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

[I]. A real or complex Hilbert space $H$ is its own dual. If $f\in H^*$ then there exists $y_f\in H$ such that the  inner product <$x,y_f$> is equal to $f(x)$ for all $x\in H.$ And $y_f$ is unique, because if $f(x)=<x, y'_f>$ for all $x,$ then $$\|y_f-y'_f\|^2=<y_f-y'_f,y_f>-<y_f-y'_f,y'_f>=f(y_f-y'_f)-f(y_f-y'_f)=0.$$ 
Also $\|f\|=\|y_f\|$ because, in the non-trivial case $f\ne 0,$ we must have $y_f\ne 0,$  so  $$(i).\quad f(y_f)/\|y_f\|=\|y_f\| \implies \|f\|\geq \|y_f\|.$$
$$(ii).\quad  \forall x\in H\;(|f(x)|=|<x,y_f>|\leq \|x\|\cdot \|y_f\|) \implies \|f\|\leq \|y_f\|.$$
[II]. For $0\ne x\in H,$ let $f(y)=$<$y,x/\|x\|$> for all $y\in H.$ Then $f(x)=\|x\|$  and $\|f\|=1.$
Suppose $g\in H^*$ with $g(x)=\|x\|$ and $\|g\|=1.$ Let $g(z)=$<$z,y_g$> for all $z\in H.$  Then $\|y_g\|=\|g\|=1.$  We have $$\|x\|=g(x)=<x,y_g>=|<x,y_g>|\leq \|x\|\cdot \|y_g\|=\|x\|.$$ 
But the Cauchy-Schwarz Inequality $|$<$u,v$>$|\leq \|u\|\cdot \|v\|$ is a strict inequality unless $u,v$ are linearly dependent. So, since $|$<$x,y_g$>$|=\|x\|=\|x\| \cdot \|y_g\|$ and $x\ne 0\ne y_g,$ there exists scalar r such that $y_g=rx.$ This implies $$0\ne \|x\|=g(x)=<x, rx>$$ so $r=1/\|x\|.$ Hence $y_g=x/\|x\|$ and $g=f.$
[III]. Appendix: To show that $H$ is its own dual: Let $E$ be an orthonormal Hilbert-space basis for $H$. Let  $0\ne f\in H^*.$ Then $f$ is uniquely determined by $\{(e,f(e):e\in E\}.$ Let $E_f=\{e\in E: f(e)\ne 0\}.$
Let $F_f$ be the set of finite non-empty subsets of $E_f.$  For $S\in F_f$ let $$x_S=\sum_{e\in S}e\;\overline {f(e)}$$ (where $\overline {z}$ denotes the complex conjugate of $z$). Then $f(x_S)=\sum_{e\in S}|f(e)|^2=\|x_S\|^2 $.  Therefore  $$\infty >\|f\|\geq \sup_{S\in F_f}\sum_{e\in S}|f(e)|^2=\sum_{e\in E_f}|f(e)|^2.$$ Therefore $y=\sum_{e\in E_f}e\; \overline {f(e)}\in H.$  The functional $g(x)=$<$x,y$> agrees with $f$ for every $x\in E,$ so $g=f.$ 
