A more intuitive way of constructing bump functions

This is a short note on creating bump functions, test functions which are 1 on the desired domain, etc. I will be working in one dimension. However, all these results can be generalized to higher dimensions by using polar coordinates.

As we know, the function f(x)= e^{-\frac{1}{x}} for x\geq 0 and 0 for x\leq 0 is a smooth function. Hence, it is an ideal candidate for constructing smooth, compactly supported functions. If we wanted to construct a smooth function that was supported on [a,b], then f(x-a)f(-x-b) is one such function.

However, the main difficulty is in constructing a bump function of the desired shape. How do we construct a bump function that is \equiv 1 on [c,d]\subset [a,b]? The idea that I had, which is different from the literature that I’ve consulted (including Lee’s “Smooth Manifolds”), is that we could consider the integrals of functions.

Consider \int_0^{\infty} f(x-a)f(-x-c)-f(x-d)f(-x-b) dx.

Basically, it we are integrating a function that is positive on [a,c], and then adding that to the integral of the negative of the same function, but now supported on [d,b].

This function will be constant on [c,d], and then decrease to 0 on [d,b]. On re-scaling (multiplying by a constant), we can obtain a bump function on [a,b] that is 1 on [c,d].

Published by ayushkhaitan3437

Hello! My name is Ayush Khaitan, and I'm a graduate student in Mathematics. I am always excited about talking to people about their research. Please please set up a meeting with me if you feel that I might have an interesting perspective to offer- https://calendly.com/ayushkhaitan/meeting-with-ayush

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