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### A simpler way to obtain smooth functions than convolutions?

Of the many mathematical concepts that I don’t understand, one of the more important ones is the convolution of functions. It is defined in the following way:

$(f*g)(x)=\int_{-\infty}^{\infty} f(x-y)g(y) dy$

Our guiding principle should be that we want to make $*"$ an abelian group action (although inverses are not always present, at least when talking about integrable functions).

However, perhaps the reason why we thought of this action in the first place was that we wanted smooth functions out of just integrable functions. For instance, given any integrable function $f$, if $\phi(x)$ is a smooth compactly supported function, $(\phi*f)(x)=\int_{-\infty}^{\infty}\phi(x-y)f(y)dy$ will be smooth (provided we can bring the derivatives under the integral sign, which is related to the Dominated Convergence Theorem).

However, why have this complicated definition? Why not just consider the function $g(x)=\int_{-\infty}^{\infty}\phi(x)f(y)dy$? Clearly, if $\phi(x)$ is smooth, so is this one.

One of the reasons why we perhaps want the more complicated definition of $(f*g)(x)=\int_{-\infty}^{\infty} f(x-y)g(y) dy$ is that there does not exist a function $\phi(x)$ such that $\int_{-\infty}^{\infty} \phi(x)f(y)dy=f(x)$ for all functions $f(x)$. Hence, there cannot exist an identity element for the set of integrable functions on the real line. Also, this definition is clearly not commutative. I’d be interested in knowing your thoughts about what other purposes convolution serves, that this simple definition does not.

### 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 $[c,d]$.