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.



the convolution of integrable functions is continuous?
Edit: sorry, there was a sign error. It should just be:


$$f(x) = g(x) = \begin{cases}
x^{-3/4}&x > 0\cr
0&x \leq 0.
\end{cases}$$
Then $\lim\_{x \to 0^+}f\*g(x) = \infty$ and $\lim\_{x \to 0^-} f\*g(x) = 0$.


