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Well-Definedness and C^Infinity Closure of Convolutions

TaylorM0192

New member
Mar 6, 2012
5
Hello,

Let me first just say, I posted this thread on mathhelpforum.com - but I read a post by Plato somewhere or another recommending here instead, since apparently the other site had some bad customer service issues... :)

I want to prove that if given two functions f and g (f is assumed continuous; g is assumed C^infinity with compact support on R), their convolution (f*g) is (a) well defined and (b) an element of C^infinity. The idea is to later use this result for some problems concerning "approximation to the identity."

Proving well-definedness is easy since (fg) is Riemann integrable and g is compactly supported, so the convolution does not diverge and is finite for all real x.

I am aware of a result back from lower-division DE class that the derivative of the convolution can be "transferred" to either f or g; but, I don't know how to prove this. If someone could lead me in that direction, I think I would be able to prove the result from there.

I also saw a proof where the Fourier transform was used; but, I want to avoid using Fourier analysis (and indeed, more sophisticated proofs involving Young/Minkowski inequalities, elements of functional analysis, measure theory, Lebesgue integration, etc.), and limit myself to just basic concepts concerning L^2 functions (i.e. mean convergence, Holder's inequality, etc.) if these concepts apply at all to any possible proofs.

Thanks!
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Hello,

Let me first just say, I posted this thread on mathhelpforum.com - but I read a post by Plato somewhere or another recommending here instead, since apparently the other site had some bad customer service issues... :)

I want to prove that if given two functions f and g (f is assumed continuous; g is assumed C^infinity with compact support on R), their convolution (f*g) is (a) well defined and (b) an element of C^infinity. The idea is to later use this result for some problems concerning "approximation to the identity."

Proving well-definedness is easy since (fg) is Riemann integrable and g is compactly supported, so the convolution does not diverge and is finite for all real x.

I am aware of a result back from lower-division DE class that the derivative of the convolution can be "transferred" to either f or g; but, I don't know how to prove this. If someone could lead me in that direction, I think I would be able to prove the result from there.

I also saw a proof where the Fourier transform was used; but, I want to avoid using Fourier analysis (and indeed, more sophisticated proofs involving Young/Minkowski inequalities, elements of functional analysis, measure theory, Lebesgue integration, etc.), and limit myself to just basic concepts concerning L^2 functions (i.e. mean convergence, Holder's inequality, etc.) if these concepts apply at all to any possible proofs.

Thanks!



Won't Leibniz's rule for differentiation an integral with variable limits do this for you? (together with the observation that the derivative a compactly supported \(C^{\infty}\) function is also a compactly supported \(C^{\infty}\) function )

CB
 
Last edited:

TaylorM0192

New member
Mar 6, 2012
5
Thanks for the hint CaptainBlack - I actually proved it this morning by simply setting up the difference quotient, combining the difference of integrals, and then interchanging the separate limits which occur (one where h->0 and the other the integral itself; this is valid once we prove the derivative converges uniformly) and then also a quick application of the Mean Value Theorem to rewrite one of the occurring terms.

If someone wants a solution, I can write it up, but it doesn't seem like there was too much interest in this problem prior to solving it. :O

Btw CB, it seems my proof is basically a proof of Leibnizs' Rule anyhow; therefore, I don't think we were allowed to assume this lol.