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Problem:
For my thesis I am trying to prove that $$ \int \nabla_\eta g(\eta)p(x-\eta)d\eta= \nabla_x \int g(\eta) p(x-\eta)d\eta $$ where $x,\eta \in \mathbb{R}^n$ (Equation (12) in https://arxiv.org/pdf/1709.03749.pdf)

My approach:
Since the $\nabla $-operator within the integral is not with respect to $x$, I cannot use Leibnitz rule directly.
Hence I tried to use rules that involve convolution:
$$ \begin{align} \int \nabla_\eta g(\eta)p(x-\eta)d\eta &= \int g'(\eta)p(x-\eta)d\eta \\ &= (g'*p)(x) \\ &= (g*p)'(x) \\ &=\nabla_x \int g(\eta) p(x-\eta)d\eta \end{align} $$ I'm very much aware that my proof isn't accurate (e.g. I ignored that things are multidimensional) but that was the best try I could come up with and it is also the way the authors suggested it on the related GIT-hub site.(https://github.com/siavashBigdeli/DMSP/issues/1)
I would very much appreciate any help with this !

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$\require{cancel}$ First off, note that

$$ \nabla_{\eta}p(x - \eta) = -\nabla_x p(x - \eta) \tag{1} $$

and now use the product rule

\begin{eqnarray} \int_V \nabla_\eta g(\eta) p(x - \eta)~{\rm d}\eta &=& \int_V\left\{\nabla_\eta[g(\eta)p(x - \eta)] - g(\eta) \nabla_\eta p(x - \eta)\right\}{\rm d}\eta \\ &=& \int_V \nabla_\eta[g(\eta)p(x - \eta)]~{\rm d}\eta - \int_V g(\eta) \nabla_\eta p(x - \eta){\rm d}\eta \\ &=& \cancelto{0}{\left. g(\eta) p(x - \eta)\right|_{S}} - \int_V g(\eta)\nabla_\eta p(x - \eta)~{\rm d}\eta \\ &\stackrel{(1)}{=}& + \int_V g(\eta)\nabla_xp(x - \eta)~{\rm d}\eta \\ &=& \nabla_x\int_V g(\eta)p(x - \eta)~{\rm d}\eta \tag{2} \end{eqnarray}

where I have assumed the product $g(\eta) p(x - \eta)$ goes to zero on the boundary $S$ of the integration volume $V$

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  • $\begingroup$ Ty, helps very much ! In this case $p(x)$ is a probability distribution and $g$ represents Gaussian distribution and corresponds to a smoothing kernel. The integral is taken over $\mathbb{R}^n$. Why is the one integral going to 0 in this case ? In 1-dimensional case it is pretty clear, but i do not know if it is also so easy in multiple dimensions. $\endgroup$
    – Dulfinator
    Aug 11 '20 at 13:43
  • $\begingroup$ @Dulfinator happy to help. Yes, the Gaussian kernel goes to zero also in multiple dimensions $\endgroup$
    – caverac
    Aug 11 '20 at 13:48

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