Questions on the real bump function and conjuction of smooth functions

Question

Before I ask you my question which I will mark in bold I will tell you what I already gathered so far.

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In a previous result I have showed that the bumpfunction is smooth. The bumpfunction is defined on some intervall $[a,b]$ with

$$\psi(x)\begin{cases}e^{-\frac{1}{x-a}}e^{-\frac{1}{b-x}}&\mbox{a<x<b}\\0&\mbox{otherwise}\end{cases}$$

For more details I refer to:

Calculate the n-th derivative for the BUMP function

I also know that the bump-function can be normalized, i.e if $a<b<c<d$ I can construct a smooth function $\psi$ with the properties

$$\phi(x)\begin{cases}=1&\mbox{}b\leq x\leq c\\\in(0,1)&\mbox{if }a<x<b\text{ or }c < x <d\\=0 &\mbox{if }x\leq a \text{ or }x\geq d\end{cases}$$

The function I came up with is:

$$\phi(x)=\psi_0(x)\begin{cases}\frac{\psi_1(x)}{\psi_2(x)+\psi_1(x)+\psi_3(x)}&\mbox{a<x<d}\\0&\mbox{otherwise}\end{cases}$$

$$\psi_1(x)\begin{cases}e^{-\frac{1}{x-a}}e^{-\frac{1}{d-x}}&\mbox{a<x<d}\\0&\mbox{otherwise}\end{cases}$$

$$\psi_2(x)\begin{cases}e^{-\frac{1}{x-a}}e^{-\frac{1}{b-x}}&\mbox{a<x<b}\\0&\mbox{otherwise}\end{cases}$$

$$\psi_3(x)\begin{cases}e^{-\frac{1}{x-c}}e^{-\frac{1}{d-x}}&\mbox{c<x<d}\\0&\mbox{otherwise}\end{cases}$$

What I have thought is if we have a value $x$ which is $a<x<b\text{ or }c < x <d$ then the nominator is bigger than the dominator i.e $\psi(x)\in(0,1)$ because all components are also positive, if we have a value which is $c\leq x \leq d$ then $\psi_2$ and $\psi_3$ are Zero and the denominator and the Nominator is the same i.e $\psi(x)=1$. Because the function is a product of smooth functions $\psi_0$ itself is smooth again.

I am however not quite sure about this when I plotted this function on Wolfram Alpha:

(e^{-1/(x+0.5)}*e^{-1/(0.5-x)})/((e^{-1/(x+0.5)}*e^{-1/(-0.25-x)})+(e^{-1/(x+0.5)}*e^{-1/(0.5-x)})+(e^{-1/(x-0.25)}*e^{-1/(0.5-x)}))

I got a bump function however the values were not normalized as originally intended.

For $a<b$ there exists also a function $\tau$ for which we have

$$\tau(x)\begin{cases}=1&\mbox{}x\geq b\\\in(0,1)&\mbox{if }a<x<b\\=0 &\mbox{if }x\leq a \end{cases}$$

Again I first define two helper-functions in this case

$$\tau_1(x)=\begin{cases} e^{-\frac{1}{x-a}} &\mbox{if } x>a\\0&\mbox{otherwise}\end{cases}$$

$$\tau_2(x)=\begin{cases} e^{-\frac{1}{x-a}}e^{-\frac{1}{b-x}} &\mbox{if } a<x<b\\0&\mbox{otherwise}\end{cases}$$

Then the desired function is

$$\tau(x)=\frac{\tau_1(x)}{\tau_1(x)+\tau_2(x)}$$

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Now my question

Given two smooth functions $f,g:\mathbb{R}\rightarrow\mathbb{R}$

There exist a smooth function $h:\mathbb{R}\rightarrow\mathbb{R}$ for which we have

$$h(x)\begin{cases}=f(x)&\mbox{, if }x\leq a\\=g(x) &\mbox{, if }x\geq b\end{cases}$$

Now if I create a function $\tau$ for the missing interval $(a,b)$ in the Fashion I did above then I have $\lim_{x\downarrow a}\tau(x)=f(a)$ and $\lim_{x\uparrow a}\tau(x)=g(b)$ as desired, but the Problem is to Show that it is smooth the function $h(x)$ must also fullfill the condition:

$$\lim_{x\downarrow a}\tau^{(n)}(x)=f^{(n)}(a)\wedge \lim_{x\uparrow b}\tau^{(n)}(x)=g^{(n)}(b)\text{ for every n-th derivative} $$

I don't know how I can prove this with the results I already have.

