Characterization of Parseval Frame Let $H$ be a Hilbert space, and let $e_j\in H$ for every $j\in\mathbb N$.
Is it possible to show that
$$\left\|f\right\|^2=\sum_{j=1}^\infty\left|\langle f,e_j\rangle\right|^2\text{ for every }f\in H\text{ if and only if }f=\sum_{j=1}^\infty\langle f,e_j\rangle e_j\text{ for every }f\in H\tag*{?}$$
I get the feeling that we need $e_j\perp e_k$ if $j\neq k$.
 A: For the reverse implication see this answer.
Note that orthogonality is not required for a Parseval Frame.
Assume that $\|f\|^2=\sum_j|\langle f,e_j\rangle|^2$ for all $f$. Given $m>n$,
\begin{align}
\Big\|\sum_{j=1}^m\langle f,e_j\rangle\,e_j-\sum_{j=1}^n\langle f,e_j\rangle e_j\Big\|^2
&=\Big\|\sum_{j=n+1}^m\langle f,e_j\rangle e_j\Big\|^2\\[0.3cm]
&=\sup\Big\{\Big|\Big\langle\sum_{j=n+1}^m\langle f,e_j\rangle e_j,h\Big\rangle\Big|:\ \|h\|=1\Big\}\\[0.3cm]
&=\sup\Big\{\Big|\sum_{j=n+1}^m\Big\langle\langle f,e_j\rangle e_j,h\Big\rangle\Big|:\ \|h\|=1\Big\}\\[0.3cm]
&\leq\sup\Big\{\sum_{j=n+1}^m|\langle f,e_j\rangle\,\langle e_j,h\rangle|:\ \|h\|=1\Big\}\\[0.3cm]
&\leq\bigg(\sum_{j=n+1}^m|\langle f,e_j\rangle|^2\bigg)^{1/2}
\sup\Big\{\bigg(\sum_{j=n+1}^m|\langle e_j,h\rangle|^2\bigg)^{1/2}:\ \|h\|=1\Big\}\\[0.3cm]
&\leq\bigg(\sum_{j=n+1}^m|\langle f,e_j\rangle|^2\bigg)^{1/2}
\sup\Big\{\bigg(\sum_{j=1}^\infty|\langle e_j,h\rangle|^2\bigg)^{1/2}:\ \|h\|=1\Big\}\\[0.3cm]
&=\bigg(\sum_{j=n+1}^m|\langle f,e_j\rangle|^2\bigg)^{1/2}.
\end{align}
The convergence of $\sum_j|\langle f,e_j\rangle|^2$ guarantees that the last series can be made arbitrarily small if $m$ and $n$ are large enough. So
$$
\sum_j\langle f,e_j\rangle e_j
$$
exists. Now we need to show that it equals $f$. We will need the Polarization Identity
$$
\langle f,g\rangle=\tfrac14\,\sum_{k=0}^3i^k\|f+i^kg\|^2,
$$
and also the particular case to complex numbers,
$$
z\overline w=\tfrac14\,\sum_{k=0}^3i^k|z+i^kw|^2.
$$
We have
\begin{align}
\langle f,g\rangle&=\tfrac14\,\sum_{k=0}^3i^k\|f+i^kg\|^2
=\tfrac14\,\sum_{k=0}^3i^k\sum_j|\langle f+i^kg,e_j\rangle|^2\\[0.3cm]
&=\sum_j\tfrac14\,\sum_{k=0}^3i^k|\langle f,e_j\rangle+i^k\langle g,e_j\rangle|^2\\[0.3cm]
&=\sum_j\langle f,e_j\rangle\langle e_j,g\rangle.
\end{align}
Then
\begin{align}
\Big\langle f-\sum_j\langle f,e_j\rangle e_j,g\Big\rangle
&=\langle f,g\rangle-\sum_j\langle f,e_j\rangle\langle e_j,g\rangle=0.
\end{align}
As this can be done for any $g$, it follows that
$$
f=\sum_j\langle f,e_j\rangle e_j. 
$$
A: No, you don't need to assume orthogonality. The claim becomes more clear in terms of the analysis operator (aka Bessel operator)
$$ 
Af = (\langle f, e_j\rangle)_j 
$$
and the synthesis operator (aka reconstruction operator)
$$
S(c_j) = \sum_j c_j e_j
$$
The claim is that $SA=I$ if and only if $A$ is an isometry into $\ell^2$. 
Proof or $\impliedby$. 
Suppose $A$ is an isometry into $\ell^2$. Then its adjoint $A^*$ is a bounded operator from $\ell^2$ to $H$, which agrees with $S$ because 
$$\langle Af, x\rangle_{\ell^2} =\sum_j \langle f, e_j\rangle_H \overline{x_j} =  
\left\langle f, \sum_j x_je_j\right\rangle_H = \langle f, Sx\rangle_H
$$
By the polarization identity, $\langle Af, Ag\rangle_{\ell^2} = \langle f, g\rangle_H $ for all $f,g\in H$. Hence $\langle A^*Af, g\rangle_H = \langle f, g\rangle_H $ for all $f,g\in H$, which implies $A^*A=I$. Since $A^*=S$, we conclude that $SA=I$.  
Proof of $\implies$. Here I have to assume that $A$ is a bounded operator into $\ell^2$; it may be possible to remove this assumption, but I don't see how; in any case it is  standard in frame theory. As above, we have $S=A^*$. Since $SA=I$, it follows that $A*Af=f$ for every $f\in H$, hence 
$$
\|Af\|^2 = \langle Af, Af\rangle_{\ell^2} = \langle A^*Af, f\rangle_H = \langle f, f\rangle _H = \|f\|^2
$$
as desired.
A: If $(e_j)$ is a sequence in $H$ with $<e_j,e_k>=\delta_{jk}$, then we have:
$(e_j)$ is an orthonormalbasis of $H$ $ \iff$
$\left\|f\right\|^2=\sum_{j=1}^\infty\left|\langle f,e_j\rangle\right|^2\text{ for every }f\in H \iff\sum_{j=1}^\infty\langle f,e_j\rangle e_j\text{ for every }f\in H\tag*{}$
