Let \(E, F \subset \mathbb{R}\) be two Lebesgue-measurable subsets of \(\mathbb{R}\), each of finite measure, and let \(\chi_E\) and \(\chi_F\) denote their respective characteristic functions.
- Prove that the convolution \(\chi_E * \chi_F\) defined by
\[\chi_E * \chi_F(x) = \int_{\mathbb{R}} \chi_E(y) \chi_F(x - y) \, dy\]
is a continuous function of \(x\).
- Show that as \(n \to \infty\)
\[n \big(\chi_E * \chi_{[0, 1/n]}\big) \longrightarrow \chi_E\]
pointwise almost everywhere.
Solution
1. The convolution \(\chi_E*\chi_F\) is uniformly continuous.
Because \(m(E),m(F)<\infty\), both characteristic functions lie in \(L^2(\mathbb R)\), with \[ \|\chi_E\|_2=m(E)^{1/2}, \qquad \|\chi_F\|_2=m(F)^{1/2}. \] Cauchy–Schwarz shows that the convolution integral exists at every \(x\), with absolute value at most \(m(E)^{1/2}m(F)^{1/2}\) [@Fol13]. Let \[ h(x)=(\chi_E*\chi_F)(x). \] For \(t,x\in\mathbb R\), \[ h(x+t)-h(x) =\int_{\mathbb R}\chi_E(y) \bigl(\chi_F(x+t-y)-\chi_F(x-y)\bigr)\,dy. \] By Cauchy–Schwarz, \[ |h(x+t)-h(x)| \le \|\chi_E\|_2 \left(\int_{\mathbb R} |\chi_F(x+t-y)-\chi_F(x-y)|^2\,dy\right)^{1/2}. \] After the change of variables \(u=x-y\), the second factor is \[ \|\tau_t\chi_F-\chi_F\|_2, \] where \((\tau_t\phi)(u)=\phi(u+t)\). Therefore \[ \sup_{x\in\mathbb R}|h(x+t)-h(x)| \le \|\chi_E\|_2\,\|\tau_t\chi_F-\chi_F\|_2. \] Translations are continuous in \(L^2(\mathbb R)\) [@Fol13], so the right-hand side tends to \(0\) as \(t\to0\). Hence \(h\) is uniformly continuous, and in particular continuous.
2. Rewrite the second convolution as a one-sided average.
For every \(x\in\mathbb R\), \[ (\chi_E*\chi_{[0,1/n]})(x) =\int_{\mathbb R}\chi_E(y)\chi_{[0,1/n]}(x-y)\,dy. \] The second characteristic function is nonzero exactly when \[ x-\frac1n\le y\le x. \] Thus \[ n(\chi_E*\chi_{[0,1/n]})(x) =n\int_{x-1/n}^{x}\chi_E(y)\,dy =\frac{1}{1/n}\int_{x-1/n}^{x}\chi_E(y)\,dy. \]
3. Apply the Lebesgue differentiation theorem.
Since \(\chi_E\in L^1_{\mathrm{loc}}(\mathbb R)\), almost every \(x\) is a Lebesgue point [@Fol13]. At such a point, \[ \left|\frac1r\int_{x-r}^{x}\chi_E(y)\,dy-\chi_E(x)\right| \leq 2\frac1{2r}\int_{x-r}^{x+r}|\chi_E(y)-\chi_E(x)|\,dy \longrightarrow0. \] Thus the one-sided averages satisfy, for almost every \(x\), \[ \lim_{r\downarrow0}\frac1r\int_{x-r}^{x}\chi_E(y)\,dy =\chi_E(x). \] Taking \(r=1/n\) yields \[ n(\chi_E*\chi_{[0,1/n]})(x) \longrightarrow \chi_E(x) \] for almost every \(x\in\mathbb R\).