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Let $H$ be a Hilbert space over $\mathbb{C}$. Let $\phi_{j} \in H$ and let $c_j$ be scalars. Prove that there exists a representation $$f=\sum_{j}c_{j}\langle f,\phi_{j}\rangle\phi_{j}, \forall f\in H$$ if and only if $$\|f\|^2=\sum_{j}c_{j}|\langle f,\phi_{j}\rangle|^2, \forall f\in H$$

My queries: For $\implies$: $\|f\|^2= \langle f,f \rangle = \sum_{i,j} c_{i} \overline{c_{j}} \langle f,\phi_{i} \rangle \overline{\langle f,\phi_{j} \rangle}\langle \phi_{i}, \phi_{j}\rangle$. It can not be simplified further due to $\phi_{j}$ are may not be orthogonal. Even if one assumes orthogonality it would become $\|f\|^2=\sum_{j}|c_{j}|^2|\langle f,\phi_{j}\rangle|^2\|\phi_{j}\|^2$ which is still not the desired form. Is there any way to apply the fact that the representation for $f$ is valid for "all" $f$ in some way to deduce the desired equality?

For sufficiency I intend to apply the polarization identity to $\langle f,\phi_{j}\rangle$ then substitute it into the desired formula for $f$, but the polarization identity implied $\langle f,\phi_{j} \rangle =0$ which is not true in general, is there any point I have missed?

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If $f=\sum_{j}c_{j}\langle f,\phi_{j}\rangle\phi_{j}$, then $$ \|f\|^2=\langle f,f\rangle=\langle \sum_{j}c_{j}\langle f,\phi_{j}\rangle\phi_{j},f\rangle =\sum_{j}c_{j}\langle f,\phi_{j}\rangle\,\langle\phi_{j},f\rangle=\sum_jc_j|\langle f,\phi_j\rangle|^2 $$ (the series distributes with respect to the inner product because we are assuming it is a norm limit).

Conversely, if $\|f\|^2=\sum_jc_j|\langle f,\phi_j\rangle|^2$, then we have $$ \langle f,f\rangle = \sum_{j}c_{j}\langle f,\phi_{j}\rangle\,\langle\phi_{j},f\rangle =\langle \sum_{j}c_{j}\langle f,\phi_{j}\rangle\phi_{j},f\rangle $$ for all $f$. Using polarization, we get, for a fixed $f$ and any $g$ $$ \langle f,g\rangle = \langle \sum_{j}c_{j}\langle f,\phi_{j}\rangle\phi_{j},g\rangle. $$ So $f=\sum_{j}c_{j}\langle f,\phi_{j}\rangle\phi_{j}$.

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