TITLE: Inner product in dual Hilbert space
QUESTION [5 upvotes]: Let $H^*$ be a dual space of a Hilbert space $H$. Then inner product is defined as
$$(f,g)=(J^{-1}f,J^{-1}g),$$
where $f,g\in H^*$ and $J\colon H\to H^*$ is the canonical isomorphism. I want to prove that 
$$\|f\|=\|J^{-1}f\|=\sqrt{(J^{-1}f,J^{-1}f)}=\sqrt{(f,f)}.$$
Any ideas on how to approach this proof?

REPLY [2 votes]: $$
\|f\|_{H^*} = \sup_{\|x\|_H\le 1} f(x) =  \sup_{\|x\|_H\le 1} (J^{-1}f,x)_H
= \sup_{\|g\|_{H^*}\le 1} (J^{-1}f,J^{-1}g)_H
=  \|J^{-1}f\|_H
$$
first equality: definition of operator norm, second: use isomorphism,
third: $J$ is isometric, fourth: Cauchy-Schwarz.