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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 .

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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. –  Haskell Curry Jan 7 '13 at 21:13

2 Answers 2

up vote 4 down vote accepted

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 $1\color{red}\leqslant p\color{red}\leqslant q\color{red}\leqslant +\infty$.

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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 . –  Theorem Jan 7 '13 at 21:22

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.

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