Convolution of indicator functions is continuous

Question

Suppose I have an indicator function on a set of measure $E$, which is a subset of $[0,1]$. Is the function of this indicator convoluted with itself a continuous function? How can I show that it is? Intuitively I believe it is..

Answer 1

Yes, such a convolution is continuous. More generally, the convolution of two square-integrable functions $f,g\in L^2(\mathbb R)$ is continuous on $\mathbb R$.

There exist sequences $f_n\to f$ and $g_n \to g$ converging in $L^2$, with $f_n, g_n$ continuous and compactly supported. Clearly, $f_n*g_n$ is continuous. Also, the difference $f*g-f_n*g_n$ uniformly tends to zero, because $$\sup| f*g-f_n *g| = \sup| (f-f_n)*g| \le \|f_n-f\|_{L^2} \|g\|_{L^2}$$ and $$\sup| f_n*g-f_n *g_n| =\sup|f_n*(g-g_n)| \le \|f_n\|_{L^2} \|g-g_n\|_{L^2}$$ (In both cases, apply the Cauchy-Schwarz inequality to the integral that defines the convolution.)

Being a uniform limit of continuous functions, $f*g$ is continuous.