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Let $\mathbb K$ denote $\mathbb R$ or $\mathbb C$. I'm trying to prove the discrete version of this result, i.e.,

Let $x= (x_1, \ldots, x_m) \in \mathbb K^m$ and $\lambda=(\lambda_1, \ldots, \lambda_m) \in \mathbb R_{\ge 0}^m$. Then $$ \lim_{p \to \infty} \bigg ( \sum_{i=1}^m \lambda_i |x_i|^p \bigg )^{1/p} = \max _{1 \leq i\leq m} |x_{i}|. $$

Could you confirm if my below attempt is correct?

Proof WLOG, we assume $x_1 = \max _{1 \leq i\leq m} |x_{i}|$. Then $$ (\lambda_1 |x_1|^p)^{1/p} \le \bigg ( \sum_{i=1}^m \lambda_i |x_i|^p \bigg )^{1/p} \le \bigg ( |x_1|^p \sum_{i=1}^m \lambda_i \bigg )^{1/p}. $$

We take the limit $p \to \infty$ and get $$ \begin{align} \liminf_{p\to\infty} (\lambda_1 |x_1|^p)^{1/p} &\le \liminf_{p\to\infty} \bigg ( \sum_{i=1}^m \lambda_i |x_i|^p \bigg )^{1/p} \le \liminf_{p\to\infty} \bigg ( |x_1|^p \sum_{i=1}^m \lambda_i \bigg )^{1/p} \\ \limsup_{p\to\infty} (\lambda_1 |x_1|^p)^{1/p} &\le \limsup_{p\to\infty} \bigg ( \sum_{i=1}^m \lambda_i |x_i|^p \bigg )^{1/p} \le \limsup_{p\to\infty} \bigg ( |x_1|^p \sum_{i=1}^m \lambda_i \bigg )^{1/p}. \end{align} $$

The claim then follows by noticing that

$$ \lim_{p\to \infty} (\lambda_1)^{1/p} =\lim_{p \to \infty}\bigg (\sum_{i=1}^m \lambda_i \bigg )^{1/p}=0. $$

This completes the proof.

Masacroso
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Akira
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    Your last line should read “$=1$” but looks ok other than that – FShrike Mar 29 '23 at 23:40
  • Since it seems there is really not much to say about your proof, I voted to close (and the answer there are somewhat similar to what you did here) – Arctic Char Mar 30 '23 at 00:10
  • @Masacroso $(\lambda_1 |x_1|^p)^{1/p} = (\lambda_1)^{1/p} |x_1|$... – Akira Mar 30 '23 at 01:10
  • Big error. You said $\lambda=(\lambda_1, \ldots, \lambda_m) \in \mathbb R_{\ge 0}^m$. But if the maximum of $|x_i|$ occurs where $\lambda_i=0$, this result fails. – GEdgar Mar 30 '23 at 01:17
  • @Akira, alright, I didn't paid enough attention that $\lambda _1|x_1|^p\leqslant \lambda _1|x_1|^p+\ldots +\lambda _n|x_n|^p$, yes, your first equality is right, sorry. However you last two equalities are wrong in general, because if $\lambda _1=0$ and $\lambda _2=1$... – Masacroso Mar 30 '23 at 01:20
  • It's a typo. I forgot to assume $\lambda_i \neq0$ for all $i$. – Akira Mar 30 '23 at 01:22
  • Any reason for a downvote...? – Akira Mar 30 '23 at 08:34

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