0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Check this. $\endgroup$ – Mhenni Benghorbal Sep 14 '14 at 19:06
  • $\begingroup$ @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? $\endgroup$ – Sanjay Sep 14 '14 at 19:37
  • $\begingroup$ Can anyone help with this? $\endgroup$ – Sanjay Sep 14 '14 at 21:17
  • $\begingroup$ I proved that there cannot be any eigenvectors, but I still don't see how to prove its compactness... $\endgroup$ – Sanjay Sep 14 '14 at 21:39
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.