2
$\begingroup$

Given an operator $T : l^p \rightarrow l^p$ ($1 \leq p < \infty$) s.t. $Ty= (\beta_1 y_1, \beta_2 y_2, \dots)$ for $y=(y_1, y_2, \dots) \in l^p$, where $\beta= (\beta_1, \beta_2, \dots)$ is bounded, show that if $T$ is compact, then $\lim \beta_n \rightarrow 0$.

I have shown this in the opposite direction, but this way is giving me some trouble. Could someone point me in the right direction?

$\endgroup$
  • 1
    $\begingroup$ I'm not clear about your notation: are you considering vectors with $n$ components or infinite sequences in $l^p$? $\endgroup$ – Rudy the Reindeer Feb 6 '17 at 15:13
  • 1
    $\begingroup$ Hint: If $e_k$ is the standard Schauder Basis of $\ell^p(\Bbb N)$ consider $x_n=e_n$. From $T$ compact it follows $T(B_1(0))$ must be precompact. If you look at $T(x_n)=\beta_n e_n$ and you suppose $\beta_n\not\to0$, what can you show? $\endgroup$ – s.harp Feb 6 '17 at 15:16
  • $\begingroup$ @RudytheReindeer They are infinite sequences. Sorry, my notation was unfortunate. I have changed it now. $\endgroup$ – Henrymerrild Feb 6 '17 at 15:50
  • $\begingroup$ @s.harp Thanks for your reply. Should I invoke that $\sup ||T(e_n)|| < \infty$ or am I heading in the wrong direction? $\endgroup$ – Henrymerrild Feb 6 '17 at 15:51
0
$\begingroup$

Let $e_n$ be the standard basis of $\ell^p(\Bbb N)$. You find $T(e_n)=\beta_n\,e_n$. Suppose that $\beta_n\not\to0$, then there is a subsequence and an $\epsilon>0$ $\beta_{n_k}$ so that $\beta_{n_k}>\epsilon>0$ for all $k$. It follows for any $k,l$ so that $k\neq l$: $$\|T(x_{n_k})-T(x_{n_l})\|=\sqrt[p]{|\beta_{n_k}|^p+|\beta_{n_l}|^{p}}≥\sqrt[p]2 \,\epsilon$$ Thus $T(x_{n_k})$ admits no subsequence that is a Cauchy sequence, and so also no convergent subsequence. But if $T$ is compact $T(\overline{B_1(0)})$ must be pre-compact, thus every sequence in it must have a convergent subsequence. Since $\|x_n\|=1$ it follows that $T(x_{n_k})$ is a sequence in $T(\overline{B_1(0)})$ that has no convergent sub-sequence, contradicting compactness of $T$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you, I had just worked out the first step of what you wrote, when you gave this answer. I appreciate it :) $\endgroup$ – Henrymerrild Feb 6 '17 at 16:28
0
$\begingroup$

$T$ is compact, so the closure of $\left \{ T\vec e_i=\beta_i\vec e_i :i\in \mathbb N\right \}$ is compact, and so contains a Cauchy subsequence,

$\left \| \beta_{m_k}\vec e_{m_k}- \beta_{n_k}\vec e_{n_k}\right \|=\sqrt[p]{|\beta_{n_k}|^p+|\beta_{m_k}|^{p}}$

and since the the LHS is as small as we want for $n,m$ large enough, it is clear that RHS$\to 0$ as $k\to \infty,\ $ which implies that $\beta_{n_k}\to 0$ as $k\to \infty.$ This means that $\left \{ \beta_i \right \}$ is not finite.

To finish, observe that $\beta_i\in P\sigma (T)$ for each $i\in \mathbb N,\ $ and so that the only possible limit point for $\left \{ \beta_i \right \}$ is $0$.

| cite | improve this answer | |
$\endgroup$

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.