Define inner product on dual space.

Question

V is a Hilbert space

By Riesz Representation Theorem: $\forall f\in V^*\exists v$ s.t $f=l_v $ where $l_v(x)=<x,v>$

and $||l_v||=||v||$(Using this fact can check that norm of dual space satisfies parallelogram law)

Using polarisation identities: we can define a norm on the dual space: $$ \langle f,g\rangle=\frac{1}{4}(\|f+g\|^2-\|f-g\|^2+i\|f+ig\|^2-i\|f-ig\|^2) $$

However, since $l_v=f$ & $l_w=g$ for some v and w: $\frac{1}{4}(\|l_v+l_w\|^2-\|l_v-l_w\|^2+i\|l_v+il_w\|^2-i\|l_v-il_w\|^2)$=$\frac{1}{4}(\|v+w\|^2-\|v-w\|^2+i\|v-iw\|^2-i\|v+iw\|^2)=<v,w>$

Hence, $<f,g>=<w,v>$.

Is this right?

Answer 1

There is one mistake. You are assuming that $l_{iv}=il_v$ but $l_{iv}=-il_v$. So finally you get $\langle f, g \rangle =\langle w, v \rangle$.