Calculate Operator norm of a given operator

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 ....

• It seems like the RHS is missing a square root. Commented Dec 6, 2020 at 18:31
• Thanks thats true.... Commented Dec 6, 2020 at 18:32
• Once you have that, you get $\|T\| \leq \max |c_n| := \lambda$. Pix $x = a_{N(\lambda)}$ where $N(\lambda)$ is such that $c_{N(\lambda)} = \max |c_n|$. Then, $\|x\| = 1$ and $\|Tx\| = \max |c_n|$. Commented Dec 6, 2020 at 18:35
• You are allowed to pull $\max |c_n|^2$ out of the sum getting something bigger.
– Ruy
Commented Dec 6, 2020 at 18:36
• Thanks a lot so far.... What if my sequence doesn't have a maximum? In this case we should find $\Vert T\Vert =\sup_{n} \vert c_{n} \vert$ I guess.... Do you know how to argue in this case? Commented Dec 6, 2020 at 19:03

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, $$\|T^*T\|=\max\{|c_n|^2:\ n\}.$$