Term by term differentiation

Question

If a function $f(x)$ is expressed as a Fourier series and we know $f'(x)$. Is it then true that if we differentiate the Fourier expression we must get $f'(x)$? E.g. if $f(x)=x^2$ for $x\in [0,\pi]$ which we extend to a sine or cosine series then can I conclude that term-by-term differentiation of the expression gives $2x$ or $2|x|$ respectively? I have a hunch that it is so, but I may be wrong. Thanks.

Answer 1

Here's an example to get you thinking about where some difficulties may lie. Let's think about the following "sawtooth function": $$ s(x) = x - 1/2\qquad (\mbox{for } 0 \le x < 1) $$ extended periodically to $\mathbb{R}$ by setting $s(x+n) = s(x)$ for all $n \in \mathbb{Z}$.

You can compute directly that $s(x)$ has the following (non-absolutely convergent) Fourier series: $$ \frac{-1}{2 \pi i} \sum_{n \ne 0} \frac{1}{n}e^{2\pi i n x}. $$ It makes sense to think about differentiating $s$ away from the integers, where we expect to get $s'(x) = 1$. But if we differentiate the above expression term-by-term, we are faced with the following expression, which doesn't really make a lot of sense: $$ -\sum_{n \ne 0} e^{2 \pi i n x} =\ ??? $$

The punch line here is that things can get rather subtle rather fast when dealing with Fourier series, and quickly leads in to things like Sobolev spaces and distributions.

It turns out that the proper hypotheses you'll want on a Fourier series $\sum c_n e^{2\pi i n x}$ to be $k$-fold termwise differentiable is for the Fourier coefficients to be "appropriately small" in the following sense: if $$ \sum |c_n|\cdot|n|^k < \infty $$ holds for some $k$, then the function represented by the Fourier series will be $k$-times differentiable, and will be differentiable termwise. If this holds for all $k$, then the function is smooth.

My reference for this is Paul Garrett's piece "Functions on Circles", which you can find on his website.