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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.

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  • $\begingroup$ 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 $\sup_{y\in B(x,r)} ||T(y-x)||= ||T||r$. $\endgroup$ – WoolierThanThou Aug 27 '19 at 21:09
  • $\begingroup$ If I believed that we'd be done, I'd just have put it as an answer. $\endgroup$ – WoolierThanThou Aug 27 '19 at 21:10
  • $\begingroup$ 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 $||Tx||$'. $\endgroup$ – WoolierThanThou Aug 27 '19 at 21:11
  • $\begingroup$ @WoolierThanThou it's trivial. if it is in the direction of $Tx$, then just flip it about $x$. $\endgroup$ – mathworker21 Aug 27 '19 at 21:36
  • $\begingroup$ @WoolierThanThou $\max(||a+b||,||a-b||)\ge(||a+b)||+||a-b||)/2\ge||a||$. $\endgroup$ – David C. Ullrich Aug 27 '19 at 23:03
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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 $||Ty|| \ge (1-\epsilon)r||T||$.

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