# Superposition of of bump functions identically equal to 1.

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.

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.