Let $T: X \to Y$ a bounded linear operator. Prove that $\|T\|=\sup\{\|T(x)\|: x\in X , \|x\|<1 \}.$
It is $||T||= \sup \{||T(x)||: x\in X ,||x|| \leq1 \}$ so $||T|| \geq \sup\{||T(x)||: x\in X , ||x||<1 \}$.
I can't prove the other inequality.
It isn't so difficult but for some reason I am stuck.
Any help?
Thanks in advance!