The generalized Parseval identity states that given $\{e_i\}_1^\infty$ a complete orthonormal system in Hilbert space $H$ then for all $g,h\in H : \langle g,h\rangle =\sum_{i=1}^{i=\infty} \langle g,e_i \rangle \langle e_i , h \rangle$. I don't know if it is true, yet I try to show that the series absolutely converges, i.e $\sum_1^\infty |\langle g,e_i\rangle||\langle e_i,h\rangle| <\infty $.
The only thing I had in mind for bounding $\sum_1^\infty |\langle g,e_i\rangle||\langle e_i,h\rangle|$ is Cauchy Schwartz inequality , i.e $\sum_1^\infty |\langle g,e_i\rangle||\langle e_i,h\rangle| \le \sum_1^\infty ||g||\cdot||e_i||\cdot||e_i|| \cdot||h|| = \sum_1^\infty ||g|| \cdot ||h||$ , but it seems as if this is too curse bound since this sum doesn't generally converge.
Best Answer
Fix $n \in \Bbb{N}$, and consider the vectors \begin{align*} u &= (|\langle g, e_1 \rangle|, \ldots, |\langle g, e_n \rangle|), \\ v &= (|\langle e_1, h \rangle|, \ldots, |\langle e_n, h \rangle|). \end{align*} Then, by Cauchy-Schwarz on $\Bbb{R}^n$, $$u \cdot v \le \|u\| \|v\| \implies \sum_{i=1}^n |\langle g, e_i \rangle| |\langle e_i, h \rangle| \le \sqrt{\sum_{i=1}^n |\langle g, e_i \rangle|^2} \sqrt{\sum_{i=1}^n |\langle e_i, h \rangle|^2}.$$ Given that the $e_i$s are orthonormal, we have \begin{align*} \sqrt{\sum_{i=1}^n |\langle g, e_i \rangle|^2} &\le \|g\| \\ \sqrt{\sum_{i=1}^n |\langle e_i, h \rangle|^2} &\le \|h\|, \end{align*} hence $$\sum_{i=1}^n |\langle g, e_i \rangle| |\langle e_i, h \rangle| \le \|g\| \|h\| < \infty.$$