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Let $f$ be a continuous but nowhere differentiable function. Is $f$ convolved with mollifier, a smooth function?

Srivatsan
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Rajesh D
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  • the derivative operation commutes over the integral (of convolution) but i get two different scenarios when i change the order of convolution, i.e fmoll or mollf ? – Rajesh D Nov 02 '10 at 04:04
  • Yes, 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. –  Nov 02 '10 at 13:29

1 Answers1

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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.

Srivatsan
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Andrew Stacey
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  • Why do we get to choose ? It isnt part of a proof. Looks like its some sort of an axiom or something ! BTW convolution is not always commutative according to statements in your answer. – Rajesh D Nov 02 '10 at 09:49
  • @Andrew a polynomial is neither compact nor in L^2.I guess you mean a polynomial multiplied with some bump function. – Rajesh D Nov 02 '10 at 10:18
  • @Rajesh: A more precise statement of the theorem is: "If $f$ and $g$ are such that $f \star g$ makes sense and $f$ is $C^1$ then $f \star g$ is $C^1$ with $D(f \star g) = (D f) \star g$.". So the ability to choose comes from the fact that convolution is commutative and that $D(f \star g)$ does not "prefer" $f$ over $g$. Regarding polynomials: you didn't specify the space in which you were working. Convolution makes sense in a much wider context that $L^2$-functions. Again, a more precise statement is: "If $f \star p$ makes sense then it is a polynomial". – Andrew Stacey Nov 02 '10 at 11:39
  • @Andrew: the notations in your statements are in sync with neither that of Question (OP) nor with your answer.Could you please be specific. – Rajesh D Nov 02 '10 at 14:16
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    @Rajesh: I don't understand that comment. There is no notation in the question (other than "f") and there's not a lot of notation in my answer either, so I'm not sure what you mean by "not in sync". – Andrew Stacey Nov 02 '10 at 20:35
  • @Andrew: In your last comment, what is 'f' in the first half and what it is in the later part, and what is it in your answer and in the question. Yes there isnt lot of notation, but whatever is there, isnt consistent. – Rajesh D Nov 03 '10 at 01:14
  • also see http://math.stackexchange.com/questions/8711/derivative-commuting-over-integral – Rajesh D Nov 03 '10 at 03:16
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    @Rajesh: Ah, I see. When I write "If $f$ ..." then I am implicitly making $f$ a local variable whose properties and so forth are only set for the scope of that statement. So the $f$s can all be different functions with different properties. I suggest that at this point you look in an introductory book on Fourier Theory. – Andrew Stacey Nov 03 '10 at 09:48
  • @Andrew: your arguments lack mathematical reasoning. It is evident from having a look at the answer given by you. Could you please consider my request to give a JUSTIFICATION for choosing from either of the integrals ! (even if you consider it below your standard) – Rajesh D Nov 07 '10 at 10:12
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    @Rajesh, all Andrew is saying is that for any f,g in certain spaces, $D(f\star g)$ is equal to $Df\star g$ OR $f\star Dg$, provided either of the latter expressions are well-defined, i.e. the derivatives exist and the integrals defining the convolution are convergent.

    For instance, in your scenario, since f is continuous and g is smooth with compact support, we have $D(f\star g)$ = $f\star Dg$. However, if f was $C^1$, we also would have $D(f\star g) = Df\star g$. This is proved by using the Lebesgue Dominated Convergence Theorem to move the derivative inside the integral.

     – Greg O. Nov 07 '10 at 16:50
  • @Rajesh: I'm sorry, but providing such full justification is beyond the purposes of this site. If you really need more than I've provided, you really should study Fourier analysis properly (ie take a course, or study from a book). I had a suspicion that you hadn't grasped the point that you could choose which to use, but it seems that you need more than that. In which case, this isn't where to find help. – Andrew Stacey Nov 07 '10 at 19:23
  • @Greg: Thank you very much. I am not just interested in the answer but a clue to probe or study further.(eg Lebesgue Dominated convergence theorem).I dont expect full explanations but clues to probe or study further on my own from books, which i think is fair ( especially me not being a math major). – Rajesh D Nov 08 '10 at 02:22
  • @Andrew: Had I known about the LDC theorem, I do'nt think i would have asked this question here ! – Rajesh D Nov 08 '10 at 02:29