Prove that $\displaystyle \lim_{p \longrightarrow \infty} |\cdot|_p = |\cdot|_{\infty}$

Question

Let $n \in \mathbb{N}$. Consider $p \in \mathbb{R}_+^*$ and the norms $|\cdot|_p, |\cdot|_{\infty}:\mathbb{C}^n \longrightarrow \mathbb{R}$ given by, for every $\xi=(\xi_1,\xi_2,\cdots,\xi_n) \in \mathbb{C}^n$,

$$|\xi|_p=(|\xi_1|^p+|\xi_2|^p+\cdots + |\xi_n|^p)^{\frac{1}{p}}$$ and $$ |\xi|_{\infty}=\sup_{j \in \{1,\cdots, n\}}|\xi_j|.$$

I want to prove that $$ \lim_{p \longrightarrow \infty} |\xi|_p = |\xi|_{\infty}.$$

But I have no idea of proceeding, I am not familiar with norms in $\mathbb{C}^n$.

https://math.stackexchange.com/questions/326172/the-l-infty-norm-is-equal-to-the-limit-of-the-lp-norms is here answer for you? — zkutch, Jun 12 '20 at 00:51
Did you mean $\displaystyle|\xi|\infty=\text{sup}{j=1,2,\dots}|\xi_j|$ ? — W. Wongcharoenbhorn, Jun 12 '20 at 01:10
So, we can use this? $$\text{sup}_{j=1,2,\dots}|\xi_j|\le|\xi|_p\le\sqrt[p]{n}\cdot \text{sup}|\xi_j|$$ — W. Wongcharoenbhorn, Jun 12 '20 at 01:31
Does this answer your question? The $ l^{\infty} $-norm is equal to the limit of the $ l^{p} $-norms. — Moishe Kohan, Jun 12 '20 at 01:34

score 1 · Accepted Answer · answered Jun 12 '20 at 01:56

What I meant to say is that, just for an alternative approach, if there is some $|\xi_j|\not = 0$ , otherwise it follows immediately, Let $|\xi_k|=\sup|\xi_j|$ we have, $$1\le \left(\dfrac{|\xi_1|^p}{|\xi_k|^p}+\dfrac{|\xi_2|^p}{|\xi_k|^p}+\dots+1+\dots+\dfrac{|\xi_n|^p}{|\xi_k|^p}\right)^{1/p}\le \left(1+1+\dots 1\right)^{1/p}=n^{1/p}$$ Then by taking $\displaystyle \lim_{p\rightarrow\infty}$ and sandwich theorem, we obtain the result $\Box$

Prove that $\displaystyle \lim_{p \longrightarrow \infty} |\cdot|_p = |\cdot|_{\infty}$

1 Answers1