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I am trying to prove $\lim_{p\rightarrow \infty}\|x\|_p = \|x\|_\infty = \sup_i |x_i|$, where $x$ is a (infinite-dim) sequence. My proof is as follows:

$$\lim_{p\rightarrow \infty}\|x\|_p = \lim_{p\rightarrow \infty} (\sum_{i=1}^\infty|x_i|^p)^{1/p}$$ $$=\lim_{p\rightarrow \infty} \ \sqrt[p]{ |M|^p\times q\times(1+ \sum_{j\neq 1}\frac{(\frac{|x_j|}{|M|})^p}{q} ) }$$ where I've assumed $|M|$ is the supremum of all $|x_i|$'s (which wlog I've set to $|x_1|$) and $q$ is the number of repetitions of $M$ in $x$. $$=\lim_{p\rightarrow \infty} \ |M| \times q^{1/p}\times\sqrt[p]{ (1+ \sum_{j\neq 1}\frac{(\frac{|x_j|}{|M|})^p}{q} ) }$$ Since $|M|>|x_j|$ for $j\neq 1$, when $p$ goes to infinity $(\frac{|x_j|}{|M|})^p$ goes to zero.

$$=\lim_{p\rightarrow \infty} |M|\times q^{1/p}=|M|$$

What am I doing wrong?

I came across the following proof which is far more complicated than mine by a reputable responder. So, I'm guessing I must be doing something wrong. https://math.stackexchange.com/a/326266/371990

P.S. I have a guess as to why my solution might not be correct. I think I'm wrongly assuming that $\max \ x_i = \sup \ x_i$ which may not be true in infinite dimensional space.

Teodorism
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    Your proof is correct, your statement is wrong. $|x|\infty=\max{i}|x_i|\not=\max_{i}x_i$. The trivial example would be $x=(-3,-2,0)$. $|x|_{\infty}=3$ and not $0$. – Behrouz Sep 19 '19 at 23:55
  • Why have they proved it here with liminf and limsup then? https://math.stackexchange.com/a/326266/371990 – Teodorism Sep 19 '19 at 23:58
  • The reason is that they prove it for the infinite dimensional space of real-valued sequences. – Behrouz Sep 20 '19 at 01:37
  • @Behrouz Mine is infinite dim as well. – Teodorism Sep 20 '19 at 15:42
  • What if $q\to \infty$? – Mostafa Ayaz Sep 20 '19 at 16:13
  • @MostafaAyaz The limit exists so the sum must be convergent. – Teodorism Sep 20 '19 at 16:18
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    How do you get rid of the sum $$\sum_{j \geq 1} \frac{(\frac{|x_j|}{M})^p}{q}$$ which apears at the end of your computation...? (Note, btw, that the supremum does not need to be attained, i.e. $q$ might be zero.) – saz Sep 20 '19 at 18:37
  • @saz $|M|>|x_j|$ for $j\neq 1$. So when $p$ goes to infinity, $(\frac{|x_j|}{|M|})^p$ goes to zero. – Teodorism Sep 20 '19 at 18:39
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    @saz I also think my prove is wrong because of the implicit assumption that sup is attained. – Teodorism Sep 20 '19 at 18:43
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    For a sequence of infinite sums, just because every term tends to zero, it need not follow that the sequence of sums tends to zero. You cannot pass a limit through an infinite sum without further justification. – Nate Eldredge Sep 20 '19 at 18:49
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    @Teodorism If it is infinite dimensional case, then your proof needs a lot of detailing: To begin with, you should start with showing $q$ is finite and the sup is achieved which are not that hard (if you assume that $(x_n)\in \ell_p$ for $p\geq p^*$, and also you should take care of the relatively serious comment of Nate Eldredge – Behrouz Sep 20 '19 at 19:19

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