## Lecture 18. The Gaussian upper bound for the heat kernel

Let $\mathbb{M}$ be a complete $n$-dimensional Riemannian manifold and, as usual, denote by $L$ its Laplace-Beltrami operator. As in the previous lecture, we will assume that the Ricci curvature of $\mathbb{M}$ is bounded from below by $-K$ with $K \ge 0$. Our purpose in this lecture is to prove a Gaussian upper bound for the heat kernel. Our main tools are the parabolic Harnack inequality proved in the previous lecture and the following integrated maximum principle:

Proposition Let $g :\mathbb{M}\times \mathbb{R}_{\ge 0} \to \mathbb{R}$ be a non positive continuous function such that, in the sense of distributions,
$\frac{\partial g}{\partial t} +\frac{1}{2} \Gamma(g) \le 0,$
then, for every $f \in L^2_\mu(\mathbb{M})$, we have
$\int_{\mathbb{M}} e^{g(y,t)} ) (P_t f)^2 (y) d\mu(y) \le \int_{\mathbb{M}} e^{g(y,0)} ) f^2 (y) d\mu(y)$.

Proof: Since
$\left(L-\frac{\partial }{\partial t}\right)(P_tf)^2 = 2 P_tf\left(L-\frac{\partial }{\partial t}\right)(P_t f) + 2 \Gamma(P_t f) = 2 \Gamma(P_tf)$, multiplying this identity by $h_n^2(y) e^{g(y,t)}$, where $h_n$ is the usual localizing sequence , and integrating by parts, we obtain
$0 = 2 \int_0^\tau \int_{\mathbb{M}} h_n^2 e^g \Gamma(P_tf) d\mu(y) dt - \int_0^\tau \int_{\mathbb{M}}h_n^2 e^g \left(L-\frac{\partial }{\partial t}\right)(P_tf)^2 d\mu(y) dt$
$= 2 \int_0^\tau \int_{\mathbb{M}} h_n^2 e^g \Gamma(P_tf) d\mu(y) dt + 4 \int_0^\tau \int_{\mathbb{M}} h_n e^g P_tf \Gamma(h_n,P_tf) d\mu(y) dt + 2 \int_0^\tau\int_{\mathbb{M}}h_n^2 e^g P_tf \Gamma(P_tf,g)d\mu(y) dt$
$- \int_0^\tau \int_{\mathbb{M}} h_n e^g (P_tf)^2 \frac{\partial g}{\partial t} d\mu(y) dt - \int_{\mathbb{M}} h_n e^g (P_tf)^2 d\mu(y)\bigg|_{t=0} + \int_{\mathbb{M}} h_n e^g (P_tf)^2 d\mu(y)\bigg|_{t=\tau}$
$\ge 2 \int_0^\tau \int_{\mathbb{M}} h_n^2 e^g \left(\Gamma(P_tf)+P_tf\Gamma(P_tf,g)+\frac{P_tf^2}{4} \Gamma(g)\right) d\mu(y) dt + 4 \int_0^\tau \int_{\mathbb{M}} h_n e^g P_tf \Gamma(h_n,P_tf) d\mu(y) dt + \int_{\mathbb{M}} h_n e^g (P_tf)^2 d\mu(y)\bigg|_{t=\tau} - \int_{\mathbb{M}} h_n e^g (P_tf)^2 d\mu(y)\bigg|_{t=0},$
where in the last inequality we have made use of the assumption on $g$. From this we conclude
$\int_{\mathbb{M}} h_n e^g (P_t f)^2 d\mu(y)\bigg|_{t=\tau} \le \int_{\mathbb{M}} h_n e^g (P_tf)^2 d\mu(y)\bigg|_{t=0} - 4 \int_0^\tau \int_{\mathbb{M}} h_n e^g P_tf \Gamma(h_n,P_tf) d\mu(y) dt.$
We now claim that
$\underset{n\to \infty}{\lim} \int_0^\tau \int_{\mathbb{M}} h_n e^g P_tf \Gamma(h_n,P_tf) d\mu(y) dt = 0.$
To see this we apply Cauchy-Schwarz inequality which gives
$\left|\int_0^\tau \int_{\mathbb{M}} h_n e^g P_tf \Gamma(h_n,P_tf) d\mu(y) dt\right|\le \left(\int_0^\tau \int_{\mathbb{M}} h_n^2 e^g (P_tf)^2 \Gamma(h_n) d\mu(y) dt\right)^{\frac{1}{2}} \left(\int_0^\tau \int_{\mathbb{M}} e^g \Gamma(P_tf) d\mu(y) dt\right)^{\frac{1}{2}}$
$\le \left(\int_0^\tau \int_{\mathbb{M}} e^g (P_tf)^2 \Gamma(h_n) d\mu(y) dt\right)^{\frac{1}{2}} \left(\int_0^\tau \int_{\mathbb{M}} e^g \Gamma(P_tf) d\mu(y) dt\right)^{\frac{1}{2}} \to 0,$
as $n\to \infty$. With the claim in hands we now let $n\to \infty$ in the above inequality
obtaining
$\int_{\mathbb{M}} e^{g(y,t)} ) (P_t f)^2 (y) d\mu(y) \le \int_{\mathbb{M}} e^{g(y,0)} ) f^2 (y) d\mu(y)$
$\square$

