$\|f\|^2 =\sum_j|\langle f, f_j \rangle|^2$ implies $\langle f, g \rangle=\sum_j \langle f, f_j\rangle\langle f_j, g\rangle$. 
Let $\mathcal{H}$ be a complex Hilbert space, let $f, g \in \mathcal{H}$ and let $(f_j)_{j \in J} \subset \mathcal{H}$, which is indexed by an index set $J \subseteq \mathbb{Z}$. Suppose $\langle f, f \rangle = \sum_{j \in J} | \langle f, f_j \rangle|^2$. Then 
  $$ \langle f, g \rangle = \sum_{j \in J} \langle f, f_j \rangle \langle f_j, g \rangle. $$

I have seen the above result several times, and it seems to be a standard result.  In general, it is mentioned that the result follows by "polarisation". However, I am not entirely sure how to apply the polarisation identity here.
Any help and/or comment is highly appreciated.
 A: Hint: (in the real case) $$\langle f,g\rangle=\frac{1}{4}(\|f+g\|^2-\|f-g\|^2)=\frac{1}{4}(\sum_j\langle f+g,f_j\rangle^2-\sum_j\langle f-g,f_j\rangle^2)$$
now expand the last brackets using linearity of the inner product. Basically
$$\langle x+y,z\rangle^2=(\langle x,z\rangle+\langle y,z\rangle)^2=\langle x,z\rangle^2+2\langle x,z\rangle\langle y,z\rangle+\langle y,z\rangle^2$$
for the complex case, see comment.
A: First notice that your sum converges absolutely because
$$
           \sum_{j}2|\langle f,f_j\rangle \langle f_j,g\rangle| \le \sum_{j}\{|\langle f,f_j\rangle|^2+|\langle g,f_j\rangle|^2\}
$$
Therefore you can define
$$
      [f,g] = \sum_j \langle f,f_j \rangle \langle f_j,g\rangle
$$
and this form has the properties of an inner product, including positivity:
$$
          [f,f] = \sum_j |\langle f,f_j\rangle|^2 = \|f\|^2=\langle f,f\rangle.
$$
Because of this, the polarization identity gives $[f,g]=\langle f,g\rangle$ for all $f,g$, which gives you identity you wanted:
$$
            \langle f,g\rangle = \sum_{j}\langle f,f_j\rangle\langle f_j,g\rangle,\;\;\; f,g\in\mathcal{H}.
$$
