Let $H$ be a separable Hilbert space, and let $A$ and $B$ be trace-class operators on $H$ such that $A^{-1/2}B$ is a Hilbert-Schmidt operator. Then is it possible to know whether the operator $B'A^{-1}B$ is bounded, or, more specifically, trace-class? ($A$ is assumed to be self-adjoint.)
The motivation is the following. According to Theorem 1.(A) of this paper, the expression $R_X^{-1/2}R_{XY} = VR_Y^{1/2}$ is valid, where $R_X, R_Y, R_{XY}$ are the variance operators of $X$ and of $Y$, and the covariance operator of $X$ and $Y$, respectively. Morevoer, the operator $VR_Y^{1/2}$ on the right-hand side is Hilbert-Schmidt because $V$ is bounded with norm less than 1 and because $R_Y^{1/2}$ is Hilbert-Schmidt. But taking adjoints of both sides of that expression is not valid because the operator $R_X^{-1/2}$ on the left-hand side is defined only on a proper dense subspace of $H$. But my conjecture is that the equality $R_{YX}R_X^{-1}R_{XY} = R_Y^{1/2}V'VR_Y^{1/2}$ might be true somehow, so that the operator $R_{YY} – R_{YX}R_X^{-1}R_{XY}$ is well-defined as a trace-class operator just as in the finite-dimensional case.
Best Answer
Consider pairwise orthogonal rank-one projections $\{P_k\}$ and define $$ A=\sum_k \frac1{k^{3/2}}\,P_K. $$ Then $A$ is trace-class. Fix an orthonormal basis $\{e_k\}$ with $P_ke_k=e_k$. Let $$ x=\sum_k\frac1{k^{3/2}}\,e_k. $$ Let $B$ be the rank-one projection onto $\mathbb C x$. Then $$ A^{-1/2}Bx=A^{-1/2}x=\sum_k\Big(\frac1{k^{3/2}}\Big)^{-1/2}\,\frac1{k^{3/2}}\,e_k=\sum_k\frac1{k^{3/4}}\,e_k. $$ Then $A^{-1/2}B$ exists and it is finite-rank, so in particular it is Hilbert-Schmidt. But $A^{-1}B$ cannot exist, for $$ A^{-1}Bx=A^{-1}x=\sum_k k^{3/2}\,\frac1{k^{3/2}}\,e_k=\sum_k \,e_k $$ More properly, the domain of $A^{-1}$ are those $y=\sum_ky_ke_k$ such that $$ \sum_k k^3\,|y_k|^2<\infty. $$ This puts the range of $B$ outside of the domain of $A^{-1}$.