For some reason I used to think that if $T$ is a linear operator on normed spaces $V \to W$ then saying $T$ is an isometry is the same as saying $\|T\|_{op} = 1$.
Well, I got stuck on a proof and subsequently looked up the definition and realised that the definition of isometry is that a linear operator $T$ is an isometry if $\|Tx\| = \|x\|$.
Now I've been wondering whether we have that $\|T\|_{op} = 1$ implies $T$ is an isometry?
The other direction holds: if $\|Tx\| = \|x\|$ for all $x$ then $\frac{\|Tx\|}{\|x\|} = 1$ for all $x \neq 0$ and hence $\sup_{\|x\|=1}\|Tx\| = \sup_{\|x\|=1} \frac{\|Tx\|}{\|x\|} = \|T\| = 1$.
Thanks for your help.