Q:

Show that $\|A\|=\sup|f(Ax)|$

Let $X$ be a linear normed space, $f\in X^*$, $A\in\mathcal{L}(X)$. Show that $\|A\|=\sup|f(Ax)|$, where the upper bound is taken from the set $\{x\in X:\|x\|=1, f\in X^*,\|f\|=1\}$

A:

Showing $\sup|fAx|\leq\|A\|$ is trivial.  For the other inequality, given $\varepsilon>0$ use the definition of the operator norm to find $x\in X$ with $\|x\|=1$ then $f\in X^*$ with $\|f\|=1$ such that $\|A\|\leq|fAx|+\varepsilon$.

