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I am trying to create a superposition of bump functions that adds identically to 1. Specifically I am looking to add two bump functions, say $f(x)$, $h(x)$ and $g(x)$ so that if $I,J,L \subset \mathbb{R}$ are the support of $f,h$ and $g$ respectively then I need $|I|=|J|=|L|$ and $I\cap L = \emptyset$. Furthermore, I need $$ f(x)+h(x)+g(x) = 1 $$ for all $x \in J$.

I guess I am getting stuck since the only bump function I know of is the usual

$$ b(x) = \exp\bigg(1-\frac{1}{1-\big(\frac{x-c}{r}\big)^2}\bigg) $$ where $c$ is the peak and $r$ is the radius. I have been trying to combine these to get what I want but I haven't been having a lot of success.

Cheers in advance for any help.

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So, what you need are some tools for creating a larger variety of bump functions. The most important tool? Convolution $f*g(x)=\int_{\mathbb{R}}f(t)g(x-t)\,dt$; if $f$ is $C^{\infty}$ and $g$ is anything at all reasonable, $f*g$ is also $C^{\infty}$. We don't need even one derivative in $f$ to make that work. A bump function is also compactly supported; for that, note that the support of $f*g$ is contained in the sum of the support of $f$ and the support of $g$ - so if $f$ is a bump function and $g$ is compactly supported, $f*g$ is a bump function.

OK, now to the problem. We want three bump functions $f,h,g$ that sum to $1$ on an interval, such that the outer two have disjoint support. Their sum $s$ is, of course, also a bump function, so let's find that first.

So, how do we get a constant out of the convolution? We convolve with a constant; $b*1(x)=\int_\mathbb{R}b$, independent of $x$. Obviously, we don't want this on the whole real line - but if, instead, we convolve with the characteristic function of an interval $U=(\rho-\gamma,\rho+\gamma)$ larger than the support $B=(r-c,r+c)$ of $b$, we'll get a bump function which is constant on some smaller interval, of length $|U|-|B|$. So then, let $$u(x)=\begin{cases}\left(\int_{\mathbb{R}}b\right)^{-1}&x\in U\\0&\text{otherwise}\end{cases}$$ and $s=b*u$. This $s$ is supported on $[r+\rho-c-\gamma,r+\rho+c+\gamma]$ and is equal to $1$ on $[r+\rho-\gamma+c,r+\rho+\gamma-c]$.

Now we need to break that $s$ up into a sum $f+h+g$, supported on three sets $I,J,L$ respectively of equal size, with $I$ and $L$ disjoint, and $J$ contained in that middle set where the sum is $1$. The obvious thing to do is to split $u$ into three equal parts, each supported on an interval of length $\frac23\gamma$. The first, leading to $f$ and $I$, will be $\left(\int_{\mathbb{R}}b\right)^{-1}$ times the characteristic function of $(\rho-\gamma,\rho-\frac13\gamma]$, so we will have $I=[r+\rho-\gamma-c,r+\rho-\frac13\gamma+c]$. Similarly, $J=[r+\rho-\frac13\gamma-c,r+\rho+\frac13\gamma+c]$ and $L=[r+\rho+\frac13\gamma-c,r+\rho+\gamma+c]$. For this all to work out, we need ($I$ and $L$ disjoint) $$r+\rho-\frac13\gamma+c < r+\rho+\frac13\gamma-c$$ $$2c<\frac23\gamma$$ and (lower endpoint of $J$ in the region where the sum is $1$) $$r+\rho-\gamma+c < r+\rho-\frac13\gamma-c$$ $$2c < \frac23\gamma$$ and another for the upper endpoint of $J$, which will simplify to the same condition as the two we already have. Basically, this three-part decomposition works as long as each of the individual pieces is large enough to have its own region where the convolution is constant.

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  • $\begingroup$ Thanks for the incredibly detailed answer, more than I thought I would get. Tell me, will this method be able to be extended to as many of these functions as I need on a given interval? Say I wanted to fill the interval (0,2pi) with functions like this each with support on an interval of length 1/k for some real k, and no function has support overlapping with more than two others? $\endgroup$ – Jandré Snyman Jan 18 at 10:08
  • $\begingroup$ Yes, the method would work for that as well. $\endgroup$ – jmerry Jan 18 at 10:11
  • $\begingroup$ Cheers mate!!! I wish I could double upvote your answer. $\endgroup$ – Jandré Snyman Jan 18 at 10:11

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