Derivative of a convolution

Remarks

Write $(f * g)(x) = \int f(y) g(x - y) \, dy$ and differentiate under the integral sign. Let $x$ range over a bounded set and let $0 < \abs{t} \le 1$. By the mean value theorem, the difference quotients $(g(x + t e_j - y) - g(x - y))/t$ are bounded by $\sup \abs{\partial_j g}$. They vanish for $y$ outside a fixed compact set $K$. Since $f$ is integrable on $K$, the dominated convergence theorem lets the limit pass inside the integral. Continuity of $f * \partial_j g$ follows in the same way from the uniform continuity of $\partial_j g$.

Example: on $\R$, take $f = \mathbf{1}_{[0, \infty)}$ and $g$ a continuously differentiable function with compact support. Then $(f * g)(x) = \int_{-\infty}^x g(u) \, du$, and the formula gives $(f * g)' = f * g' = g$, as the fundamental theorem of calculus predicts.

The derivative falls on the smooth factor; $f$ itself need not be differentiable or even continuous. Without compact support (or some decay) of $g$ the integral need not converge: for $f = 1$ and $g(x) = 1 + x^2$ the convolution is defined nowhere.

Going deeper

For a student

This is the basic step behind smoothing by convolution: by induction, $f * g$ is as differentiable as $g$. It also explains why convolution commutes with constant-coefficient differential operators $P(\partial)$: $P(\partial)(f * g) = f * P(\partial) g$. In the theory of distributions the same formula holds with $\partial(f * g) = (\partial f) * g = f * (\partial g)$ for any distributions $f, g$ one of which has compact support.