Suppose $x=(x_1,\ldots, x_n)$ is a point in $\sR^n$ , and let $\norm{x}_p$ and $\norm{x}_\infty$ be the usual $p$ -norm and $\infty$ -norm; \begin{eqnarray*} \norm{x}_p &=& \big(|x_1|^p + \cdots + |x_n|^p\big)^{1/p},\\ \norm{x}_\infty &=& \max \{|x_1|, \ldots, |x_n|\}. \end{eqnarray*} Our claim is that \begin{eqnarray} \label{claim} \lim_{p \to \infty} \norm{x}_p &=& \norm{x}_{\infty}. \end{eqnarray}In other words, for any fixed $x\in \sR^n$ , 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 $\norm{x}_\infty = |x_1|$ . Then for any real $p>0$ , we have \begin{eqnarray*} \norm{x}_\infty &=&\abs{x_1} =(\abs{x_1}^p)^{1/p} \leq \norm{x}_p \end{eqnarray*}and \begin{eqnarray*} \norm{x}_p &\le & n^{1/p} |x_1| =n^{1/p} \norm{x}_\infty. \end{eqnarray*}Taking the limit of the above inequalities (see this page) we obtain \begin{eqnarray*} \norm{x}_\infty &\le& \lim_{p\to \infty} \norm{x}_p,\\ \lim_{p\to \infty} \norm{x}_p &\le & \norm{x}_\infty, \end{eqnarray*}which combined yield the result.