Are these functions well defined Mollifiers kernels? Why? Why not?
I have found that these functions are also called test functions and smooth bump functions.
Update: Since I am particularly interested on the $f_1(x)$ case, in the answers to this other question people have proved in very different and interesting ways that $f_1(x)$ is smooth when defined piecewise in order to avoid the undefined points $x = \{-1,\ 0,\ 1\}$.
So for the current question what is left to prove (if I am not mistaken), is it if is possible to calculate the convolution integral of the Mollifier application without solving the issues at these undefined points with zero-measure (so, using $f_1(x)$ just like is defined here), or if conversely, it is required the piecewise definition used in the mentioned question.
For the other examples the same situation will follow, since $f_1(x)$ the worst or equal case scenario (again, If I am not mistaken).
Consider the following functions as $$F_i(x)=\begin{cases} \frac{f_i(x)}{I_i},\quad x<|1| \\ 0,\quad |x|\geq 1\end{cases}$$ where $I_i$ is a normalizing constant such as $\int\limits_{-\infty}^{\infty} F_i(x)\ dx=1$ and $f_i(x)$ are the following functions:
- $$f_1(x)=\frac{1}{1+\exp\left(\frac{1-2|x|}{x^2-|x|}\right)}$$ which resembles the lobes of the Fabius function and Rvachëv function (which fulfills $R'(x)=2R(2x+1)-2R(2x-1)$), as could be seen in Desmos through the approximation shown in this answer.
- $$f_2(x)=\frac{1}{1+\exp\left(\frac{\frac{\sqrt{3}}{2}\left(1-2|x|\right)}{x^2-|x|}\right)}$$ which have "almost" straight-line "sides".
- $$f_3(x)=\frac{1}{1+\exp\left(\frac{2\left(1-2|x|\right)}{x^2-|x|}\right)}$$ which look it have an "almost flat function top": I wasn't able to prove it, but by inspection they look indeed flat, and as also $f_1(x)$ both looks are not analytic at $x=0$ which is a good syntom, as can be seen in Wolfram-Alpha.
- $$f_4(x)=\exp\left(\frac{\ln(4)x^2}{|x|-1}\right)$$
- $$f_5(x)=\exp\left(\frac{\ln(8)x^2}{x^2-1}\right)$$
- $$f_6(x)=\exp\left(\frac{2(2-\sqrt{2})\ln(2)x^2}{\sqrt{|x|}-1}\right)$$
All these functions fulfills $f_i\left(\pm\frac12\right)=\frac12$, $f_i(\pm 1)\to 0$, and $f_i(0)\to 1$ ("arrows" meaning "tendency under the limit" sense).
You could see these functions here on Desmos.
I am trying to understand Mollifiers through these examples, which all of them "look smooth" but some have "avoidable issues", like undefined values but where the function is continuous (like $f_1(x)$ at $x=0$ - check this answer), or their derivatives rises Dirac's delta functions because of the absolute values but that are avoidable by defining the same functions piecewise (like in this other question), so I am trying to figure out if smoothness (being class $C^\infty$) is hold or not on these cases.
As example, $f_1(x)$ could be constructed as is shown here by choosing $f(x)=\exp\left(-\frac{1}{x}\right)\theta(x)$ with $\theta(x)$ the Heaviside step function, then making $g(x)=\frac{f(x)}{f(x)+f(1-x)}$ and then lets $f_1(x)=g\left(\frac{x-a}{b-a}\right)\cdot g\left(\frac{d-x}{d-c}\right)$ with $a=-1,\ b=c=0,\ d=1$, as could be seen in Desmos. Moreover, a simular construction is used in this answer in MSE to built an example of a smooth transition function so in the case of $f_1(x)$, in principle, the points were could be some issues should be $x=\{-1;\ 0;\ 1\}$.
I am specialy worried about the behaviour at $x\to 0$: at the edges $\partial x =\{-1;\ 1\}$ I know that since how I defined $F(x)$ some of these functions will hold smoothness there: $$\lim\limits_{x\to\partial x^\pm} \frac{d^n}{dx^n}F_i(x)=0\quad n\geq 0,\ n\in\mathbb{Z}_{+}$$ but for others as is showed on Desmos it could be some problems at the edges $\partial x$: I don't know if it is a Desmos' software issue or indeed $f_1(x)$ lost smoothness at higher derivatives as example.
Also, I am not really sure if the definition of $F(x)$ as stated is equivalent to consider $\frac{f_i(x)}{I_i}\theta(1-x^2)$, neither if all these "situations" could be avoided since Mollifiers kernels are evaluated under an integral sign in convolution operations, so in principle, a zero-measure point shouldn't have an impact in the final value, but this is not necessarily true for distributions (just think of the Dirac's Delta function example), so I am trying to understand what criterion applies on each example. As counterexample, you could check in Wolfram-Alpha that $\int\limits_{-\infty}^\infty F_1(x)\ dx=1$, even when $F_1(x=0)$ is undefined.
As example, if the problems of $f_1(x)$ at $x=0$ and at $\partial x$ are avoidable or "ficticious", meaning it is indeed smooth, then it should be a Mollifier since fulfill:
- $F_1(x)$ is compact suported
- $\int\limits_{\mathbb{R}}F_1(x)\ dx=1$ with $I_1=1$
- $\lim\limits_{\epsilon\to 0}\frac{1}{\epsilon}F_1\left(\frac{x}{\epsilon}\right)\to\delta(x)$ as you can see in Desmos.
So, finding out what is happening at these critical points $x=\{-1;\ 0;\ 1\}$ will determine if $f_1(x)$ could be used as a Mollifier or not: $$\Phi(q)(x)=q(x)\circledast F_i(x)=\int\limits_{\mathbb{R}}q(y)F_i(x-y)\ dy=\int\limits_{|x|<1}q(x-y)F_i(y)\ dy$$
Bonus track: the functions $$\lim\limits_{n\to\infty}\frac{1}{1+\exp\left(\frac{n\left(1-2|x|\right)}{x^2-|x|}\right)}$$ becomes the rectangular function: Does exist a finite $n$ where it lost its smoothness?
