Is the function $f(x)=\frac{\left(1-x^2+\sqrt{\left(1-x^2\right)^2}\right)}{2}e^{-\frac{x^2}{1-x^2}}$ a bump-function $\in C_c^\infty$? Diff. Eq.?

Question

Is the function $f(x)=\frac{\left(1-x^2+\sqrt{\left(1-x^2\right)^2}\right)}{2}e^{-\frac{x^2}{1-x^2}}$ a bump-function $\in C_c^\infty$? Which autonomous differential equation it fulfill? (note it is not defined piece-wise)

I was viewing this question on MSE and there was given here a pretty clever answer $q(x) = 1-\sqrt{x^2}+\sqrt{\left(1-\sqrt{x^2}\right)^2}$ for a compact-supported function that is not defined piece-wise, and I was trying to us it method to built a proper bump function in $C_c^\infty$ following the definition of Wikipedia (a compact-supported smooth function) that is also not defined piece-wise.

After playing a while with it on Wolfram-Alpha and Desmos, I believe that:

$$f(x)=\textstyle{\frac{\left(1-x^2+\sqrt{\left(1-x^2\right)^2}\right)}{2}}{\Large e}{\begin{array}\displaystyle{ -\frac{x^2}{1-x^2}} \\ \\ \\ \end{array}}$$

it is actually a bump function $\in C_c^\infty$ since their derivatives will be polynomials multiplied again by $f(x)$, which will keep their functions smooth and keeping smoothness at the boundaries of their non-zero values $\partial x = [-1,\,1]$, but my math skills aren´t enough to prove it, so far, I was able only to check the first 6th derivatives and it looks fine when plotted, without having discontinuities (I am worried of the hidden $\text{abs}(1-x^2)$ function which derivatives becomes Dirac's Delta functions $\delta(x \pm 1)$), neither changing their starting and ending values from zero (example).

Since the function $f(x)=0,\,|x|>1$ as intended, I think it is already a smooth function on these points so it stills being a bump-function $\in C_c^\infty$ (if I am right, the only possible analytic function of compact-support is the zero function, so I believe that piecewise zero sections are smooth given they are analytic), and within $|x|<1$ it still being smooth since the term $e^{-\frac{x^2}{1-x^2}}$ will dominate any polynomial, but in the points $x=\{-1,\,1\}$ I don't know if will dominate any derivative of the Dirac's delta function.

Also I would like to know which autonomous differential equation $f'(x) = G(f(x))$ with $G(x)$ at least $C^1$ almost-everywhere will have $f(x)$ as solution, maybe a non linear ODE, or a Delay Differential Equation (DDE) as it was shown here for another bump-function (so far is the only example I know): there is found that a bump-function which fulfill $\varphi'(t)=2\left(\varphi(2t+1)-\varphi(2t-1)\right)$ when the solution is only non-zero inside $[-1;\,1]$, but unfortunately there is no closed-form for this function, but its Fourier Transform is show to be $\hat{\varphi}(\omega) = \prod\limits_{k=0}^{\infty}\frac{\sin\left(\frac{\pi\omega}{2^k}\right)}{\frac{\pi\omega}{2^k}}$.... actually my main objective is to find a finite-duration solution to a differential equation that fulfill what is said in this paper Finite Time Differential Equations... I have the Motivation for this questions explained in this another question.

So summarizing:

• Is $f(x) \in C_c^\infty$?
• Which autonomous differential equation fulfill $f(x)$ as solution? Otherwise, proving it cannot be described through an autonomous diff. eq. is also welcome.

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My attempts so far...

I believe that the diff. eq. found by Wolfram-Alpha is wrong: $$f'(x)\left(1-x^2\right)^2+f(x)(4x-2x^3)=0,\qquad f(0)=1$$

But I have had a lot of struggle dealing with the signs, so first, to see if I am not making mistakes, I will list a few things I am using for which I am not sure if they are formally right.

