If $z$ is in an inner product space $X$, show that $f(x)=\langle x,z \rangle$ defines a bounded linear functional $f$ on $X$.

Question

If $z$ is any fixed element of an inner product space $X$, show that $f(x)=\langle x,z \rangle$ defines a bounded linear functional $f$ on $X$, of norm $||z||$. If the mapping $X\to X'$ given $z\to f$ is surjective, show that $X$ must be a Hilbert space.

There's two things to prove here and I'm not sure how to start either. Any help/starting-points would be nice.

Answer 1

Let $z\in X$. we define $f_{z}(x)=<x,z>$.

f is linear: $(f_{z}(ax+by)=<ax+by,z>= a<x,z>+b<y,z>=af_{z}(x)+bf_{z}(y)$

f is bounded: $|f_{z}(x)|=|<x,z>|\leq\|x\|\|z\|\Longrightarrow \|f\|\leq\|z\|$ (where the first inequality is the Cauchy-swartz inequality).

Let $T: X\longrightarrow X^*$ with $T(z)=f_{z}$. Observe that $T$ is surjective means that every linear functional is of the form $f_{z}$ for some $z\in X$, which is the Riesz's representation theorem.