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Let me first define some notations for convenience. Let $H$ be a Hilbert space and $\mathscr{I_\infty}(H)$ denotes the space of all compact operators on $H$. For $T\in\mathscr{I_\infty}(H)$ let $s_1(T)\ge s_2(T)\ge\cdots$ be a complete enumeration, multiplicity, of positive eigenvalues of $|T|:=\sqrt{T^*T}$. Then we have $$\displaystyle{s_n(T)=\min\limits_{\text{dim}(S)=n-1}\max\limits_{u\in S^\perp,\lVert u\rVert=1}\lVert Tu\rVert}$$

An operator $T\in\mathscr{I_\infty}(H)$ is said to be a trace class operator if $\lVert T\rVert_1:=\sum s_j(T)<\infty$ and we denote the space of all trace class operators on $H$ by $\mathscr{I_1}(H)$. We define, for $T\in\mathscr{I_1}(H)$, $\text{tr}\ T:=\sum\limits_j\langle e_j,Te_j\rangle$. I have proved that $$sup\{|\text{tr}\ TX|: X\in\mathscr{B}(H),\lVert X\rVert=1\}=\lVert T\rVert_1$$

The proof of $\mathscr{I_1}(H)$ forms a Banach space w.r.to. $\lVert\cdot\rVert_1$ goes as follows (from the book An Introduction to Quantum Stochastic Calculus by K.R. Parthasarathy): enter image description here

I understood that $T$ is trace class and the sequence of bounded linear functionals $X\mapsto\text{tr}\ T_nX$ on $\mathscr{B}(H)$ is a cauchy sequence, hence it converges to some $\lambda\in\mathscr{B}(H)^*$ i.e. $$\text{sup}\{|\text{tr}\ T_nX-\lambda(X)|:\ X\in\mathscr{B}(H),\lVert X\rVert=1\}\to0$$

Now choosing $X=|u\rangle\langle v|$ with $\lVert u\rVert=\lVert v\rVert=1$, we have $\lambda(X)=\text{tr}\ TX$. This shows that $\lambda(X)=\text{tr}\ TX$ for finite rank operator $X$, hence by continuity of $\lambda$ (with $\lVert\cdot\rVert$ on $\mathscr{B}(H)$), it follows that $\lambda(X)=\text{tr}\ TX$ for all $X\in\mathscr{I_\infty}(H)$. For the completion of the proof, this much is enough because I can prove that $$\text{sup}\{|\text{tr}\ TX|:\ X\in\mathscr{I_{\infty}}(H),\lVert X\rVert=1\}=\lVert T\rVert_1$$

Applying this, $$\lVert T_n-T\rVert_1=\text{sup}\{|\text{tr}\ (T_n-T)X|:\ X\in\mathscr{I_\infty}(H),\lVert X\rVert=1\}=\text{sup}\{|\text{tr}\ T_nX-\lambda(X)|:\ X\in\mathscr{I_\infty}(H),\lVert X\rVert=1\}\to0$$

This finishes our proof, but my question is why $\lambda(X)=\text{tr}\ TX$ for all $X\in\mathscr{B}(H)$? I know that finite rank operators are dense in $\mathscr{B}(H)$ with respect to strong operator topology. Is this $\lambda$ here continuous with respect to strong operator topology?

Can anyone help me in this regard? Thanks for your help in advance.

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  • $\begingroup$ Could you share the book that you are using? $\endgroup$
    – Akira
    May 29, 2023 at 18:42
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    $\begingroup$ An Introduction to Quantum Stochastic Calculus by K.R. Parthasarathy link.springer.com/book/10.1007/978-3-0348-0566-7 $\endgroup$
    – MathBS
    May 29, 2023 at 19:07
  • $\begingroup$ Thank you so much for your elaboration! $\endgroup$
    – Akira
    May 29, 2023 at 19:14
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    $\begingroup$ I should have mentioned it earlier, sorry I missed. I have modified it in the original post as well. $\endgroup$
    – MathBS
    May 29, 2023 at 19:23

1 Answer 1

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Like you said, you don't need this to complete the proof, so ignore it. It's enough to conclude that $\lambda(X)=\operatorname{tr}TX$ for all $X\in \mathscr{I}_\infty(H)$ and then use $$\sup\{|\operatorname{tr}SX|:X\in\mathscr{I}_\infty(H),\|X\|=1\}=\|S\|_1$$ for all $S\in \mathscr{I}_1(H)$.

The reason that it's true is that from this and possibly further material $$\mathscr{I}_\infty(H)^* \simeq \mathscr{I}_1(H)$$ $$\mathscr{I}_1(H)^* \simeq \mathscr{B}(H)$$ via the pairing $(T,X)\mapsto\operatorname{tr}(TX)$. So $$\mathscr{I}_1(H)^{**} \simeq \mathscr{B}(H)^*$$ Since $T_n\rightarrow T$ in the norm of $\mathscr{I}_1(H)$, and a Banach space is canonically identified with a closed subspace of its double-dual, it follows that $T_n\rightarrow T$ when they are viewed as functionals on $\mathscr{B}(H)$ by the action $X\mapsto\operatorname{tr}(TX)$.

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  • $\begingroup$ Thank you so much for your elaboration $\endgroup$
    – MathBS
    May 30, 2023 at 7:34

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