1. First, to show explicitly where I have risks for discontinuities, I have made transparent that $\sqrt{(1-x^2)^2} \equiv |1-x^2|$ an absolute value function, so $f(x)$ acquire the form $f(x) = e^{-\frac{-x^2}{1-x^2}}\frac{(1-x^2+|1-x^2|)}{2}$
2. Second, for keeping tractable the sign issues and maintain unchanged the solutions of the diff. equation, I have found I have to keep $\frac{1}{\text{sgn}(a)} \neq \text{sgn}(a)$, even when both plotted look the same (honestly for easy-hand-calculation I was using that a lot before).
3. When working with the function $s(x) = \log\left(1-x^2+|1-x^2|\right)$, I will have that $s'(x) = -\frac{2x}{|1-x^2|}$, but when integrating $-\frac{2x}{|1-x^2|}$ on Wolfram-Alpha it shows a completely different thing, and even so, when derivating the W-A result it didn´t shows to be $s'(x)$ (even the results has Dirac's delta functions involved). So hereinafter, I will be using $ \int -\frac{2x}{|1-x^2|} dx = \log\left(1-x^2+|1-x^2|\right) + \mathcal{c}$
4. Also, noting that $1-x^2+|1-x^2| \equiv |1-x^2|\left(1+\text{sgn}(1-x^2)\right)$
5. In all the manipulations I am ignoring possible problems because of dividing by $0$ at the points $x=\{-1,\,1\}$ (they will arise on the topic later).

Using these things, I believe that "a true" diff. equation for $f(x)$ (but not-autonomous) is: $$f'(x)\left(1-x^2\right)^2+2x(1+|1-x^2|)f(x) = 0\qquad \text{Eq. 1}$$

From here, this is equivalent to: $$\frac{f'(x)}{f(x)}=\frac{-2x(1+|1-x^2|)}{\left(1-x^2\right)^2}\qquad \text{Eq. 2}$$

Since $f'(x)/f(x) = \frac{d}{dx}\left(\log(f(x))\right)$, I tried to integrate the left-hand-side of Eq. 2 on Wolfram-Alpha but it was unable to find and antiderivative, but noting that:

$$\frac{-2x(1+|1-x^2|)}{\left(1-x^2\right)^2} = \frac{-2x}{\left(1-x^2\right)^2}+\frac{-2x|1-x^2|}{\left(1-x^2\right)^2} =\frac{-2x}{\left(1-x^2\right)^2}+\frac{-2x}{|1-x^2|}$$ integrating both fractions and using point (3) I will have that: $$\Rightarrow \log(f(x)) = \frac{-x^2}{(1-x^2)}+\log(1-x^2+|1-x^2|)+\mathcal{c} $$ From where it can be seen that by applying both side the $\exp()$ function I will recover $f(x)$ for $c=-\log(2)$, so I believe Eq. 1 is right.

As @blamethelag reccomend on the answer I tried to find a recurrence relation for Eq. 2 using the General Leibniz formula: $$\left(fg\right)^{(n)} = \sum\limits_{k=0}^{n} {n \choose k}f^{(k)}g^{(n-k)} $$ But I wasn´t able to find and useful formula (I get stuck).

But using this formula jointly with the Faà di Bruno's formula for a function composed with an exponential: $$\left(e^{g}\right)^{(n)} = e^{g}B_n(g',g'',\cdots, g^{(n)})$$ where $B_n()$ is the nth complete exponential Bell polynomial.