I actually have an idea I take two $\tau$ functions one $\tau_1$ goes to $0$ as $x$ goes to $b$ and to $1$ as $x$ tends to $a$ and another one with the exact opposite behaviour $\tau_2$. Then $\tau_1(x) f(x) + \tau_2(x) g(x)$ is the function I Need but I Need help to prove that the derivatives of the $\tau$ have the same behaviour on the border. However the Problem is that when applying the productrule to calculate for example the first derivative I would get $\tau_1'f+\tau_1f'+\tau_2'g+\tau_2g'$. Assuming that the derivatives of $\tau$ have the same convergencebehaviour I would get for $\lim_{x\uparrow b}h'(x)= g(x)+g'(x)$. But I want only $g'(x)$.

Really hope somebody can help me how I can start

EDIT:

I am also Looking for a solution that does not involve integrals

Answer 1

The $\tau$ defined in the question seems to be $0/0$ for $x<a.$ Perhaps you mean to take: $$\tau_1(x)=\begin{cases} e^{-\frac{1}{x-a}} &\mbox{if } x>a\\0&\mbox{otherwise}\end{cases}$$ $$\tau_2(x)=\begin{cases} e^{-\frac{1}{b-x}} &\mbox{if } x<b\\0&\mbox{otherwise}\end{cases}$$

Then $\tau_1+\tau_2$ is strictly positive everywhere, so the ratio

$$\tau(x)=\frac{\tau_1(x)}{\tau_1(x)+\tau_2(x)}$$ defines a smooth non-decreasing function with $f^{-1}(\{0\})=(-\infty,a]$ and $f^{-1}(\{1\})=[b,\infty).$

Given smooth $f,g:\mathbb R\to\mathbb R$ define $h=(1-\tau) f+\tau g$ with the $\tau$ defined above. This is smooth because the set of smooth functions is closed under addition and multiplication. We get $h(x)=f(x)$ for $x<a,$ so $h^{(n)}(x)=f^{(n)}(x)$ for $x<a,$ and by continuity of the derivatives, $h^{(n)}(x)=f^{(n)}(x)$ for $x\leq a.$ Similarly $h^{(n)}(x)=g^{(n)}(x)$ for $x\geq b.$ There is no need to take one-sided limits because all the derivatives are continuous.

If you want to try the product rule: $\tau^{(n)}(x)=0$ for $n\geq 1$ and $x\in(\infty,a]$ because $\tau$ is constant on $(\infty,a)$ and $\tau^{(n)}$ is continuous. Similarly $\tau^{(n)}(x)=0$ for $n\geq 1$ and $x\in[b,\infty).$

Answer 2

A few comments on your constructions at the beginning: Your $\phi$ has problems. For $x\in (a,b),$

$$\phi(x) = \frac{\Psi_1(x)}{\Psi_2(x)+\Psi_1(x)} = \frac{1}{\Psi_2(x)/\Psi_1(x)+1}.$$

However,

$$\Psi_2(x)/\Psi_1(x) \to e^{-1/(b-a)}/e^{-1/(d-a)}$$

as $x\to a^+.$ Thus $\phi(x)$ does not approach $0$ as $x\to a^+.$ There is a similar problem for $\phi$ when $x\to d^-.$

Your function $\tau$ has a similar problem: As $x\to a^+,$ $\tau(x) \to e^{-1/(b-a)}.$

Now to the $f,g,[a,b]$ problem you posed: Take your $\Psi_2$ and multiply it by a positive constant $c$ so that $\int_a^b (c \Psi_2) = 1.$ For $x\in \mathbb R,$ define

$$\phi(x)=\int_{-\infty}^x (c \Psi_2).$$

Then $\phi$ is smooth. We have

$$\begin{cases} \phi = 0&\text{ on } (-\infty,a]\\0<\phi < 1 &\text{ on } (a, b)\\ \phi=1&\text{ on } [b,\infty)\end{cases}.$$

Verify that $1-\phi$ does the above in reverse. I.e.,

$$\begin{cases} 1-\phi = 1&\text{ on } (-\infty,a]\\0<\phi < 1 &\text{ on } (a, b)\\ \phi=0&\text{ on } [b,\infty)\end{cases}.$$

Now given smooth $f,g$ on $\mathbb R,$ we can proceed much as @Dap did: Define $h=(1-\phi)f + \phi g.$ Then $h$ is smooth on $\mathbb R,$ $h=f$ on $(-\infty,a],$ and $h=g$ on $[b,\infty).$ This is the desired conclusion.

I'll end by pointing out another problem along these lines: Suppose $f\in C^\infty([a,b]).$ Show there exists $F\in C^\infty(\mathbb R)$ such that $F=f$ on $[a,b].$ This is a harder problem. I solved it using a nice result of Borel: Given $x_0\in \mathbb R,$ and numbers $a_0,a_1, \dots \in \mathbb R,$ there exists $f\in C^\infty(\mathbb R)$ such that $D^nf(x_0) = a_n$ for all $n.$ See the Corollary in

Connecting smooth functions in a smooth way