note to myself

• For $2L$ periodic functions, $\displaystyle a_n = \frac{1}{L}\int_{-L}^Lf(x)\cos nx\,dx$ for $\displaystyle n\ge 1$ and $\displaystyle b_n=\frac{1}{L}\int_{-L}^L f(x)\sin nx\,dx$ for $\displaystyle n\ge 1$.


• $\displaystyle\hat f(0)=\frac{a_0}{2}$ and for $\displaystyle n\ge 1$, $\displaystyle \hat f(n) = \frac{a_n- i b_n}{2}$, $\displaystyle \hat f(-n) = \frac{a_n+i b_n}{2}$, which is easily shown by expanding $\displaystyle \frac{a_0}{2} + \sum_{k=1}^n (a_k\cos kx + b_k\sin kx)$.

These two translate the general theory to real-valued function. For example, in the theory of fourier series, we know that for each $f\in L^2(\mathbb T)$, we have $$\|f\|^2:=(f,f)=\frac{1}{2\pi} \int_0^{2\pi} |f|^2 = \sum_{n\in \mathbb Z}|\hat f(n)|^2=: \|\hat f\|^2.$$

where $\displaystyle \hat f := (\hat f(0),\hat f(1),\hat f(-1),\hat f(2),\hat f(-2),\dots)\in \ell^2$ has the same norm as that of $\displaystyle f$. Thus we say that the linear map $\displaystyle f\mapsto \hat f:L^2(\mathbb T)\to \ell^2$ is isometric. Ok, let's prove that this is equivalent to (for real $f\in L^2[-L,L]$) $$\displaystyle\frac{1}{2L} \int_{-L}^{L} |f|^2 = \frac{(a_0)^2}{4}+\sum_{n=1}^\infty \frac{a_n^2+b_n^2}{2}.$$