Show that if \(f, g\) are continuous and compactly supported, then so is \(f\ast g\).
Show that if \(f\in L^1\) and \(g\) is bounded, then \(f\ast g\) is bounded and uniformly continuous.
If \(f, g\) are compactly supported, is it necessarily the case that \(f\ast g\) is compactly supported?
Show that under any of the following assumptions, \(f\ast g\) vanishes at infinity:
\(f, g\in L^1\) are both bounded.
\(f, g\in L^1\) with just \(g\) bounded.
\(f, g\) smooth and compactly supported (and in fact \(f\ast g\) is smooth)
\(f\in L^1\) and \(g\) smooth and compactly supported (and in fact \(f\ast g\) is smooth)
Show that if \(f\in L^1\) and \(g'\) exists with \(\dd{g}{x_i}\) all bounded, then \[\dd{}{x_i}(f\ast g) = f \ast \dd{g}{x_i}\]
Solution
1. If \(f, g\) are continuous and compactly supported, then \(f \ast g\) is continuous and compactly supported.
\(|f\ast g(x + h) - f\ast g(x)| \le \|g\|_1 \sup_z|f(z + h) - f(z)|\), which tends to \(0\) because \(f\) is uniformly continuous. \(f\ast g\) vanishes off the compact set \(\supp f + \supp g\), the image of \(\supp f \times \supp g\) under addition. See E-Y4GZM.
2. If \(f \in L^1\) and \(|g| \leq M\), then \(|f \ast g| \le M\|f\|_1\) and \(f\ast g\) is uniformly continuous.
\(|f\ast g(x)| \le \int |f(x-y)||g(y)|\,dy \le M\|f\|_1\). Substituting \(u = x - y\) gives \(|f \ast g(x + h) - f \ast g(x)| \le M\int|f(u + h) - f(u)|\,du\), which does not depend on \(x\) and tends to \(0\) as \(h \to 0\) by continuity of translation in \(L^1\). See E-E4H2J.
3. If \(f, g\) are compactly supported, then \(f \ast g\) is compactly supported.
The argument of step 1 uses no continuity: \(f\ast g(x) = 0\) for \(x \notin \supp f + \supp g\), which is compact.
4. \(f \ast g\) vanishes at infinity under each of the four listed assumptions.
4.1. The cases \(f, g \in L^1\) both bounded, \(f, g \in L^1\) with \(g\) bounded, and \(f, g\) smooth and compactly supported.
3. See E-LYXHE.
3.2. If \(f \in L^1\) and \(g\) is smooth with compact support, then \(f\ast g\) is smooth and vanishes at infinity.
2. Choose \(R\) with \(\supp g \subseteq \theset{|y| \le R}\). For \(|x| \ge 2R\) and \(|y| \le R\), \(|x - y| \ge |x|/2\), so\[ |f\ast g(x)| \le \|g\|_\infty\int_{|y| \le R}|f(x-y)|\,dy \le \|g\|_\infty\int_{|u| \ge |x|/2}|f(u)|\,du, \]which tends to \(0\) as \(|x| \to \infty\) because \(f \in L^1\).
2.3. Q.E.D.
Steps 1 and 2.
5. If \(f \in L^1\) and \(g\) is differentiable with \(\dd{g}{x_i}\) bounded, then \(\dd{}{x_i}(f \ast g) = f \ast \dd{g}{x_i}\).
The difference quotient of \(f\ast g\) in the direction \(e_i\) is \(\int f(x-y)\,\frac{g(y + h e_i) - g(y)}{h}\,dy\). By the mean value theorem the quotient of \(g\) is bounded by \(\sup|\dd{g}{x_i}|\), and it converges pointwise to \(\dd{g}{x_i}\), so dominated convergence with dominating function \(\sup|\dd{g}{x_i}|\,|f(x-\cdot)|\) applies. See E-GQEMZ.
6. Q.E.D.
Steps <1>1–<1>5 treat the parts in order.