weak operator topology convergence and the trace of spectral projections

Question

Let $B(H)$ be the algebra of all bounded self-adjoint linear operators on $H$. Let $\{T_i\}$ be a uniformly bounded net in $B(H)$ converging to $T$ in the weak operator topology. Let $Tr$ be the regular trace on $B(H)$. Is it true that $$sup_i Tr (e^{T_i}(\frac{\varepsilon}{2},\infty)) \ge Tr (e^{T}(\varepsilon,\infty)) $$ for any $\varepsilon>0$? Here $e^{T}(\varepsilon,\infty)$ denotes the spectral projection.

Answer 1

As requested in the comments, here is a proof in the case of strong instead of weak convergence (and for $\epsilon\geq 0$).

First note that the characteristic function of $(\epsilon,\infty)$ is lower semicontinuous, that is, there exists a sequence of positive continuous functions $f_k$ with compact support such that $f_k\nearrow 1_{(\epsilon,\infty)}$ pointwise. Since the functional calculus is continuous for such functions (Takesaki, Lemma II.4.3), it follows that $f_k(T_i)\to f_k(T)$ strongly for all $k\in\mathbb{N}$. As $\mathrm{Tr}$ is weakly lower semicontinuous on bounded sets, this convergence implies $$ \mathrm{Tr}(f_k(T))\leq \liminf_{i}\mathrm{Tr}(f_k(T_i))\leq \liminf_i \mathrm{Tr}(1_{(\epsilon,\infty)}(T_i)). $$ Finally, $f_k\nearrow 1_{(\epsilon,\infty)}$ implies $1_{(\epsilon,\infty)}(T)=\sup_k f_k(T)$. Thus we can use the normality of $\mathrm{Tr}$ to see that $$ \mathrm{Tr}(1_{(\epsilon,\infty)}(T))=\sup_{k\in\mathbb{N}}\mathrm{Tr}(f_k(T))\leq \liminf_i \mathrm{Tr}(1_{(\epsilon,\infty)}(T_i)). $$ Of course, if $\epsilon\geq 0$, we have $1_{(\epsilon,\infty)}\leq 1_{(\epsilon/2,\infty)}$, so this inequality implies the one in the original question.

This argument relies on the continuity of the functional calculus with respect to the strong operator topology. This continuity fails for the weak operator topology and I don't know how to proceed in this case (or if the inequality is even true).