With these, I tried to expand the nth derivative of f(x): $$\begin{array}{r c l} \frac{d^n}{dx^n} f(x) & = & \sum\limits_{\begin{smallmatrix}k=0 \\m=n-k\end{smallmatrix}}^n {n \choose k} \frac{d^k}{dx^k}\left(e^{-\frac{x^2}{1-x^2}} \right)\frac{d^m}{dx^m}\left(1-x^2+|1-x^2| \right)\\ & = & \sum\limits_{\begin{smallmatrix}k=0 \\m=n-k\end{smallmatrix}}^n {n \choose k} e^{-\frac{x^2}{1-x^2}}B_k\left(\frac{d}{dx}\left(\frac{-x^2}{1-x^2}\right),\frac{d^2}{dx^2}\left(\frac{-x^2}{1-x^2}\right),\cdots,\frac{d^k}{dx^k}\left(\frac{-x^2}{1-x^2}\right) \right) \frac{d^m}{dx^m}\left(1-x^2+|1-x^2| \right)\\ & = & e^{-\frac{x^2}{1-x^2}} \sum\limits_{\begin{smallmatrix}k=0 \\m=n-k\end{smallmatrix}}^n \underbrace{{n \choose k} B_k\left(\frac{d}{dx}\left(\frac{-x^2}{1-x^2}\right),\frac{d^2}{dx^2}\left(\frac{-x^2}{1-x^2}\right),\cdots,\frac{d^k}{dx^k}\left(\frac{-x^2}{1-x^2}\right) \right)}_{\mathbb{P}_{n,k}(x)} \frac{d^m}{dx^m}\left(1-x^2+|1-x^2| \right)\\ & = & e^{-\frac{x^2}{1-x^2}} \sum\limits_{\begin{smallmatrix}k=0 \\m=n-k\end{smallmatrix}}^n \mathbb{P}_{n,k}(x) \left( \frac{d^m}{dx^m}\left(1\right)+\frac{d^m}{dx^m}\left(-x^2\right)+\frac{d^m}{dx^m}\left(|1-x^2|\right) \right) \qquad \text{Eq.3} \end{array}$$

Here, I believe that the term $e^{\frac{-x^2}{1-x^2}}$ is going to "dominate" every possible polynomial $\mathbb{P}_{n,k}(x)$ since it already done it for the classical example of a bump function $\in C_c^\infty$ given by $r(x) = \begin{cases} e^{\frac{-x^2}{1-x^2}},\,|x|\leq 1\\ 0,\,\text{otherwise}\end{cases}$

But I am worried about the other part of the equation, since $\frac{d^m}{dx^m}\left(|1-x^2|\right)$ will make appear Dirac's Delta functions and derivatives of it (the following derivatives were solved by wolfram-alpha, by hand I have differences so maybe I am doing something wrong):

$$\begin{array}{l c l} m = 0 & \rightarrow & |1-x^2| \\ m = 1 & \rightarrow & -2x\,\text{sgn}(1-x^2)\\ m = 2 & \rightarrow & -2\,\text{sgn}(1-x^2)+4\delta(x+1)+4\delta(x-1) \\ m = 3 & \rightarrow & 4\delta(x+1)+4\delta(x-1)+4\delta'(x+1)+4\delta'(x-1)\\ m = p\geq 4 & \rightarrow & 4 \left(\frac{d^{p-3}}{dx^{p-3}}+\frac{d^{p-2}}{dx^{p-2}}\right)\Big(\delta(x+1)+\delta(x-1)\Big)\\ \end{array}$$

And I don´t know if these Dirac's delta functions and its derivatives are actually helping by "killing" things outside $x=\{-1,\,1\}$, or instead are ruining the smoothness of $r(x)$ on the points $x=\{-1,\,1\}$, discarding the hypothesis of the smoothness of $f(x)$ (and it is not seen on the plots because happens only in two zero-measure points $x=\{-1,\,1\}$).

This is why I worried about the smoothness, since for every other points I think is granted by $r(x)$.

At least for the case $m=3$, where the terms $\delta()$ and $\delta'()$, Wolfram-Alpha is able to take the limits $x \to 1$ giving zero value, and it match with their right and left side limits $x \to 1^{\pm}$ in Wolfram-Alpha here and here.

By expanding the terms of Eq. 3 you will have things of the form: $$e^{-x^2/(1-x^2)}\mathbb{P}(x)\delta^{(m)}(x\pm 1)$$ where since $w(x)=e^{-x^2/(1-x^2)}$ will "win" any polynomial on $x \to \pm 1$, I will have that $w(x)\mathbb{P}(x)=0,\,x\to\pm 1$, so by calling $u(x \pm 1) = w(x \pm 1)\mathbb{P}(x\pm 1)$ and $z = x \pm 1$, the "problematic" terms will look like $u(z)\delta^{(m)}(z)$ with $u(0) = 0$.

