# A differentiation under the integral sign.

I know that

if $f\colon \mathbb{R}^n \to \mathbb{R}$ is integrable on any measurable, according to the usual $n$-dimensional Lebesgue measure $\mu_y$, and bounded subset of $\mathbb{R}^n$, and if $g \in C^k(\mathbb{R}^n)$ is compactly supported, then the function$$h:x\mapsto \int_{\mathbb{R}^n} f(x-y)g(y)\,d\mu_y$$belongs to $C^k(\mathbb{R}^n)$, and its partial derivatives of order $\leqslant k$ are given by

$$D^{\alpha} h(x) = \int_{\mathbb{R}^n} f(x-y)D^{\alpha} g(y)\,d\mu_y$$

whose proof, whose author I thank again, is found here.

I was wondering whether this result can extend to any $g \in C^k(\mathbb{R}^n)\cap L^{\infty}(\mathbb{R}^n)$ with $f\in L^1(\mathbb{R}^n)$, but I cannot adapt the linked proof because $\chi_{L-x_0}$ can no longer be used.

## 1 Answer

Suppose that $f\in L^1$ and that $g$ is continuously differentiable with $g$ and all first order derivatives $g_1, g_2, g_3,\cdots g_n$ in $L^{\infty}$. Then $$h_j(x) = \int f(x-y)g_j(y)dy = \int f(y)g_j(x-y)dy$$ are continuous functions of $x$.

Fubini's Theorem allows you to interchange orders of integration: \begin{align} \int_{x_{j,a}}^{x_{j,b}}\left(\int f(x-y)g_j(y)dy\right)dx_j & = \int_{x_j,a}^{x_j,b}\left(\int f(y)g_j(x-y)dy\right)dx_j \\ & = \int f(y)\left(\int_{x_{j,a}}^{x_{j,b}}g_j(x-y)dx_j\right)dy \\ & = \left.\int f(y)g(x-y)\right|_{x_j=x_{j,a}}^{x_j=x_{j,b}}dy \\ & = \left.\int f(y)g(x-y)dy\right|_{x_j=x_{j,a}}^{x_j=x_{j,b}} \\ & = \left.\int f(x-y)g(y)dy\right|_{x_j=x_{j,a}}^{x_j=x_{j,b}} \end{align} The integrand of the first integral in $x_j$ is continuous in $x_j$. Therefore, by the Fundamental Theorem of Calculus, the first integral is differentiable in $x_{j,b}$ and the derivative with respect to $x_{j,b}$ is equal to the integrand. That means that the last integral on the right must also be differentiable with respect to $x_{j,b}$, and one has the equality $$\frac{\partial}{\partial x_j}\int f(x-y)g(y)dy = \int f(x-y)g_j(y)dy.$$ The expression on the right is continuous in $y$. Therefore, $\int f(x-y)g(y)dy$ is continuously (jointly) differentiable with derivative $$D \int f(x-y)g(y)dy = \int f(x-y)(Dg)(y)dy.$$ This is easily extended to higher order derivatives by induction.

• Very interesting! Thank you so much! A thing isn't clear to me: how to se that $h,h_j\in C(\mathbb{R}^n)$? – Self-teaching worker Mar 10 '16 at 20:45
• @Self-teachingworker : I changed the assumptions. Assume that $g, g_j$ are in $L^{\infty}$. – DisintegratingByParts Mar 10 '16 at 22:25
• So, for $g_j$, $j=0,...,n$ ($g_0:=g$) I see that $|h_j(x)-h_j(x_0)|$ $\le \sup_{\mathbb{R}^n}|g|\int_{\mathbb{R}^3}|f(x-y)-f(x_0-y)|d\mu_y$ but how to see that the integral approaches $0$ as $x\to x_0$? I heartily thank you. (I also understand that $f\in L^1(\mathbb{R}^n)$, right? not only locally summable, so that it cannot be used to prove such a derivation under the integral for $f(z)=\|z\|^{-1}$ of the Helmholtz theorem; anyhow, interesting result per se: thanks a lot!) – Self-teaching worker Mar 11 '16 at 11:49
• @Self-teachingworker : The function $g_j$ is continuous and, therefore, $\lim_{x\rightarrow x_0}g(x-y)=g(x_0-y)$ pointwise everywhere. So dominated convergence gives you what you want. – DisintegratingByParts Mar 12 '16 at 3:29
• Oh, thank you: you use $(\sup_{\mathbb{R}^n}g) f$ as the $g$ and $y\mapsto fg(x_n-y)$, with $x_n\to x$, as the $f_n$ of this, supposing that $f\in L^1(\mathbb{R}^n)$! – Self-teaching worker Mar 12 '16 at 9:53