• $(H,\langle\;\cdot\;,\;\cdot\;\rangle)$ 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.

  • 2
    $\begingroup$ It seems likely that they're identifying $x\otimes y$ with the operator $z\mapsto\langle z,x\rangle y$. $\endgroup$ – David C. Ullrich May 12 '16 at 18:22
  • $\begingroup$ @DavidC.Ullrich : are we considering this : en.wikipedia.org/wiki/Hilbert%E2%80%93Schmidt_operator ? $\endgroup$ – reuns May 12 '16 at 18:34
  • $\begingroup$ I don't see why - I'd assume we're talking about trace-class operators en.wikipedia.org/wiki/Trace_class In any case, all I was trying to contribute was a conjecture on how $H\otimes H$ can be regarded as a space of operators on $H$... $\endgroup$ – David C. Ullrich May 12 '16 at 18:43
  • $\begingroup$ @DavidC.Ullrich That's clear to me, but what about the other way around? Can we identify each element of $\mathfrak L(H)$ with an element of $H\otimes H$? $\endgroup$ – 0xbadf00d May 12 '16 at 18:56
  • 1
    $\begingroup$ it seems Hilbert-Schmidt is about the $\|.\|_2$ trace class norm, while you are considering here the $\|.\|_1$ trace class norm : for a given $Tr$ operator, $\|L\|_1 = tr (L^* L)^{1/2}$ while $\|L\|_2 = tr(L^* L)$ is an inner product norm, but the concept of closure of $H \otimes H$ (or $H^* \otimes H$) for some norm seems to be the same. $\endgroup$ – reuns May 12 '16 at 19:05

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.


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 $$||T-T_N||_1=\sum_{n=N+1}^\infty\lambda_n\to0\quad(N\to\infty).$$


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 $$\begin{aligned}\sum\left|\ip{Te_n}{f_n}\right| &=\sum_n\left|\ip{\sum_j\lambda_j\ip{e_n}{v_j}u_j}{f_n}\right| \\&\le\sum_j\lambda_j\sum_n\left|\ip{e_n}{v_j}\ip{u_j}{f_n}\right| \\&\le\sum_j\lambda_j\left(\sum_n\left|\ip{e_n}{v_j}\right|^2\right)^{1/2} \left(\sum_n\left|\ip{u_j}{f_n}\right|^2\right)^{1/2} \\&=\sum_j\lambda_j||v_j||\,||u_j|| \\&\le\sum_j\lambda_j \\&=||T||_1. \end{aligned}$$

  • $\begingroup$ Your Lemma is essentially the Hilbert-Schmidt theorem. Note that a bounded, linear operator $L$ on $H$ is called trace class, if $L$ is nonnegative and symmetric with finite trace. While it's clear that $L:=TT^\ast$ (with the $T$ defined in your answer) is nonnegative and symmetric, it won't have finite trace, since $$\operatorname{tr}L=\left\|T\right\|_{\operatorname{HS}(H)}^2\;,$$ where $\left\|\;\cdot\;\right\|_{\operatorname{HS}(H)}$ denotes the Hilbert-Schmidt norm on $\mathfrak L(H)$. $\endgroup$ – 0xbadf00d May 16 '16 at 12:11
  • $\begingroup$ So, while we know that $L^{1/2}$ does exist (and is nonnegative and symmetric too), I'm not sure whether or not we can show that $\operatorname{tr}\left(L^{1/2}\right)<\infty$. However, if you change your assumption on $T$, say that $T$ is Hilbert-Schmidt, then $\operatorname{tr}L<\infty$ and we can show that $$\left\|L^{1/2}\right\|_{\operatorname{HS}(H)}^2=\operatorname{tr}L\;,$$ i.e. that $L^{1/2}$ is Hilbert-Schmidt (and hence trace class) too. But that doesn't help for the question. If I've overlooked something, please correct me. $\endgroup$ – 0xbadf00d May 16 '16 at 12:11
  • $\begingroup$ @0xbadf00d Well yes, it appears to me that you've overlooked more or less everything, starting with the definition of trace class! Which is curious - a few days ago you asked if I was talking about H-S, I said no, I was talking about trace-class operators, and gave you a link to the definition - from your comment it seems like you didn't read the definition. In detail: (i) I've been talking about $L=T^*T$, not $TT^*$. (ii) I never said or implied or hinted that $tr(L)<\infty$. (iii) The biggie: You give the definition of trace class for positive operators. [...] $\endgroup$ – David C. Ullrich May 16 '16 at 12:35
  • $\begingroup$ @0xbadf00d The actual definition is much more general. By definition $T$ is trace class if $tr(L^{1/2})<\infty$. See the first two sentences of en.wikipedia.org/wiki/Trace_class . Or don't, but that's the definition of trace class I'm using above, exactly as I said a few days ago... $\endgroup$ – David C. Ullrich May 16 '16 at 12:38
  • $\begingroup$ @0xbadf00d I meant the first two sentences in the section titled "Definition". As long as I'm typing, a suggestion: Instead of saying my lemma is the same as that other lemma and then if we change the definition, etc, why not comment on what I actually wrote? Like what part of the proof of the lemma that I wrote seems wrong, or what part of the derivation that I wrote of the result from the lemma seems wrong... $\endgroup$ – David C. Ullrich May 16 '16 at 13:29

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:

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


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