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Let $a=\lbrace a_n \rbrace_{n=1}^\infty$ be a sequence of real numbers. We define $\| a\|_p = \left( \sum_{n=1}^\infty a_k^p \right)^{1/p}$ for $0<p<\infty$ and $\| a\|_\infty = \sup_k |a_k|$. If the series is finite we say that $a$ belongs to $\ell^p$. I want to show that $\ell^{p_1}\subset \ell^{p_2}$ when $0<p_1<p_2\leq \infty$. It is easy to see that $\ell^p\subset \ell^\infty$ because if $a$ is $\ell^p$ then the sequence $\lbrace a_n \rbrace_{n=1}^\infty$ tends to zero and consists of real numbers, so there is a maximum and it is attained. I am having a real problem showing the inclusion for general $p_1<p_2$, however.

This is what I have: Since $\sum a_k^{p_1}$ converges, $\lim_{k\to\infty} a_k^{p_1} = 0$. Thus choose $N\in\mathbb{N}$ where for all $n\geq N$ we have that $a_n^{p_1}<1$. Observe that $p_2/p_1 >1$. Then we have that $a_n^{p_2}=(a_n^{p1})^{p_2/p_1}<a_n^{p_1}$. By [some comparison theorem here] we may conclude that $\sum a_k^{p_2}$ converges.

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    $\begingroup$ BTW, you can also use \ell^p ($\ell^p$) instead of l^p ($l^p$) to write this space in LaTeX. $\endgroup$ – 6005 May 8 '14 at 3:32
  • $\begingroup$ That is much more fancy. $\endgroup$ – Prototank May 8 '14 at 3:46
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The key point to use is that ($p < \infty$ here) $$a \in \ell^p \implies a_n \to 0 \implies |a_n| < 1 \text{ for all large $n$}$$

Then since $p_2 > p_1$, it's true that $|a_n|^{p_2} < |a_n|^{p_1}$, so we can just compare directly. The finitely many terms that are large don't affect convergence, so we're done.

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  • $\begingroup$ This is much simpler. I went a little off track. Thank you :) $\endgroup$ – Prototank May 8 '14 at 1:17
  • $\begingroup$ You're very welcome. $\endgroup$ – user61527 May 8 '14 at 1:21

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