Now, for $m=3$, the terms with issues are of the form $u(z)\delta(z)$ and $u(z)\delta'(z)$ using the properties of the Dirac's delta function shown on the Spanish version page of Wikipedia:

• $f(x)\delta'(x)=-f'(x)\delta(x)$
• $x^n\delta(x) = 0,\,\forall n>0,\,x\in\mathbb{R}$
• $h(x)\delta(x-a)=h(a)\delta(x-a)$
• $h(x)\delta'(x-a)=h(a)\delta'(x-a)-h'(a)\delta(x-a)$

I believe it could be seen that the terms will vanish since it will behave as terms of the form $\{g(z)\to 0\}\cdot \delta(z)$: this because of the property $z^n\delta(z)=0$, and in this case $g(z)\to 0$ even faster than every possible $z^n$, and if some terms $g(z)\delta(z)\delta(z)\cdots\delta(z)$ arises, they will also become zero since every left-side multiplication will become the term zero for the remaining terms.

Unfortunately, I don´t know how to extend this for $m \geq 4$ since I don't know how to work with higher derivatives of the Dirac's delta function, which at least on Wikipedia are reviewed through Distribution Theory scope, for which I am completely ignorant.

But, if the first property of Wikipedia is right, I believe that every product by the derivatives of the Dirac's delta function could be manipulated into standard Dirac's delta functions, as example:

$$\begin{array}{r c l} u(x)\delta'(x) & = & -u'(x)\delta(x) \iff \delta'(x) = -\frac{u'(x)}{u(x)}\delta(x) \\ \Rightarrow u(x)\delta''(x) & = & u(x)\frac{d}{dx}\left(-\frac{u'(x)}{u(x)}\delta(x)\right) \\ & = & u(x)\frac{d}{dx}\left(-\frac{u'(x)}{u(x)}\right)\delta(x)-u'(x)\delta'(x) \\ & = & u(x)\frac{d}{dx}\left(-\frac{u'(x)}{u(x)}\right)\delta(x)+u'(x)\frac{u'(x)}{u(x)}\delta(x) \end{array}$$

So noting that $\frac{u^{(m)}(x)}{u(x)}\equiv \mathbb{P}(x)$ some polynomial on the variable $x$, all the terms of these derivatives will be of the form $w(x)\mathbb{P(x)}\delta(x)$ for other polynomials $\mathbb{P}(x)$ (I am abusing of the notation), and the same procedure could be extended by construction to higher derivatives of $u(x)\delta^{(m)}(x)$, so If this is right, the function $f(x)$ is keeping is smoothness on the points $x =\{-1,\,1\}$, meaning $f(x)\in C_c^\infty$ inherited by the function $\exp\left(\frac{-x^2}{1-x^2}\right)$.

But since I cannot formally proving it better as I already explained (which is more an intuitive prove than a proper one - there are too many "I believe"), I hope someone could confirm this through some theorem or valid method.

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2nd thing added

If the triangular function is defined as: $$\Lambda(x) = \frac{1}{2}\left(|1+x|+|1-x|-2|x| \right)$$

It looks like for bump functions $b(x)$ defined piecewise in $x \in [-1,\,1]$, their domain could be extended by using $g(x) = b(x)\Lambda(x^n)$ with integer $n \geq 2$. Above I have used $\Lambda(x^2) = 1-x^2+|1-x^2|$. Also positive powers of $\Lambda^m(x^n), m\in\mathbb{Z}^+$ will work.

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3rd added later - discussion about the definition of f(t)

Due to enriching comments,answers, and explanations by chat, I have a better idea of what the issues are with the proposed function. Since I would like to focus the answers with the problem with the derivatives of the Dirac's Delta function $\delta^{(m)}()$, in this section I do a brief review of the problem with the definition of the function on the points $x=\{-1,\,1\}$.

From what I have seen here, commonly bump functions are defines in open intervals as the example: $$q(x) = \begin{cases} e^{-\frac{x^2}{1-x^2}}, & |x|<1 \\ 0, & |x| \geq 1 \end{cases}$$ where in the points of the "edges" $\partial x = \{-1,\,1\}$, since the function is matching the zero constant, to keep smoothness it also has to happen that $\lim\limits_{x \to \partial x^{\pm}} \frac{d^n}{dx^n}q(x) = 0,\,\forall n \leq 0,\,n \in \mathbb{z}$, in other words, all its right and left side derivatives at the edges must match and been equal to $0$.

