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?!
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?!
On
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.
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:
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.