# Extending the legitimacy of a type of differentiation under the integral to a larger class of functions

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

Since many result concerning compactly supported functions also extend to rapidly decreasing functions, I was wondering whether this result extends to any $g \in C^k(\mathbb{R}^n)$ such that$$\forall\alpha,\beta\in\mathbb{N}\quad\exists C>0:\forall x\in\mathbb{R}^n\quad|x^\beta D^\alpha g(x)|<C$$where I use the usual multi-index notation if $f$ is Lebesgue integrable on any bounded measurable subset, but I cannot prove it. Is it true and, if it is, how can we prove it? I have tried to adapt the linked proof, but the $\chi_{L-x_0}$ characteristic function cannot be used here. I heartily thank any answerer.

You'll want $f$ to be bounded by a polynomial, I think. Then you can use Lebesgue Dominated Convergence Theorem on the limit of difference quotients
$$\dfrac{h(x+tv) - h(x)}{t} = \int_{{\mathbb R}^n} f(x - y) \dfrac{g(y+tv)-g(y)}{t}\; d\mu_y$$
For a counterexample where $f$ is not bounded by a polynomial, try $f(x) = \exp(x^2)$ and $g(x) = \exp(-x^2)$.
• I think I know the corollary to the Lebesgue dominated convergence theorem you refer to and I think we should find a $\varphi\in L^1(\mathbb{R}^n)$ such that for almost all $y\in\mathbb{R}^n$ and for $t$ small enough $\left|f(x-y)\frac{g(y-tv)-g(y)}{t}\right|\le\varphi(y)$. It will be sufficient to find a $\varphi$ such that $2\max_{\mathbb{R}^n}|g|$ $\cdot|t^{-1}f(x-y)|\le\varphi(y)$ and I think that the polynomial bound is a key too that, but I've got some problem to find it... I heartily thank you again! – Self-teaching worker Mar 18 '16 at 11:07