We define p-norm in this way: $\|x\|_p = \{\sum ^N_j=_1|x_j|^p\}^ {1\over p}$
We know that It change to $\|x\|_p = \max|x_j| $ when $ p \to \infty $
How can I prove this ?
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Hint: Chebyshev's inequality – Prahlad Vaidyanathan Sep 27 '13 at 11:17
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I don't know really Chebyshev's inequality.no idea plz help me more! I could prove that $||x||_p$ Is smaller than $ Max|x_j| + 1$ with Triangle_inequality but I want more! – jack Sep 27 '13 at 11:27
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This question is an infinite-dimensional version of this: The $ l^{\infty} $-norm is equal to the limit of the $ l^{p} $-norms. – Martin Sleziak Sep 27 '13 at 12:43
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What we have to prove is that if $x_1,\dots,x_K$ are real numbers such that $0\leqslant x_k\lt 1$, then $$\lim_{p\to \infty}\left(1+\sum_{k=1}^Kx_k^p\right)^{1/p}=1.$$ It's equivalent to show that $$\lim_{p\to \infty}\frac 1p\log\left(1+\sum_{k=1}^Kx_k^p\right)=0.$$ It's the case, since for $t\geqslant 0$, $0\leqslant\log(1+t)\leqslant t$.
Davide Giraudo
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Giraudo thx for your attention .but I can't understand something in your proof.why $ x_k $ should be $0 \le x_k \lt 1$ – jack Sep 27 '13 at 12:58
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Assume that the maximum of the $|x_j|$ is reached on the first coordinate. So the $p$-norm is $|x_1|$ times a term like in the first displayed equation of the answer. – Davide Giraudo Sep 27 '13 at 13:06