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Let $X=\mathbb{R}^n$.

The space $(X,d_p)$, where $d_p$ is the metric on $X$ defined as

$$d_p(x,y):=\bigg[\sum_{i=1}^n|x_i-y_i|^p\bigg]^{1/p}$$

is the space $\ell^n_p$.

And,

The space $(X,d_{\infty})$, where $d_{\infty}$ is the metric on $X$ defined as

$$d_{\infty}(x,y):=\max_{1 \leq i\leq n}\{|x_i-y_i|\}$$

is the space $\ell^n_{\infty}$.

[In both, $x=(x_1,\dots,x_n), y=(y_1,\dots,y_n)$]

Now, my question is, why the following is true

$$d_{\infty}(x,y)=\max_{1 \leq i\leq n}\{|x_i-y_i|\}=\lim_{p\to\infty}\bigg[\sum_{i=1}^n|x_i-y_i|^p\bigg]^{1/p}$$

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We have $$\bigg[\sum_{i=1}^n|x_i-y_i|^p\bigg]^{1/p}\leqslant n^{1/p}\max|x_i-y_i|$$

On the other hand, for each $i$; $$|x_i-y_i|\leqslant \bigg[\sum_{i=1}^n|x_i-y_i|^p\bigg]^{1/p}$$ so certainly $$\max|x_i-y_i|\leqslant \bigg[\sum_{i=1}^n|x_i-y_i|^p\bigg]^{1/p}$$

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  • $\begingroup$ @Mike Hey! Drop by the chat! $\endgroup$ – Pedro Tamaroff Dec 26 '13 at 3:03
  • $\begingroup$ @SalechAlhasov No problem. $\endgroup$ – Pedro Tamaroff Dec 26 '13 at 3:05

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