\(f\in L^1\) 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
Let \(f \in L^1(\RR^n)\), let \(g\) be differentiable with \(\dd{g}{x_i}\) bounded, and let \(e_i\) be the \(i\)th standard basis vector.
1. For \(h \neq 0\), \(\frac{(f\ast g)(x + h e_i) - (f\ast g)(x)}{h} = \int f(x-y)\, \frac{g(y + h e_i) - g(y)}{h}\,dy\).
\((f\ast g)(x+he_i) = \int f(x+he_i-y)g(y)\,dy\); the change of variables \(y \mapsto y + h e_i\) turns this into \(\int f(x-y)g(y+he_i)\,dy\). Subtract \((f\ast g)(x) = \int f(x-y)g(y)\,dy\) and divide by \(h\).
2. As \(h \to 0\), the integrands in step 1 converge pointwise to \(f(x-y)\,\dd{g}{y_i}(y)\) and are bounded by \(|f(x-y)| \sup_z |\dd{g}{z_i}(z)|\), which is integrable in \(y\).
Pointwise convergence is differentiability of \(g\) in the direction \(e_i\). The bound on the difference quotient is the mean value theorem applied to \(t \mapsto g(y + te_i)\). The dominating function is integrable because \(f \in L^1\).
3. \(\dd{}{x_i}(f\ast g) = f \ast \dd{g}{x_i}\).
By step 2 the dominated convergence theorem lets \(h \to 0\) pass under the integral in step 1.
4. If \(g\) is smooth and compactly supported, then \(f \ast g\) is smooth.
Every partial derivative \(D^\alpha g\) of a smooth compactly supported \(g\) is bounded, so step 3 applied repeatedly gives \(D^\alpha(f\ast g) = f \ast D^\alpha g\) for every multi-index \(\alpha\). Each \(f \ast D^\alpha g\) is the convolution of an \(L^1\) function with a bounded function, hence continuous, so \(f \ast g\) has continuous partial derivatives of all orders.
5. Q.E.D.
Step 3 is the differentiation formula, and step 4 is smoothness of \(f\ast g\) for smooth compactly supported \(g\).