# Diffusion semigroup generated by Laplacian

Let $M$ be a complete Riemannian manifold and $\Delta$ denote the Laplacian on it. Also assume that the spectrum of $-\Delta$ lies inside $[a, \infty)$. Let $P_t, t > 0$ denote the diffusion semigroup generated by $\Delta$, and pick a function $f \in L^2(M)$. I am looking for a proof of the following $$\Vert P_t f\Vert_{L^2} \leq Ce^{-ta}\Vert f\Vert_{L^1}, t \geq 1$$ A reference would also be appreciated.

Could we also say something like $\Vert \nabla P_t f\Vert_{L^2} \leq Ce^{-ta}\Vert f\Vert_{L^1}, t \geq 1$?

• I think you may be able to find this in E.B. Davies' book Heat Kernels and Spectral Theory. Roughly, it should be equivalent to the ultracontractivity of the semigroup $P_t$ and/or the boundedness of the heat kernel. Your second question should follow from the first, integrating by parts and using the spectral theorem which says that $\Delta P_t$ is bounded on $L^2$. I will try to look at my copy of Davies tomorrow and fill in details. – Nate Eldredge Apr 13 '15 at 0:49

## 1 Answer

This is essentially equivalent to boundedness of the heat kernel. I don't think you can expect it to be true for an arbitrary manifold, but it is true if the manifold has Ricci curvature bounded below.

First, note that if your inequality holds with $t=1$ then it holds with any $t \ge 1$. (Since the spectrum of $-\Delta$ is bounded below by $a$, we have $\|P_t f\|_2 \le e^{-ta} \|f\|_2$. Replacing $f$ by $P_1 f$ and $t$ by $t-1$ and using the semigroup property, if your inequality holds for $t=1$ we get $$\|P_t f\|_2 \le e^{-a(t-1)} \|P_1 f\|_2 \le C e^{-at} \|f\|_1.$$ So we are asking whether $P_1$ is a bounded operator from $L^1$ to $L^2$.

Since $P_1$ is symmetric on $L^2$, by taking adjoints, this is equivalent to asking whether $P_1$ is a bounded operator from $L^2$ to $L^\infty$. If we have $P_t$ bounded from $L^2$ to $L^\infty$ for all $t > 0$, we say $P_t$ is ultracontractive. In Lemma 2.1.2 of E.B. Davies, Heat Kernels and Spectral Theory, you can find the sketch of a proof that ultracontractivity implies the existence of a bounded heat kernel. But we care more about the converse, which is really trivial. Suppose $P_t$ admits an integral kernel $p(t,x,y)$ such that for each $t$, $p(t,\cdot,\cdot)$ is bounded. Then the desired inequality clearly follows with $C = e^{a} \sup_{x,y \in M} p(1,x,y)$.

A famous result of Li and Yau says that a complete Riemannian manifold with Ricci curvature bounded below has a heat kernel that satisfies Gaussian upper bounds. In particular, for each $t > 0$, $p(t,\cdot,\cdot)$ is bounded. This may be overkill but it certainly works.

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

Your second statement essentially follows from the first, at least for $t > 1$. The spectral theorem says that $\Delta P_t$ is bounded from $L^2$ to $L^2$ for each $t > 0$. (Essentially, this follows from the fact that $-\Delta$ is a positive operator, and $x e^{-x}$ is a bounded function on $[0, \infty)$.) Now integrating by parts gives \begin{align*} \|\nabla P_{1+\epsilon} f\|_2 &= \|\nabla P_{1+\epsilon} f\|_2 \\ &= \sqrt{-\langle P_{1+\epsilon} f, \Delta P_\epsilon P_1 f\rangle} \\ &\le \sqrt{\|P_\epsilon\|_{2 \to 2} \|\Delta P_\epsilon\|_{2 \to 2}} \|P_1 f\|_2\end{align*} and then use the previous statement.