How can we prove that the space of trace class operators on a Hilbert space $H$ is the closure of $H\otimes H$ with respect to the trace norm? 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.
 A: 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 $$\begin{aligned}\sum\left|\ip{Te_n}{f_n}\right|
&=\sum_n\left|\ip{\sum_j\lambda_j\ip{e_n}{v_j}u_j}{f_n}\right|
\\&\le\sum_j\lambda_j\sum_n\left|\ip{e_n}{v_j}\ip{u_j}{f_n}\right|
\\&\le\sum_j\lambda_j\left(\sum_n\left|\ip{e_n}{v_j}\right|^2\right)^{1/2}
\left(\sum_n\left|\ip{u_j}{f_n}\right|^2\right)^{1/2}
\\&=\sum_j\lambda_j||v_j||\,||u_j||
\\&\le\sum_j\lambda_j
\\&=||T||_1.
\end{aligned}$$
A: The underlying Hilbert space $\mathsf{H}$ may be real or complex, hence let $\mathbb K$ denote the scalars. Assume $\mathsf{H}$ to be infinite-dimensional, not necessarily separable.

It is shown that
  $$\mathfrak L^1(\mathsf{H})\;\cong\;\mathsf{H}\,\hat{\otimes}_\pi\,\mathsf{H'}\tag{$\pi$}$$
  holds, where $\mathsf{H'}$ is the topological dual and the decorated tensor sign refers to the Banach space completion w.r.to the projective tensor norm.

Consider a rank one operator $F=v\,\langle\,\cdot\,|w\rangle = \mu e_v\langle\,\cdot\,|e_w\rangle\,$ with $v,e_v,w,e_w\in\mathsf{H}$, where $e_v,e_w$ are unit vectors and $\,\mu\in\mathbb K\,$. We mention two central points in answering the OP, note the intermediary role of the inner product in both of them:


*

*The trace of $\,F\,$
$$\operatorname{tr}(F) = \langle v|w\rangle = \mu\langle e_v|e_w\rangle$$
results from evaluating the inner product.

*$F$ may be represented as the simple tensor $v\otimes\langle\,\cdot\,|w\rangle\in \mathsf{H}\otimes\mathsf{H'}$. This gives rise to the algebraic isomorphism
$$\begin{eqnarray}
\mathsf{H}\otimes \mathsf{H'}\quad & \overset{\cong}{\longrightarrow} &\;\mathfrak{F}(H)\tag{1} \\[0.5ex]
\sum_{\text{finite}}v_k\otimes\langle\,\cdot\,|w_k\rangle\; &
\mapsto &
\; \left(u\mapsto\sum\nolimits_{\text{finite}}v_k\langle u|w_k\rangle\right)
\end{eqnarray}
$$
where $\mathfrak{F}(\mathsf{H})$ are the finite rank operators. Cf also Operator norm and tensor norms in this context.
It is a fundamental property of Hilbert spaces that $\mathsf{H}$ and $\mathsf{H'}$ are isometrically isomorphic via $u\mapsto\langle\,\cdot\,|u\rangle$, and the isomorphism is antilinear if $\,\mathbb K=\mathbb C$.


Next the trace class is spotted among the compact operators $\mathfrak K(\mathsf{H})$:
A positive $\,A\in\mathfrak K(\mathsf{H})\,$ has the spectral decomposition $\,A=\sum_{n=1}^N\mu_n e_n\langle\,\cdot\,|e_n\rangle\,$ where


*

*$(\mu_n)$ is the null sequence of eigenvalues of $A$, counting multiplicities and decreasingly arranged,
hence all $\,\mu_n>0\,$ and $\,\mu_1=\|A\|\,$,

*$\left\{e_n\mid 1\le n\le N=\operatorname{dim}_\mathsf{H}\overline{\operatorname{Im}A}\right\}$ is an orthonormal system of eigenvectors.


The expansion converges in operator norm. For general $\,L\in\mathfrak K(\mathsf{H})\,$ exploit its polar decomposition $\,L=V|L|\,$ to get 
$L=\sum_{n=1}^N\mu_n\, f_n\langle\,\cdot\,|e_n\rangle\,$
with $\,f_n= Ve_n$. The $\mu_n$ are called singular values then, and $L$ is a trace class operator if
$$\|L\|_1\,=\,\operatorname{tr}|L|\,=\,\sum_n\mu_n<\infty\,.$$
Then $\,\operatorname{tr}(L)=\sum_{n=1}^N\mu_n\langle\, f_n|e_n\rangle\,$ is finite.
Equip the LHS of $(1)$ with the projective tensor norm:
For Banach spaces $\mathsf X,\mathsf Y$ and $t\in\mathsf X\otimes\mathsf Y$ the norm is given as
$$\|t\|_\pi = \inf\left\{ \sum\nolimits_{k=1}^n \|x_k\|_{\mathsf X} \,\|y_k\|_{\mathsf Y}\Big\vert\;t=\sum\nolimits_{k=1}^n x_k\otimes y_k
\right\}$$
The completed tensor product $\mathsf X\hat{\otimes}_\pi\mathsf Y$ has the universal property that every
jointly continuous bilinear map $B:\mathsf X\times\mathsf Y\rightarrow\mathsf Z$, where $\mathsf Z$ is another Banach space, uniquely factors through the canonical bilinear map $\kappa$
\begin{array}{ccc}
\mathsf X\times\mathsf Y & \xrightarrow{\kappa} & \mathsf X\hat{\otimes}_\pi \mathsf Y  \\
 & \searrow{B} &  \swarrow{\exists !b} \\
 & \qquad\mathsf Z
\end{array}
thus $B=b\circ\kappa$ and $b$ is a continuous linear map.
Passing to completions in $(1)$ yields an isomorphism:
When going from right to left, continuity holds because
if $L=\sum_{\text{finite}}\mu_k f_k\langle\,\cdot\,|e_k\rangle\in\,\mathfrak{F}(H)$ then $\|\sum\mu_k\,f_k\otimes e_k\|_\pi
\le\sum|\mu_k|\|f_k\|\|e_k\|\le\|L\|_1$. Completion of the target space and extension by continuity yields the map $\mathfrak L^1(\mathsf H)\to\mathsf{H}\,\hat{\otimes}_\pi\,\mathsf{H'}$.
Continuity in the opposite direction results from the universal property applied to the (obvious) bilinear map
$\mathsf H\times\mathsf{H'}\to\mathfrak L^1(\mathsf H)$.
This completes the proof of ($\pi$).
