An identity of inner product

Question

I am trying to prove the following identity $$\|\psi\|^2=\|A(A+iaI)^{-1}\psi\|^2+a^2\|(A+iaI)^{-1}\psi\|^2$$ where $A$ is self-adjoint and $a$ is a real number.

My approach is to rewrite $A(A+iaI)^{-1}$ as $$(A+iaI-iaI)(A+iaI)^{-1}=I-ia(A+iaI)^{-1}$$ and so $$\|A(A+iaI)^{-1}\psi\|^2=\langle\psi-ia(A+iaI)^{-1}\psi,\psi-ia(A+iaI)^{-1}\psi\rangle$$ $$=\|\psi\|^2+a^2\|(A+iaI)^{-1}\psi\|^2+\langle\psi,-ia(A+iaI)^{-1}\psi\rangle+\langle-ia(A+iaI)^{-1}\psi,\psi\rangle$$

Then I get stuck here. I cannot make the last two terms to vanish, and it seems that, except for the cross-terms, there is a difference in the minus sign.

Can anyone give a hint about it?

Answer 1

Suppose first that the Hilbert space is $L^2(\Omega, {\mathcal B}, \mu)$ and $A$ is multiplication by a real-valued bounded measurable function $f(x)$. For every $x \in \Omega$, $$\left|\frac{f(x)\cdot\psi(x) }{f(x)+ia} \right|^2+a^2\left|\frac{\psi(x)}{f(x)+ia} \right|^2=\frac{f(x)^2|\psi(x)|^2}{f(x)^2 +a^2}+\frac{f(x)^2|\psi(x)|^2}{f(x)^2 +a^2}=|\psi(x)|^2 \,,$$ so the identity holds in this case by integrating both sides over $\Omega$ with respect to $\mu$.

By the spectral theorem (see [1]) every self-adjoint operator is unitarily equivalent to a multiplication operator, so the general case follows.

[1]http://homepages.math.uic.edu/~furman/4students/halmos.pdf

Answer 2

Hint: Rearrange your identity first as

$$I= A(A+iaI)^{-1} + ia(A+iaI)^{-1}$$

and then apply your calculation to $\|\psi\|^2$ and use the self-adjointness of A and $a \in \mathbb{R}$.