# $\partial_{x_j}\partial_{x_i}\int_{\mathbb{R}^n}k(x,y)d\mu_y=\int_{\mathbb{R}^n}\partial_{x_j}\partial_{x_i}k(x,y)d\mu_y$ under some assumptions

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 $x$.

I see, by using the corollary to the dominated convergence theorem quoted in this question, 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 $\int_{[a,t]}\int_{\mathbb{R}^n}\partial_{x_j}\partial_{x_i}k(x,y) \,d\mu_y \,d\mu_{x_j}$.

• You may prove the statement by applying your corollary twice. In regards to your second footnote. The answer is no. In general, if you want derivatives to live in your $L^p$ space, you have to define them to exist. These are called Sobolev Spaces, usually denoted $W^{r,p}$ which means it's $L^p$ where all functions have $r$ derivatives that live in $L^p$ as well. – Jeb Mar 17 '16 at 15:02
• @Jeb Thank you very much for your comment! In order to apply that corollary twice, I should know that $\exists g\in L^1(\mathbb{R}^n)$ : for almost all $y\in\mathbb{R}^n$ and $\forall x\in\mathcal{O}\quad$ $|\partial_{x_i} k(x,y)|\le g(y)$, which I don't know how to prove... (my note 2 meant: can the fact that $k(x,\cdot)\in L^1(\mathbb{R}^n)$ together with the stated assumptions on $\partial_{x_j}\partial_{x_i} k(\cdot,y)$ be used to prove that?) – Self-teaching worker Mar 17 '16 at 15:19

## 1 Answer

I have been able to contact the author of the paper, who has told me that the assumption $$\exists G\in L^1(\mathbb{R}^n):|\partial_{x_i}k(x,y)|\le G(y),\quad\forall x\in\mathcal{O},\quad\text{for almost every }y\in\mathbb{R}^n$$made for (b) is implicitly made.