Annihilator of a von Neumann algebra is closed with respect to trace class norm

Question

Let $A$ be a von Neumann algebra on a Hilbert space $H$ and define $$A^{\perp}:=\{v \in L^1(H)~|~\text{tr}(uv)=0~(u \in A)\}.$$ I want to show that $A^\perp$ is closed subspace of $L^1(H)$ with respect to trace class norm. Here $L^1(H)$ is the space of trace-class operators on $H$.
My approach, subspace is clear but about the closedness, let $(v_\lambda)_\lambda$ be a net in $A^\perp$ converging to $v$ in trace class norm, where trace class norm is defined as: If $E$ is an orthonormal basis of $H$ then $$\|v\|_1:=\sum_{x\in E} \langle |v|(x),x\rangle,~\text{ where }~|v|=(v^*v)^{1/2}.$$ Then we have, $\|v_\lambda -v\|_1 \to 0$ and since $v_\lambda \in A^\perp$ we have $\text{tr}(uv_\lambda)=0$ for $u\in A,$ and for all $\lambda$. I want to show that $\text{tr}(uv)=0,$ where trace of trace class operator $v$ is defined as $$\text{tr}(v)=\sum_{x\in E} \langle v(x),x\rangle,~\text{where $E$ is any orthonormal basis of $H$}.$$ Now note that, $$\|v_\lambda -v\|_1 \to 0 \implies \lim_{\lambda} \sum_{x\in E} \langle |v_\lambda-v|(x),x\rangle=0.$$ Can you please help me how to proceed from here if I am approaching in right direction. Thank you for your time.

Answer 1

Your approach is fine. What you need to do next is use this version of Hölder's inequality: $$ |\operatorname{tr}(vu)|\leq\|v\|_1\,\|u\|. $$ Proof here.