So far so good, but, seeing the Wikipedia page for Compact support it is said that for a function defined in an open interval $(-1,\,1)$, its support it still being $[-1,\,1]$, so I think is like being "cheating": the function domain have two points where is undefined by definition, so is discontinuous there (I think), instead of being defined on $[-1,\,1]$ and having the issue of being undefined because of dividing by zero on the exponent.

However, about this forbidden division (which is mentioned by @blamethelag), there is an issue: following Wolfram-Alpha, the limit of $q(x)$ at the edges does not exist, and neither are equal their left and right side limits (solved because of the open interval definition I believe), but the same analysis for the function $f(x)$ shows that actually it is not only having identical right and left hand sides limits (which is the standard way of extending a function), following Wolfram-Alpha the limits at exactly the edges exists and are also zero, so If W-A is right, actually the function $f(x)$ is fulfilling the continuity definition $\lim\limits_{x \to c} f(x) = f(c)$, so if this is right, the function $f(t)$ should be properly defined as a function.

But since it must also fulfill the existence of all its derivatives to be a smooth function, here is where the derivatives of the Dirac's Delta function $\delta^{(m)}()$ could be doing a mess, and its where I am worried about.

Also another discussion was an opinion received in other question here: @CalvinKhor has correctly noted that the function $f(x)$ is still defined piecewise since the absolute value function by definition is defined piecewise. There is no way to refute this, but I believe that in the spirit of the question it is still a function not defined piecewise because of the following: as @blamethelag noted, if I work with the piecewise section within $[-1,\,1]$, the differential equation for the function $f(x)$ will be defined by: $$\frac{f'(x)}{f(x)}=\frac{-2x}{(1-x^2)^2}$$, which solution will be behaving very different from being zero outside the interval $[-1,\,1]$. But differently from it, using the definition of $f(x)$ presented here I was able to found the differential equation: $$\frac{f'(x)}{f(x)}=\frac{-2x}{(1-x^2)^2}+\frac{-2x}{|1-x^2|}$$ which, If I am right, it will be describing a function that actually behaves as the zero function outside the interval $[-1,\,1]$.

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Note aside

If I have a function $y(x) = e^{p(x)}(1-x^2+|1-x^2|)$, using point (3) its trivial differential equation will be given by: $$\frac{y'(x)}{y(x)} = \frac{d}{dx}p(x)-\frac{2x}{|1-x^2|}$$ So I was trying to found some $p(x)$ that fulfills: $$y(x)\left(\frac{d}{dx}p(x)-\frac{2x}{|1-x^2|}\right) = 2y(2x+1)-2y(2x-1)$$ for matching the solution with an already known result, this unsuccessfully, but maybe someone else have an idea of how to made the matching...

Or maybe found some $p(x)$ so the differential equation take an autonomous form $f'(x) = G(f(x))$ with $G(x)$ at least $C^1$ almost-everywhere.

English language is not native for me, so probably this have a lot of mistakes: my apologies in advance

Answer 1

Your question is too vague, unclear and does not respect this site standards. However my flag did not do anything, as you made some efforts in the direction I suggested in comment and explained that you are not acquaintanced with analysis I will try to give an answer.

