Let $f$ be a continuous but nowhere differentiable function. Is $f$ convolved with mollifier, a smooth function?
Let $f$ be a continuous but nowhere differentiable function. Is $f$ convolved with mollifier, a smooth function?
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$\begingroup$
analysis
convolution
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0Yes, this is actually a very important property of mollifiers (maybe the most important?). Do you know what an approximate identity is? You can approximate your function with the mollified version which often has much nicer properties. – 2010-11-02
1 Answers
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Yes.
The key is that when you (as you say in the comments) get the two scenarios:
$ f \star (D g) = D(f \star g) = (D f) \star g $
then you get to choose which!
So if $D f$ doesn't make sense, then you can ignore it and choose to use the identity $D(f \star g) = f \star (D g)$.
Taking this to the extreme, you get the - bizarre, in my opinion - result that if $p$ is a polynomial, then $f \star p$ is always a polynomial.
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0@Andrew: Had I known about the LDC theorem, I do'nt think i would have asked this question here ! – 2010-11-08