Functional in Hilbert space Let $H$ be a Hilbert space and $0\neq x\in H$. I want to prove that there is an unique $f\in H^*$, such that $\|f\|=1$ and $f(x)=\|x\|$.
Any ideas on how to approach this problem

Let $K=\Bbb R\text{ or } \Bbb C$ the field of scalars. Let $V$ be the subspace of $H$ defined by $$V=\{\lambda x\}_{\lambda\in K}$$ Consider the continuous form $\phi\in V^*$ defined by $$\phi(\lambda x)=\lambda\space ||x||$$
It is clear that $\phi( x)= ||x||$.
Besides, for all $\lambda x\in V$ one has
$$|\phi(\lambda x)|=|\lambda|\space ||x||=||\lambda x||=||x||$$ Hence
$$||\phi||_V=1$$Now by Hahn-Banach theorem we can extend $\phi$ to the whole $H$ with preservation of the norm.
(We get this way a form $f\in H^*$ such that the restriction $f_{|V}=\phi$ and $||f||=||\phi||=1$)
