Differentiating twice under the integral (with assumptions)

[DavideGenoa]

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$^1$ $$\partial_{x_j}\partial_{x_i}\int_{\mathbb{R}^n}k(x,y)d\mu_y=\int_{\mathbb{R}^n}\partial_{x_j}\partial_{x_i}k(x,y)d\mu_y.$$

I see, by using a known corollary to the dominated convergence theorem (quoted here), 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, although I suspect it may have to do with the rather strong assumption that $\partial_{x_j}\partial_{x_i} k(\cdot,y)\in C(\mathcal{O})$...

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 $\infty$-ly thank any answerer!$^1$When I read a text I always assume that it is prefectly worded. Nevertheless, after many fruitless trials, I am beginning to suspect that this result is intended to hold provided that the condition (b) here holds.

$^2$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 $\int_{[a,t]}\int_{\mathbb{R}^n}\partial_{x_j}\partial_{x_i}k(x_t,y) d\mu_y d\mu_{x_j}$.

 

[jvicens]

I do not know how to conclude from this, since I do not know how to prove that this integral is $\int_{\mathbb{R}^n}\partial_{x_j}\partial_{x_i}k(x,y)d\mu_y$.
Thank you for your interesting question. I will try to address your concerns and provide a proof for the desired result.

First, let me clarify the notation in the statement of the problem. In the notation $\partial_{x_j}\partial_{x_i} k(\cdot,y)\in C(\mathcal{O})$, the symbol $\cdot$ refers to the first argument of the function. In other words, we are considering the function $x\mapsto \partial_{x_j}\partial_{x_i} k(x,y)$, which is continuous on the open set $\mathcal{O}$ for almost all $y\in\mathbb{R}^n$. This is a common notation in multivariable calculus and is known as partial differentiation.

Now, let us proceed with the proof. We will use the notation $f(y) = \int_{\mathbb{R}^n} k(x,y) d\mu_y$ for the function that we are interested in. By the assumptions given, we have that $f$ is well-defined and integrable on $\mathcal{O}$. We will also use the notation $g(y) = \partial_{x_j}\partial_{x_i} k(x,y)$ for the continuous function on $\mathcal{O}$.

We begin by using the dominated convergence theorem. Since $\partial_{x_i}k(x,\cdot)\in L^1(\mathbb{R}^n)$ for all $x\in\mathcal{O}$, we can choose a sequence of functions $\{g_n\}$ in $L^1(\mathbb{R}^n)$ such that $g_n(y) \to g(y)$ for almost all $y\in\mathbb{R}^n$. Then, by the dominated convergence theorem, we have
$$
\frac{d}{dx_j} \int_{\mathbb{R}^n} \partial_{x_i} k(x,y) d\mu_y = \lim_{n