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Let $T: X \to Y$ be a bounded linear map between normed spaces. The operator norm is defined by $$\sup_{\|x\| = 1} \|T(x)\|$$

Is this equivalent to

$$\sup_{x \in B(0, 1)} \|T(x)\|$$

where $B(0, 1)$ is the open unit ball?

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    $\begingroup$ And this. $\endgroup$ – user228113 Dec 19 '16 at 11:05
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Yes. Call $A$ the first expression and $B$ the second. It is clear that $A\le B$. Now, if $0<\|x\|<1$ we have $$ \|T(x)\|=\|x\|\,\Bigl\|T\Bigl(\frac{x}{\|x\|}\Bigr)\Bigr\|\le A\,\|x\|\le A. $$ Take the sup to get $B\le A$.

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  • $\begingroup$ I'm sorry, could you elaborate slightly more on why "It is clear that A \leq B"? $\endgroup$ – user71487 Mar 31 at 18:46
  • $\begingroup$ Since $T$ is continuous, $$\sup_{x \in B(0, 1)} \|T(x)\|=\sup_{x \in \bar B(0, 1)}\|T(x)\|.$$ $\endgroup$ – Julián Aguirre Mar 31 at 20:20
  • $\begingroup$ What I'd like to understand is precisely why $$ \sup_{x \in B(0,1)} ||T(x)|| \geq \sup_{x \in \overline{B}(0,1)} ||T(x)|| $$ $\endgroup$ – user71487 Mar 31 at 20:36
  • $\begingroup$ If $\|x\|=1$, then there is a sequence $\{x_n\}\subset B(0,1)$ such that $x_n\to x$. Since $T$ is continuous, $Tx_n\to Tx$. $\endgroup$ – Julián Aguirre Mar 31 at 21:43

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