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It seems that a large class of periodic functions (e.g. continuously differentiable functions) can be represented as a trigonometric series, at least almost everywhere. But is there any function (less pathological the better) that CANNOT be represented as a trigonometric series almost everywhere? I know there are functions whose Fourier series does not converge, but that does not imply that it cannot be represented by a trigonometric series, does it? Many results in Fourier analysis assume at least integrability of functions, so if the function is not integrable, there seem to be little I can say about its trigonometric representation.

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  • $\begingroup$ What precisely do you mean by "represent" in the case of a series that does not converge almost everywhere? $\endgroup$ – Robert Israel May 5 '17 at 1:04
  • $\begingroup$ By "representing a function by a trigonometric series," I meant that there is trigonometric series which converges to the function pointwise. $\endgroup$ – ashpool May 5 '17 at 1:08
  • $\begingroup$ This question as stated includes non-measurable functions. The basic analysis results I'm aware of just apply to measurable functions. So maybe it is harder than it looks? $\endgroup$ – Reinstate Monica May 5 '17 at 1:13
  • $\begingroup$ Here is a discussion and extension of a famous result that there is $f\in L^1(0,2\pi)$ whose Fourier series diverges almost everywhere: akademiai.com/doi/abs/10.1007/BF01904870?journalCode=10473 . Is this the type of thing you were wondering about? $\endgroup$ – DisintegratingByParts May 5 '17 at 2:26
  • $\begingroup$ It turns out the more natural notion of convergence for such series is $L^2$ convergence - that is $\int_{0}^{P} |f(x)-f_n(x)|^2\,dx \to 0$, where $P$ is the period. $\endgroup$ – Thomas Andrews May 5 '17 at 15:51
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It depends what you mean by "can be represented."

It is perhaps natural to take it to mean that for all $\epsilon>0$ there is an $n(\epsilon)$ such that the series truncated at $m$ terms for any $m\geq n(\epsilon)$ has the property of always being within $\epsilon$ of the function being represented. If that is the meaning, then I'm pretty sure that continuous differentiation is necessary and sufficient.

Note that "always" here means at every point, using an $n(\epsilon)$ that does not depend on the selected point.

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