Distribution theory problem I need some help with this problem related with distributions:
With $\cal{D}(\Omega)$ we denote de set of the functions of class $C^{\infty}$ in $\Omega$ and compact support.
Let N=3. We consider $\Phi\in \cal{D}(\mathbb{R}^3)$, with $\Phi(0,0,0)=0$. Let $\psi,\phi,$ and $\varphi$ in $\cal{D}(\mathbb{R})$, with $\int_{\mathbb{R}}\psi=\int_{\mathbb{R}}\phi=\int_{\mathbb{R}}\varphi=1$. Show that:
$$\lim_{n\to\infty}\displaystyle\int_{\mathbb{R}^3}(n^3\phi(nx)\psi(ny)\varphi(nz)\Phi(-x,-y,-z)-\Phi(x,y,z))\;dx\,dy\,dz=0$$
Thanks a lot for any help.
Edited. The correct statement is: $$\lim_{n\to\infty}\displaystyle\int_{\mathbb{R}^3}n^3\phi(nx)\psi(ny)\varphi(nz)(\Phi(-x,-y,-z)-\Phi(x,y,z))\;dx\,dy\,dz=0$$
 A: Hint: define $x':=nx$, $y':=ny$ and $z':=nz$, and use a dominated convergence argument. 
A: Let's see if I am correct:
First, with the change of variable you indicated (which has jacobian $\frac{1}{n^3}$), we get that our expression becomes: $\int_{R^3} \phi(x')\psi(y')\varphi(z')(\Phi(-x'/n,-y'/n,-z'/n)-\Phi(x'/n,y'/n,z'/n) )dx'dy'dz'$. 
We want to change limit with integral. So have to dominate our integrand by an integrable function, in order to apply dominated convergence theorem. The function $|\phi\psi\varphi|2*sup|\Phi|$ dominates the integrand, and it's integrable, because $\int_{\mathbb{R}^3}|\phi\psi\varphi|2*sup|\Phi|=2*sup|\Phi|\int_{\mathbb{R}}\phi\int_{\mathbb{R}}\psi\int_{\mathbb{R}}\varphi=2*sup|\Phi|<\infty$, (we could separate the integrals because each one of the three functions depends only on one variable).
Then, $\lim_{n\to\infty} \int_{\mathbb{R}^3} \phi(x')\psi(y')\varphi(z')(\Phi(-x'/n,-y'/n,-z'/n)-\Phi(x'/n,y'/n,z'/n) )dx'dy'dz'=$
$=\int_{\mathbb{R}^3}\lim_{n\to\infty} \phi(x')\psi(y')\varphi(z')(\Phi(-x'/n,-y'/n,-z'/n)-\Phi(x'/n,y'/n,z'/n) )dx'dy'dz'=$
$=\int_{\mathbb{R}^3}\phi(x')\psi(y')\varphi(z')\lim_{n\to\infty}(\Phi(-x'/n,-y'/n,-z'/n)-\Phi(x'/n,y'/n,z'/n) )dx'dy'dz'=$
$=\int_{\mathbb{R}^3}\phi(x')\psi(y')\varphi(z')((\Phi(-0,0,0)-\Phi(0,0,0) ))=1*(0-0)=0$
