Well-Definedness and C^Infinity Closure of Convolutions

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In summary, the conversation discusses proving the well-definedness and smoothness of the convolution of two functions, without using Fourier analysis or more advanced concepts. The use of Leibniz's rule for differentiation and integration with variable limits is suggested, and the original poster eventually solves the problem by applying the Mean Value Theorem. It is mentioned that this proof is essentially a proof of Leibniz's Rule, which was not allowed in the original problem.
  • #1
TaylorM0192
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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!
 
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  • #2
TaylorM0192 said:
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
 
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  • #3
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.
 

FAQ: Well-Definedness and C^Infinity Closure of Convolutions

What is the definition of "well-definedness" in the context of convolutions?

Well-definedness refers to the property of a mathematical operation, such as convolution, being uniquely defined and independent of the chosen representation or variables. In the context of convolutions, this means that the result of convolution is the same regardless of the choice of functions or variables used.

How is well-definedness related to the C^infinity closure of convolutions?

The C^infinity closure of convolutions ensures that the result of convolution is a smooth function, meaning that it has derivatives of all orders. This is important for well-definedness because if the functions used in the convolution are not smooth, the result may not be well-defined.

Can convolutions be well-defined with non-smooth functions?

No, convolutions are only well-defined when the functions involved are smooth, meaning they have derivatives of all orders. If one or both of the functions are not smooth, the result of convolution may not be well-defined.

How is the C^infinity closure of convolutions achieved?

The C^infinity closure of convolutions is achieved by taking the convolution of two smooth functions and then taking the limit as the smoothness of the functions increases. This ensures that the resulting function is also smooth.

Why is well-definedness important in mathematics?

Well-definedness is important in mathematics because it ensures consistency and avoids ambiguity in mathematical operations. Without well-definedness, the result of an operation may depend on the representation or variables chosen, leading to incorrect or inconsistent results.

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