- Show that if \(f\in L^1\) and \(g\) is bounded, then \(f\ast g\) is bounded and uniformly continuous.
Solution
Let \(f \in L^1(\RR)\) and let \(g\) be measurable with \(|g| \leq M\) a.e., where \(M > 0\).
1. \(f\ast g\) is bounded, with \(|f\ast g(x)| \leq M\norm{f}_1\) for every \(x\).
\(|f\ast g(x)| = |\int f(x-y)g(y)\,dy| \leq \int |f(x-y)|\,|g(y)|\,dy \leq M \int |f(x-y)|\,dy = M\norm{f}_1\).
2. For \(h \in \RR\) let \(\tau_h f(y) \coloneqq f(y-h)\). For every \(x, h \in \RR\), \(|f\ast g(x+h) - f\ast g(x)| \leq M\, \norm{\tau_h f - f}_1\).
\(f\ast g(x+h) = \int f(x+h-y)g(y)\,dy = \int f(x - (y-h))g(y)\,dy\), so \(f\ast g(x+h) - f\ast g(x) = \int \big(f(x+h-y) - f(x-y)\big)g(y)\,dy\), and \(|f\ast g(x+h) - f\ast g(x)| \leq M \int |f(x+h-y) - f(x-y)|\,dy = M\norm{\tau_h f - f}_1\), the equality by the change of variables \(u = x - y + h\).
3. \(\lim_{h \to 0} \norm{\tau_h f - f}_1 = 0\).
For a compactly supported continuous \(\varphi\) the claim holds by uniform continuity of \(\varphi\) and dominated convergence on a fixed compact set. Given \(\eps > 0\), choose such \(\varphi\) with \(\norm{f - \varphi}_1 < \eps/3\), which exists by density of \(C_c(\RR)\) in \(L^1(\RR)\). Then \(\norm{\tau_h f - f}_1 \leq \norm{\tau_h(f - \varphi)}_1 + \norm{\tau_h\varphi - \varphi}_1 + \norm{\varphi - f}_1 < \eps\) for \(|h|\) small, since \(\norm{\tau_h(f-\varphi)}_1 = \norm{f-\varphi}_1\).
4. \(f\ast g\) is uniformly continuous.
Given \(\eps > 0\), step 3 gives \(\delta > 0\) with \(\norm{\tau_h f - f}_1 < \eps/M\) for \(|h| < \delta\). Then step 2 gives \(|f\ast g(x+h) - f\ast g(x)| < \eps\) for all \(x \in \RR\) and \(|h| < \delta\).
5. Q.E.D.
Step 1 gives boundedness and step 4 gives uniform continuity.