I am working on this problem: In a complex inner product space $$\langle x,y \rangle = \frac{1}{2\pi}\int^{2\pi}_0 \|x+e^{it}y\|^2e^{it} \ dt.$$
My thought was to write out $y$ in a basis $y=\sum \alpha_ie_i$ giving $\langle x,y \rangle$ as a weighted sum of Fourier coefficients $\sum\alpha_i\hat{x}_i$ and establishing some relationship between this sum and the integral. Am I crazy?
Just compute. Observe \begin{align} \frac{1}{2\pi } \int^{2\pi}_0 \langle x+e^{it}y, x+e^{it}y\rangle e^{it} \ dt =&\ \frac{1}{2\pi } \int^{2\pi}_0 [\langle x, x\rangle e^{it}+ \langle x, y\rangle +\langle y, x\rangle e^{2it} + \langle y, y\rangle e^{it}] \ dt\\ =&\ \frac{1}{2\pi} \int^{2\pi}_0 \langle x, y\rangle\ dt = \langle x, y\rangle . \end{align}
Note: I used the fact that $\langle \cdot, \cdot \rangle$ is conjugate linear in the first component and linear in the second component.