Is there a reasonable/canonical way to mollify a Borel probability measure without changing its marginals. Let $\pi \in \mathcal{P}(\mathbb{R}^2)$ with marginals $\mu,\nu$. I want to smooth out $\pi$ up to scale $\varepsilon$, which is equivalent to cutting it off at frequency $k\approx \varepsilon^{1}$ in Fourier space, but I do not want to change is marginals. The resulting measure should have a smooth density. It goes without saying that the usual procedure (convolution with a smooth, positive, bump function) does not work. Of course, you can assume $\mu,\nu$ are absolutely continuous with smooth densities.

$\begingroup$ Whenever this can be done $\mu$ and $\nu$ must have densities. So, you need to impose some conditions on $\mu$ and $\nu$. $\endgroup$– Iosif PinelisCommented Aug 20, 2023 at 13:44

$\begingroup$ This goes without saying. I have edited the question to indicate this. $\endgroup$– almosteverywhereCommented Aug 21, 2023 at 13:27

1$\begingroup$ I guess you will need a case where the marginals $\mu,\nu$ are unchanged when they are mollified. $\endgroup$– Gerald EdgarCommented Aug 21, 2023 at 13:57

$\begingroup$ @GeraldEdgar Yes, I agree. What I had in mind is some smoothing operation that is $\mu,\nu$dependent with the natural property that it leaves $\mu,\nu$ invariant. Maybe this is asking for too much. $\endgroup$– almosteverywhereCommented Aug 22, 2023 at 7:52
2 Answers
$\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\tX}{\tilde X}\newcommand{\tY}{\tilde Y}\newcommand{\tpi}{\tilde\pi} $Here is how this can be done in an explicit way, at least when the densities (say $p$ and $q$) of $\mu$ and $\nu$ are everywhere $>0$.
Let $(X,Y)$ be a random point in $\R^2$ with distribution $\pi$ and thus marginals $\mu$ and $\nu$.
Let $F$ and $G$ be the c.d.f.'s of $X$ and $Y$ (respectively).
Let $\pi_\ep$ be a distribution over $\R^2$ with a smooth density such that $\pi_\ep$ is close to $\pi$. Let $(X_\ep,Y_\ep)$ be a random point in $\R^2$ with distribution $\pi_\ep$. Let $F_\ep$ and $G_\ep$ be the c.d.f.'s of $X_\ep$ and $Y_\ep$ (respectively), so that $F_\ep$ and $G_\ep$ are close to $F$ and $G$, and hence $F^{1}\circ F_\ep$ and $G^{1}\circ G_\ep$ are smooth maps each close to the identity map.
Let $\tX:=F^{1}(F_\ep(X_\ep))$ and $\tY:=G^{1}(G_\ep(Y_\ep))$. Then the distribution (say $\tpi$) of the random point $(\tX,\tY)$ in $\R^2$ will have a smooth density and marginals $\mu$ and $\nu$, and $\tpi$ will be close to $\pi$.

$\begingroup$ Thanks! Any idea how one can generalise this to higher dimensions? $\endgroup$ Commented Aug 26, 2023 at 10:11

$\begingroup$ @almosteverywhere : I suggest you ask the question about higher dimensions in another post. For higher dimensions, I guess one could try to use Dirk's idea. However, the problem that I then envision is to even show that a feasible distribution $\pi_\epsilon$ with a finite $\Phi(\pi_\epsilon)$ exists (let alone a minimizer). $\endgroup$ Commented Aug 27, 2023 at 1:34
You can always formulate this problem as an optimization/feasibility problem and look for solutions. The conditions on the marginals gives two linear constraints: $$P_1\pi_\epsilon = \mu,\quad P_2\pi_\epsilon = \nu.$$
The closeness to a given $\pi$ can be formulated as another constraints (for example) $$\\pi\pi_\epsilon\\leq\epsilon.$$ For smoothness you can now optimize your favorite smoothness measure $\Phi$ over the constraint set, i.e. you consider $$\min \Phi(\pi_\epsilon)\quad \text{s.t.}\quad P_1\pi_\epsilon = \mu,\quad P_2\pi_\epsilon = \nu,\quad \\pi\pi_\epsilon\\leq\epsilon.$$