# Confusion regarding the proof of $\mathscr{I}_1(H)$ (trace class operators on $H$) is Banach space with respect to $\lVert\cdot\rVert_1$.

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):

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.

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

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 $$X\mapsto\operatorname{tr}(TX)$$.