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Let $f:\mathbb{R}^n\to\mathbb{R}$ be a function Lebesgue summable on all $\mu$-measurable and bounded subsets of $\mathbb{R}^n$, where $\mu$ is the usual Lebesgue measure defined on $\mathbb{R}^n$, and let $g:\mathbb{R}^n\to\mathbb{R}$ be a function of class $C^k(\mathbb{R}^n)$ whose support is contained in the compact subset $V\subset\mathbb{R}^n$. That implies that all the derivatives of $g$, up to the $k$-th, are bounded and supported within $V$.

I suspect that these conditions are enough to guarantee that the function $$h(\boldsymbol{x}):=\int_V f(\boldsymbol{y}-\boldsymbol{x})g(\boldsymbol{y})\,d\mu_{\boldsymbol{y}}$$is of class $C^k(\mathbb{R}^n)$. In fact, I notice, if I am not wrong, that $$h(\boldsymbol{x})=\int_{V-\boldsymbol{x}} f(\boldsymbol{y})\bar{g}(\boldsymbol{y}+\boldsymbol{x})\,d\mu_{\boldsymbol{y}}=\int_{\mathbb{R}^n} f(\boldsymbol{y})\bar{g}(\boldsymbol{y}+\boldsymbol{x})\,d\mu_{\boldsymbol{y}}$$where $$\bar{g}(\boldsymbol{y}) := \begin{cases} g(\boldsymbol{y}), & \boldsymbol{y}\in V \\ 0, & \boldsymbol{y}\in\mathbb{R}^n\setminus V \end{cases}$$and $V-\boldsymbol{x}=\{\boldsymbol{y}\in\mathbb{R}^n:\boldsymbol{y}+\boldsymbol{x}\in V\}$, and I suppose that the conditions on $g$ may be enough to allow us to differentiate under the integral sign.

Is my intuition that $h\in C^k(\mathbb{R}^n)$ correct and, if it is, how can we prove it?


A trial of mine: I know a corollary of Lebesgue's dominated convergence theorem that

- if $f:V\times [a,b]\to \mathbb{R}$, $(\boldsymbol{x},t)\mapsto f(\boldsymbol{x},t)$ with $V$ measurable is such that $\forall t\in[a,b]\quad f(-,t)\in L^1(V)$, i.e. the function $\boldsymbol{x}\mapsto f(\boldsymbol{x},t) $ is Lebesgue summable on $V$,

- and if there is a neighbourhood $B(t_0,\delta)$ of $t_0$ such that, for almost all $\boldsymbol{x}\in V$ and for all $t\in B(t_0,\delta)$, $\left|\frac{\partial f(\boldsymbol{x},t)}{\partial t}\right|\le\varphi(\boldsymbol{x})$, where $\varphi\in L^1(V) $, then $$\frac{d}{dt}\int_V f(\boldsymbol{x},t) d\mu_{\boldsymbol{x}}\bigg|_{t=t_0}=\int_V\frac{\partial f(\boldsymbol{x},t_0)}{\partial t}d\mu_{\boldsymbol{x}}$$but I am not able to find a proper $\varphi$ to use in this context. Nevertheless I would not be amazed if there were an even more straightforward method to prove that the integral and derivative sign can be commutated...

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    $\begingroup$ As written, the premises don't really make sense. First you write $f\colon \mathbb{R}^n \to \mathbb{R}$, then afterwards you say $f$ is defined (and integrable) on $V$. I suspect the intended premise is that $f$ is locally integrable, that is, the restriction of $f$ to every bounded Lebesgue-measurable $V\subset \mathbb{R}^n$ belongs to $L^1(V)$. Is that correct? $\endgroup$ Commented Feb 28, 2016 at 10:39
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    $\begingroup$ If you only assume that $f$ is summable on a specific $V$, then you can't even be sure that $h(x)$ is defined for any $x\neq 0$. To be sure that $h$ is defined on all of $\mathbb{R}^n$, you need local integrability. (That is, $f$ should be integrable over every bounded measurable subset of $\mathbb{R}^n$, or, equivalently one can say that $f$ should be integrable over every compact subset of $\mathbb{R}^n$.) If you assume that $f$ is locally integrable, everything works out nicely. $\endgroup$ Commented Feb 28, 2016 at 11:19

1 Answer 1

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Your intuition is correct, if $f\colon \mathbb{R}^n \to \mathbb{R}$ is locally integrable, and if $g \in C_c^k(\mathbb{R}^n)$, then the function $h$ given by

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

belongs to $C^k(\mathbb{R}^n)$, and its partial derivatives of order $\leqslant k$ are given by

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

The change of coordinates $z = y-x$ gives us

$$h(x) = \int_{\mathbb{R}^n} f(z)g(z+x)\,d\mu_z,\tag{$\ast$}$$

and in that form we can apply the dominated convergence theorem to justify differentiation under the integral.

