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Jan
26
awarded  Popular Question
Jan
19
comment Determinant of discrete Laplacian
By user35593's comment (which I didn't check but looks reasonable to me) you have lower and upper bounds on $ |det|$ in terms of what you expect, as soon as your partition is "approximately equidistant" $0<C_1\leq \frac{\Delta_i}{\Delta_j}\leq C_2$. However the scalings are different in the OP's 2 matrices: in order to stay consistent the first matrix should be multiplied by the uniform mesh $\Delta$, otherwise it doesn't make sense. Also, these matrices are the discretization of the positive operator "negative laplacian", not of the laplacian.
Jan
5
comment Monotonicity of the Hellinger integral/distance
Why don't you simply differentiate w.r.t. $t\geq 0$?
Dec
28
answered Chain rule for weakly differentiable functions
Nov
25
comment Lower semi-continuity of the Hellinger-Fisher-Rao distance
Yes that works, thank you gew. Unfortunately I realized since I first posted that I really the need weak-* part. I think I have a proof, but it's much more involved and uses a dynamic representation of H through some Riemannian structure hidden in there. I'll triple check and post it if correct.
Nov
23
revised Lower semi-continuity of the Hellinger-Fisher-Rao distance
edited title
Nov
23
asked Lower semi-continuity of the Hellinger-Fisher-Rao distance
Sep
17
comment Lebesgue differentiation theorem holds on locally doubling space?
I would guess so, since Lebesgue differentiation theorem is essentially local in nature.
Jul
19
comment Existence of the solution of a Dirichlet type differential equation
You can show smoothness by local interior regularity: since $f_{\epsilon}$ is bounded you have $u\in W^{2,p}_{loc}$ for any large $p$, thus by Morrey's inequality $u\in C^{1,\alpha}_{loc}$. Bootstrapping on Schauder's estimates you immediately get $u\in \mathcal{C}^{\infty}$ (of course not all the way up to the boundary). As far as I can tell the $u(0)=\epsilon$ is not imposed, so it's not a constraint (cf the "suppose" in their theorem 1.2) but only a normalization (which they explain just below the theorem).
Jun
26
comment continuity of the Boltzmann entropy in the Wasserstein metric
Well, sure, if the relative H is lsc then of course the original one is lsc too. But do you have a reference?
Jun
26
comment continuity of the Boltzmann entropy in the Wasserstein metric
what about lower semi-continuity then? after thinking on it for a while this should be enough for my purpose, contrarily to what I said in my post
Jun
26
revised continuity of the Boltzmann entropy in the Wasserstein metric
fixed typos
Jun
26
accepted continuity of the Boltzmann entropy in the Wasserstein metric
Jun
26
comment continuity of the Boltzmann entropy in the Wasserstein metric
yes you're perfectly right, very nice counterexample. Yet another naive belief goes to the bin...
Jun
26
asked continuity of the Boltzmann entropy in the Wasserstein metric
May
6
awarded  Yearling
Apr
13
comment monotone parabolic systems, convex variational structure and Legendre transform
yep, user5678's answer below answers my question 1 (should have thought of it!), and the reference helps for question 2.
Apr
12
accepted monotone parabolic systems, convex variational structure and Legendre transform
Mar
21
comment if $u_\epsilon \rightarrow u$ weakly in $L^2$ then also $\partial_t u_\epsilon \rightarrow \partial_t u $ weakly in $L^2$
By $u_\epsilon \in L^{\infty}(I,H^1(M))\cap Lip(I,L^2(M))$ do you mean that you have uniform bounds (in $\epsilon$), or that for any fixed $\epsilon$ the function $u_{\epsilon}$ lies in this space? if you have uniform bounds then the statement is trivial, since then $\partial_tu_{\epsilon}$ is bounded in $L^{\infty}(I,L^2(M))\subset L^2(I,L^2(M))$ so in by the Banach-Alaoglu-Bourbaki theorem $\partial_tu_{\epsilon}\rightharpoonup v$ weakly in $L^2(I,L^2(M))$ for some $v$ and then by continuity $v=\partial_t u$. Otherwise I think the statement is false.
Dec
17
awarded  Popular Question