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Is the following true?

The convolution of two infinitely differentiable as well as integrable real functions can be nowhere continuous.

A reference/proof idea would be very helpful.

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    $\begingroup$ First of all, if you are considering the convolution of two non-integrable functions, how are you defining this? $\endgroup$
    – Yemon Choi
    Commented Aug 16, 2010 at 22:58
  • $\begingroup$ The functions have to be in $L^1[\mathbf{R}]$ $\endgroup$
    – Ashutosh
    Commented Aug 16, 2010 at 23:10
  • $\begingroup$ "can be nowhere continuous." I'm confused; are you asking for an examples of two smooth, integrable functions that have a continuous convolution? Because Nate gives an example. Sure you don't mean "nowhere continuous"? $\endgroup$
    – Dylan Wilson
    Commented Aug 16, 2010 at 23:51
  • $\begingroup$ I'm sorry. I meant nowhere continuous. $\endgroup$
    – Ashutosh
    Commented Aug 16, 2010 at 23:56
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    $\begingroup$ Here's a sort of example I don't want: Let f_n be a smooth approximation to n times the characteristic function of [n, n + (1/n)^3]. Let f be the sum of f_n's where n runs over all positive integers. Let g be an even function which matches f on positive numbers. Then g is an infinitely differentiable L1 function whose convolution with itself at 0 is infinite and therefore discontinuous at 0. But this is not something I want. I want (f * g) to be finite everywhere and continuous nowhere and I don't know how to do this. $\endgroup$
    – Ashutosh
    Commented Aug 17, 2010 at 20:48

2 Answers 2

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I believe the answer is yes. though I don't quite have a precise construction. The basic idea is to find a series $f = \sum_n f_n$ of increasingly narrow bump functions $f_n$ whose $L^1$ norms decay very quickly (e.g. exponentially fast in $n$), but such that the series converges to a nowhere continuous function $f$; I think some sort of "typewriter function" construction will do this. If the bump functions $f_n$ are reasonable, then they should be expressible as (or modifiable to) a convolution $f_n = g_n * h_n$, where the $g_n$ are also narrow (but perhaps tall) bump functions whose $L^1$ norms also decay very quickly. If one then sets $g(x) := \sum_n g_n(x-x_n)$ and $h(x) := \sum_n h_n(x+x_n)$ where $x_n$ goes to infinity extremely quickly with $n$ (e.g. $x_n = 2^{2^n}$) then I think $g*h$ should differ from $f$ by a continuous function and thus also be nowhere continuous, despite $g, h$ being infinitely smooth and integrable.

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  • $\begingroup$ If $f_n$ is a sequence of continuous functions whose sum converges pointwise to $f$ then isn't it true that the set of points of continuity of f is dense in reals? $\endgroup$
    – Ashutosh
    Commented Sep 1, 2010 at 0:40
  • $\begingroup$ Well, one only needs the series here to converge almost everywhere, rather than pointwise, so I think we can escape the Baire class. The construction I had in mind had the f_n being concentrated near multiples of 1, then multiples of 1/2 (perhaps with the opposite sign to keep the partial sums bounded), then multiples of 1/4, and so forth, while becoming very rapidly narrower as one progresses, which should create a lot of oscillation at every scale, while still being absolutely summable in L^1 and thus convergent pointwise a.e. $\endgroup$
    – Terry Tao
    Commented Sep 1, 2010 at 2:47
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I believe we get the stronger statement that the convolution of an infinitely differentiable integrable function $f$ with any integrable function $g$ will result in $f*g$ being infinitely differentiable. In fact, $(f*g)' = f'*g$ (This can be found in The Fourier Transform and Its Applications, Bracewell) Thus the convolution of two infinitely differentiable, integrable functions will necessarily be (infinitely) differentiable and so also continuous.

EDIT: Also, if my math is correct, we can check this formula via the Fourier transform:

$\widehat{((f*g)')}(r) = ir\widehat{f*g}(r) = ir\widehat{f}(r)\widehat{g}(r) = \widehat{f'}(r)\widehat{g}(r) = \widehat{(f'*g)}(r)$

Apologies, this only applies in the case that $f$ is compactly supported. Back to the drawing board...

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    $\begingroup$ What if f' is not integrable? $\endgroup$
    – senti_today
    Commented Aug 17, 2010 at 1:14
  • $\begingroup$ Please double-check the hypotheses of the theorems you're using. The proofs I'm familiar with require $f'$ to be integrable, which is not assumed here. $\endgroup$
    – Nate Eldredge
    Commented Aug 17, 2010 at 1:19
  • $\begingroup$ I think this problem is more subtle than it looks, and this answer is incorrect as written (without a lot of extra explanation). For example, I'm fairly sure that $f'$ need not make sense as a tempered distribution in general, so how can you take the Fourier transform? $f'$ can be very badly behaved! $\endgroup$
    – Zen Harper
    Commented Aug 17, 2010 at 1:36
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    $\begingroup$ Also, the space of Fourier transforms of continuous $L^1$ functions (as well as the Fourier transforms of $L^1$ itself) doesn't have any nice characterisation as far as I know, so even if we could take the Fourier transform, it wouldn't necessarily be able to help us...consider $e^{-x^2} sin(\exp(\exp(\exp(x))))$, which is $C^\infty$ and $L^1$, but with very nasty derivatives!! $\endgroup$
    – Zen Harper
    Commented Aug 17, 2010 at 4:49
  • $\begingroup$ My mistake, you are right - I have been working with $\mathbb{T}$ for so long I forgot I require $f$ to be compactly supported. $\endgroup$
    – Vince
    Commented Aug 17, 2010 at 15:38

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