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A comparison principle for parabolic equation
No problem, you're very welcome! 
Apr 21 
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A comparison principle for parabolic equation
One last comment: the "$a=\frac{F(u)F(v)}{uv}$" trick is preciesly what makes the duality method work for very weak solutions, see the proof of theorem 6.5 (with the extra difficulty that one first has to truncate to avoid the degenerate levelset $u=0$). 
Apr 21 
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A comparison principle for parabolic equation
I updated my previous answer 
Apr 21 
revised 
A comparison principle for parabolic equation
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Apr 21 
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A comparison principle for parabolic equation
So there is a PDE in the end, and you were basically trying to compare a subsolution and a supersolution of your PDE. Your computation must have gone wrong at some point: the weak formulation shouldn't involve time derivatives on the boundary, but only in the interior $\Omega\times(0,T)$ of the parabolic domain. 
Apr 21 
answered  A comparison principle for parabolic equation 
Apr 21 
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Convex interaction energy
The OP wanted absolutely continuous measures (w.r.t Lebesgue, I guess). But taking any of one's favourite approximation should do the trick. 
Apr 20 
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Convex interaction energy
It is well known that this interaction energy is convex with respect to geodesic interpolation in the quadratic wasserstein distance, so I guess this is quite unlikely. Have you tried an explicit computation with $\mu_0,\mu_1$ two uniform densities on concentric balls but with different radii? 
Apr 16 
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A comparison principle for parabolic equation
What is the PDE, then? it seems you have time derivatives on the boundary, which I've never seen before... 
Apr 16 
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A comparison principle for parabolic equation
Is $p\in R$ really negative, or is it just a typo? In which sense do you understand $\varphi '$ (I guess $\partial \varphi/\partial t$) if you only have $\varphi\in \mathcal{C}^1(0,T;L^2(\Omega))$ as a time regularity? Of course with your assumptions you also have $\varphi\in L^{2}(0,T;H^1(\Omega))\hookrightarrow L^{2}(0,T;H^{1/2}(\partial\Omega))\hookrightarrow L^{2}(0,T;L^2(\partial\Omega))$, but this still doesn't make sense of the time derivative on the boundary. I'm a little surprised by your $\partial\Omega$ boundary terms, can you tell us what is your (degenerate) parabolic PDE? 
Apr 15 
answered  reference request: trace/lifting operator for $L^{\infty}$ data in bounded $\Omega\subset R^d$ 
Apr 13 
asked  reference request: trace/lifting operator for $L^{\infty}$ data in bounded $\Omega\subset R^d$ 
Apr 12 
revised 
Existence of Minimizer of $h(\rho) = c \\rho\_{3}  \int_{\mathbb R^3} \, dx \frac{\rho(x)}{x} $
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Apr 12 
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Existence of Minimizer of $h(\rho) = c \\rho\_{3}  \int_{\mathbb R^3} \, dx \frac{\rho(x)}{x} $
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Apr 12 
revised 
Existence of Minimizer of $h(\rho) = c \\rho\_{3}  \int_{\mathbb R^3} \, dx \frac{\rho(x)}{x} $
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Apr 12 
revised 
Existence of Minimizer of $h(\rho) = c \\rho\_{3}  \int_{\mathbb R^3} \, dx \frac{\rho(x)}{x} $
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Apr 12 
answered  Existence of Minimizer of $h(\rho) = c \\rho\_{3}  \int_{\mathbb R^3} \, dx \frac{\rho(x)}{x} $ 
Apr 9 
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Existence of Minimizer of $h(\rho) = c \\rho\_{3}  \int_{\mathbb R^3} \, dx \frac{\rho(x)}{x} $
I wasn't arguing about the criticality, only pointing out that your function $\rho_m$ is not in $L^3(R^3)$ (the problem is integrability at the origin). Again: it is radial and behaves as $1/r$ at the origin, but $1/r\notin L^{3}$ in any neighborhood of the origin (in dimension 3). As a consequence it doesn't even make sense to look at the criticality, since you have $h(\rho_{m,R})=\infty$ for any $R>0$. 
Apr 8 
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Existence of Minimizer of $h(\rho) = c \\rho\_{3}  \int_{\mathbb R^3} \, dx \frac{\rho(x)}{x} $
you should be really careful for several reasons: 1) your function $\rho_m$ does not belong to $L^3$ (in dimension 3 with $dx\approx r^2dr$ this would require $\frac{1}{r^3}r^2dr$ to be integrable near $r=0$, which is obviously false). 2) for random $\eta$ your perturbation $\rho_m+t\eta$ may not be an admissible candidate (you may loose positivity/nonnegativity, or violate the constraint $\rho_{L^1}=1$). So even convexity may not be enough and you have to prove your claim "by hand". I'm working on it, I believe the result holds. 
Apr 6 
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Existence of Minimizer of $h(\rho) = c \\rho\_{3}  \int_{\mathbb R^3} \, dx \frac{\rho(x)}{x} $
I guess by stationary function you mean critical point? In which case of course you absolutely cannot conclude anything, your $\rho_m$ may be a saddle point... 