If $H$ is a Hilbert space then integral version of polarisation identity.

Question

If $H$ is a Hilbert space, then $\langle x, y\rangle = \frac{1}{2\pi}\int_0^{2\pi}e^{\imath\theta}\lVert x + e^{\imath\theta}y\rVert^2 d\theta$.

I have no idea where to start. So far I had only little troubles solving problems in functional analysis, but now I'm stuck, so I guess I'm just missing something, because it should be quite easy problem.

Answer 1

Here's another method: \begin{align*} \frac{1}{2\pi}\int_0^{2\pi}e^{i\theta}\|x+e^{i\theta}y\|^2d\theta &=\frac{1}{2\pi}\int_0^{2\pi}e^{i\theta}\left(\|x\|^2+e^{-i\theta}\langle x,y\rangle+e^{i\theta}\langle y,x\rangle+\|y\|^2\right)d\theta\\ &=\langle x,y\rangle+\frac{1}{2\pi}\left(\|x\|^2+\|y\|^2\right)\int_0^{2\pi}e^{i\theta}d\theta+\frac{1}{2\pi}\langle y,x\rangle\int_0^{2\pi}e^{2i\theta} d\theta\\ &=\langle x,y\rangle. \end{align*}

Answer 2

I'll just write out Daniel Fischer's comment: \begin{align} \frac{1}{2\pi}\int_{0}^{2\pi}e^{i\theta}\| x+e^{i\theta}y\|^{2}d\theta &= \frac{1}{2\pi}\int_{0}^{2\pi}e^{i\theta}\langle x+e^{i\theta}y,x+e^{i\theta}y\rangle d\theta \\ &= \frac{1}{2\pi}\int_{0}^{2\pi}e^{i\theta}\left(\|x\|^{2}+2 \text{Re}\langle x,e^{i\theta}y\rangle+\|y\|^{2} \right)d\theta \\ &=\frac{1}{2\pi}\int_{0}^{2\pi}2\left(\text{Re}\langle x,y \rangle e^{i\theta}\cos \theta + \text{Im}\langle x,y \rangle e^{i\theta}\sin \theta \right)d\theta \\ &=\frac{1}{2\pi}\left(\text{Re}\langle x,y \rangle\int_{0}^{2\pi}2e^{i\theta}\cos\theta d\theta+\text{Im}\langle x,y \rangle \int_{0}^{2\pi}2e^{i\theta}\sin\theta d\theta \right) \\ &=\frac{1}{2\pi}\left(2\pi\text{Re}\langle x,y \rangle +2i\pi \text{Im}\langle x,y \rangle\right) \\ &=\langle x,y \rangle. \end{align}