Characterization of Parseval Frame

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 $$ f=\sum_j\langle f,e_j\rangle e_j. $$