How can I show that for a Fourier series u(x) and it's associated partial sum $u_J(x)=\sum\limits_{k=-J}^J u_ke^{2\pi i kx}\,dx$
that $\int_0^1u(x)u_J(x)\,dx = \sum\limits_{k=-J}^J |u_J|^2$
I have proved the result when integrating $|u_J(x)|^2$ but cannot do this one.
Assuming $u(x)$ is real: $$\begin{aligned} \int_0^1 u(x) u_J(x)\ dx &= \int_0^1 u(x) \sum_{k=-J}^{J}u_k e^{2\pi i k x}\ dx \\ &= \sum_{k=-J}^J u_k \int_0^1 u(x) e^{2\pi i k x}\ dx \\ &= \sum_{k=-J}^J u_k \overline{\int_0^1 \overline{u(x)} e^{-2\pi i k x}\ dx} \\ &= \sum_{k=-J}^J u_k \overline{\int_0^1 u(x) e^{-2\pi i k x}\ dx} \\ &= \sum_{k=-J}^J u_k \overline{u_k} \\ &= \sum_{k=-J}^J |u_k|^2 \end{aligned}$$