Prove $f(x) = \int_{\Bbb {R}} \chi_E (y) \chi_E (y-x)dy$ is continuous, where $E$ is a subset of $\Bbb R$ with finite Lebesgue measure.

Question

Prove $f(x) = \int_{\Bbb {R}} \chi_E (y) \chi_E (y-x)dy$ is continuous, where $E$ is a subset of $\Bbb R$ with finite Lebesgue measure.

This looks like the convolution equation, so I consider $f$ as a linear functional of a translation of indicator function. Since the indicator function here is bounded, the linear functional is continuous. Should I prove the translation of indicator function is continuous? How?

Answer 1

Let $g=\chi_E$. We have $0\le g \le 1$ and $g$ is integrable. Thus, for any $\epsilon > 0$ there exists a continuous function $g_\epsilon$ with $0\le g_\epsilon \le 1$ and $\| g - g_\epsilon \|_1 < \epsilon$. Let $f_\epsilon = g_\epsilon * g_\epsilon$ (convolution).

Notice that $f_\epsilon$ is continuous and for each $x\in\mathbb R$ we have \begin{align} |f_\epsilon(x) - f(x)| &\le \int |g_\epsilon(y) - g(y)||g_\epsilon(x-y)| \; dy \\ &+ \int |g(y)||g_\epsilon(x-y) - g(x-y)| \; dy < 2 \epsilon. \end{align} Now, let $x_n\in\mathbb R$ converge to $x \in \mathbb R$. Then, for each $\epsilon > 0$ we have \begin{align} |f(x_n) - f(x)| &\le |f_\epsilon(x_n) - f_\epsilon(x)| \\ &+ |f(x_n) - f_\epsilon(x_n)| + |f(x) - f_\epsilon(x)| \\ &< |f_\epsilon(x_n) - f_\epsilon(x)| + 4\epsilon \to 4\epsilon. \end{align} Thus, $f(x_n)\to f(x)$ and $f$ is continuous.