Inequality involving operator norm

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.

• Your idea is right, but you're forgetting where the supremum comes in. Hence, letting $y$ range over all of $B(x,r)$, you will get $\sup_{y\in B(x,r)} ||T((y-x)/r)||= ||T||,$ implying, by continuity and linearity, that $\sup_{y\in B(x,r)} ||T(y-x)||= ||T||r$. – WoolierThanThou Aug 27 '19 at 21:09
• If I believed that we'd be done, I'd just have put it as an answer. – WoolierThanThou Aug 27 '19 at 21:10
• However, now, the task is reduced to proving that the supremum can be attained 'in a direction that doesn't interfere with the norm of $||Tx||$'. – WoolierThanThou Aug 27 '19 at 21:11
• @WoolierThanThou it's trivial. if it is in the direction of $Tx$, then just flip it about $x$. – mathworker21 Aug 27 '19 at 21:36
• @WoolierThanThou $\max(||a+b||,||a-b||)\ge(||a+b)||+||a-b||)/2\ge||a||$. – David C. Ullrich Aug 27 '19 at 23:03

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 $$||Ty|| \ge (1-\epsilon)r||T||$$.