Proving Plancherel Identity for complex inner product Given that $V$ is an infinite dimensional complex inner product space and that $\phi_k\in V$ is a complete orthonormal system, I am to prove Plancherel's identity, $$||f||^2=\sum_{k=1}^\infty \langle f,\phi_k\rangle^2$$
My first thoughts were to use the definitions of $||f||^2$ and $\langle f,\phi_k\rangle$ to try and show $$\frac{1}{2\pi}\int_{-\pi}^\pi |f(x)|^2- \frac{1}{4\pi^2}\sum_{k=1}^\infty \left[\int_{-\pi}^\pi f(x)\overline{\phi_k(x)}dx\int_{-\pi}^\pi f(x)\overline{\phi_k(x)}dx\right]=0$$ but this has been fruitless. The hint provided was to use the following identities:
$$\langle f,g\rangle= \frac{1}{4}\left(||f+g||^2-||f-g||^2+i||f+ig||^2-i||f-ig||^2\right)$$
$$||f + g||^2 =||f||^2+ 2\Re \langle f,g\rangle +||g||^2$$
However, I'm not sure how to make use of them, particularly the first, which seems like it would be the most helpful for this problem. I would love some help. Thanks very much.
 A: It's Parseval's identity, and you are missing the absolute value:
$$\tag1
\|f\|^2=\sum_{k=1}^\infty| \langle f,\phi_k\rangle|^2.
$$
What you do, is first note that
\begin{align}
0&\leq\Big\|f-\sum_{k=1}^N\langle f, \phi_k\rangle\,\phi_k\Big\|^2=
\|f\|^2+\sum_{k=1}^N |\langle f,\phi_k\rangle|^2-2\operatorname{Re}\langle f,\sum_{k=1}^N\langle f,\phi_k\rangle\,\phi_k\rangle\\[0.3cm]
&=\|f\|^2-\sum_{k=1}^N |\langle f,\phi_k\rangle|^2.
\end{align}
It follows that $\sum_{k=1}^N |\langle f,\phi_k\rangle|^2\leq\|f\|^2$ for all $N$, and so
$$\tag2
\sum_{k=1}^\infty |\langle f,\phi_k\rangle|^2\leq\|f\|^2.
$$
This is Bessel's Inequality. As the coefficients are square-summable, the element $$ \sum_{k=1}^\infty \langle f,\phi_k\rangle\,\phi_k$$ exists in $V$ (this requires that $V$ is complete).
Using that the inner product is continuous,
$$
\Big\langle f-\sum_{k=1}^\infty \langle f,\phi_k\rangle\,\phi_k,\phi_j\Big\rangle=\langle f,\phi_j\rangle-\langle f,\phi_j\rangle=0.
$$
As $\{\phi_k\}$ is complete, we get that
$$\tag3
f=\sum_{k=1}^\infty \langle f,\phi_k\rangle\,\phi_k.
$$
Using $(3)$ and the continuity of the inner product,
$$
\|f\|^2=\sum_{k,j}\langle f,\phi_k\rangle\,\overline{\langle \phi_j,f\rangle}\,\langle\phi_k,\phi_j\rangle=\sum_{k=1}^\infty|\langle f,\phi_k\rangle|^2.
$$
