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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