# Prove that for a bounded operator $A$, then $\sum_{n=1}^\infty |a_{mn}|^2 \leq \|A\|^2$.

Suppose $A$ is some bounded operator on the Hilbert space of $\ell^2(\mathbb{N})$, with a representing matrix $(a_{mn})_{m,n=1}^\infty$. How can I prove that $$\sum_{n=1}^\infty |a_{mn}|^2 \leq \|A\|^2$$

I have started by noting that since $A$ is a bounded operator, then $\|A\| = \|A^\ast\|$ (the adjoint operator of $A$). The representing matrix of such an adjoint would have entries $((\overline{a_{mn}})_{m,n=1}^\infty)^t$. Could I perhaps use these two ideas in order to prove the result. Any help is appreciated!

• If $e_m$ is the m-th basis vector of the standard ONB of $l^2(\mathbb{N})$, look at $\| Ae_m\|^2$. – agb Apr 20 '17 at 10:04

Let $(e_n)_n$ the orthogonal basis of $l^2(\mathbb{N})$ we have by definition: $$A(e_n)=\sum_{j\geq 1} a_{nj}e_j\qquad \forall j\in \mathbb{N}$$ So $$\|A(e_n)\|^2=\left\|\sum_{j\geq 1} a_{nj}e_j \right\|^2=\sum_{j\geq 1} \left\|a_{nj}e_j \right\|^2=\sum_{j\geq 1} |a_{nj}|^2$$ $$\|A(e_n)\|^2\leq \|A\|^2 \|e_n\|^2=\|A\|^2$$ and