The norm of Fourier coefficients

Question

Let's consider the space $\mathcal{R}$ of integrable functions with inner product given by: $$<f,g> = \frac{1}{2 \pi} \int \limits_{0}^{2 \pi} f(\theta) \overline{g(\theta)} \mbox{d} \theta$$ and norm $\left\lVert f \right\rVert$ by $$\left\lVert f \right\rVert = <f, f> = \frac{1}{2 \pi} \int \limits_{0}^{2 \pi} |f(\theta)|^2 \mbox{d} \theta.$$ I've defined a family $$\{e_n \}_{n \in \mathbb{Z}} = e^{in \theta}.$$ It's easy to show that the family above is orthogonal (even orthonormal in fact).
Now I define Fourier coefficients in a following way: $$a_n = <f, e_n>.$$ It is written in my book - Fourier Analysis An Introduction (Princeton Lectures) that $$\left\lVert \sum_{|n| \le N} a_ne_n \right\rVert ^2 = \sum_{|n| \le N} |a_n|^2.$$ I'm afraid I can't see how it was obtained, because: $$\left\lVert \sum_{|n| \le N} a_ne_n \right\rVert ^2 = \big< \sum_{|n| \le N} a_ne_n, \sum_{|n| \le N} a_ne_n \big> = \frac{1}{2 \pi} \int \limits_{0}^{2 \pi} \big| \sum_{|n| \le N} a_ne_n \big |^2 \mbox{d} \theta \le \frac{1}{2 \pi} \int \limits_{0}^{2 \pi} \big( \sum_{|n| \le N} \big| a_ne_n \big | \big) ^2 \mbox{d} \theta$$

Answer 1

By bilinearity of the inner product, $$\left\langle\sum_{|n| \le N} a_n e_n, \sum_{|m| \le N} a_m e_m\right\rangle = \sum_{|n| \le N} \sum_{|m| \le N} a_n \overline{a_m} \langle{e_n, e_m \rangle}.$$ Now use the fact that the family $\{e_n\}$ is orthonormal.