TITLE: the convolution of integrable functions is continuous?
QUESTION [10 upvotes]: The question is simple but I still can't prove it or contradict it. Here it goes:

Suppose $f$ and $g$ are defined on the circle
  (or, equivalently, $2\pi$ periodic functions) and Lebesgue integrable, 
  is their convolution $(f*g)(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x-y) g(y) dy $ continuous?

In the case when two functions are bounded, it is proved in Elias Stein's 
Fourier Analysis (page 47) that their convolution is truly continuous.
However, for unbounded functions, I have tried tools in real analysis, say, Lusin's theorem, transition continuity of $L_1$ functions, etc., but can't figure it out.

REPLY [15 votes]: A less explicit answer: Salem and Zygmund proved that convolution $L^1(\mathbb T) \times L^1(\mathbb T) \to L^1(\mathbb T)$ is onto.
This was shown to hold for all locally compact groups by Paul Cohen in 1959. This result was the starting point of an entire industry establishing "factorization theorems".
A nice survey on this topic is Jan Kisynski, On Cohen's proof of the Factorization Theorem, Annales Polonici Mathematici 75, 2 (2000), 177-192.