Timeline for heat kernel $p_t(x_0,y) \in D(\Delta) \cap L^\infty$ for a manifold with Ricci curvature bounded below?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 1, 2014 at 19:10 | comment | added | Fabrice Baudoin | @NateEldredge $Ricci \ge 0$ alone is not enough to get a uniform bound $p_t(x,y) \le C(t)$ on the heat kernel. The non-collapsing condition $\inf_{x \in \mathbb{M}} \mu (B(x,1)) >0$ is also needed. | |
| Nov 4, 2013 at 5:59 | comment | added | wang mu | You proved $P_t$ maps $L^1$ to $L^\infty$. Is $P_t f(x)$ uniformly bounded above for every t>0 and $x \in X$ for a function $f \in L^\infty(X)$? | |
| Nov 1, 2013 at 17:44 | comment | added | Nate Eldredge | @wangmu: Oops, I didn't save the edit. But the idea is that the $L^1$ norm of $P_t h$ for $h \ge 0$ is the sup of $\int (P_t h) f$ over all $0 \le f \le 1$. (Restrict both $h$ and $f$ to also lie in $L^2$ if necessary.) Since $\int (P_t h)f = \int h P_t f$ and $0 \le P_t f \le 1$ we get the result. | |
| Nov 1, 2013 at 17:42 | comment | added | Nate Eldredge | @wangmu: I edited to explain why $P_t$ maps $L^1$ to $L^1$. I am not sure in general whether $P_t$ maps $L^\infty$ to Lipschitz functions. And yes, your argument is essentially the reason why $P_t$ maps $L^2$ into $W^{1,2}$. | |
| Nov 1, 2013 at 14:09 | comment | added | wang mu | For every Dirichlet form, suppose $f \in L^2(X)$ but $\int_X |\nabla f|^2=\infty$. You still can get $P_t f \in W^{1,2}(X) $because $\int_X |\nabla P_tf|^2=-\int_X \langle \Delta P_t f,f \rangle<\infty$? | |
| Nov 1, 2013 at 13:46 | comment | added | wang mu | :Your answer is quite helpful. 1 why "sine $P_t$ maps $L^\infty$ to $L^\infty$ and is symmetric, it also maps $L^1$ to $L^1$"? 2 $P_t$ maps $L^\infty$ to Lipschitz function? | |
| Oct 31, 2013 at 22:22 | history | answered | Nate Eldredge | CC BY-SA 3.0 |