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Let $V_t(x)=x^2+t\phi(x)$ where $t>0$ and $\phi\in C^\infty_c(\mathbb{R})$.

My question is what can be said about the continuity of the (unique) minimizer (among probability measures) of the following lower semi continuous (w.r.t weak topology) strictly convex functional $$E(\mu)=\int_\mathbb{R}V_t(x)d\mu(x)-P\iint_{\mathbb{R}^2}\log|x-y|d\mu(x)d\mu(y)+\int_\mathbb{R}\log\Big(\dfrac{d\mu}{dx}\Big)d\mu(x)$$

with respect to $t$ and $P>0$. The minimizer $\mu_t$ of this equation with p.d.f $\rho_t$ is characterized by the following equation:

There exists a constant $C_t^P\in\mathbb{R}$ such that for all $x\in\mathbb{R}$ $$V_t(x)-2P\int_\mathbb{R}\log|x-y|\rho_t(y)dy+\log\rho_t(x)=C_t^P$$.

We can differentiate w.r.t $x$ and find the relation $O[\rho_t]=-V'_t$ where $O$ is defined by $$O[f]=\dfrac{f'}{f}+2P\mathcal{H}[f]$$

where $\mathcal{H}[f](x):=p.v\displaystyle\int_\mathbb{R}\dfrac{f(t)}{t-x}dt$.

Is there a small chance that $\rho_t$ will be continuous with respect to $t$ ? I think it would be easier to invert the non-linear operator $O$ and show continuity of the inverse with small $P$.

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  • $\begingroup$ Since $\log|x-y|$ is just the Green's function in 2D your optimality condition can also be expressed as $-2P\Delta\rho +\log\rho -V=cst$ (up to signs and constants, which I probably messed up). In this respect the constant is the unique one such that this is actually solvable, e.g. $\log\rho \pm V$ should integrate to zero (otherwise one should not expect the PDE to be solvable). Maybe that helps? $\endgroup$ Commented Oct 8, 2023 at 20:50
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    $\begingroup$ I would also guess that the monotonicity of the $\log$ should go in the right direction, i-e the nonlinear operator $[-2P\Delta +\log](\cdot)$ should satisfy some comparison principle? This posisbly helps in proving continuity w.r.t. parameters. $\endgroup$ Commented Oct 8, 2023 at 20:51
  • $\begingroup$ I'll try to investigate this possible comparison principle. Thanks! $\endgroup$ Commented Oct 9, 2023 at 13:52

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