Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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 $n$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.