $(H,\langle\;\cdot\;,\;\cdot\;\rangle)$ be a separable Hilbert space over $\mathbb R$
- $\mathfrak L^1(H)$ be the space of trace class operators on $H$ and $$\operatorname{tr}L:=\sum_{n\in\mathbb N}\langle Le_n,e_n\rangle\;\;\;\text{for }L\in\mathfrak L^1(H)$$ for some orthonormal basis $(e_n)_{n\in\mathbb N}$ of $H$
As you know, $\operatorname{tr}L$ is called the trace of $L\in\mathfrak L(H)$ and its value is finite and independent of the choice of $(e_n)_{n\in\mathbb N}$. I've read that
the closure of the tensor product $H\otimes H$ with respect to the trace norm $$\operatorname{tr}|L|:=\sum_{n\in\mathbb N}\langle\left(L^\ast L\right)^{\frac 12}e_n,e_n\rangle\;\;\;\text{for }L\in\mathfrak L^1(H)$$ equals $\mathfrak L^1(H)$.
How can we prove this statement rigorously? I suppose there is some identification going on here, cause otherwise it wouldn't make much sense to talk about the trace norm of a tensor.
Best Answer
Edit: First we recall the definition: $T$ is a trace-class operator if $$||T||_1:=tr((T^*T)^{1/2})<\infty.$$We show that $H\otimes H$ is dense in the trace class, with respect to that norm.
Note that the definition is really all we're going to use about the trace class. In particular: It's not clear to the OP why a trace-class operator actually has a finite trace. That's totally irrelevant to the proof below, showing that $H\otimes H$ is dense. But it wasn't clear to me either for the longest time; we give a proof in the bonus section below.
Original:
I know nothing about operator theory. I concocted a lemma; if the lemma is correct it's awesomely standard - you may want to try to check that out. The result is immediate from the lemma.
Lemma If $A$ is a bounded positive-definite trace-class operator then there is an orthonormal basis for $H$ consisting of eigenvectors for $A$.
Proof By my favorite version of the Spectral Theorem we may assume that $H=L^2(\mu)$ where $\mu$ is a measure on $X$, and $A$ is a multiplication operator $$Af=mf$$for some $m\in L^\infty(\mu)$; of course $A$ positive definite implies $m\ge0$ almost everywhere.
Let $$F=\{m=0\},$$ $$E_n=\{2^{-n}\le m<2^{-n+1}\}.$$Let $B^*$ be an orthonormal basis for $L^2(F)$ and let $B_n$ be an orthonormal basis for $L^2(E_n)$ (so $B_n=\emptyset$ if $\mu(E_n)=0$). Then $B=B_*\cup\bigcup_{n\in\Bbb Z}B_n$ is an orthonormal basis for $L^2(\mu)$.
So $$\int_Xm\sum_{f\in B}|f|^2\,d\mu=\sum_{f\in B}\langle Af,f\rangle<\infty.$$Since $m$ is bounded away from $0$ on $E_n$ we have $$|B_n|=\int_{E_n}\sum_{f\in B_n}|f|^2\,d\mu<\infty.$$In particular $L^2(E_n)$ is finite-dimensional, so $E_n$ is the union of finitely many (perhaps zero) disjoint atoms. Let $B_n'$ be the orthonomal basis for $L^2(E_n)$ consisting of the normalized indicator functions of those atoms.
Then $B'=B_*\cup\bigcup_{n\in\Bbb Z}B_n'$ is an orthonormal basis for $L^2(\mu)$ consisting of eigenvectors for $A$. QED.
Now say $T$ is a trace-class operator. Let $$A=(T^*T)^{1/2},$$and let $(e_n)$ be an orthonormal basis for $H$ such that $$Ae_n=\lambda_ne_n.$$So $\sum\lambda_n=||T||_1<\infty$.
Let $P_N$ be the orthogonal projection onto the span of $e_1,\dots,e_N$: $$P_Ne_n=\begin{cases}e_n,&(1\le n\le N), \\0,&(n>N).\end{cases}$$Define $$T_N=TP_n,$$ $$A_N=AP_N=P_NA.$$Now $$T_Nx=\sum_{n=1}^N\langle x,e_n\rangle T_Ne_n,$$so $T_N$ lies in (the space of operators corresponding to) $H\otimes H$. And now a miracle happens: $$(T-T_N)^*(T-T_N)=(I-P_N)AA(I-P_N)=(A-A_N)^2.$$That is, $\left((T-T_N)^*(T-T_N)\right)^{1/2}=A-A_N,$ so $$||T-T_N||_1=\sum_{n=N+1}^\infty\lambda_n\to0\quad(N\to\infty).$$
$\newcommand\ip[2]{\langle#1,#2\rangle}$
Bonus: Say $A=(T^*T)^{1/2}$ as above. For some time I was stuck on why $tr(A)<\infty$ should imply that $\sum|\ip{Te_n}{e_n}|<\infty$ for any orthonormal basis $(e_n)$. Turns out I was trying to prove too little! A stronger statement is easier to prove, because it's clear that this or that can't work for the stronger statement.
Proposition If $T$ is a trace-class operator then $$\sum\left|\ip{Te_n}{f_n}\right|\le||T||_1$$for any two orthonormal bases $(e_n)$ and $(f_n)$.
Proof: Say $A=(T^*T)^{1/2}$ as always. It's clear that $$||Tx||=||Ax||;$$hence there exists $U:H\to H$ such that $$T=UA$$and $$||Ux||\le||x||\quad(x\in H).$$(In fact we can take $U$ to be a "partial isometry": $||Ux||=||x||$ for $x$ in the range of $A$ and $Ux=0$ for $x$ in the orthogonal complement of the range of $A$.)
Let $(v_n)$ be an orthonormal basis with $$Av_n=\lambda_nv_n,$$and let $u_n=Uv_n$. Now $x=\sum\ip{x}{v_n}v_n$ implies that $$Ax=\sum\lambda_n \ip{x}{v_n}v_n,$$so$$Tx=\sum\lambda_n \ip{x}{v_n}u_n.$$So $$\begin{aligned}\sum\left|\ip{Te_n}{f_n}\right| &=\sum_n\left|\ip{\sum_j\lambda_j\ip{e_n}{v_j}u_j}{f_n}\right| \\&\le\sum_j\lambda_j\sum_n\left|\ip{e_n}{v_j}\ip{u_j}{f_n}\right| \\&\le\sum_j\lambda_j\left(\sum_n\left|\ip{e_n}{v_j}\right|^2\right)^{1/2} \left(\sum_n\left|\ip{u_j}{f_n}\right|^2\right)^{1/2} \\&=\sum_j\lambda_j||v_j||\,||u_j|| \\&\le\sum_j\lambda_j \\&=||T||_1. \end{aligned}$$