Suppose $\mu_t : [0, + \infty) \to \mathcal{P}(\mathbb{R}^n)$ is a curve of probability measures on $\mathbb{R}^n$. For each $\nu \in \mathcal{P}(\mathbb{R}^n)$ I define $\nu \ast \rho$ as the absolutely continuous measure w.r.t. the Lebesgue measure on $\mathbb{R}^n$ which density is given by $$ \int_{\mathbb{R^n}} \rho(x-y) d \nu(y)$$ where $\rho$ is a never vanishing smooth function integrating to $1$ w.r.t. the Lebesgue measure. Moreover I define $\frac{d}{dt} \mu_t$ as the functional on $C^{\infty}_c( [0, +\infty) \times \mathbb{R}^n)$ defined as $$ \langle \frac{d}{dt} \mu_t, \varphi \rangle = -\int_0^{+\infty} \int_{\mathbb{R}^n} \dot{\varphi}(x,t) d \mu_t(x) dt $$

I would like to prove that $$ \frac{d}{dt} (\mu_t \ast \rho) = \biggl ( \frac{d}{dt} \mu_t \biggr ) \ast \rho $$ but I am non sure about the meaning/definition of the RHS.

Any help would be really appreciated!

EDIT: I think I have understood it. If I use the definition of convolution of a distribution times a function I have that $$ \langle \biggl ( \frac{d}{dt} \mu_t \biggr ) \ast \rho , \varphi \rangle = \langle \frac{d}{dt} \mu_t, \tilde{\rho} \ast \varphi \rangle $$ for every $\varphi \in C^{\infty}_c( [0, +\infty) \times \mathbb{R}^n)$ where $$ \tilde{\rho}(x) := \rho(-x) \quad x \in \mathbb{R}^n $$ and $\rho \ast \varphi$ is the function defined as $$ (\tilde{\rho} \ast \varphi)(x,t) = \int_{\mathbb{R}^n} \tilde{\rho}(x-y) \varphi(x,t) dx = \int_{\mathbb{R}^n} \rho(y-x) \varphi(x,t) dx \quad x \in \mathbb{R}^n \, \, , \, \, t \in [0, +\infty)$$

This should work.

  • $\begingroup$ I think your definition of $\frac{d}{dt}\mu_t$ differs from the usual one by a minus sign $\endgroup$ Jun 3, 2019 at 10:42
  • $\begingroup$ Yes, I forgot it! Thanks! $\endgroup$
    – Bremen000
    Jun 3, 2019 at 10:45
  • $\begingroup$ I missed it earlier but also if you're using e.g. $C^\infty_c([0,\infty))$ instead of $(0,\infty)$ then there's a boundary term $\langle f',g \rangle = - \langle f, g' \rangle + f(0)g(0)$ (...but this is not really the point of your question, and it seems you have the answer) $\endgroup$ Jun 3, 2019 at 11:14


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