We are now ready for the main result of this lecture:

Theorem: For any $0 < \epsilon < 1$ there exist positive constants $C_1=C_1(\epsilon)$ and $C_2=C_2 (n,\epsilon)$, such that for every $x,y\in \mathbb{M}$ and $t > 0$ one has
$p(x,y,t)\le \frac{C_1}{\mu(B(x,\sqrt t))^{\frac{1}{2}}\mu(B(y,\sqrt t))^{\frac{1}{2}}} \exp \left(C_2Kt-\frac{d(x,y)^2}{(4+\epsilon)t}\right).$

Proof: Given $T > 0$, and $\alpha > 0$ we fix $0 < \tau \le (1+\alpha)T$. For a function $\psi\in C^\infty_0(\mathbb{M})$, with $\psi \ge 0$, in $\mathbb{M} \times (0,\tau)$ we consider the function
$f(y,t) = \int_{\mathbb{M}} p(y,z,t) p(x,z,T) \psi(z) d\mu(z),\ \ \ x\in \mathbb{M}.$
Since $f = P_t(p(x,\cdot,T)\psi)$, it satisfies the Cauchy problem
$\begin{cases} Lf - f_t = 0 \ \ \ \ \text{in}\ \mathbb{M} \times (0,\tau), \\ f(z,0) = p(x,z,T)\psi(z),\ \ \ z\in \mathbb{M}. \end{cases}$
Let $g :\mathbb{M}\times [0,\tau] \to \mathbb{R}$ be a non positive continuous function such that, in the sense of distributions,
$\frac{\partial g}{\partial t} +\frac{1}{2} \Gamma(g) \le 0.$
From the previous lemma, we know that:
$\int_{\mathbb{M}} e^{g(y,\tau)} f^2(y,\tau) d\mu(y) \le \int_{\mathbb{M}} e^{g(y,0)} f^2(y,0) d\mu(y).$
At this point we fix $x\in \mathbb{M}$ and for $0 < t \le\tau$ consider the indicator function $\mathbf 1_{B(x,\sqrt t)}$ of the ball $B(x,\sqrt t)$. Let $\psi_k\in C^\infty_0(\mathbb{M})$, $\psi_k \ge 0$, be a sequence such that $\psi_k \to \mathbf 1_{B(x,\sqrt t)}$ in $L^2(\mathbb{M})$, with supp$\ \psi_k\subset B(x,100\sqrt t)$. Slightly abusing the notation we now set $f(y,s) = P_s(p(x,\cdot,T)\mathbf{1}_{B(x,\sqrt t)})(y) = \int_{B(x,\sqrt t)} p(y,z,s) p(x,z,T) d\mu(z).$ Thanks to the symmetry of $p(x,y,s) = p(y,x,s)$, we have
$f(x,T) = \int_{B(x,\sqrt t)} p(x,z,T)^2 d\mu(z).$

