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I am trying to prove that for $p\to \infty$ the p-norm:

$$\|x\|_{p}=\Bigl(\sum_{i=1}^{n}|x_i|^{p}\Bigr)^{\frac{1}{p}}$$ we get:

$$\|x\|_{\infty}=\max\{|x_1|\}$$

$$|x_i|\leftarrow_{\infty}(|x_i|^{p})^{\frac{1}{p}}\leq\Bigl(\sum_{i=1}^{n}|x_i|^{p}\Bigr)^{\frac{1}{p}}\leq (p|x_i|^{p})^{\frac{1}{p}}\rightarrow_{\infty}|x_{i}|$$

It is for an arbitrary $i$ so we take the $\max\{|x_1|,|x_2|,...,|x_n|\}?$

gbox
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  • I don't understand, why $\bigl( \sum_{i=1}^n |x_i|^p \bigr)^{1/p}$ is equal to $(n \cdot |x_i|^p)^{1/p}$, seems like $i$'s in indices are different (one for summation index, the other for "arbitrary $i$"). – G. Strukov Oct 31 '17 at 20:56
  • @G.Strukov correct, Edited – gbox Oct 31 '17 at 21:25
  • Now it's true more frequently, but still not for all $i$. Understanding for which $i$ inequality $\bigl(\sum_{j=1}^n |x_j|^p)^{1/p} \leq (p|x_i|)^{1/p}$ holds for all $p$ might be useful for finishing the proof. – G. Strukov Oct 31 '17 at 21:31
  • @G.Strukov $|x_{j}|\leq |x_{i}||$ – gbox Oct 31 '17 at 21:50

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