We let $K := \operatorname{supp} g$, and define $L = \{x \in \mathbb{R}^n : \operatorname{dist}(x,K) \leqslant 1\}$. Then $L$ is also compact, hence of finite Lebesgue measure. Now we fix an arbitrary $x_0 \in \mathbb{R}$ and show that $h$ is continuously differentiable on $B_1(x_0)$. The integrand in $(\ast)$ is only nonzero for $z$ such that $z+x \in K$, and rearranging gives $z \in K + (x_0 - x) - x_0$. Since we only look at $x$ with $\lVert x-x_0\rVert < 1$, we have $K + (x_0 - x) \subseteq L$, so

$$f(z)g(z+x) \neq 0 \implies z \in L - x_0$$

for $x\in B_1(x_0)$. As a continuous function with compact support, $g$ is bounded, say $\lvert g(y)\rvert \leqslant M$ for all $y$. Then we have

$$\lvert f(z)g(z+x)\rvert \leqslant M\cdot \lvert f(z)\rvert \cdot \chi_{L - x_0}(z)$$

for all $z \in \mathbb{R}^n$, and choosing $d(z) := M\cdot \lvert f(z)\rvert \cdot \chi_{L - x_0}(z)$ as the dominating function yields the continuity of $h$ on $B_1(x_0)$. If $k \geqslant 1$, then the partial derivatives of $g$ are all continuous functions with compact support, and hence bounded. We may assume that $\lvert \partial_{i}g(y)\rvert \leqslant M$ for all $y\in \mathbb{R}^n$ and $1 \leqslant i \leqslant n$. Then $d$ is also a dominating function for

$$f(z)\partial_i g(z+x),$$

and another application of the dominated convergence theorem - repsectively of the mentioned corollary - yields the continuous partial differentiability of $h$ on $B_1(x_0)$, and the formula

$$ \frac{\partial h}{\partial x_i} (x) = \int_{\mathbb{R}^n} f(z) \partial_i g(z+x)\,d\mu_z.$$

If $k > 1$, we can iterate the argument to obtain the existence and continuity of the higher-order partial derivatives of $h$ on $B_1(x_0)$. Since $x_0$ was arbitrary, it follows that $h \in C^k(\mathbb{R}^n)$.

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  • $\begingroup$ Very clear and explicative answer, as yours always are. I also see, by substituting $f$ with $z\mapsto f(-z)$ that $$D_x^\alpha\int_{\mathbb{R}^n}f(x-y)g(y)d\mu_y=\int_{\mathbb{R}^n}f(x-y)D_{y}^{\alpha} g(y)d\mu_y$$also holds, which is a result that I think to be useful in the theory of convolution... Thank you so much! $\endgroup$ Commented Feb 29, 2016 at 10:47
  • $\begingroup$ Forgive me if I'm bothering you: does something analogous hold with $h(x):=\int_{\mathbb{R}^n}f(x-y)g(y,t-\alpha\|x-y\|)d\mu_y$ and opportune assumptions on $g$, for ex. $g\in C_c^k(\mathbb{R}^{n+1})$ and? I ask this question because I'm trying to prove to myself that the Laplacian of the electric retarded potential $V(x,t):=\frac{1}{4\pi\varepsilon_0}\int_{\mathbb{R}^n} \frac{g(y,t-c^{-1}\|x-y\|)}{\|x-y\|}d\mu_y$ satsfies the Lorenz gauge condition. I heartily thank you! $\endgroup$ Commented Aug 1, 2017 at 12:19
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    $\begingroup$ Substituting $z = x-y$, we get $$h(x) = \int_{\mathbb{R}^n} f(z) g(x-z, t - \alpha \lVert z\rVert)\,d\mu_z,$$ and under appropriate conditions, we can differentiate under the integral (and then substitute back). $\endgroup$ Commented Aug 1, 2017 at 12:29
  • $\begingroup$ Thank you very much! Well, I would say that (I use $D^\alpha$ for the derivatives w.r.t. $\tau$, too), if (1) $g\in C^k(\mathbb{R}^{n+1})$, $(\xi,\tau)\mapsto g(\xi,\tau)$; (2) there is a compact set $K\subset\mathbb{R}^n$ such that $\forall\tau\in\mathbb{R}$ the support of $\xi\mapsto D^\alpha g(\xi,\tau)$ is contained in $K$; (3) $\exists M>0:\forall(\xi,\tau)\in\mathbb{R}^{n+1}\quad|D^\alpha g(\xi,\tau)|\le M$ (these three conditions are met if $g\in C_c^k(\mathbb{R}^{n+1})$) ... $\endgroup$ Commented Aug 3, 2017 at 20:11
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    $\begingroup$ Yes. Assuming of course that $f$ is locally integrable. $\endgroup$ Commented Aug 3, 2017 at 20:17

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