I would like to share a problem that has been bugging me for a bit, which is on the general topic of bump functions.
The setting is that I have one piecewise function $p$, that comprises of a constant leg (suppose this is $g$) and one that looks parabolic (suppose this is $h$).
For reference, \begin{equation} p(t) = \begin{cases} g(t) \\ h(t) \end{cases} = \begin{cases} 1\text{, for }t<0, \\ \frac{1}{2} + \frac{1}{10} \sqrt{-80t^2+60t+25}, \end{cases} \end{equation}
Let us consider $p$ inside $[-1,1]$ for the sake of simplicity.
They connect at $0$, and my goal is to smooth this out. The idea is to extract a stripe of length 2$\epsilon$, for some $\epsilon$ that is relatively small -- like $0.1$ for example, but not important at the moment -- and replace that part with a bump function that glues together the pieces into one, smooth function.
If I understand correctly, I should pick a bump function $f$, i.e. \begin{equation} f = \begin{cases} \exp{{(-1/x^2)}}\text{, for } x>0\\ 0\text{, otherwise,} \end{cases} \end{equation}
consider its integral (suppose it's $F$ but is still not clear to me why I need its integral). Then, in the $(-\epsilon, \epsilon)$ interval, I would replace said functions with $gF+h(1-F)$ and voila?
The issue is that this doesn't work so far. I have experimented with a few different variations for bump functions, and still something seems to evade me.
I would appreciate any insight on this. Ultimately, I want to understand the process exactly, in order to replicate this again in different points, using different functions.
All other similar questions simply provide the bump equivalent of the step function and I cannot completely see the connection.
The function $g$ is already very nice, so we just have to change $p$ in an interval $(0,\epsilon)$. The goal is to redefine it in that interval, so that the resulting function $\widetilde{p}$ is $C^\infty$ in $[-\infty,t_*)$, where $t_*$ is the smallest positive root of $h$.
Let $\varphi$ be a positive bump function in $(0,\epsilon)$, that vanishes with all its derivatives outside $(0,\epsilon)$, and has integral $1$. For example, we can take $$\forall x \in (0,\epsilon), \quad \varphi(x)=c_\epsilon \exp\Bigl(\frac{-1}{x(\epsilon-x)}\Bigr) \: \quad \text{and} \: \quad\forall x \notin (0,\epsilon), \quad \varphi(x)=0\,,$$
where $c_\epsilon$ is chosen to ensure that $$\Phi(x):=\int_0^x \varphi(t) \,dt $$ satisfies $\Phi(\epsilon)=1$. Define $$\widetilde{p}(x)=1+(h-1)\cdot \Phi(x)$$ for $x \in (0,\epsilon)$ and $\widetilde{p}(x)=p(x)$ for $x \notin (0,\epsilon)$.
Clearly $\widetilde{p}(0)=1=g(0)$ and $\widetilde{p}(\epsilon)=p(\epsilon)$.
By the Leibniz formula for the $n$'th derivative of a product (https://en.wikipedia.org/wiki/General_Leibniz_rule) we see that $$\forall n \ge 1, \quad \widetilde{p}^{(n)}(0)=0$$ and $$\forall n \ge 1, \quad \widetilde{p}^{(n)}(\epsilon)=h^{(n)}(\epsilon) \,,$$ as required.