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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...

An old man in the sea.
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  • $x$ as a periodic function is not differentiable (not even continuous) at the period ends ( here $\pm \pi$) so the fourier series doesn't converge uniformly; even if it did there is no guarantee the result would be differentiable – Conrad Jan 13 '22 at 16:54
  • @Conrad In light of the statements here, wouldn't uniform convergence of the piecewise differentiated series imply that the piecewise differentiated series would converge to the correct derivative? – FShrike Jan 13 '22 at 18:17
  • @FShrike If the derivative series converges uniformly the result is true but here the derivative series converges nowhere pointwise (terms do not go to zero) so the question is moot; the original function is not continuous so in particular not piecewise differentiable; my point was that even if the original series converges uniformly ( hence to a continuous function) the fourier series may not be differentiable (equivalently the continuous function to which it converges is not absolutely continuous and actually may not be even of bounded variation) – Conrad Jan 13 '22 at 18:25
  • @Conrad your first comment is very interesting... I don't see how it's not diff at the end points. Also, why is it necessary for us to have the function diff at the end points? Sorry for all these questions. Thanks for all the help. – An old man in the sea. Jan 15 '22 at 21:10
  • Fourier series require periodic functions so something like $f(x)=x$ which is not periodic, needs a specified $2\pi$ interval and the periodic extension from it to get a Fourier series; various choices of the interval give different Fourier series and the standard ones are $[-\pi, \pi]$ with the series you mention or $[0,2\pi]$ for which the series is obtained by a substitution $x\to \pi -x$ so it is just all negative sum above; in the first case obviously the limit at $\pi$ from the left is $\pi$ but periodic extension means that the limit from the right is $-\pi$ so there is a jump – Conrad Jan 16 '22 at 01:39
  • Note that in general if $\sum_{n \ge 1} f_n=f, f_n,f$ are differentiable on some interval (even infinitely differentiable) and the convergence is uniform, it doesn't follow that $\sum f_n'$ exists or is $f'$ so the differentiation term by term theorem for Fourier series is a strong result that requires all the hypothesis given, in particular full piecewise differentiability of $f$ so excludes examples as above – Conrad Jan 16 '22 at 01:51
  • @Conrad I think I understand now... I think I was misled (+ lazy/distracted) by the phrasing of a theorem in the book. The author states that $f$ must be piecewise differentiable in $(-\pi, \pi)$. I think he meant along the whole real numbers... – An old man in the sea. Jan 17 '22 at 22:31
  • @Conrad Hum... On the same page, the author also says that we can take derivatives term by term for the $x^2$ fourier series representations. How is $x^2$ piecewise differentiable? – An old man in the sea. Jan 17 '22 at 22:47
  • The piecewise differentiable condition is a very strong sufficient condition but it is not necessary; for $x^2$ (which is continuous but not differentiable at the ends) we use a different theorem, namely that a series of functions for which the differentiated series converges uniformly is indeed differentiable term by term; note also that the same caveat about the ends applies here too as the fourier series of $x^2$ satisfies the condition above on any compact interval included in $(-\pi,\pi)$ so the differentiation theorem applies only there (as convergence goes) – Conrad Jan 17 '22 at 23:27
  • More generally regarding a fourier series as an object associated to an integrable periodic function $f$ (so ignoring convergence issues), one can easily show that if the trigonometric series obtained by formally differentiating the original fourier series term by term is also a fourier series associated to some integrable function $g$, then $f$ is absolutely continuous (hence differentiable ae) and $f'=g$ ae and conversely with the condition that the fourierr series of $g$ has no free term (otherwise the result applies to $g-c$ giving $f-cx$ absolutely continuous) – Conrad Jan 17 '22 at 23:52
  • So for $x^2$ you can go two ways - compute the fourier series and show that its derivative term by term is a fourier series, or use the fact that the series of $2x$ on $[-\pi, \pi]$ has no free term and apply the result above; note that if we look at $x^2,x$ on $[0,2\pi]$ the series of $2x$ has a free term $2\pi$ so the right function which has a term by term differentiable fourier series is not $x^2$ but $x^2-2\pi x$ – Conrad Jan 18 '22 at 00:05
  • @Conrad, join all of your comments in an answer, and I'll accept it and give you plus one. – An old man in the sea. Jan 18 '22 at 09:49

2 Answers2

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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 $1$...)

paul garrett
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  • Paul, thanks for the comment. In the book, the author states that Fourier series representations of piecewise cont, and periodic functions, on the discontinuity points, converge to the middle point between the side-limits, i.e. $ \frac{f(-\pi_+)+f(\pi_-}{2}$.

    I think my doubt is more for the interior of the interval... Which condition does the identity function $f(x)=x$ not satisfy so that the fourier series of the derivative not being deduced from taking the derivative of the fourier series term by term?

     – An old man in the sea. Jan 15 '22 at 21:05
  • @Anoldmaninthesea., the behavior of the Fourier series is not "purely local", that is, pointwise non-convergence of the term-wise derivative at points where the function is smooth occurs despite that smoothness... and is due to the jump discontinuity of the periodicized version of the function... even though that jump is some distance away from interior points. – paul garrett Jan 15 '22 at 21:25
  • Very interesting! :) – An old man in the sea. Jan 17 '22 at 22:33
  • @Anoldmaninthesea. it is indeed a funny business! :) – paul garrett Jan 17 '22 at 22:46
  • Paul, in the book, the author also states that x^2( with a 2pi period) has a valid fourier series, and one which converging uniformly can be differentiated term by term. In this case, x^2 is also discontinuous at the end points, no? Why is it then possible to do diff term by term? – An old man in the sea. Jan 17 '22 at 22:49
  • Since $x^2$ is even, the $(-\pi,\pi)$-periodicized version of it is continuous at the endpoints (since $(-\pi)^2=)+\pi)^2)$). :) If we periodicized the $(0,2\pi)$ version of it, the Fourier coefficients would be different, and the resulting Fourier series would not converge as well, due to the jump discontinuity. – paul garrett Jan 17 '22 at 22:54
  • Many thanks Paul, I think I got it! – An old man in the sea. Jan 18 '22 at 09:45
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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.

enter image description here

FShrike
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  • The series above sums to $1$ except at $\pm \pi$ by various summability methods (Abel,Caesaro) but it cannot converge anywhere since the terms do not go to zero – Conrad Jan 13 '22 at 18:33