Question

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.

Answer 1

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.

Answer 2

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...