Applying the integrated maximum principle to $f_k(y,s) = P_s(p(x,\cdot,T)\psi_k)(y)$, we find
$\int_{\mathbb{M}} e^{g(y,\tau)} f^2_k(y,\tau) d\mu(y) \le \int_{\mathbb{M}} e^{g(y,0)} f^2_k(y,0) d\mu(y).$
At this point we observe that as $k\to \infty$
$\left|\int_{\mathbb{M}} e^{g(y,\tau)} f^2_k(y,\tau) d\mu(y) - \int_{\mathbb{M}} e^{g(y,\tau)} f^2(y,\tau) d\mu(y)\right|$
$\le 2 ||e^{g(\cdot,\tau)}||_{L^\infty(\mathbb{M})} ||p(x,\cdot,T)||_{L^2(\mathbb{M})} ||p(x,\cdot,\tau)||_{L^\infty(B(x,110 \sqrt t))} ||\psi_k - \mathbf 1_{B(x,\sqrt t)}||_{L^2(\mathbb{M})} \to 0.$
By similar considerations we find
$\left|\int_{\mathbb{M}} e^{g(y,0)} f^2_k(y,0) d\mu(y) - \int_{\mathbb{M}} e^{g(y,0)} f^2(y,0) d\mu(y)\right|$
$\le 2 ||e^{g(\cdot,0)}||_{L^\infty(\mathbb{M})} ||p(x,\cdot,T)||_{L^\infty(B(x,110 \sqrt t))} ||\psi_k - \mathbf 1_{B(x,\sqrt t)}||_{L^2(\mathbb{M})} \to 0.$
Letting $k\to \infty$ we thus conclude that the same inequality holds with $f_k$ replaced by $f(y,s) = P_s(p(x,\cdot,T)1_{B(x,\sqrt t)})(y)$. This implies in particular the basic estimate
$\underset{z\in B(x,\sqrt t)}{\inf}\ e^{g(z,\tau)} \int_{B(x,\sqrt t)} f^2(z,\tau) d\mu(z)$
$\le \int_{B(x,\sqrt t)} e^{g(z,\tau)} f^2(z,\tau) d\mu(z) \le \int_{\mathbb{M}} e^{g(z,\tau)} f^2(z,\tau) d\mu(z)$
$\le \int_{\mathbb{M}} e^{g(z,0)} f^2(z,0) d\mu(z) = \int_{B(y,\sqrt t)} e^{g(z,0)} p(x,z,T)^2 d\mu(z)$
$\le \underset{z\in B(y,\sqrt t)}{\sup}\ e^{g(z,0)} \int_{B(y,\sqrt t)} p(x,z,T)^2 d\mu(z)$.

