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I Differentiating twice under the integral (with assumptions)

  1. Mar 16, 2016 #1
    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}##.
     
  2. jcsd
  3. Mar 21, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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