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Suppose $\Sigma = \mathcal{P}(\Omega) < \infty$ and $\mu$ is the counting measure. I'm seeking to show that $l^q \subset l^p$ for $1 \leq q\leq p \leq \infty$. The main obstacle to a proof is showing $$ \left(\sum_{\omega \in \Omega} |\omega|^q\right)^{1/q} \leq \left(\sum_{\omega \in \Omega} |\omega|^p\right)^{1/p}$$. Could someone point me in the right direction? Thanks in advance!

anegligibleperson
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2 Answers2

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Try proving it first in the case where $\left(\sum_{\omega \in \Omega} |\omega|^p\right)^{1/p} = 1$. Can you see how the general case follows from that?

Mikko Korhonen
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  • Thanks, but how does one show the case when the right hand side is equal to 1? After that the general result follows from homogeneity of the norm. – anegligibleperson Nov 24 '13 at 09:21
  • In that case you can see that $|\omega| \leq 1$ for all $\omega \in \Omega$ (since $|\omega|^p \leq 1$), hence $|\omega|^q \leq |\omega|^p$ for all $\omega \in \Omega$ because $q \leq p$. – Mikko Korhonen Nov 24 '13 at 09:26
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If $(w_n) \in \ell^q$, then $|w_n| < 1$ must hold eventually, hence you may assume that $|w_n| < 1$ for all $n$, whence $$ |w_n|^p \leq |w_n|^q $$ So $\sum |w_n|^p < \infty$ as well.

Prahlad Vaidyanathan
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