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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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 $B\le A$.
Julián Aguirre
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I'm sorry, could you elaborate slightly more on why "It is clear that A \leq B"? – user71487 Mar 31 '19 at 18:46
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Since $T$ is continuous, $$\sup_{x \in B(0, 1)} |T(x)|=\sup_{x \in \bar B(0, 1)}|T(x)|.$$ – Julián Aguirre Mar 31 '19 at 20:20
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What I'd like to understand is precisely why $$ \sup_{x \in B(0,1)} ||T(x)|| \geq \sup_{x \in \overline{B}(0,1)} ||T(x)|| $$ – user71487 Mar 31 '19 at 20:36
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If $|x|=1$, then there is a sequence ${x_n}\subset B(0,1)$ such that $x_n\to x$. Since $T$ is continuous, $Tx_n\to Tx$. – Julián Aguirre Mar 31 '19 at 21:43