At this point we choose $g(y,t) = g_x(y,t) = - \frac{d(x,y)^2}{2((1+2\alpha) T - t)}.$ Using the fact that $\Gamma(d)\le 1$, one can easily check that $g$ satisfies
$\frac{\partial g}{\partial t} +\frac{1}{2} \Gamma(g) \le 0.$
Taking into account that
$\inf_{z\in B(x,\sqrt t)} e^{g_x(z,\tau)} = \inf_{z\in B(x,\sqrt t)} e^{-\frac{d(x,z)^2} {2((1+2\alpha)T- \tau)}} \ge e^{\frac{-t}{2((1+2\alpha)T- \tau)}},$
if we now choose $\tau = (1+\alpha)T$, then from the previous inequality we conclude that
$\int_{B(x,\sqrt t)} f^2(z,(1+\alpha)T) d\mu(z) \le \left(\underset{z\in B(y,\sqrt t)}{\sup}\ e^{-\frac{d(x,z)^2}{2(1+2\alpha)T} + \frac{t}{2\alpha T}}\right) \int_{B(y,\sqrt t)} p(x,z,T)^2 d\mu(z).$
We now apply the Harnack inequality which gives for every $z\in B(x,\sqrt t)$
$f(x,T)^2 \le f(z,(1+\alpha)T)^2(1+\alpha)^{n} e^{\frac{t}{2\alpha T}+\frac{Kt}{3} +\frac{nK\alpha T}{2 }}.$
Integrating this inequality on $B(x,\sqrt t)$ we find
$\left(\int_{B(y,\sqrt t)} p(x,z,T)^2 d\mu(z)\right)^2 = f(x,T)^2$
$\le \frac{(1+\alpha)^{n} e^{\frac{t}{2\alpha T}+\frac{Kt}{3} +\frac{nK\alpha T}{2 }}}{\mu(B(x,\sqrt t)) } \int_{B(x,\sqrt t)} f^2(z,(1+\alpha)T) d\mu(z).$
Thus, we obtain
$\int_{B(y,\sqrt t)} p(x,z,T)^2 d\mu(z) \le \frac{(1+\alpha)^{n} e^{\frac{t}{2\alpha T}+\frac{Kt}{3} +\frac{nK\alpha T}{2 }}}{\mu(B(x,\sqrt t)) } \left(\underset{z\in B(y,\sqrt t)}{\sup}\ e^{-\frac{d(x,z)^2}{2(1+2\alpha)T} + \frac{t}{2\alpha T}}\right).$
Choosing $T = (1+\alpha)t$ in this inequality we find
$\int_{B(y,\sqrt t)} p(x,z,(1+\alpha)t)^2 d\mu(z) \le \frac{(1+\alpha)^{n} e^{\frac{Kt}{3} +\frac{nK}{2} \alpha (1+\alpha)t+ \frac{1}{2\alpha (1+\alpha)}}}{\mu(B(x,\sqrt t))} \left(\underset{z\in B(y,\sqrt t)}{\sup}\ e^{-\frac{d(x,z)^2}{2(1+2\alpha)(1+\alpha)t} + \frac{1}{2\alpha (1+\alpha)}}\right).$
We now apply again the Harnack inequality to obtain for every $z\in B(y,\sqrt t)$
$p(x,y,t)^2 \le p(x,z,(1+\alpha)t)^2 (1+\alpha)^{n} \exp\left(\frac{1}{2\alpha }+\frac{Kt}{3}+\frac{nK\alpha t}{4} \right).$
Integrating this inequality in $z\in B(y,\sqrt t)$, we have
$\mu(B(y,\sqrt t)) p(x,y,t)^2 \le (1+\alpha)^{n} \exp\left(\frac{1}{2\alpha }+\frac{Kt}{3}+\frac{nK\alpha t}{4} \right) \int_{B(y,\sqrt t)} p(x,z,(1+\alpha)t)^2 d\mu(z).$
Combining this inequality with the above inequality we conclude
$p(x,y,t) \le \frac{(1+\alpha)^{n} e^{\frac{3+\alpha}{4\alpha(1+\alpha)} +\frac{Kt}{3} +\frac{nKt}{4} \left( \alpha^2 +\frac{3}{2} \alpha \right) }}{\mu(B(x,\sqrt t))^{\frac{1}{2}}\mu(B(y,\sqrt t))^{\frac{1} {2}}}\left(\underset{z\in B(y,\sqrt t)}{\sup}\ e^{-\frac{d(x,z)^2}{4(1+2\alpha)(1+\alpha)t}}\right).$
If now $x\in B(y,\sqrt t)$, then
$d(x,z)^2 \ge (d(x,y) - \sqrt t)^2 > d(x,y)^2 - t,$
and therefore
$\underset{z\in B(y,\sqrt t)}{\sup}\ e^{-\frac{d(x,z)^2}{4(1+2\alpha)(1+\alpha)t}} \le e^{\frac{1}{4(1+2\alpha)(1+\alpha)}} e^{-\frac{d(x,y)^2}{4(1+2\alpha)(1+\alpha)t}}.$
If instead $x\not\in B(y,\sqrt t)$, then for every $\delta > 0$ we have $d(x,z)^2 \ge (1-\delta) d(x,y)^2 - (1+ \delta^{-1}) t$. Choosing $\delta = \alpha/(\alpha+1)$ we find $d(x,z)^2 \ge \frac{d(x,y)^2}{1+\alpha} - (2 + \alpha^{-1}) t,$ and therefore
$\underset{z\in B(y,\sqrt t)}{\sup}\ e^{-\frac{d(x,z)^2}{4(1+2\alpha)(1+\alpha)t}} \le e^{-\frac{d(x,y)^2} {4(1+2\alpha)(1+\alpha)^2 t} + \frac{2 + \alpha^{-1}}{4(1+2\alpha)(1+\alpha)}}$
For any $\epsilon > 0$ we now choose $\alpha > 0$ such that $4(1+2\alpha)(1+\alpha)^2 = 4+\epsilon$ to reach the desired conclusion $\square$

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