Consider a function which has only jump singularities of the form of the function itself or one of its derivatives jumping. Now let $\hat{f}(k)$ be its Fourier transform/series. We know the decay of the magnitude of the Fourier series/transform depend on how many times $f$ is continuously differentiable.
But now what I am asking is are there any such similar relations which not only tells how many times $f$ is continuously differentiable, but also gives complete information about all the jump singularities of $f$, i.e., the location of the jump $x_i$, the order of the derivative jumping $k_i$, and the amount of the jump $D_i$.
Is it unintuitive to ask such a question? Is such a thing not intuitively possible? I have been sincerely working hard for the past two years trying find an answer to this question, and I would sincerely appreciate suggestions and insights on this question.
EDIT (In view of comment by fedja)
We assume we have complete information about all the Fourier series/transform coefficients both magnitude and phase of all the infinite coefficients. I have mentioned the relation to be somewhat like asymptotic with the only reason that no 'finite number of Fourier coefficients can give any information about the singularities. That's why I expect the relations which give information about singularities need to involve all the infinite coefficients, and in this sense I used the word asymptotic, but we essentially assume we can utilize the full information about all the Fourier coefficients.
If we assume we do not have information about a finite number of Fourier coefficients, it would still not make any difference to the problem as these finite missing coefficients do not carry any info about the singularities, as they would form addition of a trignometric polynomial which is smooth.
EDIT 2
I am not interested in recontrsucting the function. There could be a lot of functions which could possess the same singularities as the given function but different from the given function. I want a property of the Fourier transform/coefficients which is obeyed by all such functions.
EDIT 3
To make my point more precise, may I add that the method should be amenable to an algorithm with infinite computations. Would it be a legitimate thing? I may be not be quite right here, but let me try to express my interest. I assume we were given all (except some finite) the Fourier coefficients, but not in any closed form expression. Say they were supplied to us as real numbers data. Now I'd like know how to do computations on them to determine the singularities.
EDIT 4 (after the answers by Igor Khavkine and Paul Garret)
I am seeking for something as rigorous as this theorem. Let the function $f(\theta)$ be a BV and periodic with period $2\pi$. Let $s_n(\theta)$ be the $n^{th}$ Fourier series partial sum then, we have $$\lim_{n\to\infty}\frac{{s'_n}(\theta)}{n} = \frac{1}{\pi}(f(\theta^+)-f(\theta^-))$$. Not exactly same or similar to this, but it should be as rigorous as this theorem.
Best Answer
Consider functions of the form $s_k(x) = \Theta(x) x^k$, where $\Theta(x)$ is the Heaviside step function. Then any function of the kind you describe can be written as $f(x) = g(x) + \sum_{i,k} (A_{ik} s_k(x-y_i))$, where the constants $A_{i,k}$ and $y_i$ determine the magnitudes of some finite number of discontinuities in derivatives of order less than $r$ and their locations, and $g(x)$ is $C^r$ regular. Obviously the same decomposition will be true for the Fourier transform $\hat{f}(\xi)$. It is straightforward to compute the Fourier transforms $\hat{s}_k(\xi)$ and match them to the asymptotics of $\hat{f}(\xi)$ for large $\xi$. Essentially, they will decay as powers of $\xi$ and $\hat{s}_k(x)$ will always dominate $\hat{g}(\xi)$ for any $g(x)$ that is $C^r$ regular, with $r>k$.
Then, $\hat{f}(\xi) = \sum_k F_k(\xi) \xi^{-k-1}$, where each $F_k(\xi)$ is a finite linear combination of the form $\sum_i A_{ik} \exp(i\xi y_i)$. It should be straight forward to extract the locations $y_i$ and the jump magnitudes $A_{ik}$ if the $F_k(\xi)$ are known.