# Pathological functions as Fourier series

A sufficient condition for a periodic function to be representable as a Fourier series are the so called Dirichlet conditions:

$f$ must be absolutely integrable over a period.
$f$ must have a finite number of extrema in any given bounded interval, i.e. there must be a finite number of maxima and minima in the interval.
$f$ must have a finite number of discontinuities in any given bounded interval, however the discontinuity cannot be infinite.

But these conditions are known not to be necessary. What is an example function violating these conditions, which still has a Fourier series representation?

• Provided you use the Lebesgue integral, the characteristic function of the rationals would work. Even with the Riemann integral, the blancmange function should work. Then there's the Weierstrass function, of course. – Chappers Oct 8 '17 at 17:01
• So you meant $f(x) = \sum_{n=-\infty}^\infty c_n e^{2i\pi nx}$ pointwise. Then $g(x) = x -\lfloor x \rfloor$, $g(\mathbb{Z}) = 1/2$ has a Fourier series and if I'm not wrong so does $f(x) = \sum_{k=0}^\infty \frac{1}{k!} g(x 2^k)$ (infinitely many discontinuities) – reuns Oct 8 '17 at 18:16

So an easy answer to your question is: take a continuous function with an infinite number of extrema but of bounded variation. For instance, $$f(x) = \sin x\, \sin \log x\qquad 0\le x\le 2\pi$$ extended periodically. It looks reasonable:
but there are infinitely many extrema near $0$, as one can see from the derivative
$$f'(x) = \frac{\sin x}{x}\cos \log x + \cos x\,\sin \log x\approx \cos \log x + \sin \log x$$ which changes sign infinitely often as $x\to 0^+$.
The derivative is bounded, and so $\int_0^{2\pi} |f'(x)|\,dx$ is finite, which implies $f$ is a BV function.
• You can decompose a function $f$ of bounded variation into the difference of monotone functions that are continuous where $f$ is continuous. So the oscillations are not necessarily an impediment to applying the classical result. Too many discontinuities could be a problem for the classical result, though. – DisintegratingByParts Oct 9 '17 at 9:18
• Yes, one doesn't have to work hard to prove convergence for this function, but it still qualifies as an example in terms of OP's question. To make it discontinuous, one can throw in $\operatorname{sign}$ function: $\sin x \operatorname{sign} \sin \log x$ – user357151 Oct 9 '17 at 15:43