the norm of a linear operator

Question

In in the demonstration of Lax-Milgramm lemma, they use a linear operator $A:V\to V$, where $V$ is a Hilbert space; My basic problem is how to prove that

$$\|Au\|_V=\sup_{v\in V^*}\frac{(Au,v)_V}{\|v\|_V},\ \forall u\in V$$

Till now I have only one inequality, $$\sup_{v\in V^*}\frac{(Au,v)_V}{\|v\|_V}\leq \|Au\|_V$$

what about the other inequality ? thank you for your time.

Answer 1

You found $$ \sup_{v\in V^{\ast }}\frac{(Au,v)_{V}}{\parallel v\parallel _{V}}\leqslant \parallel Au\parallel _{V} $$ Take $$ v=Au $$ Then $$ \frac{(Au,Au)_{V}}{\parallel Au\parallel _{V}}\leqslant \sup_{v\in V^{\ast }}% \frac{(Au,v)_{V}}{\parallel v\parallel _{V}}\leqslant \parallel Au\parallel _{V} $$ But $$ \frac{(Au,Au)_{V}}{\parallel Au\parallel _{V}}=\frac{\parallel Au\parallel _{V}^{2}}{\parallel Au\parallel _{V}}=\parallel Au\parallel _{V} $$