singular value decomposition of sum

Question

Let $A,B$ be positive, linear trace class operators on some Hilbert space. I would like to know if the following trace inequality for some $\mu>0$ is true $$ \mathrm{Tr}\!\left(A\,(A+B+\mu I)^{-1}\right)\leq \mathrm{Tr}\!\left(A\,(A+\mu I)^{-1}\right). $$ I am not sure if the pseudo inverse in this case is actually well defined? I am worrying abound finding a SVD where $A$ and $B$ have both positive eigenvalues. Is this possible in general?

Answer 1

There is no need to do a singular value decomposition here, it is sufficient to multiply and divide by $(A+\mu\,I)$ to get $$ \mathrm{Tr}\!\left(A\,(A+B+\mu I)^{-1}\right) = \mathrm{Tr}\!\left(A\,(A+\mu I)^{-1}\,(A+\mu I)\,(A+B+\mu I)^{-1}\right) \\ = \mathrm{Tr}\!\left(A\,(A+\mu I)^{-1}\,(A+\mu I)^{1/2}\,(A+B+\mu I)^{-1}\,\,(A+\mu I)^{1/2}\right) \\ \leq \mathrm{Tr}\!\left(|A\,(A+\mu I)^{-1}|\right) \|(A+\mu I)^{1/2}\,(A+B+\mu I)^{-1}\,(A+\mu I)^{1/2}\|_\infty $$ where $\|\cdot\|_{\infty}$ denotes the operator norm, $|C| = \sqrt{C^*C}$ and I used the cyclicity of the trace and the fact that operators that are functions of $A$ commute with each other. Since $A\geq 0$, so is $A\,(A+\mu I)^{-1}$ and so one can remove the absolute value from the last trace. On the other hand, since $A$ and $B$ are positive operators, it holds $$ 0 \leq A+\mu I \leq A+B+\mu I $$ in the sense of operators. Equivalently, for any $\varphi$ in your Hilbert space $$ \langle \varphi, (A+\mu I)\varphi\rangle = \|(A+\mu I)^{1/2}\varphi\| \leq \|(A+B+\mu I)^{1/2}\varphi\| $$ Taking $\psi = (A+B+\mu I)^{-1/2}\varphi $ one deduces that $\|(A+\mu I)^{1/2}\,(A+B+\mu I)^{-1/2}\|_\infty \leq 1$. The same being true for the adjoint, one deduce finally that $$ \|(A+\mu I)^{1/2}\,(A+B+\mu I)^{-1}\,(A+\mu I)^{1/2}\|_\infty \\ \leq \|(A+\mu I)^{1/2}\,(A+B+\mu I)^{-1/2}\|_\infty\, \|(A+B+\mu I)^{-1/2}\,(A+\mu I)^{1/2}\|_\infty $$ is smaller than $1$, concluding the proof.