Show that convolution of two $L^1(\mathbb{R})$ functions is continuous

Question

Suppose $f, g \in L^1(\mathbb{R})$. I want to show that their convolution is continuous.

I can show continuity if one of the functions were in $L^\infty(\mathbb{R})$.

I have tried to approximate one of the functions, say $f$ with $f_M = f\cdot \mathbb{I}\{ |f(x)| \leq M\}$. But I have problem bounding the residual, which would be of form:

$$ \int_\mathbb{R} |f(x-z) - f_M(x-z)|\cdot |g(z)| dz$$

Any help/hint would be greatly appreciated!

Answer 1

This is not true. The counterexample I know is:

$f(x) = g(x) = \begin{cases} x^{-3/4}&0<x< 1\cr 0& \text{otherwise} \end{cases}$

then $\lim_{x \to 0^+}f*f(x) = \infty$ and $\lim_{x \to 0^-}f*f(x)=0$.

I guess you need at least one of $f$ or $g$ to be $L^{\infty}.$

Answer 2

$f*g$ is not a well defined function from $\mathbb R$ to itself, in general. For example, if $f(x)=g(x)=\frac 1 {\sqrt {|x|}}$for $0<|x| \leq1 $ and $0$ elsewhere then $(f*g)(0)=\infty$.