I was encountered with a problem about convolution but I think there must be something wrong.
Problem: Suppose that $f,g\in L^{1}(T=\mathbb{R}/\mathbb{Z})$, prove that $$\|\int _T f(x-y)g(y)dy\|_{\infty}\leq\|f\|_1\|g\|_1.$$
But I tried to integrate this convolution on T and get $$\int_T\int _T |f(x-y)g(y)|dydx=\int _T |g(y)|dy\int_T|f(x-y)|dx=\|f\|_1\|g\|_1.$$
If the problem is wrong, how can we get the continuity of the convoluion?
if $f,g\in L^1(T)$, can we prove that $$\int_Tf(x-y)g(y)dy$$ is a continuous function of x?
I know it is true if $f\in L^1$ and $g\in L^{\infty}$.
Suppose $f(y)=1/\sqrt y$ for $0<y<1,$ $f=0$ elsewhere. Then $f\in L^1.$ Degine $g=f.$ For small $x<0,$ we have
$$f*g(x) = \int_0^{1+x}\frac{1}{\sqrt{y-x}}\frac{1}{\sqrt{y}}\,dy.$$
As $x\to 0^-,$ the last integral converges to
$$ \int_0^{1}\frac{1}{y}\,dy=\infty$$
by the monotone convergence theorem. Hence the convolution is not bounded. This also shows the convolution is not continuous.