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

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 norms stay invariant if we exchange two components in $x$, we can arrange things such 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 (see this 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. $\mathrm{\square }$

Title	${lim}_{p\to \mathrm{\infty }}69645069x{86422285}_p=69645069x{86422285}_{\mathrm{\infty }}$
Canonical name	limptoinftylVertXrVertplVertXrVertinfty
Date of creation	2013-03-22 14:02:53
Last modified on	2013-03-22 14:02:53
Owner	Koro (127)
Last modified by	Koro (127)
Numerical id	12
Author	Koro (127)
Entry type	Result
Classification	msc 46B20
Related topic	PowerMean