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Disclaimer: due mi ignorance about the topic I change the question to look for what I was really traying to ask, so there are a few answers that were right as it were described the original thing I write (I assumed that every bump function were smooth, which is false). Also I messed it up with requiring to be piecewise in just one compact-suported interval and $0$ outside, which I tried to fix it later (hope now it is understandable). Please be considered and don't downvote them, since was my fault. And also they were indeed useful since through them I realized I was asking something different to was I were intended to ask.

I am looking for the simplest cases possible of one-variable closed-form smooth 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|< 1 $ is a valid one, since is defined picewise at most in two pieces: its compact support on one piece defined with just one function, and $0$ outside.

If possible, an answer 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", like infinite sums of other simpler functions).

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.

Later I understood thanks to @CalvinKhor that its equivalent to define them piecewise, but since are continuous under the limits-based definition of continuity, I like them most since are simple to manipulate within differential equations.

Joako
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    A “bump functions” tag is probably unnecessary. – littleO Oct 18 '21 at 20:48
  • @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. – Joako Oct 18 '21 at 21:14
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    Perhaps one of the constructions in Are there other kinds of bump functions than $e^\frac1{x^2-1}$? could work. – projectilemotion Oct 18 '21 at 21:15
  • @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). – Joako Oct 18 '21 at 21:19
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    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. – projectilemotion Oct 18 '21 at 21:23
  • @projectilemotion thanks for the info, but actually the problem is that the function $\varphi(t)$ doesn't have a closed form so its the same problem (thinking about $\hat{\varphi}(t)$ as the function which don't have a closed form Fourier transform). – Joako Oct 18 '21 at 21:33
  • Is there a particular reason you want its Fourier transform in closed form? Often with bump functions you don’t need to explicitly calculate things. Instead you usually just estimate – JackT Oct 19 '21 at 00:29
  • @JackT yes it is, I am trying to understand something and it could be a good "example" to review how a Bump functions behaves under the problem I am interested in (is another question I am still working about here). – Joako Oct 19 '21 at 01:23
  • If you relax the assumption from $C^\infty_c$ to $C^1_c$ or even $C^{100000000}_c$ and allow a finite number of piecewise definitions, then there are examples. If you are waiting for a proof that the functions you want do not exist, (well, if they don't exist, but I would bet on it) I guess it would require some fancy abstract algebra not too different from that used to prove the Risch algorithm works, which is IMO far too complicated for an answer on Math.SE – Calvin Khor Dec 15 '21 at 03:26
  • (I forgot to mention that the examples I have in mind are piecewise made up of polynomials, hence easy to plot, and their Fourier transforms are explicit, hence very easy to plot) – Calvin Khor Dec 15 '21 at 03:49
  • @CalvinKhor sorry but I don´t understand your comment because I don´t know the things you have mention... Do you have any reference to share to see what you are talking about? – Joako Dec 15 '21 at 04:58
  • Which things? If you mean risch search for it on Wikipedia or google – Calvin Khor Dec 15 '21 at 05:00

2 Answers2

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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.

Calvin Khor
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  • 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. – Joako Oct 19 '21 at 12:20
  • @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 https://en.wikipedia.org/wiki/Piecewise – Calvin Khor Oct 19 '21 at 14:04
  • 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. – Joako Oct 19 '21 at 14:13
  • 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. – Calvin Khor Dec 15 '21 at 07:42
  • Calvin Khor: I believe I found a function that dismiss your hypothesis here: If I am right, things of the form $f(x) = \frac{(1-x^2+|1-x^2|)}{2}\exp\left(-\frac{x^2}{1-x^2}\right)$ are bump functions $\in C_c^\infty$ that are not defined piecewise. Hope you can check if I am not mistaken about it. – Joako Feb 22 '22 at 19:15
  • No, you just took the normal bump function and tried to hide the piecewise nature with $\max(x,0)=(x+|x|)/2$. This trick is eg used to write “the Batman equation” in one line. Maybe this is what you wanted. But $|x|$ can only be defined piecewise. If you allow this, then you can transform all piecewise functions into ‘non-piecewise’ functions. And I don’t see anything about the Fourier transform (sorry, on mobile) What hypothesis of mine are you talking about? @Joako – Calvin Khor Feb 23 '22 at 00:28
  • @Joako also as explained by blamethelag your function is strictly speaking not defined when $x=\pm1$, even if I can guess what you want to piecewise define it at those points to be. – Calvin Khor Feb 23 '22 at 00:33
  • Calvin: Maybe I have misunderstood your initial answer, but what I understand from it is that in the whole real line is impossible to present a smooth function with compact support only with wide-known-things -and-operations since it will become analytical, and I believe with that tricks it shows it is actually false, maybe the trick is wrong but I believe in the other question is prove to work. In the question I understand you are talking for the functions in the domain of time/space and not in the Fourier domain (where the transform is going to be analytical for sure). – Joako Feb 23 '22 at 00:35
  • If you are talking about my answer here, i was precise about the collection of functions allowed and I specifically did not include $|x|$. The trick shows that the conclusion is false if you include this function. But also, as I said, including |x| means “piecewise” loses all meaning! – Calvin Khor Feb 23 '22 at 00:39
  • I am not saying it is true, is what I am asking on the other question, but since $e^{\frac{-x^2}{1-x^2}},,|x|\leq 1$ is already a bump-function $\in C_c^\infty$ (from what I have found here is widely used), having the same issue mention by @blamethelag, I think that somehow its inherited to the example I give there.... when using the abs() function as a non-piecewise I believe it is not a violation to that since it stills keeping smoothness - but is what I believe, closed forms includes the nth roots says Wiki, but it don´t mention abs() fn... – Joako Feb 23 '22 at 00:51
  • ...and defining the bump-function instead by $|x| <1$ left the domain as an open interval, and I don´t know if that works against the compactness of the support... I have been learning through Wikipedia and questions so maybe I have understand things mistakenly. – Joako Feb 23 '22 at 00:54
  • I need to retract my comment above- I would say the trick still doesn’t work because you still require it to be piecewise to define it at those two points. In fact I already allowed $|x|=\sqrt{x^2}$ but if your resulting function is smooth you either never passed through $0$ of |x| or you need to extend via limits or whatever ie violate being ‘non-piecewise’. – Calvin Khor Feb 23 '22 at 01:00
  • These discussion is for the other question, but the function is proven to have limits at $\pm 1$ identical to its right and left limits on $\pm 1^{\pm}$ equal zero, fulfilling the continuity definition $\lim_{x \to c} f(x) = f(c)$ (which is inherited from $\exp(-x^2/(1-x^2))$... also a thing I don´t understand is defining bump functions as $|x|<1$ since its support it still being $[-1,,1]$ so it is actually defining a discontinuous function at the points $x={-1;,1}$.... About the trick, I like it since it allow me to find differential equations, thing I can´t do with the piecewise versions – Joako Feb 23 '22 at 01:20
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    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 – Calvin Khor Feb 23 '22 at 02:34
  • Thanks for the invitation. I moved to the chat room, but I am not really sure how they work so I left you a few messages. – Joako Feb 23 '22 at 02:57
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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 $[-1,1]$.

Greg Martin
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    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. – Joako Oct 18 '21 at 20:41
  • 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. – Joako Oct 18 '21 at 22:29
  • 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. – Greg Martin Oct 18 '21 at 23:34
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    I added it know, thanks. – Joako Oct 18 '21 at 23:37