# Operator norm. Alternative definition

Let $T\colon X\to Y$ be a linear operator with norm $$\|T\|=\sup_{\|x\|=1}\|Tx\|.$$ Prove that $$\|T\|=\sup_{\|x\|\leq 1}\|Tx\|.$$

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This should be easy. $\{x\in X; \|x\|=1\}\subseteq \{x\in X; \|x\|\le 1\}$ should help you establish one inequality. To get the other one, try to use $\|cx\|=|c|\|x\|$ . – Martin Sleziak Dec 18 '11 at 17:38
Also you need to assume that $X\neq\{0\}$ – Norbert Dec 18 '11 at 19:02

Put $s_1:=\sup_{||x||=1}||Tx||$, and $s_2:=\sup_{||x||\leq 1}||Tx||$. As Norbert says, we have to assume $X\neq\{0\}$, otherwise $s_1=-\infty$ whereas $s_2=0$. Since for $x\in X, ||x||=1\Rightarrow ||x||\leq 1$, we have $s_1\leq s_2$. Let $x\in X$ such that $||x||\leq 1$ and $x\neq 0$. Then $\lVert ||x||^{-1}x\rVert=1$ and $$||T(x)||=||x||\cdot ||T(||x||^{-1}x)||\leq ||x||s_1\leq s_1.$$ Since this inequality is true for $x=0$, we get $s_2\leq s_1$ and finally $s_1=s_2$ if these $\sup$ are finite. If $s_2=+\infty$, then so is $s_1$, since for all $n$ we can find $x_n$, $||x_n||\leq 1$ such that $||T(x_n)||\geq n$. Hence $s_1\geq ||T(||x_n||^{-1}x_n)||=||x_n||^{-1}n\geq n$ for all $n$.