Timeline for Minimizing some $H^{-1}$ functional over (a subset of) probability densities in $R^d$
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
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| Jun 11, 2014 at 11:28 | comment | added | leo monsaingeon | My bad: it will hold of course by density of $\mathcal{C}^{\infty}_c$, since the standard cutoff/mollification $f_n=\zeta_n (f*\rho_n)\to f$ in any $L^q$ as long as $f\in L^q$, while $\nabla f_n\to \nabla f$ in any $L^p$ as long as $f\in L^p$. So your agument works just fine, thanks Mark! | |
| Jun 11, 2014 at 9:31 | comment | added | leo monsaingeon | (previous comment, continued) For example given $f\in L^{q}$ with $q\in (2,2^*)$ and $\nabla f\in L^2$ does the Sobolev inequality $|f|_{L^{2^*}}\leq C |\nabla f|_{L^2}$ hold? If $q\leq 2$ this is clear, but if $q>2$? | |
| Jun 11, 2014 at 9:26 | comment | added | leo monsaingeon | I guess you meant $\Psi_n$ bounded in $L^{d^*}=L^{2d/(d-2)}$ (critical Sobolev exponent), right? I tried this approach by cutting of $\Psi_{n,k}=(\Psi_n\vee -k)\wedge k\in L^q$ for all $q>p=d/(d-2)$ and $|\nabla \Psi_{n,k}|_{L^2}\leq |\nabla \Psi_n|_{L^2}\leq C$ to apply the standard Sobolev inequality and conclude that $|\Psi_{n,k}|_{L^{2d/(d-2)}}\leq C |\nabla\Psi_{n,k}|\leq C$, hence $|\Psi_n|_{L^{2d/(d-2)}}\leq C$ by letting $k\to\infty$. However in dimension $d=3$ the exponent $\frac{d}{d-2}=3> 2$ so the classical Sobolev inequality is not clear to me (see next comment). | |
| Jun 11, 2014 at 2:58 | comment | added | Mark Peletier | I just realized that your case is slightly better than the standard $L^1$-right-hand-side case, since you have bounded $\|\nabla \Psi_n\|_2$ by construction. By the Sobolev inequality that implies that $\Psi_n$ is bounded in $L^{d/(d-2)}$ and you can extract a weakly converging subsequence. Then you can pass to the limit in your equality $\Psi_n = G*u_n$ after integrating against a test function. | |
| Jun 11, 2014 at 2:44 | comment | added | Mark Peletier | 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). | |
| Jun 9, 2014 at 23:00 | comment | added | leo monsaingeon | @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 $\Psi\in L^{\frac{d}{d-2}}_{weak}$? | |
| Jun 9, 2014 at 22:59 | comment | added | leo monsaingeon | @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? | |
| Jun 9, 2014 at 21:53 | comment | added | Mark Peletier | Actually, @piero-dancona says it much nicer as an answer to your other question at mathoverflow.net/a/159500/16530. | |
| Jun 9, 2014 at 21:38 | comment | added | Mark Peletier | 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? | |
| Jun 9, 2014 at 10:28 | history | edited | leo monsaingeon | CC BY-SA 3.0 |
added 4 characters in body
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| Jun 9, 2014 at 10:16 | history | asked | leo monsaingeon | CC BY-SA 3.0 |