# Equivalent definition operator norm [duplicate]

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?

• And this. – user228113 Dec 19 '16 at 11:05

## 1 Answer

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 $B\le A$.

• I'm sorry, could you elaborate slightly more on why "It is clear that A \leq B"? – user71487 Mar 31 at 18:46
• 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 at 20:20
• 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 at 20:36
• 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 at 21:43