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.
Best Answer
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 $\lim_{x \to 0^-} f*g(x) = 0$.