Deducing an operator inequality over a complex Hilbert space

Question

Let $0\ne x\in H$ be a vector in a complex Hilbert space and define $w=(\sigma-A)^{-1}x$, where $A\in B(H)$ is a continuous linear operator on $H$ and $\sigma$ is another vector. Let $\nu$ be a complex scalar, then, how does one deduce the equality

$$ \frac{1}{2\pi}\left(\langle\nu w,(\sigma-A)w\rangle+\langle\bar{\nu}(\sigma-A)w,w\rangle\right)=\frac{\|w\|^2}{\pi}\operatorname{Re}\left[\nu\left(\bar{\sigma}-\frac{\langle w,Aw\rangle}{\|w\|^2}\right)\right] $$

Answer 1

Notice that $$\langle \nu w, (\sigma-A)w\rangle = \nu\langle w, (\sigma-A)w\rangle = \overline{\overline{\nu}\langle(\sigma-A)w, w\rangle } = \overline{\langle\overline{\nu}(\sigma-A)w, w\rangle}$$ so

\begin{align} \frac1{2\pi}\Big(\langle \nu w, (\sigma-A)w\rangle + \langle\overline{\nu}(\sigma-A)w, w\rangle\Big) &= \frac1{2\pi} \cdot2\operatorname{Re}\langle \nu w, (\sigma - A)w\rangle\\ &= \frac1\pi\operatorname{Re}\Big[\nu\langle w, \sigma w - Aw\rangle \Big]\\ &= \frac1\pi\operatorname{Re}\Big[\nu\left(\overline{\sigma}\langle w, w\rangle - \langle w, Aw\rangle \right)\Big]\\ &= \frac1\pi\operatorname{Re}\Big[\nu\left(\overline{\sigma}\|w\|^2 - \langle w, Aw\rangle \right)\Big]\\ &= \frac{\|w\|^2}{\pi}\operatorname{Re}\left[\nu\left(\overline{\sigma} - \frac{\langle w, Aw\rangle}{\|w\|^2} \right)\right]\\ \end{align}

Note that $w \ne 0$ because $x \ne 0$ and $\sigma - A$ is bijective.