Prove that T is compact

Question

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.

Answer 1

A convenient to show that an operator is compact is to show that it is the limit (for the operator norm) of operators of finite rank.

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.