Can one express $f'(x)$ with the same basis as one uses for $f$?

Question

If I have an orthonormal basis $\{\phi_n\}_1^\infty$ in space $L^2(a,b)$ and the generalized Fourier series expansion for $f$ would be:

$$f= \sum \langle f, \phi_n\rangle\phi_n,$$

then can one use the same basis $\{\phi_n\}_1^\infty$ to give a generalized Fourier series expansion for $f'$? That is can one do:

$$f'= \sum \langle f', \phi_n\rangle\phi_n,$$

Is this true or false? Is it true in some cases, but generally not?

Thank you for any help! =)

Answer 1

The formula $f' = \sum \langle f', \phi_{n}\rangle \phi_{n}$ is just the ordinary $L^{2}$ expansion of $f'$, assuming $f'$ is in the same $L^{2}$ space as $f$ itself.

The formula $f' = \sum \langle f, \phi_{n}\rangle \phi_{n}'$ holds assuming $\{\phi_{n}\}$ is the "standard" exponential/trigonometric basis $\phi_{n}(x) = e^{inx}$ of $L^{2}\bigl([-\pi, \pi]\bigr)$: $$ \langle f', \phi_{n}\rangle\phi_{n} = -\langle f, \phi_{n}'\rangle\phi_{n} = in\langle f, \phi_{n}\rangle\phi_{n} = \langle f, \phi_{n}\rangle\phi_{n}'. $$ This chain of equality uses the fact that $\phi_{n}$ is an eigenfunction of the derivative.