Prove that $||a||_{l^p} \rightarrow \sup_n|a_n|$ as ${p \rightarrow \infty}$

Asked Oct 28 '17 at 16:03

Active Oct 28 '17 at 16:18

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Let $a_n$ be an infinite bounded sequence. Prove that $||a||_{l^p} \rightarrow \sup_n|a_n|$ as ${p \rightarrow \infty}$.

I've read proofs for that, but I want to ask about the following:

For any finite $N \in \mathbb{N}$, we have:

$$\lim_{p \rightarrow \infty}(|a_1|^p + ... + |a_N|^p)^{1/p} = \sup_{n \in N} |a_n| $$

Can I just take the limit $N \rightarrow \infty $? What is the justification for that?

edited Oct 28 '17 at 16:10

asked Oct 28 '17 at 16:03

catch22

1

You can't just take $N \to \infty$, but the basic idea is good. Construct a suitable $\delta# argument to finish the proof. – Hans Engler Oct 28 '17 at 16:16
The sequence $(1,1,1\dots)$ is a counterexample. You need to assume in addition that there exists $p<\infty$ with $||a||_p<\infty$. (No, you can't just take $N\to\infty$ and say you're done - the problem is that $\lim_N\lim_p$ need not be the same as $\lim_p\lim_N$.) – David C. Ullrich Oct 28 '17 at 17:11

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