# $s \in L^{1}(H)$ $\iff$ $s=\sum_{i=0}^\infty x_{i} \otimes y_{i}$ [closed]

Let $H$ be a separable Hilbert space, and let $L^1(H)$ be the space of trace-class operators on $H$. I'd like to prove that $s\in L^{1}(H)$ if and only if there exists $\{ x_{i} \} , \{ y_{i} \} \subseteq H$ such that $\sum_{i=0}^\infty || y_{i}||^{2} < \infty$ , $\sum_{i=0}^\infty || x_{i}||^{2} < \infty$ and for which $s=\sum_{i=0}^\infty x_{i} \otimes y_{i}$ (where $x \otimes y$ denotes the operator $z \mapsto \langle z,y\rangle x$). How do I do this?

• What is your question? – Ahaan S. Rungta Dec 17 '14 at 16:03
• Can you explain what your notation means? What is $H$? What is $L^1(H)$? If you can explain your notation and what precisely you're asking, we'll be able to help. – user98602 Dec 17 '14 at 16:56
• It is clear from the context that $H$ is a separable Hilbert space, $L^1 (H)$ are the trace-class operators, and $x\otimes y$ is the rank-one operator $z\mapsto \langle z,y\rangle\,x$. – Martin Argerami Dec 18 '14 at 1:52
• One direction is clear- if $s$ is of the given form, then $s$ is trace class. For the other direction, I think you can approximate $s$ by finite rank operators, and see what that gives you. – voldemort Dec 20 '14 at 19:19

I think that Proposition B.17 here might answer your question. To sum up: if $(e_n)_{n\geq 1}$ is an orthonormal basis for your Hilbert space $H$, then for all trace class operators $T\in L^1(H)$, we consider the functional $$\varphi_T\colon S\mapsto\mathrm{tr}(TS)=\sum_{n=1}^\infty\langle TSe_n,e_n\rangle,\quad S\in B(H).$$ The functional $\mathrm{tr}$ is called the trace, and it satisfies $\mathrm{tr}(BC)=\mathrm{tr}(CB)$ for all Hilbert-Schmidt operators $B,C\in B(H)$, namely operators satisfying $\sum_{n=1}^\infty\|Be_n\|^2<\infty$.
Any trace class operator $T$ is the product of two Hilbert-Schmidt operators, say, $T=BA^*$ (Prop. B.8). Since $B$ and $A^*S$ are both Hilbert-Schmidt for all $S\in B(H)$ (since the set of Hilbert-Schmidt operators is an ideal, see Prop. B.7) we have $$\varphi_T(S)=\mathrm{tr}(TS)=\mathrm{tr}(BA^*S)=\mathrm{tr}(A^*SB)=\sum_{n=1}^\infty\langle SBe_n,Ae_n\rangle.$$ Note also that $$\varphi_{Be_i\otimes Ae_i}(S)=\sum_{n=1}^\infty\langle (Be_i\otimes Ae_i)Se_n,e_n\rangle=\sum_{n=1}^\infty\langle\langle Se_n,Ae_i\rangle Be_i,e_n\rangle=\sum_{n=1}^\infty\langle Be_i,\langle Ae_i,Se_n\rangle e_n\rangle=\left\langle Be_i,\sum_{n=1}^\infty\langle S^*Ae_i,e_n\rangle e_n\right\rangle=\langle Be_i,S^*Ae_i\rangle=\langle SBe_i,Ae_i\rangle,$$ so we have $$\varphi_T(S)=\sum_{i=1}^\infty\varphi_{Be_i\otimes Ae_i}(S).$$ Here comes the tricky part: the map $\varphi\colon T\mapsto\varphi_T$ of $L^1(H)$ into $B(H)^*$ is actually an isometry when $L^1(H)$ is equipped with the trace norm $\|T\|_1=\mathrm{tr}|T|$ -- therefore it is injective. Moreover, $\sum_{i=1}^\infty Be_i\otimes Ae_i$ converges in the trace-norm, since $$\sum_{i=1}^\infty\|Be_i\otimes Ae_i\|_1=\sum_{i=1}^\infty\|Be_i\|\|Ae_i\|\leq\left(\sum_{i=1}^\infty\|Be_i\|^2\right)^{1/2}\left(\sum_{i=1}^\infty\|Ae_i\|^2\right)^{1/2}<\infty$$ by Cauchy-Schwarz, so we actually have $$\varphi_T=\sum_{i=1}^\infty\varphi_{Be_i\otimes Ae_i}=\varphi_{\sum_{i=1}^\infty Be_i\otimes Ae_i}$$ by continuity of $\varphi$ and thus $T=\sum_{i=1}^\infty Be_i\otimes Ae_i$ by injectivity of $\varphi$.
Conversely, if $(x_n)_{n\geq 1}$ and $(y_n)_{n\geq 1}$ are sequences such that $\sum_{n=1}^\infty\|x_n\|^2<\infty$ and $\sum_{n=1}^\infty\|y_n\|^2<\infty$, then the partial sums of $\sum_{n=1}^\infty x_n\otimes y_n$ are finite rank and hence trace class, and as above, the series converges absolutely in the trace norm (i.e., $\sum_{n=1}^\infty\|x_n\otimes y_n\|_1<\infty$), so it must also be trace class. Recall that the trace norm $\|\cdot\|_1$ majorizes the operator norm $\|\cdot\|$, so that the series also defines a bounded operator.