TITLE: Discontinuous convolutions
QUESTION [7 upvotes]: 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.

REPLY [5 votes]: 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.