# Prove that T is compact

If $H$ is a Hilbert space with basis $\{\varphi_{k}\}^{\infty}_{k=1}$, how do I show that the operator $T$ defined by $T(\varphi_{k})=\frac{1}{k}\varphi_{k+1}$ is compact and has no eigenvectors? Thanks.

• Check this. – Mhenni Benghorbal Sep 14 '14 at 19:06
• @MhenniBenghorbal I looked at your link, but I am still kind of unsure how to proceed. Could you give me a hint on how I should proceed? – Sanjay Sep 14 '14 at 19:37
• Can anyone help with this? – Sanjay Sep 14 '14 at 21:17
• I proved that there cannot be any eigenvectors, but I still don't see how to prove its compactness... – Sanjay Sep 14 '14 at 21:39

Define the operator $T_n$ by $T_n(\phi_k) = \frac 1k \phi_{k+1}$ if $k \leq n$ and $T_n(\phi_k) = 0$ for $k > n$. Let $x = \sum x_k \phi_k$ be an element in $H$. We want to evaluate $\|(T-T_n)x\|$. This is given by $$(T-T_n)x = \sum_{k \geq n+1} \frac {x_k}k \phi_{k+1}.$$
Use Cauchy-Schwarz inequality to obtain an estimate on $\|T-T_n\|$. This will show that this operator norm tends to $0$ when $n$ tends to infinity. Since the $T_n$ are of finite-rank by construction, this concludes the proof.