Cesàro summation in Hilbert space

Question

Let $H$ be Hilbert space, $\{e_n\}_{n\in \mathbb{Z}}$ be complete orthonormal basis, and bounded linear operator $U$ with $Ue_n=e_{n+1}$. Then, for any comact operator $T$, $$\lim _{n \to \infty } ||\frac{1}{n+1} \sum _{k=0} ^{n} U^k T U^{-k} ||=0$$

It is enough to prove $\lim _{k\to \infty }||U^k T U^{-k} ||=0$. $\{e_n\}$ is weak convergence sequence to $0$ by Bessel's inequality, so ${Te_n}$ is strong convergence sequence to $0$. I tried to use it, but $||U^k T U^{-k} || \leq \sum _{m \in \mathbb{Z}} ||Te_{m-k}||$ appear, so it is not convergent to $0$ as $k\to \infty$.

Someone have an idea to prove ?

Answer 1

Since Unitary operator, $||U^k TU^{-k}||=||T||$, so it is not convergent to $0$. By using spectrum, it is more easy to show

Answer 2

$\newcommand\s{\frac1n\sum_{k=1}^n}$ $\newcommand\ss[1]{\s U^k#1U^{-k}}$

Edit: The reason this makes no sense is it's an answer to a totally different question. The question was edited; the new question has nothing whatever to do with the original.

Original question, I think, based on the answer: Say $H$ is a Hilbert space, $U$ is a unitary operator on $H$ and $T$ is a compact operator on $H$. Show that $$\lim_{n\to\infty}\left|\left|\frac 1n\sum_{j=1}^n U^jTU^{-j}\right|\right|=0.$$

Comment in the original question: This would follow if $||U^j TU^{-j}||\to0$. How do we show that?

Answer:

Since $U$ is an isometric isomorphism it's clear that $||U^k TU^{-k}||=||T||$, so your plan won't work.

Since $$\left|\left|\ss T-\ss{T'}\right|\right|\le||T-T'||\quad(*)$$you can assume that $T$ has finite rank. So $T$ is a finite sum of operators of rank one, hence you can assume that $T$ has rank one: $$Tx=u<x,v>.$$So $$U^kTU^{-k}x=U^k u<U^{-k}x,v>.$$Now again $(*)$ shows that you can assume that $u$ is a finite linear combination of basis elements, and then that you can assume the same of $v$. So you can assume that $u=e_j$ and $v=e_m$, and now $$\begin{align}\left|\left|\ss Tx\right|\right|^2&=\frac1{n^2}\sum_{k=1}^n|<U^{-k},x>|^2 \\&=\frac1{n^2}\sum_{k=1}^n|<x,U^k v>|^2 \\&\le\frac1{n^2}||x||^2.\end{align}$$