Given T a bounded linear operator $T:X \to Y$ where $X$ and $Y$ are normed spaces,
I'm trying to show that $\|T\|_*=\sup\{ \frac{\|Tx\|_Y}{\|x\|_X},x\neq0\}$ equals to $\sup\{\|T(x)\|_Y\,:\, \|x\|_X< 1\}$.
I can show that $\|T\|_*=\sup\{\|T(x)\|_Y\,:\, \|x\|_X\le 1\}$, but I can't go further.
Do you have any suggestions?
Best Answer
Right so what you can do is find an $x \in\overline{B_1(0)}$ that is $\epsilon >0$ close to the supremum. i.e. $$\| Tx\| \geq \|T\| - \epsilon.$$ Then we can find $y \in B_1(0)$ such that $y = (1-\epsilon)x$ and so by linearity it follows that, $$\| Ty\| \geq (1-\epsilon)(\|T\| - \epsilon)$$ and as $\epsilon$ was arbitrary we can conclude that, $$\sup \{\|Tx\| : \|x\| \leq 1\} \leq \sup \{\|Tx\| : \|x\| < 1\}.$$ The other inequality should be clear.