3
$\begingroup$

This question just struck me, is it true that if $1 <p <q <\infty$ , is the inclusion map

$$\ell^p \mathbb N \subset \ell^q \mathbb N$$compact ?

Hölders inequality gives us that the inclusion is continuous . But for compactness it doesn't seem very direct .

Thank you for ur help .

$\endgroup$
  • 1
    $\begingroup$ The answer is ‘no’, as shown below. However, any bounded linear operator from $ {\ell^{q}}(\mathbb{N}) $ into $ {\ell^{p}}(\mathbb{N}) $ must be compact. This is Pitt's Theorem. $\endgroup$ – Haskell Curry Jan 7 '13 at 21:13
4
$\begingroup$

No: let $e^{(n)}_k:=\delta_{nk}$. Then the sequence $\{e^{(n)}\}$ is bounded in $\ell^p$, for $1\leqslant p\leqslant \infty$. If $n_1\neq n_2$, then $$\lVert e^{n_1}-e^{n_2}\rVert_q=\begin{cases} 2^{1/q},&\mbox{if }1\leqslant q<\infty\\\ 1&\mbox{if }q=+\infty, \end{cases}$$

which proves that there is no convergent subsequence in $\ell^q$. So we actually can take $1\color{red}\leqslant p\color{red}\leqslant q\color{red}\leqslant +\infty$.

$\endgroup$
  • $\begingroup$ Thanks ! I was thinking the other way , by using the fact that if the map is compact , then if $x_k \to x$ weakly , then $Tx_k \to Tx$ in norm, but i couldn't think of a functional from $\ell^{p'}$ which would satisfy my claim . $\endgroup$ – Theorem Jan 7 '13 at 21:22
4
$\begingroup$

The sequence of standard basis vectors $(1,0,\ldots),(0,1,0,\ldots),(0,0,1,0,\ldots),\ldots$ is bounded in every $\ell^p$, but has no convergent subsequence in any of these spaces.

$\endgroup$

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.