Prove that if ,$||f||^2 = A\sum_{j}|<f, \phi_j>|^2 $ then $f = \sum_{j}<f, \phi_j> \phi_j$

Question

Let $\phi_k$ be some sequence of real functions in an infinite Hilbert space $H$ such that there exists $A \in \mathbb{R}$ such that for all $f \in H$ ,$||f||^2 = A\sum_{j}|<f, \phi_j>|^2 $

Prove that $f = \sum_{j}<f, \phi_j> \phi_j$.

There is a hint to use polarization, but I failed to prove that.

Answer 1

First of all, note that $${\left\| f \right\|^2} = \left\langle {f,f} \right\rangle $$ then, by the linearity of inner product, you also have $$A\sum\limits_j {{{\left\langle {f,{\phi _j}} \right\rangle }^2}} = \sum\limits_j {\left\langle {f,{\phi _j}} \right\rangle A\left\langle {f,{\phi _j}} \right\rangle } = \sum\limits_j {\left\langle {f,A\left\langle {f,{\phi _j}} \right\rangle {\phi _j}} \right\rangle } $$using these two equations in the given equality (and again by linearity of the inner product), we get $$\left\langle {f,f - \sum\limits_j {A{\phi _j}\left\langle {f,{\phi _j}} \right\rangle } } \right\rangle = 0$$Now since this equation holds for any function, we should have $$f = \sum\limits_j {A{\phi _j}\left\langle {f,{\phi _j}} \right\rangle } $$hope it helps ;)