4
$\begingroup$

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...

$\endgroup$
  • 2
    $\begingroup$ 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? $\endgroup$ – Daniel Fischer Feb 28 '16 at 10:39
  • 2
    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Feb 28 '16 at 11:19
2
$\begingroup$

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 $h \in C^k(\mathbb{R}^n)$.

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – Self-teaching worker Feb 29 '16 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! $\endgroup$ – Self-teaching worker Aug 1 '17 at 12:19
  • 1
    $\begingroup$ 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). $\endgroup$ – Daniel Fischer Aug 1 '17 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 $g\in C_c^k(\mathbb{R}^{n+1})$) ... $\endgroup$ – Self-teaching worker Aug 3 '17 at 20:11
  • 1
    $\begingroup$ Yes. Assuming of course that $f$ is locally integrable. $\endgroup$ – Daniel Fischer Aug 3 '17 at 20:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.