Show that $f(x)$ is orthogonal to $f'(x)$ in $L^2(-\pi, \pi)$

Question

I have the following problem:

Suppose $f$ is of class $C^{(1)}$, $\;2\pi$-periodic, and real-valued. Show that $f'$is orthogonal to $f$ in $L^2(-\pi, \pi)$ by

a) expanding $f$ in (generalized) Fourier series $f = \sum \langle f, \phi_n\rangle \phi_n$ and using the following theorem:

Suppose $\{\phi_n\}_1^\infty$ is an orthonormal basis for $L^2(a,b)$. Then for any $f, g \in L^2(a,b)$ we have

$$\langle f, g\rangle = \sum_n \langle f, \phi_n \rangle \overline{\langle g, \phi_n \rangle}.$$

b)

By using the fact that $(f^2)' = 2ff'$.

I think I got $\textbf{b)}$ solved:

$$\langle f, f'\rangle = \int_{-\pi}^\pi f(x)f'(x)\;dx = \frac{1}{2}\int_{-\pi}^\pi \frac{d}{dx} f(x)^2\;dx = \frac{1}{2}\left[ f(x)^2\right]_{-\pi}^\pi = 0.$$

I tried a) but it seemed more problematic...how should I solve a)?

Answer 1

Assuming that you are expanding with respect to the standard basis $\{\sin (nx), \cos (nx)\}_n$, you compute $$ f' = \sum \langle f,\phi_n \rangle \phi'_n, $$ and hence $$ \langle f,f' \rangle = \sum \langle f,\phi_n \rangle \langle \phi_n,\phi'_n \rangle = 0 $$ because $\int_{-\pi}^\pi \sin (nx) \cos (nx)\, dx =0$ for every $n$.