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Question. Let $f$ be an arbitrary $2\pi$-periodic integrable function on reals. Does there exists any sequence $\{t_n\}$ with $f(\theta)=\sum t_ne^{in\theta}$?

If NOT, what about if we replace integrable function with a continuous one?!

ABB
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  • Your second part does not make sense. If the result is not true for continuous fucntions it cannot be true for integrable functions. – Kavi Rama Murthy Oct 31 '19 at 08:09
  • you need more for pointwise convergence, say Holder regularity. If you assume $f$ is $L^2$ then you have convergence of the series in $L^2$. – Calvin Khor Oct 31 '19 at 08:19
  • @ Calvin Khor: Any counterexample? – ABB Oct 31 '19 at 08:59
  • Yes, there is a classical "construction" that uses the Uniform Boundedness Principle. In fact I recently asked a question about this construction here - https://math.stackexchange.com/questions/3406915/what-is-the-baire-category-argument-here-divergence-of-many-fourier-series-at – Calvin Khor Oct 31 '19 at 09:10
  • @Calvin Khor: It seems you assume that $f(\theta)=\sum t_ne^{in\theta}$ implies $\hat{f}(n)=t_n$!! Why? – ABB Oct 31 '19 at 12:08
  • well to finish that to full rigour, you're going to need to define that equal sign very very precisely. That is, what do you mean when you say $$ f(\theta)=\sum t_ne^{in\theta}$$ – Calvin Khor Oct 31 '19 at 12:43
  • I mean that does there exists a sequence ${t_n}$ such that $f$ is represented by $f(\theta)=\sum t_ne^{in\theta}$ pointwisely? – ABB Oct 31 '19 at 12:51
  • Could you please write more about why pointwise convergence of the series $\sum t_ne^{in\theta}$ implies uniform convergence? – ABB Oct 31 '19 at 13:29
  • I don't know if its true and i shouldn't comment while busy, sorry – Calvin Khor Oct 31 '19 at 13:50
  • It is clear you are asking a more subtle question that involves the uniqueness of the trigonometric series representing a function, and are not asking a basic question on Fourier series, especially since you already know about the book of Katznelson. Please edit the question to make this more clear for other people – Calvin Khor Nov 01 '19 at 02:08

2 Answers2

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There is the following theorem of de la Vallée-Poussin, which I quote from page 880 of J. Marshall Ash's Uniqueness of Representation by Trigonometric Series:

If $S=\sum_n t_n e^{inx}$ converges to $f$ at each $x$, and if $f$ is finite at each $x$ and if $\int_{\mathbb T} |f(x)| dx< \infty$, then $S$ is the Fourier series of $f$.

This theorem is to be found in Zygmund's Trigonometric Series, Vol I, page 326. The beginning of Ash's paper proves the special case for $f=0$, but it seems that the proof of the general case spans several pages, so I will leave it at that.

For your question: consider a continuous function $f$ constructed via the Uniform Boundedness Principle (see for example, the note of Paul Garrett linked in this question). Suppose that $S$ converged pointwise to $f$. By de la Vallée-Poussin's theorem, $S$ must be the Fourier series of $f$. But the Fourier series of $f$ does not converge at every point, which is a contradiction.

Calvin Khor
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If a sequence $\{t_n|n\in\Bbb Z\}$ exists for which $f$ satisfies the identity$$f(\theta)=\sum_{n\in\Bbb Z}t_ne^{in\theta},$$applying $g\mapsto\int_0^{2\pi}g(\theta)e^{-ik\theta}d\theta$ with $k\in\Bbb Z$ gives $\int_0^{2\pi}f(\theta)e^{-ik\theta}d\theta=2\pi t_k$, since in terms of the Kronecker delta $\int_0^{2\pi}e^{i(n-k)\theta}d\theta=2\pi\delta_{nk}$ for integers $n,\,k$. In other words, if a solution exists it is$$f(\theta)=\sum_{n\in\Bbb Z}\left(\frac{1}{2\pi}\int_0^{2\pi}f(\theta^\prime)e^{-in\theta^\prime}d\theta^\prime\right)e^{in\theta}.$$If such an $f$ is modified at countably many values of $\theta$, the integrals in the second display-line equation are unchanged, so the function obtained from such a sum is the "original" function. Then the display-line equations are each true of the "new" $f$, except at countably many points.

J.G.
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  • Seems you assume $\int( \sum_{n\in \mathbb{Z}}t_ne^{in\theta})e^{ik\theta}=\sum_{\in \mathbb{Z}}\int t_ne^{in\theta}e^{ik\theta}$? Why can the infinite sum and integration be interchanged? We only have the point-wise convergent. – ABB Oct 31 '19 at 14:09