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I was recently experimenting in wolfram alpha to "make" smooth Bump functions, and I found some interesting attempts which have all the real line as domain, but I believe are compact supported since its values are non-zero only for a tight interval:

  1. $f(x) = e^{-2 \cdot x^{2n} \cdot e^{x^2}}$ for integer $n \geq 1$ are non-zero between (-1; 1), have max value 1, and increasing n gives them a flat-top making them, I think, non-analytical. I believe they could be nice transitions/window functions.

As example: https://www.wolframalpha.com/input/?i=e%5E%28-2*x%5E8*e%5E%28x%5E2%29%29+for+x%3D-1.2+to+1.2

  1. $f(x) = e^{-(2x)^{2n} \cdot e^{(2x)^2}}$ for integer $n \geq 1$ are non-zero between (-1/2; 1/2), have max value 1, and increasing n gives them a flat-top making them, I think, non-analytical. Also increasing n make them really squared so I believe they could be in the limit a representation of the standard rectangular function (wolfram-alpha calculate the area only up n=200 and it was almost 1).

As example: https://www.wolframalpha.com/input/?i=e%5E%28-%282*x%29%5E100*e%5E%28%282*x%29%5E100%29%29+for+x%3D-1.2+to+1.2

  1. $f(x) = e^{-(n+1)! \cdot x^{2n} \cdot e^{x^2}}$ for integer $n$ between [1; 4] are non-zero between (-1; 1), have max value 1, increasing n gives them a flat-top making them, I think, non-analytical, and their area under the curve integrates approximately 1, so I believe they could be interesting mollifiers for numerical calculations.

I don't have enough mathematical background to probe if they are Bump functions, so I will be happy to receive your opinions about them, hoping they will be useful for anybody else.

Xander Henderson
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Joako
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    Bump functions have compact support. None of these do. – Rushabh Mehta Sep 27 '21 at 01:17
  • so, for becomes a bump functions they have to be defined as $g(x) = f(x)$ if $|x| \leq 1$ and $g(x) = 0$ if $|x| > 1$ ??? its a necessary condition? – Joako Sep 27 '21 at 01:27
  • @Joako If you define it that way, the function is not continuous. – Rushy Sep 27 '21 at 01:29
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    For a function $f$ to be a bump function you need it to be smooth and compactly supported. – Rushy Sep 27 '21 at 01:31
  • I am following the example shown on wikipedia https://en.wikipedia.org/wiki/Bump_function, which is also defined similarly (by parts), so I hope you can explain why mine is wrong. – Joako Sep 27 '21 at 01:35
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    If you define it by parts as you did, your function is compactly supported, however it is not smooth. The reason for this is that it has a jump discontinuity at $x = -1$ and $x = 1$. For example taking your first function $f$, the limit $\lim_{x \to 1^{-}} g(x) = e^{-2e}$ while $\lim_{x \to 1^{+}} g(x) = 0$. – Rushy Sep 27 '21 at 01:41
  • Thanks, I better understand it now, I was thinking that because the $\lim f(x) = 0$ and $\lim df(x)/dx =0$ when $x \rightarrow \pm \infty$ it was comppact suported in the real line. – Joako Sep 27 '21 at 01:52
  • If for integer $n>1$ I defined the function: $$f(x) = e^{1-1/(1-(2x)^{2n})} \cdot (\theta(x+1/2) - \theta(x-1/2))$$, with $\theta(x)$ the unitary step function, I will have a family with $\max_x{f(x)} = 1$, with $f(x) \geq 0$ $\forall x$, $f(x) = 0 \text{ if } |x|>1/2$, flat-top for $n>1$, and also $\lim_{x \rightarrow \pm 1/2^{\pm}} f(x)=0$ ¿Is $f(x)$ a Bump function?.

    Also $f(x) = 1/2 \rightarrow x^=\pm 1/2 \cdot (\frac{\log(2)}{1+\log(2)})^{\frac{1}{2n}}$ so $\lim_{n \rightarrow \infty} x^ = \pm 1/2$, so, Could be $\lim_{n \rightarrow \infty} f(x) = $ rectangular function $\Pi(x)$??

