I'm trying to prove the $|\cdot|_{p}$ norm will become the maximum norm when $p \to \infty$.

Let $\mathbb K$ denote $\mathbb R$ or $\mathbb C$, and $x= (x_1, \ldots, x_m) \in \mathbb K^m$. Then $$\lim_{p \to \infty} \left ( \sum_{i=1}^m |x_i|^p \right )^{1/p} = \max _{1 \leq i\leq m} |x_{i}|$$

Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!

My attempt:

It suffices to prove the statement in case $x \in \mathbb {(R^+)}^{m}$, where it becomes $$\lim_{p \to \infty} \left ( \sum_{i=1}^m (x_i)^p \right )^{1/p} = \max _{1 \leq i\leq m} x_{i}$$

Let $l:= \max _{1 \leq i\leq m} x_{i}$. We have $$l = (l^p)^{1/p} \le \left ( \sum_{i=1}^m (x_i)^p \right )^{1/p} \le (ml^p)^{1/p} = m^{1/p}l$$

Then $$l = \lim_{p \to \infty} l \le \lim_{p \to \infty} \left ( \sum_{i=1}^m (x_i)^p \right )^{1/p} \le \lim_{p \to \infty} (m^{1/p}l) = l$$

and thus by squeeze theorem $$\lim_{p \to \infty} \left ( \sum_{i=1}^m (x_i)^p \right )^{1/p} = l$$

This completes the proof.

  • $\begingroup$ Hi @AlexR., Please elaborate on why we should subscript $l_m$. I think I well defined $l$ by $l:= \max _{1 \leq i\leq m} x_{i}$. $\endgroup$ – LAD Aug 7 '19 at 4:56
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    $\begingroup$ @LeAnhDung I think Alex's comment is bad and that part of your proof is fine. I do, however, think your proof is flawed in that it works with limits before establishing the limit exists. You should be working with liminf's and limsup's $\endgroup$ – mathworker21 Aug 7 '19 at 5:02
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    $\begingroup$ @LeAnhDung I mean, it comes down to how you phrase it. Once you have $l \le (\sum x_i^p)^{1/p} \le m^{1/p}l$, you can say: since $\lim_{p \to \infty} l = l$ and $\lim_{p \to \infty} m^{1/p}l = l$, the squeeze theorem implies $\lim_{p \to \infty} (\sum x_i^p)^{1/p} = l$. That's what the squeeze theorem says. My objection in my previous comment was that you can't immediately go from$ l \le (\sum x_i^p)^{1/p} \le m^{1/p}l$ to $\lim_{p \to \infty} l \le \lim_{p \to \infty} (\sum x_i^p)^{1/p} \le \lim_{p \to \infty} m^{1/p}l$, since you don't know if the limits exist (yet). $\endgroup$ – mathworker21 Aug 7 '19 at 9:37
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    $\begingroup$ Yes it's correct. Good job! However, that is not what I suggested. I clearly suggested to use liminf's and limsup's. You should go from $l \le (\sum x_i^p)^{1/p} \le m^{1/p}l$ to $l = \liminf_p l \le \liminf_p (\sum x_i^p)^{1/p} \le \limsup_p (\sum x_i^p)^{1/p} \le \limsup_p m^{1/p}l = l$, so $\liminf_p (\sum x_i^p)^{1/p} = \limsup_p (\sum x_i^p)^{1/p}$ and thus the limit exists (and is then equal to $l$). $\endgroup$ – mathworker21 Aug 8 '19 at 13:41
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    $\begingroup$ @LeAnhDung no problem. upvoted your answer :) $\endgroup$ – mathworker21 Aug 8 '19 at 14:53

On the basis of @mathworker21's suggestion, I added a proof that $\lim_{p \to \infty} \left ( \sum_{i=1}^m (x_i)^p \right )^{1/p}$ exists here. It would be great if someone helps me verify my attempt.

My attempt:

Lemma: $$\left( \sum_{i=1}^m (x_{i})^{p} \right)^{q} \ge \left( \sum_{i=1}^m (x_{i})^{q} \right)^{p}, \quad (x_1, \ldots,x_m) \in {(\mathbb R^+)}^m, \quad p,q \in \mathbb R^+, \quad p \le q$$

$$\text{Let } l:= \max _{1 \leq i\leq m} x_{i}.\ \text{Then } \left ( \sum_{i=1}^m (x_i)^p \right )^{1/p} \ge (l^p)^{1/p} = l.\ \text{As such, the sequence}$$

$$\left \langle \left ( \sum_{i=1}^m (x_i)^p \right )^{1/p} \right \rangle_{p \in \mathbb N}$$ is bounded from below. Next we prove that this sequence is decreasing by showing $$\left ( \sum_{i=1}^m (x_i)^p \right )^{1/p} \ge \left ( \sum_{i=1}^m (x_i)^{p+1} \right )^{1/(p+1)}$$ or equivalently $$\left ( \sum_{i=1}^m (x_i)^p \right )^{p+1} \ge \left ( \sum_{i=1}^m (x_i)^{p+1} \right )^{p}$$, which is true by our Lemma. This completes the proof.

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