Inequality involving operator norm

Question

T is a bounded operator from a normed space $X$ to a normed space $Y$. I need to prove that for every $x\in X$ and $r>0\,$ one has $\sup\limits_{y\in B(x,r)}\|Ty\|\geqslant\|T\|r$.

For each $y\in B(x,r)$ we have $\frac{y-x}{r}\in B(0,1)$, so that $\|T\big(\frac{y-x}{r}\big)\| < \|T\|$. But this is not helping me to get the desired inequality.

Any hint would be appreciated.

Answer 1

Fix $\epsilon > 0$. Take $\Delta \in B(0,r)$ with $||T\Delta|| \ge (1-\epsilon)r||T||$. Then $||Tx+T\Delta|| < (1-\epsilon)r||T||$ and $||Tx-T\Delta|| < (1-\epsilon)r||T||$ imply $||2T\Delta|| < 2(1-\epsilon)r||T||$. So, there is some $y \in B(x,r)$ with $||Ty|| \ge (1-\epsilon)r||T||$.