Let $\mathscr{H}$ be a separable complex Hilbert space and $A$ be a self-adjoint bounded operator on $\mathscr{H}$. Let $h:\mathbb{R} \rightarrow [0,\infty)$ be convex. Then is it true that $$ \operatorname{tr}{h(A)}=\sup \sum_n h(\langle x_n|A x_n\rangle) $$ where the supremum is taken over othonormal-families $x_n$.
This is trivial for finite-dim $\mathscr{H}$, but what about infinite-dim? I would hope that this is true regardless of $\operatorname{tr}{h(A)} =\infty$ or $<\infty$. However, I would be glad with just proving that this is true if $h(A)$ is in trace-class.
On another note, are there any good references that deal with infinite-dim trace inequalities?
Note that $h$ is continuous. By functional calculus (or the Spectral Theorem), the operator $B=h(A)$ is positive. We have $$\tag1 \operatorname{tr}(B)=\sum_n\langle x_n,Bx_n\rangle $$ for any orthonormal basis $\{x_n\}$. Now if $\{y_n\}$ is an orthonormal set, choose $\{z_n\}$ such that $\{y_n\}\cup\{z_n\}$ is an orthonormal basis. Then, since $\langle x,Bx\rangle\geq0$ for all $x$, $$\tag2 \sum_n\langle y_n,By_n\rangle\leq \sum_n\langle y_n,By_n\rangle+\sum_n\langle z_n,Bz_n\rangle=\operatorname{tr}(B). $$