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Inclusion of $L^p$ spaces for functions has been discussed here.
Does this apply to $l^p$ space of sequences similarly?

I tried to show the following: For $1\leq p<q<\infty$, $l^q\subset l^p$ By using Hölder inequality but it doesn't seem to work.

My question is that is this true? If yes, what's the right way to prove it and what's a good counter example for showing $l^p\subset l^q$ is not true? Thanks.

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Spock
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2 Answers2

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If $\sum_n |x_n|^p < \infty$, then $|x_n| \leq 1$ definitely. Therefore, if $q>p$, then $|x_n|^q \leq |x_n|^p$, and we conclude that $\ell^p \subset \ell^q$.

Now the opposite embedding can't be true, otherwise $\ell^p \simeq \ell^q$ for every pair $(p,q)$, and this is obviously false.

Siminore
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  • $|x_n|^q \leq |x_n|^p$ is true. but how do we control their powers since we have $(\sum_n |x_n|^p)^{1/p}$? – Spock Feb 09 '14 at 09:20
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    Also, how is $|x_n| \leq 1$ if we have a sequence with $2$ in the first term and infinite $0$s? – Spock Feb 09 '14 at 09:29
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    Come on, you can do better that this! I wrote definitely, i.e. beyond some large index. You should not care of the first terms, since they can be always bounded by some fixed constant. Moreover, $|x_n|^q \leq |x_n|^p$ implies that $|x_n|_q^q \leq |x_n|_p^p$ by summation. Hence the injection $\imath \colon \ell^p \to \ell^q$ is continuous. – Siminore Feb 09 '14 at 13:29
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    "can't be true because it's obviously false" does not make a good argument. You also commit a logical fallacy saying "for every pair". Maybe it's true for some pairs, not for all of them. –  Nov 15 '15 at 07:45
  • " I wrote definitely, i.e. beyond some large index": I think you meant eventually, not "definitely". And "$\ell^p \simeq \ell^q$" should be $\ell^p=\ell^q$. – Anne Bauval Jan 04 '23 at 23:35
  • why does the inclusion follow from $|x_{n}|^{q}≤|x_{n}|^{p} $? – hzm Jan 07 '23 at 17:41
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    @AnneBauval Of course, typical false friend for italians (we say "definitivamente" to mean "eventually"). – Siminore Jan 10 '23 at 16:40
  • @Siminore How funny, for us french, "éventuellement" means "possibly", and "définitivement" means "fixed for ever". – Anne Bauval Jan 10 '23 at 16:46
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For completeness, a counterexample for $p>q$: pick $s\in (1/p,1/q)$ and consider the sequence $x_n=1/n^s$. This sequence is in $\ell^p$ but not in $\ell^q$.