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

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What have you tried in order to prove this? –  Mariano Suárez-Alvarez Nov 2 '10 at 3:58
    
the derivative operation commutes over the integral (of convolution) but i get two different scenarios when i change the order of convolution, i.e f*moll or moll*f ? –  Rajesh D Nov 2 '10 at 4:04
    
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. –  Jonas Teuwen Nov 2 '10 at 13:29
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1 Answer

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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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 2 '10 at 9:49
    
@Andrew a polynomial is neither compact nor in L^2.I guess you mean a polynomial multiplied with some bump function. –  Rajesh D Nov 2 '10 at 10:18
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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 2 '10 at 20:35
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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 3 '10 at 9:48
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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 7 '10 at 16:50
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