A double differentiation under the integral sign

[DavideGenoa]

Hello, friends!
Let $k:\mathcal{O}\times\mathbb{R}^n\to\mathbb{R}$, with $\mathcal{O}\subset\mathbb{R}^m$ open, be such that $\forall x\in\mathcal{O}\quad k(x,\cdot)\in L^1(\mathbb{R}^n) $, i.e. the function $y\mapsto k(x,y)$ is Lebesgue summable on $\mathbb{R}^n$, according to the usual $n$-dimensional Lebesgue measure.

I read (theorem 1.d here, p. 2) that

if $\partial_{x_j}\partial_{x_i} k(\cdot,y)\in C(\mathcal{O})$ for almost all $y\in\mathbb{R}^n$ and there exists $ g\in L^1(\mathbb{R}^n)$ such that $$|\partial_{x_j}\partial_{x_i} k(x,y)|\le g(y)\quad\forall x\in\mathcal{O}\quad\text{for almost all }y\in\mathbb{R}^n,$$then$^1$ $$\partial_{x_j}\partial_{x_i}\int_{\mathbb{R}^n}k(x,y)\,d\mu_y=\int_{\mathbb{R}^n}\partial_{x_j}\partial_{x_i}k(x,y)\,d\mu_y$$where $x_i,x_j$ are components of $x$.

A corollary to the dominated convergence theorem says 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 $V$,</sub>
- 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
$$\frac{d}{dt}\int_V f(\boldsymbol{x},t) d\mu_{\boldsymbol{x}}\bigg|_{t=t_0}=\int_V\frac{\partial f(\boldsymbol{x},t_0)}{\partial t}d\mu_{\boldsymbol{x}}$$

and therefore I see that, provided that $\forall x\in\mathcal{O}\quad \partial_{x_i}k(x,\cdot)\in L^1(\mathbb{R}^n) $ (which I do not know how to prove$^2$), then$$\partial_{x_j}\int_{\mathbb{R}^n}\partial_{x_i}k(x,y)\,d\mu_y=\int_{\mathbb{R}^n}\partial_{x_j}\partial_{x_i}k(x,y)\,d\mu_y$$but then I do not know how we can "move" $\partial_{x_i}$ outside the integral.

How can we prove that $\partial_{x_j}\partial_{x_i}\int_{\mathbb{R}^n}k(x,y)\,d\mu_y=\int_{\mathbb{R}^n}\partial_{x_j}\partial_{x_i}k(x,y)\,d\mu_y$? I thank any answerer very much.

Notes:
$^1$When I read a text I always assume that it is prefectly worded. Nevertheless, after many fruitless trials, I am not excluding that this result is intended to hold provided that the condition (b) here holds. I would be very grateful to any answerer confirming or denying that.
$^2$By using Fubini's theorem I only see that, once chosen an arbitrary interval $[a,t]$, the function $\partial_{x_i}k(x_t,\cdot)-\partial_{x_i}k(x_a,\cdot)$, where $x_t$ has $t$ as its $j$-th component, is summable and its integral is $\int_{[a,t]}\int_{\mathbb{R}^n}\partial_{x_j}\partial_{x_i}k(x,y) \,d\mu_y \,d\mu_{x_j}$.

 

[jvicens]

Hello,

Thank you for sharing your findings on this topic. It seems that you have done a thorough investigation and have come across some interesting results. I am not an expert in this specific field, but I would suggest reaching out to other researchers or consulting with a mentor or colleague for further guidance on this problem. It may also be helpful to review other literature on this topic to see if other researchers have addressed this issue or have proposed alternative solutions. I wish you the best of luck in your research.