Let me consider the following subset of probability measures in $R^d$ $$ \mathcal{K}_M=\left\{0\leq u(x)\in L^1(R^d):\quad \int u(x)dx =1,\,\int|x|^2u(x)dx\leq M,\,\int u(x)|\log u(x)|dx\leq M\right\} $$ for some fixed $M>0$. For given $u\in \mathcal{K}$ let me define $\Psi=(-\Delta)^{-1}u$ the Newton potential, i-e up to multiplicative constants $$ \Psi(x)=G*u(x)=\int \frac{1}{|x-y|^{d-2}}u(y)dy\qquad \text{a.e. }x\in R^d $$ where $G(z)=\frac{1}{|z|^{d-2}}$ is the usual Poisson kernel. On this admissible set the functional is defined as $$ u\in \mathcal{K}_M\mapsto F[u]=\int |\nabla \Psi(x)|^2dx, $$ where possibly $F[u]=\infty$ if $\nabla \Psi\notin L^2$. This is roughly speaking a squared $H^{-1}$ norm of $u$.

Goal: show that there is a minimizer $u^*\in \mathcal{K}_M$ $$ F[u^*]=\min\limits_{u\in \mathcal{K}_M}F[u]. $$

What I have so far: since my functional is bounded from below there is a minimizing sequence $\{u_n\}_n\subset \mathcal{K}_M$. By the Dunford-Pettis theorem and easy arguments that I shall not detail here, it is possible to show that there is some $u^*\in \mathcal{K}_M$ such that $u_n\rightharpoonup u^*$ in $L^1(R^d)$. Now by definition of my functional we have that $\nabla\Psi_n\rightharpoonup \Phi^*$ in $L^2(R^d)$ for some $\Phi^*\in L^2$ with $|\Phi^*|^2_{L^2}\leq \liminf |\nabla\Psi_n|^2_{L^2}=\inf\limits_{u\in \mathcal{K}_M}F[u]$. So if I could show that $\Psi_n\to\Psi^*$ in some sense and if uniqueness for solutions to the Poisson equation $-\Delta \Psi=u$ holds, I may conclude that $\Phi^*=\nabla\Psi^*$ with $\Psi^*=(-\Delta)^{-1}u^*=G*(u^*)$ and I would be done.

The problem: from this other post it seems that $u\in L^1$ is unfortunately a borderline case for the Poisson equation: if $u$ is only in $L^1$ then $\Psi$ is not in $L^p(R^d)$ for any $p\geq 1$ (the problem seems to be the decay at infinity). In order to get compactness of $\{\Psi_n\}$ I cannot hope to use a naive argument like: "$|u_n|_{L^1}=1\Rightarrow |\Psi_n|_{L^p}\leq C$, then extract a weakly $L^p$ converging subsequence with $\Phi^*=\nabla\Psi$, and conclude by uniqueness that $\Psi=G*(u^*)$". Since $|u_n|_{L^1}=1$ some extra information is provided by the Hardy-Littlewood-Sobolev lemma, which guarantees here that the sequence $\Psi_n$ is bounded in the Lorentz/weak $L^p$ space $L^{\frac{d}{d-2},\infty}(R^d)=L^{\frac{d}{d-2}}_{weak}(R^d)$. So everything essentially boils down to compactness in $L^{\frac{d}{d-2}}_{weak}(R^d)$ and uniqueness for Poisson equation within this class of solutions for $L^1(R^d)$ data. Let me point out that by the very setting of the problem $\nabla \Psi_n$ is bounded in $L^2$, which should help (is there any kind of Sobolev/interpolation inequality for $\Psi$ in Lorentz spaces and $\nabla \Psi$ in classical $L^q$ spaces?). Also I actually have that $\{u_n\}$ is bounded in $L^1\log L^1$, which is very slightly better than just $L^1$.

This question is also related to these post 1, post 2, post 3. I've been unsuccessfully trying to solve this problem for the last couple of weeks, so any help is much appreciated!

  • $\begingroup$ Why is uniqueness for $-\Delta \Psi = u$ with $u\in L^1(R^d)$ an issue? If I have two solutions $\Psi_1$ and $\Psi_2$ of the equation, both in the sense of distributions (so $\Psi_{1,2}$ only need be $L^1_{\mathrm{loc}}$), then the difference $\Psi=\Psi_1-\Psi_2$ satisfies $\Delta\Psi=0$ in the sense of distributions. Then $\Psi$ is $C^\infty$, as can be seen by regularizing $\Psi$ by convolution. If $\Psi$ is bounded and harmonic, then it is constant ... Maybe the boundedness of $\Psi$ is the issue? $\endgroup$ – Mark Peletier Jun 9 '14 at 21:38
  • $\begingroup$ Actually, @piero-dancona says it much nicer as an answer to your other question at mathoverflow.net/a/159500/16530. $\endgroup$ – Mark Peletier Jun 9 '14 at 21:53
  • $\begingroup$ @Mark: yes, I remember Piero's nice answer. This decay at infinity condition was my first thought too. I would totally agree with you if I had "decay at infinity" in the form $\Psi_{1,2}\in L^p$ for some $p\geq 1$. However here the $\Psi$'s are not in $L^p(R^d)$ for any $p$, but only in weak-$L^p$/Lorentz as in my OP. I'm not familiar at all with those spaces and I'm certainly missing an obvious point, but let me ask 2 questions nonetheless (probably trivial): 1) does $\Psi\in L^p_{weak}$ imply any "decay at infinity", and 2) if so is that decay condition sufficient to guarantee uniqueness? $\endgroup$ – leo monsaingeon Jun 9 '14 at 22:59
  • $\begingroup$ @Mark: also, this would only settle half of the problem, namely uniqueness. But what about compactness, which would be here $\Psi_n\to\Psi$ in some sense for some $\Psi\in L^{\frac{d}{d-2}}_{weak}$? $\endgroup$ – leo monsaingeon Jun 9 '14 at 23:00
  • $\begingroup$ Yes, if a function $f$ is in $L^p_{weak}(R^d)$, then any truncation $f$ to a finite range (e.g. $(f\vee -1)\wedge 1$) is in $L^q(R^d)$ for $q>p$. This follows from the property that the distribution function satisfies $\lambda_f(t) \leq \|f\|_{p,weak}^p t^{-p}$ and the layer-cake principle (e.g. Lieb-Loss Sec 1.13). $\endgroup$ – Mark Peletier Jun 11 '14 at 2:44

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