$\mathop{lim}\displaylimits_{p\to\infty}\delimiter 69645069x\delimiter 86422285% _{p}=\delimiter 69645069x\delimiter 86422285_{\infty}$

Suppose $x=(x_{1},\ldots,x_{n})$ is a point in $\mathbb{R}^{n}$, and let $\lVert x\rVert_{p}$ and $\lVert x\rVert_{\infty}$ be the usual $p$-norm and $\infty$-norm;

 $\displaystyle\lVert x\rVert_{p}$ $\displaystyle=$ $\displaystyle\big{(}|x_{1}|^{p}+\cdots+|x_{n}|^{p}\big{)}^{1/p},$ $\displaystyle\lVert x\rVert_{\infty}$ $\displaystyle=$ $\displaystyle\max\{|x_{1}|,\ldots,|x_{n}|\}.$

Our claim is that

 $\displaystyle\lim_{p\to\infty}\lVert x\rVert_{p}$ $\displaystyle=$ $\displaystyle\lVert x\rVert_{\infty}.$ (1)

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

Proof. Since both norms stay invariant if we exchange two components in $x$, we can arrange things such that $\lVert x\rVert_{\infty}=|x_{1}|$. Then for any real $p>0$, we have

 $\displaystyle\lVert x\rVert_{\infty}$ $\displaystyle=$ $\displaystyle|x_{1}|=(|x_{1}|^{p})^{1/p}\leq\lVert x\rVert_{p}$

and

 $\displaystyle\lVert x\rVert_{p}$ $\displaystyle\leq$ $\displaystyle n^{1/p}|x_{1}|=n^{1/p}\lVert x\rVert_{\infty}.$

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

 $\displaystyle\lVert x\rVert_{\infty}$ $\displaystyle\leq$ $\displaystyle\lim_{p\to\infty}\lVert x\rVert_{p},$ $\displaystyle\lim_{p\to\infty}\lVert x\rVert_{p}$ $\displaystyle\leq$ $\displaystyle\lVert x\rVert_{\infty},$

which combined yield the result. $\Box$

Title $\mathop{lim}\displaylimits_{p\to\infty}\delimiter 69645069x\delimiter 86422285% _{p}=\delimiter 69645069x\delimiter 86422285_{\infty}$ limptoinftylVertXrVertplVertXrVertinfty 2013-03-22 14:02:53 2013-03-22 14:02:53 Koro (127) Koro (127) 12 Koro (127) Result msc 46B20 PowerMean