Why is this inner product equal to $0$?

Question

Why is / how can I show that

$$ \Big\langle f - \sum_{k=-N}^N\langle f, e_k\rangle, \sum_{k=-N}^N(\langle f, e_k\rangle-a_k)e_k\Big\rangle = 0?$$

Here the $e_k$ are functions $e_k(x)=e^{2\pi i k x}$, $f\in L_p([0, 1]), a_k\in \mathbb{C}$, and the inner product is defined as $$\langle f, g\rangle = \int_{[0, 1]}f\bar g$$

Answer 1

Let $S_N=\sum_{k=-N}^N \langle f,e_k\rangle e_k$, $$\left\langle f-S_N,S_N-\sum_{k=-N}^N a_ke_k\right\rangle=\langle f,S_N\rangle-\sum_{k=-N}^N a_k\langle f,e_k\rangle-\langle S_N,S_N\rangle+\sum_{k=-N}^Na_k\langle S_N,e_k\rangle$$ But $\langle S_N,e_k\rangle=\langle f,e_k\rangle$ thus $$ \left\langle f-S_N,S_N-\sum_{k=-N}^N a_ke_k\right\rangle=\langle f,S_N\rangle-\langle S_N,S_N\rangle $$ and $$ \langle S_N,S_N\rangle=\sum_{k=-N}^N \langle f,e_k\rangle^2=\langle f,S_N\rangle $$ Finally $\left\langle f-S_N,S_N-\sum_{k=-N}^N a_ke_k\right\rangle=0$

Answer 2

Assuming

$$ \Big\langle f - \sum_{k=-N}^N\langle f, e_k\rangle e_k, \sum_{k=-N}^N(\langle f, e_k\rangle-a_k)e_k\Big\rangle $$

we have

$$ \langle f,\sum \langle f,e_k\rangle e_k\rangle - \langle f,\sum a_k e_k\rangle -\langle\sum\langle f,e_k\rangle e_k,\sum\langle f,e_k\rangle e_k\rangle + \langle\sum \langle f,e_k\rangle e_k,a_k e_k\rangle $$

or

$$ \sum \langle f,e_k\rangle^2-\sum a_k\langle f,e_k\rangle -\sum \langle f,e_k\rangle^2+\sum a_k\langle f,e_k\rangle = 0 $$