${lim}_{p\to \mathrm{\infty }}69645069x{86422285}_p=69645069x{86422285}_{\mathrm{\infty }}$

Suppose $x=(x_1,\mathrm{\dots },x_n)$ is a point in ${ℝ}^n$, and let ${\parallel x\parallel }_p$and ${\parallel x\parallel }_{\mathrm{\infty }}$ be the usual $p$-norm and $\mathrm{\infty }$-norm;

${\parallel x\parallel }_p$$=$${\left({|x_1|}^p+\mathrm{\cdots }+{|x_n|}^p\right)}^{1/p},$

${\parallel x\parallel }_{\mathrm{\infty }}$$=$$\max \{|x_1|,\mathrm{\dots },|x_n|\}.$

Our claim is that

$\underset{p\to \mathrm{\infty }}{lim}{\parallel x\parallel }_p$$=$${\parallel x\parallel }_{\mathrm{\infty }}.$(1)

In other words, for any fixed $x\in {ℝ}^n$, the above limit holds.This, or course, justifies the notation for the $\mathrm{\infty }$-norm.

Proof.Since both normsstay invariant if we exchange two components in $x$, we can arrange thingssuch that ${\parallel x\parallel }_{\mathrm{\infty }}=|x_1|$. Then for any real $p>0$, we have

${\parallel x\parallel }_{\mathrm{\infty }}$$=$$|x_1|={({|x_1|}^p)}^{1/p}\le {\parallel x\parallel }_p$

and

${\parallel x\parallel }_p$$\le$$n^{1/p}|x_1|=n^{1/p}{\parallel x\parallel }_{\mathrm{\infty }}.$

Taking the limit of the above inequalities (seethis page (http://planetmath.org/InequalityForRealNumbers))we obtain

${\parallel x\parallel }_{\mathrm{\infty }}$$\le$$\underset{p\to \mathrm{\infty }}{lim}{\parallel x\parallel }_p,$

$\underset{p\to \mathrm{\infty }}{lim}{\parallel x\parallel }_p$$\le$${\parallel x\parallel }_{\mathrm{\infty }},$

which combined yield the result. $\mathrm{\square }$