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.

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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