# Are there any non-constant "smooth bump functions" in "closed form" whose Fourier Transforms are also in closed form?

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.

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.

• A “bump functions” tag is probably unnecessary. Commented Oct 18, 2021 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. Commented Oct 18, 2021 at 21:14
• Perhaps one of the constructions in Are there other kinds of bump functions than $e^\frac1{x^2-1}$? could work. Commented Oct 18, 2021 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). Commented Oct 18, 2021 at 21:19
• 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. Commented Oct 18, 2021 at 21:23

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.

• 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. Commented Oct 19, 2021 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 en.wikipedia.org/wiki/Piecewise Commented Oct 19, 2021 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. Commented Oct 19, 2021 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. Commented Dec 15, 2021 at 7:42
• 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 Commented Feb 23, 2022 at 2:34

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 $$[-1,1]$$.

• 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. Commented Oct 18, 2021 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. Commented Oct 18, 2021 at 22:29
• I added it know, thanks. Commented Oct 18, 2021 at 23:37
• @Joako smooth is a very vague term unfortunately though I don’t think I have seen a book point it out. many papers or books say ‘sufficiently smooth’ which means ‘assume as many derivatives exist as needed to make what I’m saying true :)’. I believe Do Carmo’s geometry book says differentiable to actually mean infinitely differentiable? So unfortunately the ship has long since sailed :) Commented Jul 22 at 14:14
• @CalvinKhor You are right. But I think it is worthwile to rise awareness... in my case I was not even aware the difference existed neither how important was until I found on MSE about non-analytc smooth functions and smooth bump functions, and it matter of fact it is so unknow that I only find advanced abstract math without simple explicit examples... I did find one using was where on Wikipedia and add it, and result it has been useful for many others as you can see here. Unfortunately my attempt was erased due downvotes, but at least I tried. Commented Jul 22 at 15:08