Dual of Hilbert space : induced norm vs. operator norm

Question

Let $\mathfrak{H}$ be a Hilbert space. Is the operator norm on the dual $\mathfrak{H}^*$ induced by a inner product ?

Answer 1

Let $H$ be a Hilbert space and $R: H^*\rightarrow H$ be the Riesz izomorphism. We define an inner product in $H^*$ by using the Riesz map:

$(f^*,g^*)_{H^* \times H^*}:=(Rf^*,Rg^*)_{H\times H}=(f,g)_{H\times H}$ where $f$ and $g$ are the images of the functionals $f^*$ and $g^*$ under $R$. Therefore, the induced norm by this inner product is

$\|f^*\|^2=(f^*,f^*)_{H^*\times H^*}=(Rf^*,Rf^*)_{H\times H}=(f,f)_{H\times H}=\|f\|^2\quad\quad$ (1)

So it is left to show that $\|f^*\|_{op}=\|f^*\|$, i.e the operator norm is equal to the induced from the inner product norm. But the Riesz map $R$ is an isometry $\Rightarrow \|f^*\|_{op}=\|Rf^*\|=\|f\|$, which by (1) is exactly $\|f^*\|$ .