Let $A, B$ be a trace class operators on separable Hilbet space $\mathcal{H}$ such that $0\leq A\leq B\leq 1$ and $A \log A, B \log B$ be in the trace class.
Is it true that $\mbox{tr}(\overline{A \log B}) \ge \mbox{tr}(A \log A)$ ?
$\overline{A}-$ is a close of the operator $A.$
I know a proof for finite dimensional Hilbert space. Sketch of a proof:
$\log B \geq \log A \Rightarrow A^{1/2} (\log B) A^{1/2}\geq A^{1/2} (\log A) A^{1/2} \Rightarrow \mbox{tr}(A \log B) \ge \mbox{tr}(A \log A).$
I don't have an idea in infinite dimensional case.
I explain what I mean by $\log A:$
-For $x\in \overline{A(\mathcal{H})}$ define $(\log A)x:=(\log A_{|\overline{A(\mathcal{H})}})x.$
-For $x\in (\overline{A(\mathcal{H})})^{\perp}$ define $(\log A)x:=0.$
The operator $A\log B$ isn't defined on the whole space but the dense subspace of $\mathcal{H}$ so we can't compute the trace. I think $A\log B$ is bounded (I don't have a proof) then we can close it and $\overline{A\log B}$ will be bounded defined on the whole space.
The operator $A\log A$ is a standard defined by the spectral theorem $A\log A=\int_{[0,\| A \|]} (x \log x) dE_A (x),$ where $E_A$ is the spectral measure of $A.$ This is the correct definition because the function $x\mapsto x\log x$ is defined on the interval $[0,\infty).$ From this definition $A\log A$ is bounded and defined on the whole space $\mathcal{H}$ (because $x\log x$ is bounded).