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I am looking for the simplest cases possible of one-variable bump functions $\in C_c^\infty$ [1] with known Fourier transforms in "closed form" (also the function itself). This means it can be described by commonly known functions (exponentials, polynomials, trigonometric, logarithmic, etc.), not defined by more than two piece-wise "steps" (within and outside the compact-supported domain), so things like: $$f(x) = \begin{cases} f_1(x),\,x_0 \leq x < x_1; \\ f_2(x),\,x_1 \leq x < x_2; \\ \,\,\,\,\,\vdots \\ f_n(x),\,x_{n-1} \leq x \leq x_n \end{cases}$$ are not allowed for $n \geq 3$.

As example, $f(x) = e^{\frac{1}{x^2-1}},\,|x|\leq 1 $ is a valid one, since is defined picewise at most in two pieces: its compact support on one piece, and $0$ outside.

If possible, with domain in $[-1;\,1]$, and also if possible, the functions and its Fourier transforms made by functions that can be described "shortly" (since I want to plot them on Wolfam-Alpha and it didn't recognize functions that are "too long").

Beforehand, thanks you very much.


added later

I believe here are show a way to made bump functions $\in C_c^\infty$ that are not defined piecewise, but unfortunately, I don´t think its helps to find their Fourier Transforms.

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    $\begingroup$ A “bump functions” tag is probably unnecessary. $\endgroup$
    – littleO
    Oct 18, 2021 at 20:48
  • $\begingroup$ @littleO I have seen a lot of questions related, and also a lot a confusion about how are they defined (myself included), especially since they have to be "smooth" at the boundaries of its domain ($\partial t$), so is not only required that $f(\partial t) = f'(\partial t) = 0$ so it rises and decline "softly" to $0$ (the value outside their domain), also have to fulfill that $\lim_{t \to \partial t} \frac{d^n}{dt^n}f(t) = 0, \forall n \geq 0 \in \mathbb{Z}$ so every derivative is continuous on $\partial t$. As example, the only answer, given by an expert, is wrong because of derivative issues. $\endgroup$
    – Joako
    Oct 18, 2021 at 21:14
  • $\begingroup$ Perhaps one of the constructions in Are there other kinds of bump functions than $e^\frac1{x^2-1}$? could work. $\endgroup$ Oct 18, 2021 at 21:15
  • $\begingroup$ @projectilemotion I have tried almost every one of bump functions listed there in Wolfram Alpha but neither ones gives a result in closed form for its Fourier Transform (saddly). $\endgroup$
    – Joako
    Oct 18, 2021 at 21:19
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    $\begingroup$ If you allow infinite products to be a closed form (which it isn't by the usual definition), then by this answer and the attached paper, the Fourier transform of the bump function $\varphi$ satisfying the given conditions is given by $$\widehat{\varphi}(z)=\prod_{h=0}^{\infty} \frac{\sin\left(\frac{\pi z}{2^h}\right)}{\frac{\pi z}{2^h}}.$$ There are also other equivalent expressions given in the paper. $\endgroup$ Oct 18, 2021 at 21:23

2 Answers 2

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If your smooth function is defined on the whole real line only using $+,-,\times,\div$, and finitely many of those functions (polynomials, exponentials, trigs, and their inverses) and without a piecewise definition, and without an infinite sum or integral or whatever, then your function is better than smooth: it’s analytic. But analytic and compact support implies identically zero by the identity theorem. So the only such function is the trivial function.

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  • $\begingroup$ thanks for answering. I believe I explained myself mistakenly so I extend the requirement in the questions. I am thinking in things like $f(x) = e^{1/(x^2-1)},\,|x|\leq 1$ as simple defined ones, but more than one step is none (not considering the obvious step of $0$ value outside the compact-support domain). This precise example is a bump function defined with a simple exponential function, that is smooth in the boundaries, and is not analytic since outside its support is zero for adjacent points, so its Taylor series there is not definable. $\endgroup$
    – Joako
    Oct 19, 2021 at 12:20
  • $\begingroup$ @Joako I do not understand your clarification. If you allow multiplying by indicator functions then a finite number must be allowed, surely? e.g. $f(x) = a, |x|<1, b, |x-1|<1, c, |x-2|<1,\dots$ The Fourier transform is a linear transformation, after all. And it is very easy to encode all sorts of sets into this type of notation, e.g. instead of $|x|\le 1$ you could try $sin(x)+1/2 \le 1$. (Indeed, $|x|\le1$ is a simple way to write two inequalities at once, $x\ge-1$ and $x\le 1$) I suggest you stick to standard words cf en.wikipedia.org/wiki/Piecewise $\endgroup$ Oct 19, 2021 at 14:04
  • $\begingroup$ I am not saying that is not possible to have bumps functions defined picewise, I am just looking for a simple case that could be defined "at most" in two pieces: its compact support on one piece, an $0$ otherwise. As the example I give. $\endgroup$
    – Joako
    Oct 19, 2021 at 14:13
  • $\begingroup$ Regarding the recent downvote- My answer was written for a prior version of the question, which OP has changed; disregarding that Questions usually shouldn't be changed after getting an Answer, I don't believe the current formulation has a reasonable chance of getting a definitive answer, but the moment i'm proven wrong I will remove this answer. $\endgroup$ Dec 15, 2021 at 7:42
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    $\begingroup$ The fact that there is an obvious continuous extension does not mean that you do not need to extend your function for it to be defined at $\pm1$. Regarding the support/finding an ODE: the way you define a function, so long as your definition is equivalent, is irrelevant to its properties. If you have more to discuss please move the discussion to chat or otherwise you can try reaching me in this chat room $\endgroup$ Feb 23 at 2:34
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The triangular function $T(x)=1-|x|$ on the domain $[-1,1]$ has the simple Fourier transform $$ \hat T(\xi) = \frac{\sin^2(\pi\xi)}{\pi^2\xi^2}. $$ Indeed one can even write $$ T(x) = \tfrac{1}{2} \bigl(|x-1|+|x+1|\bigr)-2|x| $$ which explicitly evaluates to $0$ outside $[-1,1]$.

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    $\begingroup$ I don't believe the triangular function is a "Bump function" since its not differentiable on every point, so it can't be a $C^\infty$ function. $\endgroup$
    – Joako
    Oct 18, 2021 at 20:41
  • $\begingroup$ Please note that for avoid the confusion about "bump" being any function like a "bump", I explicitly left in the question that it has to be in $C_c^\infty$ as the bump function definitions showed here. $\endgroup$
    – Joako
    Oct 18, 2021 at 22:29
  • $\begingroup$ Ah ... it would help if you included the $C_c^\infty$ condition in the body of the question (not just the title), and also the link to the definition you're working with. $\endgroup$ Oct 18, 2021 at 23:34
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    $\begingroup$ I added it know, thanks. $\endgroup$
    – Joako
    Oct 18, 2021 at 23:37

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