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..

  • $\begingroup$ I don't think so. Consider the convolution of $I_{\mathbb{Q}\cap [0,1]}$ with itself. I have a hard time believing that won't have countably infinitely many discontinuities (but then again, I haven't done the work, either) $\endgroup$
    – nomen
    Commented Dec 6, 2013 at 5:02
  • $\begingroup$ @nomen The convolution of $I_{\mathbb{Q}\cap [0,1]}$ with itself is identically zero, hence continuous. $\endgroup$ Commented Jan 29, 2014 at 5:04

1 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.


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