Inner product in dual Hilbert space

Question

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?

Answer 1

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