I have the following operator on a Hilbert space:
$$T=\sum_{n=0}^{\infty}c_{n}\langle a_{n}\mid \cdot\rangle b_{n},$$
where the $c_{n}$ is a bounded sequence of complex numbers and $a_{n},b_{n}$ are two orthonormal families.
I have to show that the operator norm of $T$ is the maximum of the $c_{n}$.
Using the definition of the operator norm I have after some steps:
$$\Vert T\Vert=...=\sup_{\Vert x\Vert =1}\sqrt{\sum_{n=0}^{\infty}\vert c_{n}\vert^{2}\vert\langle a_{n}\mid x\rangle\vert^{2}}$$
I don`t know how to proceed. I should find that $\Vert T\Vert = \max_{n\in\mathbb{N}}\vert c_{n}\vert$. So I had the idea to show $\leq$ and $\geq$. Using the Bessel's inequality I get $\Vert T\Vert\leq \max_{n\in\mathbb{N}}\vert c_{n}\vert$, but I don't how to show the other direction ....
You have that $\|T\|^2=\|T^*T\|$. And $$ T^*Tx=\sum_{n,m}c_n\overline{c_m} \overline{\langle a_n,x\rangle}\,\langle b_n,b_m \rangle\, a_n =\sum_n|c_n|^2\,\langle x,a_n\rangle\,a_n. $$ From this you see that, since the maps $\langle\cdot,a_n\rangle a_n$ are a family of pairwise orthogonal projections, $$ \|T^*T\|=\max\{|c_n|^2:\ n\}. $$