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)
Solution
1. If \(f\) and \(g\) are compactly supported, then \(f \ast g\) is compactly supported.
1.1. \(\supp(f \ast g) \subseteq \supp f + \supp g\).
If \(x \notin \supp f + \supp g\), then \((x - \supp g) \cap \supp f = \emptyset\), so \(f(x-y) = 0\) for every \(y \in \supp g\) and \(f \ast g(x) = \int f(x-y)g(y)\,dy = 0\). Since \(\supp f + \supp g\) is closed by step 2, the support of \(f\ast g\), the closure of \(\theset{f\ast g \neq 0}\), is contained in it.
1.2. \(\supp f + \supp g\) is compact.
It is the image of the compact set \(\supp f \times \supp g\) under the continuous map \((a,b) \mapsto a+b\).
1.3. Q.E.D.
A closed subset of a compact set is compact; apply steps 1 and 2.
2. If \(f, g \in L^1\) are both bounded, then \(f \ast g\) vanishes at infinity.
2.1. \(\abs{\int_{|y| \le |x|/2} f(x-y)g(y)\,dy} \leq \|g\|_\infty \int_{|z| \ge |x|/2} |f(z)|\,dz\), which tends to \(0\) as \(|x| \to \infty\).
On \(\theset{|y| \le |x|/2}\), \(|x-y| \ge |x|/2\). Substitute \(z = x - y\) and use \(|g| \le \|g\|_\infty\). The tail integral of \(|f| \in L^1\) tends to \(0\) by dominated convergence.
2.2. \(\abs{\int_{|y| > |x|/2} f(x-y)g(y)\,dy} \leq \|f\|_\infty \int_{|y| > |x|/2} |g(y)|\,dy\), which tends to \(0\) as \(|x| \to \infty\).
Use \(|f| \le \|f\|_\infty\); the tail integral of \(|g| \in L^1\) tends to \(0\).
2.3. Q.E.D.
\(f\ast g(x)\) is the sum of the two integrals in steps 1 and 2, and both tend to \(0\) as \(|x| \to \infty\).
3. If \(f, g \in L^1\) with only \(g\) bounded, then \(f \ast g\) vanishes at infinity.
3.1. For \(M > 0\) write \(f = f_M + r_M\) with \(f_M = f \chi_{\theset{|f| \le M}}\). Then \(|f_M| \le M\) and \(\|r_M\|_1 \to 0\) as \(M \to \infty\).
\(\int |r_M| = \int_{\theset{|f| > M}} |f| \to 0\) by dominated convergence, since \(|f| \in L^1\) and \(\chi_{\theset{|f| > M}} \to 0\) a.e.
3.2. \(\|r_M \ast g\|_\infty \le \|r_M\|_1 \|g\|_\infty\).
\(|r_M \ast g(x)| \le \int |r_M(x-y)||g(y)|\,dy \le \|g\|_\infty \|r_M\|_1\).
3.3. \(f_M \ast g\) vanishes at infinity.
\(f_M\) and \(g\) are bounded and in \(L^1\), so step 2 applies.
3.4. Q.E.D.
3.3. For \(|x| > R\), \(|f \ast g(x)| \le |f_M \ast g(x)| + |r_M \ast g(x)| < \eps\).
4. If \(f, g\) are smooth and compactly supported, then \(f \ast g\) is smooth and vanishes at infinity.
4.1. \(f \ast g\) is \(C^\infty\), with \(D^\alpha(f \ast g) = (D^\alpha f) \ast g\) for every multi-index \(\alpha\).
Each \(D^\alpha f\) is continuous and compactly supported, hence bounded, and \(g\) is integrable, so dominated convergence justifies differentiating under the integral in \(f\ast g(x) = \int f(x-y)g(y)\,dy\). Each \((D^\alpha f) \ast g\) is continuous.
4.2. \(f \ast g\) vanishes at infinity.
By step 1, \(f \ast g\) is zero outside a compact set.
4.3. Q.E.D.
Steps 1 and 2.
5. Q.E.D.
Step 1 answers the first question affirmatively; steps 2, 3, and 4 prove vanishing at infinity under the three listed assumptions.