How can we show that $\langle (LQ)(LQ)^\ast h,g\rangle_H=\sum_{n\in\mathbb N}\langle h,LQf_n\rangle_H\langle g,LQf_n\rangle_H$?

Question

Let

• $U,H$ be separable $\mathbb R$-Hilbert spaces
• $(f_n)_{n\in\mathbb N}$ be an orthonormal basis of $U$
• $L$ be a bounded linear operator from $U$ to $H$
• $Q$ be a bounded linear operator on $U$

How can we show that $$\langle (LQ)(LQ)^\ast h,g\rangle_H=\sum_{n\in\mathbb N}\langle h,LQf_n\rangle_H\langle g,LQf_n\rangle_H\tag1$$ for all $h,g\in H$? I guess this is easy, but I don't get how we obtain it.

Answer 1

Any vector $y$ in $U$ satisfies $$ y=\sum_n\langle y,f_n\rangle f_n $$ (precisely because $\{f_n\}$ is orthonormal. Then coefficients are $\ell^2$, so series manipulations are ok. Then \begin{align} \langle LQ(LQ)^*h,g\rangle&=\langle (LQ)^*h,(LQ)^*g\rangle =\sum_{m,n}\langle (LQ)^*h,f_m\rangle\,\overline{\langle (LQ)^*g,f_n\rangle}\,\langle f_m,f_n\rangle\\ \ \\ &=\sum_{n}\langle (LQ)^*h,f_n\rangle\,\overline{\langle (LQ)^*g,f_n\rangle}\\ \ \\ &=\sum_{n}\langle h,LQf_n\rangle\,\overline{\langle g,LQf_n\rangle}. \end{align}