- Show that if \(f, g\) are continuous and compactly supported, then so is \(f\ast g\).
Solution
1. \(f \ast g\) is uniformly continuous.
1.1. \(|f \ast g(x + h) - f \ast g(x)| \le \|g\|_1 \sup_{z}|f(z + h) - f(z)|\) for all \(x, h\).
\(|f\ast g(x+h) - f\ast g(x)| = \left|\int (f(x + h - y) - f(x - y))g(y)\,dy\right| \le \int |f(x+h-y) - f(x-y)|\,|g(y)|\,dy\), and the first factor of the integrand is at most \(\sup_z|f(z+h) - f(z)|\).
1.2. \(\sup_z|f(z + h) - f(z)| \to 0\) as \(h \to 0\).
A continuous function with compact support is uniformly continuous.
1.3. Q.E.D.
Steps 1 and 2 give \(\sup_x|f\ast g(x + h) - f\ast g(x)| \to 0\) as \(h \to 0\).
2. \(f \ast g\) is compactly supported.
2.1. \(\supp f + \supp g\) is compact.
It is the image of the compact set \(\supp f \times \supp g\) under the continuous map \((a,b) \mapsto a + b\).
2.2. Q.E.D.
If \(x \notin \supp f + \supp g\), then \((x - \supp g) \cap \supp f = \emptyset\), so \(f(x - y)g(y) = 0\) for every \(y\) and \(f\ast g(x) = 0\). So \(\theset{f\ast g \neq 0}\) lies in the closed set \(\supp f + \supp g\), and so does its closure \(\supp(f\ast g)\), which is compact by step 1.
3. Q.E.D.
Steps 1 and 2.