A doubt on the derivative of Fourier Series for $x$.

Question

I'm reading a book on Fourier Series, and the author states that if $f$ is piecewise differentiable ( derivative exists and is piecewise continuous) in $(-\pi,\pi) $, and if f can be written in the following way

$$f\left( x \right) = \frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {\left( {{a_n}\cos nx + {b_n}\sin nx} \right)}$$

then, he states, that the derivative is just

$f'\left( x \right) = \sum\limits_{n = 1}^\infty {\left( {n{b_n}\cos nx - n{a_n}\sin nx} \right)} $. This is just as if one would take the derivatives term by term.

Afterwards, he also states that the Fourier series for $x$,considered in the range $[-\pi, \pi]$ $$\left(\sum_{n\geq 1} \frac{2}{n} (-1)^{n+1}\sin(nx)\right)$$ is not uniformly convergent... and thus there's no point in taking the derivatives term by term. In fact he states that if we did, we would get some nonsense.

However, aren't Fourier Series uniformly convergent? Also, isn't $f(x)=x$ piecewise diff?

Why can't we take derivatives on the fourier representation of $x$?

I'm probably missing something here...

Answer 1

Not an answer, but just a visual demonstration: the red curve is the Fourier series of $f(x)=x$ summed to $1000$ terms; the blue... blob... is Desmos's attempt after two minutes of calculation to plot $f'(x)$. Some nonsense indeed.

More formally, consider the piecewise differentiation:

$$2\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\sin(nx)\mapsto2\sum_{n=1}^\infty(-1)^{n+1}\cos(nx)$$

Where we note the distinct lack of a dividing $1/n$ term! This means the resulting series simply doesn't settle down, as the amplitudes are never regulated, hence the blobby nonsense below.

Answer 2

The assertion in the book is approximately true, but there is a probably-unexpected trap: "differentiability" or "continuity" of a function expressed as standard Fourier series requires that the end-point values agree. So $f$ is "continuous" for these purposes if and only $f$ is continuous on $(-\pi,\pi)$, and $f$ has a right limit at $-\pi$ and a left limit at $+\pi$ and these two one-sided limits are the same value.

Similarly with differentiability.

This "extra" constraint arises because the sines and cosines (or, equivalently, exponentials) have that property...

So, the sawtooth function is continuous (and differentiable) in the interior of the interval, but not continuous at the endpoint: there's a jump.

Due to this, the Fourier series will not converge very well (it cannot converge uniformly, because then its limit would be continuous ... in this "periodic" sense, but it's not).

(Still, if we need to differentiate that Fourier series as a distribution, it does make sense, and converges perfectly well in a Sobolev space... though certainly not pointwise. This possibility is useful in computations often even when "the answer" does not directly mention distributions. In the case at hand, differentiating gives $-2\pi$ times the Dirac comb (periodic $\delta$) plus $1$...)