Parseval's identity type result for Hilbert spaces Parseval's identity for Hilbert spaces state that if $H$ is a Hilbert space and $(e_i)_{i \in I}$ is its orthonormal basis, then for all $x \in H$ we have
$$
\lVert x \rVert_H^2 = \sum_{i \in I} \left\lvert \langle x, e_i \rangle_H \right\rvert^2.
$$
Is the following generalization of this result true?
For all $x,y \in H$ we have
$$
\langle x, y \rangle_H = \sum_{i \in I} \left\lvert \langle x, e_i \rangle_H \langle y, e_i \rangle_H \right\rvert.
$$
It is clearly true for Euclidean spaces, but I am at a loss when it comes to general Hilbert spaces.
This question's accepted answer seems hand-wavy since I don't see how $\langle x, y \rangle$ equals $\sum_i x_i y_i$.
 A: If $\ (e_i)_{i\in I}\ $ is an orthonormal basis, then
\begin{align}
x&=\sum_{i\in I}x_ie_i\ \text{, and}\\
y&=\sum_{j\in I}y_je_j\ ,
\end{align}
where $\ x_i=\langle x,e_i\rangle\ $ and $\ y_j=\langle y,e_j\rangle\ $.  Therefore,
\begin{align}
\langle x,y\rangle&=\Big\langle\sum_{i\in I}x_ie_i,\sum_{j\in I}y_je_j\Big\rangle\\
&=\sum_{i\in I}x_i\Big\langle e_i,\sum_{j\in I}y_je_j\Big\rangle\\
&=\sum_{i\in I}x_i\sum_{j\in I}\overline{y}_j\langle e_i,e_j\rangle\\
&=\sum_{i\in I}x_i\overline{y}_i\ ,
\end{align}
by the orthonormality of $\ (e_i)_{i\in I}\ $.  Note that the question whose answer you cited was about real Hilbert spaces, for which the above identity reduces to
$$
\langle x,y\rangle=\sum_{i\in I}x_iy_i=\sum_{i\in I}\langle x,e_i\rangle\langle y,e_i\rangle\ .
$$
For a general Hilbert space, however, the conjugation must remain in the identity:
$$
\langle x,y\rangle=\sum_{i\in I}x_i\overline{y}_i=\sum_{i\in I}\langle x,e_i\rangle\overline{\langle y,e_i\rangle}=\sum_{i\in I}\langle x,e_i\rangle\langle e_i,y\rangle\ ,
$$
as given by Kavi Rama Murthy in his comment.
