Let $H$ be an Hilbert space, $\{u_n\}$ an orthogonal set , $\{v_n\}$ an orthonomal set and $k_n$ a sequence of complex numbers such that $\|k_nu_n\|<1.$ Consider the operator defined by
$$Tx=\sum_{n}k_n\langle x,u_n\rangle v_n.$$
I have already checked that this is well-defined and is a bounded operator thanks to the Bessel's Inequality. Now what I am not being able to see is that its spectrum lies inside the unit circle of $\mathbb{C}$, and I am a bit confused about this because I don't know if this explicitly means that $|\lambda|\leq 1$ or $|\lambda|=1$. I have tried seeing that $\langle Tx,x\rangle $ has always norm $1$, I have tried constructing explicit inverses for when $|\lambda|\neq 1$, but so far I got nothing, also it is kinda of annoying the fact that neither the $v_n$ or the $u_n$ is a basis for the whole space. I am bit out of ideas, so any help is appreciated just a hint to get me going.
Best Answer
So after some discussion here's my attempt at solving this
We will have that $||Tx||^2=<\sum\limits_{n=1}^{\infty}k_n<x,u_n>v_n,\sum_\limits{i=1}^{\infty}k_i<x,u_i>v_i> = \sum_{n=1}^{\infty}|k_n|^2|<x,u_n>|^2$, and if we consider the elements $u_n'=\frac{u_n}{||u_n||}$, we will have that this last series is equal to $\sum_{n=1}^{\infty}||k_nu_n||^2|<x,u_n'>|^2 \leq ||x||^2$ , using Bessel's Inequality and the fact that $||k_nu_n||<1$ for any $n$, and so we will have that $||T||\leq 1$ and so we get that the operator is well defined, continuous and that the spectrum lies inside the unit circle.