Exercise Methods of modern mathematical physics Reed Simon

Question

I encountered this exercise in Methods of modern mathematical physics by Reed & Simon. Exercise 22 Chapter 2 Volume 1

Let $\{ \eta_n \}_{n=-\infty}^\infty$ be a set of vectors in the Hilbert space, $l_2(\mathbb{Z})$, so that $a_{nm}= \lvert \langle \eta_n, \eta_m \rangle \rvert$ is the matrix (in natural basis) of an operator $A$ on $l_2(\mathbb{Z})$. Prove that $$ \sum_{n=-\infty}^\infty \| \langle f, \eta_n \rangle \|^2 \leq \lVert A \rVert \lVert f \rVert^2 $$ for any $f \in l_2(\mathbb{Z})$.

Anyone willing to give a push in the right direction?

Answer 1

The answer can be seen to follow from the case in the comment.

Let $A$, $\eta_n$ be as in the question. Define $C$ to be the operator with matrix coefficients $c_{nm}=(\eta_n,\eta_m)$, this is still a bounded operator and we have seen $\sum_n |(f,\eta_n)|^2≤\|C\|\,\|f\|^2$. We want to see $\|C\|≤\|A\|$.

Let $f$ be so that $\|Cf\|≥\|C\|\,\|f\|-\epsilon$. Define $f^*:=\sum_n |f_n|\,e_n$, note $\|f\|=\|f^*\|$ and $|c_{nm}|=a_{nm}$. One has: $$\|Cf\|^2=\sum_n \left|\sum_m c_{nm}\,f_m\right|^2≤\sum_n\left(\sum_m |c_{nm}\,f_m|\right)^2=\|Af^*\|^2$$ it follows: $$\|Af^*\|≥\|C\|\,\|f\|-\epsilon = \|C\|\,\|f^*\|-\epsilon$$ this holds for any $\epsilon$, so $\|A\|≥\|C\|$.