A doubt regarding derivative of convolution!!

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In the following calculation:

$\int_{\mathbb R^{d}} u_{o \epsilon} div (\phi) dx = \int_{\mathbb R^{d}} (u_{o} * \psi_{\epsilon}) div(\phi) dx = \sum_{i=1}^{d} \int_{\mathbb R^{d}} ( u_{o} * \psi_{\epsilon}) \phi_{x_{i}} dx$

After this step;

Can I shift the derivative to obtain: $\sum_{i=1}^{d} \int_{\mathbb R^{d}} ( u_{o} * \psi_{\epsilon}) \phi_{x_{i}} dx = \int_{\mathbb R^{d}} u_{o} div( \phi * \psi_{\epsilon}) dx $ ; I mean how to obtain the term "$div (\phi * \psi_{\epsilon})" ???$

if this information is somehow useful: THEN

[" here: $\phi \in C_{0}^{1} (\mathbb R^{d})^{d}$ with $|| \phi ||_{L^{\infty}(\mathbb R^{d})} \leq 1$ " ]

AND

[$\psi_{\epsilon}(x) = \frac{1}{\epsilon} \psi(\frac {x}{\epsilon}) $ ; where: $\psi \in C_{o}^{\infty}(\mathbb R^{d}) $ be a cut-off function with the following properties:

a) the function $\psi$ is non-negative & its support is contained in the unit ball of $\mathbb R^{d}$ ;

b) $\int_{\mathbb R^{d}} \psi(x) dx = 1$ ;

c) $\psi (-x) = \psi (x)$ .]