• Is $f$ a bump function? As you changed your definition of $f$ I will give it here for the sack of clarity and in the prevision of further changes $$ f(x) = \frac{1-x^2+ |1-x^2|}{2}e^{- \frac{x^2}{1-x^2}}. $$ First thing to do is to check $f$ is well defined as a function. The only illegal operation you are doing is to divide something by $0$ at $x = \pm 1$ so $f$ is not defined at $\pm 1$. A common way to overcome this difficulty is to extend the function at those points by its limit, provided it exist, so that the extension is continuous. So you have to check the limit of $f$ at $x = \pm 1$, $f$ is null outside $[-1,1]$ and $$ f(x) \underset{1^-}\sim (1-x^2)e^{-\frac{1}{1-x^2}} \longrightarrow 0 $$ and by symmetry of $f$ you get the same limit in $-1^+$ and the right to extend $f$ as a continuous function. We know that the support of $f$ is compact so we just have to check $f$ is $C^\infty$, this is a long but easy work so I will just tell how to do it. Notice $f$ is null outside $[-1,1]$ whence $C^\infty$ with null derivatives. Inside $(-1,1)$ you have $f$ infinitely differentiable because the usual operations preserves differentiability. Moreover, in $(-1,1)$ $$ f(x) = (1-x^2)e^{- \frac{x^2}{1-x^2}} $$ and you show by induction that forall $n$, $f^{(n)}(x) = \frac{P_n(x)}{Q_n(x)}e^{- \frac{x^2}{1-x^2}}$ with $P_n, Q_n$ polynomials. Then use comparison theorem to discover $$ f^{(n)}(x) \underset{x \rightarrow \pm 1,~~|x| < 1}{\longrightarrow} 0 $$ and use a well know analysis theorem telling you that $f^{(n)}$ is differentiable at $\pm 1$ with null derivative. All this work shows that $f$ is indeed a bump function, I encourage you to check the details. Some comments on this: I don't understand why you want a bump function that has an analytical expression, those objects are usually used in the theory to prove the existence of something or to pass a property from local to global. The question you refer to talks about computational problems with transcendental functions but you can already do a lot with classical bump functions and in general there is no closed form for the solution of a differential equation.

• Which properties does it fulfill? This question is to broad to be answered, you should either work yourself on the function or specify your question.

• "the only possible analytic function of compact-support is the zero function, so I believe that piecewise zero sections are smooth given they are analytic". I think you are confusing the notion of an analytical form and the notion of an analytical function. The first thing is to have a closed formula, i.e. your object is directly built via a finite number of operations from commonly known objects. For instance the tangent is obtained from the sine and the cosine but the roots of a $5$th order polynomial does not have a closed formula from the coefficients of the polynomial. The second notion is to locally being equal to its Taylor series, which prohibits to be of compact support and not null. It is written on the Wikipedia page you gave in the question: there is no bump function that is analytic.

• Which non trivial differential equation does $f$ is solution to? In my opinion this is not a good question because I don't know what is a non trivial differential equation. A better question would be: Is there an interesting differential equation which has $f$ as a solution? For this question the only thing you can do is to differentiate $f$ and hope to get $f$ again: $$ f'(x) = -2 \frac{e^{-\frac{x^2}{1-x^2}}x(x^2-2)}{x^2-1} = 2\frac{x(x^2-2)}{(1-x^2)^2}f(x) $$ Notice that this ODE has solutions that are not null outside $[-1,1]$ so I don't know if it is really characterizing $f$ (I suppose this is your interest in knowing an ODE solved by $f$).

• Can we compute $\hat f$? I don't see any smart way to compute it by hand but considering the Fast Fourier Transform, you can actually compute it with a computer and plot it.

• Could that equation be used to figure out the Fourier Transform? In what we have found , $f' = g f$ so taking the Fourier yields $i\xi \hat f (\xi) = \widehat{fg}(\xi)$. The usual trick to get rid of the product with $g$ in the Fourier is to use the formula $\widehat{f * g} = \hat{f} \hat{g}$ where "$*$" stands for the convolution product. But to benefit from it you write $$ \widehat{fg} = \hat f * \hat g $$ which involves computing $\hat g$, maybe feasible with some contour integral, and to solve the equation with the convolution, that is impossible in my knowledge. The convolution is not even helping for the numerical computations so this way is to avoid, to me.

Answer 2

The only points where smoothness is not clear are $\{-1,1\}$. However, the singularities as a result of your compactly supported function are of the form $|1-x|)^{-R}$ for some $R>0$. However, around $x=1$, the exponential decay completely dominates any polynomial blow up since $\lim_{x\to\infty} x^R e^{-x}=0$ for all $R>0$. Thus, your function is smooth and those points and a member of the set $C_c^\infty(\mathbb{R})$.