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X is an n-dim Riemannian manifold with the Dirichlet form $$ \varepsilon (u,v) =-\int_X \langle \nabla u,\nabla v \rangle $$

for $u,v \in W^{1,2}(X)$. Let $P_t$ and $p_t(x,y)$ be the associate semigroup and heat kernel respectively.

If there exists $f \in L^2(x)$ such that $$ \varepsilon (u,v) =-\int_X \langle \nabla u,\nabla v \rangle= \int_X f\phi dvol $$ Then we say $\Delta u=f$. Denote $$D(\Delta)=\{u:u \in W^{1,2}(X),\Delta u \in L^2(X)\}$$(Obviously, the definition of $\Delta$ and $D(\Delta)$ coincide with the generator and the domain of the generator of the Dirichlet form).

Suppose $Ric(X) \ge -(n-1)$, then for any $h \in L^2(X)$, can we get $P_t h \in D(\Delta) \cap L^\infty$?

Fix $x=x_0$, then $p_t(x_0,y) \in D(\Delta) \cap L^\infty$ and $\Delta p_t(x_0,y) \in D(\Delta) \cap L^\infty$?

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Yes.

It follows from the spectral theorem that for any $h \in L^2$, we have $P_t h \in D(\Delta)$.

I think this is easiest to see from the multiplication operator form of the spectral theorem; i.e., there is a measure space $(\hat{X}, \mu)$, a measurable function $g : \hat{X} \to \mathbb{R}$, and a unitary operator $U : L^2(X) \to L^2(\hat{X})$ such that $\Delta = U^{-1} M_g U$. Here $M_g$ is the multiplication operator on $L^2(\hat{X})$ defined by $M_g v = gv$ with domain $D(M_g) = \{ v \in L^2(\hat{X}) : gv \in L^2(\hat{X})\}$. Since $\Delta$ is a negative definite operator, $g \le 0$ almost everywhere.

Now it is easy to check that $P_t = U^{-1} M_{e^{tg}} U$. But notice that $g e^{tg}$ is a bounded function. Thus for any $v \in L^2(\hat{X})$, we have $M_{e^{tg}} v \in D(M_g)$. Using $U$ to move back to $L^2(X)$, we see that for any $h \in L^2(X)$, $P_t h \in D(\Delta)$.

A similar argument shows that $P_t h \in D(\Delta^\infty)$.

To show $P_t h \in L^\infty$, one could use the heat kernel bounds shown by Li and Yau:

Li, Peter; Yau, Shing-Tung. On the parabolic kernel of the Schrödinger operator. Acta Math. 156 (1986), no. 3-4, 153–201.

They show, among other things, that in a complete Riemannian manifold with Ricci curvature bounded from below, we have Gaussian upper estimates for the heat kernel. In particular, for each $t$, $p_t$ is bounded. From this it is obvious that $P_t$ maps $L^1$ into $L^\infty$.

Now $P_t$ also maps $L^\infty$ into $L^\infty$, so it maps $L^1 + L^\infty$ into $L^\infty$; in particular it maps $L^2$ into $L^\infty$.

Since $P_t$ maps $L^\infty$ into $L^\infty$ and is symmetric, it also maps $L^1$ into $L^1$. So it maps $L^1$ into $L^1 \cap L^\infty$; in particular it maps $L^1$ into $L^2$. In particular, since $p_t(x_0, \cdot) = P_{t/2} [p_{t/2}(x_0, \cdot)]$, where $p_{t/2}(x_0, \cdot) \in L^1$, we have $p_t(x_0, \cdot) \in L^2$. Then by using $p_t(x_0, \cdot) = P_{t/2} [p_{t/2}(x_0, \cdot)]$ again, we get $p_t(x_0, \cdot) \in D(\Delta)$.

Finally, we note that for $h \in D(\Delta)$, we have $\Delta P_t h = P_t \Delta h$. So we have $$\Delta [p_t(x_0, \cdot)] = \Delta P_{t/2} [p_{t/2}(x_0, \cdot)] = P_{t/2} \Delta [p_{t/2}(x_0, \cdot)].$$ Since $p_{t/2}(x_0, \cdot) \in D(\Delta)$ as previously argued, we have $\Delta p_{t/2}(x_0, \cdot) \in L^2$. Since $P_{t/2}$ maps $L^2$ into $L^\infty \cap D(\Delta)$, we have $\Delta [p_t(x_0, \cdot)] \in L^\infty \cap D(\Delta)$ as desired.

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  • $\begingroup$ :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? $\endgroup$ – wang mu Nov 1 '13 at 13:46
  • $\begingroup$ 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 $\int_X |\nabla P_tf|^2=-\int_X \langle \Delta P_t f,f \rangle<\infty$? $\endgroup$ – wang mu Nov 1 '13 at 14:09
  • $\begingroup$ @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 $W^{1,2}$. $\endgroup$ – Nate Eldredge Nov 1 '13 at 17:42
  • $\begingroup$ @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. $\endgroup$ – Nate Eldredge Nov 1 '13 at 17:44
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    $\begingroup$ @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. $\endgroup$ – Fabrice Baudoin Jul 1 '14 at 19:10

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