     – Joako Oct 01 '21 at 19:11

2 Answers2

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Any function $f \in C_C^{\infty}$ can be expressed as $$ f(x) = \begin{cases} \phi(x), & x \in (a,b) \\ 0, & x\in \mathbb{R}\setminus (a,b)\end{cases} $$

where, $\phi \in C^{\infty}(a,b)$ and the one-sided derivatives of all orders exist are are null at $x = a,b$. For example,

$$ f(x)=\begin{cases} e^{-\frac{1}{(x+1)^2}} \cdot e^{-\frac{1}{(x-1)^2}}, & |x| < 1\\ 0, & |x|\ge 1\end{cases} $$

fullfills the requirements.

PierreCarre
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Actually I was wrong and neither one is a "Bump function", or equivalently, kind $C_c^\infty$.

Since they value is zero outside its domain, at the domain boundaries $\partial x$ the value of the function must be $0$, for example, if the function $f(x)$ is defined on $x_0 < x < x_F$ (or equivalently, $\partial x = [x_0;\,x_F]$), then must be true that $f(x_0)=f(x_F)=0$ to preserve continuity.

Also, its derivative must be zero at the boundaries $\partial x$, since the function within the domain have to become a constant (zero) outside the domain, so $\lim_{x \to \partial x^\pm}\frac{d}{dx}f(x) = 0$ to preserve continuity of the derivative.

Actually fulfilling these both requirements will make any function to "looks like" it rises and finished "softly" from the $0$ value, and this is why I make a mistake, since for a function to be "smooth" (or class $C^\infty$), it also has to have every superior derivative continuous in the whole domain of the function.

To see what this "means", lets see the definition of the second derivative: $$f''(x) = \lim_{h \to 0}\frac{f'(x+h)-f'(x)}{h} = \lim_{h \to 0}\frac{f(x+h)-2\,f(x)+f(x-h)}{h^2}$$ Since the first derivative is already $f'(\partial x)=f(\partial x)= 0$ in a neighborhood of the boundaries, its easy to see that also the second derivative will be zero on the boundaries $f''(\partial x)=0$. This is not a rigorous demonstration, but a more intuitive explanation.

Given that the second derivative $f''(\partial x)=0$ on the boundaries, by induction, the same argument can be applied for the third derivative, and so on, so for having smooth infinite order derivatives it must to fulfill at the boundaries of its domain that: $$\lim_{x \to \partial x^\pm} \frac{d^n}{dx^n}f(x) = 0, \,\forall n \geq 0,\, n \in \mathbb{Z}$$

Note that is not saying that every function that fulfill $f(x)=f'(x)=0$ will have then $\frac{d^n}{dx^n} f(x) = 0$, which is false. What I trying to explain is that for $f(x)$ to be in $C_c^\infty$, it have to satisfy $\lim_{x \to \partial x^\pm} \frac{d^n}{dx^n}f(x) = 0, \,\forall n \geq 0,\, n \in \mathbb{Z}$ (this is why I stated before is not a rigurous explanation, but a intuitive one).

It can be seen in this way: the function $f(x)$ have to tie with the $0$ constant outside its domain, and as a constant function all this derivatives must be $0$.

This is actually a huge requirement, since the boundaries acts like an "absorbing" point for the derivatives. I have found a few bump functions in here, but unfortunately, I didn't found any with known Fourier transform as is asked on here and here.

Joako
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    It is generally not true that zero one side derivative at the "boundary" implies a null second derivative. if this was true, you would just have to come up a a function whose derivative vanishes at the "boundary" and this does not work, as you noted. – PierreCarre Oct 19 '21 at 14:23
  • You are right, maybe I explain myself wrong. What I want to say is that you don´t just have $f(\partial x)=f'(\partial x)=0$ is not enough, you also need that the function $d^nf(\partial x)/dx^n = 0, \forall n \in \mathbb{Z}, n \geq 0$. – Joako Oct 19 '21 at 16:59