Norm of a linear operator and Hahn-Banach Theorem

Question

Let $X$ and $Y$ be normed spaces and $T$ $\in$ $L(X,Y)$. Prove that $\| T\|=\sup\{|\varphi (Tx)|:x \in X, \|x\|\leq 1, \varphi \in Y^*, \|\varphi\| \leq 1 \}$. I think that I should use Hahn-Banach Theorem, could you help me?

Answer 1

Let the above definition of norm be denoted by $\|T\|_*$. Then we have for any linear functional $F$ and any $x$ satisfying above constraints that $$ |F(Ax)| \le \|F\|\|Ax\| \le \|Ax\| \le \|A\|\|x\| \le \|A\|$$ which proves one direction.

For the other way, let $x^*$ be the point that achieves $\|Ax^*\| = \|A\|$ (To be more rigorous, by definition, given $\varepsilon >0$, we can find a $x^*$ such that $\|Ax^*\| \ge \|A\| - \varepsilon$). As you pointed out, use Hahn Banach theorem (HBT) with the sub-linear function $\|x\|$. In particular, HBT states that there is a bounded linear functional $\Psi$ such that when applied to the vector $Ax^*$, you get $\Psi(Ax^*) = \|Ax^*\|$. This functional will achieve equality or the $\varepsilon$ version I mentioned above.