Identity for complex inner product spaces

Question

Suppose $V$ is a complex inner product space. I want to prove that for any $x,y \in V$, we have the following identities:

$\langle x, y\rangle = \frac{1}{2\pi} \int_{0}^{2\pi} \|x + e^{it}y\|^2 e^{it}dt$

$\langle x, y\rangle = \frac{1}{N} \sum_{k=1}^{N} \|x + e^\frac{2\pi ik}{N}y\|^2 e^\frac{2\pi ik}{N},$ where $N \geq 3.$

I tried to use the polarization identity (and I still thnk is is the way to approach the problem), but I couldn't figure out the trick, so help would be highly appreciated.

Answer 1

\begin{align} &\frac{1}{N} \sum_{k=1}^{N}\|x + e^\frac{2\pi ik}{N}y\|^2 e^\frac{2\pi ik}{N}\\ &=\frac{1}{N} \sum_{k=1}^{N}\left(\|x\|^2+\|e^\frac{2\pi ik}{N}y\|^2+\langle x, e^\frac{2\pi ik}{N} y\rangle+\langle e^\frac{2\pi ik}{N} y,x\rangle\right)e^\frac{2\pi ik}{N}\\ &=\frac{1}{N} \sum_{k=1}^{N}(\|x\|^2+\|y\|^2)e^\frac{2\pi ik}{N}+\frac1N\sum_{k=1}^{N}\langle x, y\rangle e^\frac{-2\pi ik}{N}e^\frac{2\pi ik}{N}+\frac1N\sum_{k=1}^{N}\langle y, x\rangle e^\frac{2\pi ik}{N}e^\frac{2\pi ik}{N}\\ &=0+\frac1N\sum_{k=1}^{N}\langle x, y\rangle+0\\ &=\langle x, y\rangle \end{align} Some observations:

• The sum identity is a generalization of polarization identity in complex Hilbert space ($n=4$). $$\langle x,y\rangle ={\frac {1}{4}}\left(\|x+y\|^{2}-\|x-y\|^{2}+i\|x+iy\|^{2}-i\|x-iy\|^{2}\right)$$
• $\{\frac{2\pi k}n:k,1,\dots,N\}$ is a partition of $[0,2\pi]$, so the sum is a Riemann sum of the integral. $$\frac{1}{N} \sum_{k=1}^{N} \|x + e^\frac{2\pi ik}{N}y\|^2 e^\frac{2\pi ik}{N}\longrightarrow\frac{1}{2\pi} \int_{0}^{2\pi} \|x + e^{it}y\|^2 e^{it}dt$$
• The integral is path integral of $\|x+zy\|^2$ on the circle $|z|=1$. $$\int_{0}^{2\pi} \|x + e^{it}y\|^2 ie^{it}dt=\oint_{|z|=1} \|x + zy\|^2 dz$$