Unbounded Operator not bounded on any ball

Question

I came across a small proof where the following implication was used:

Let $X$ be a normed v.s. and $T \in X'$ an unbounded operator ( $X'$ denotes the dual of $X$ ), i.e.

$$ \| T (x) \| \gt M \| x \| \qquad \forall M >0, x \in X $$

then

$$ \sup_{ \| x \| \le \epsilon } \| T(x) \| \gt M \qquad \forall \epsilon >0, M >0 $$

How do I see this? I tried using linearity to reshape the supremum condition, but I cannot clearly reason why this implication is true.

Thanks in advance!

Answer 1

If $\sup_{\|x\|\le\epsilon}\|T(x)\|=M<\infty$, since $T(x)=T(\frac{1}{\epsilon} \epsilon x) = \frac{1}{\epsilon} T(\epsilon x)$, whenever $\|x\|\le 1$, we must have $\|\epsilon x\|\le \epsilon$, $\|T(x)\|\le \frac{1}{\epsilon} M<\infty$.