Let be a complete -dimensional Riemannian manifold and, as usual, denote by its Laplace-Beltrami operator. As in the previous lecture, we will assume that the Ricci curvature of is bounded from below by with . 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 be a non positive continuous function such that, in the sense of distributions,
then, for every , we have
.
*

**Proof:** Since

, multiplying this identity by , where is the usual localizing sequence , and integrating by parts, we obtain

where in the last inequality we have made use of the assumption on . From this we conclude

We now claim that

To see this we apply Cauchy-Schwarz inequality which gives

as . With the claim in hands we now let in the above inequality

obtaining

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

**Theorem:** * For any there exist positive constants and , such that for every and one has
*

**Proof:** Given , and we fix . For a function , with , in we consider the function

Since , it satisfies the Cauchy problem

Let be a non positive continuous function such that, in the sense of distributions,

From the previous lemma, we know that:

At this point we fix and for consider the indicator function of the ball . Let , , be a sequence such that in , with supp. Slightly abusing the notation we now set Thanks to the symmetry of , we have

Applying the integrated maximum principle to , we find

At this point we observe that as

By similar considerations we find

Letting we thus conclude that the same inequality holds with replaced by . This implies in particular the basic estimate

.

At this point we choose Using the fact that , one can easily check that satisfies

Taking into account that

if we now choose , then from the previous inequality we conclude that

We now apply the Harnack inequality which gives for every

Integrating this inequality on we find

Thus, we obtain

Choosing in this inequality we find

We now apply again the Harnack inequality to obtain for every

Integrating this inequality in , we have

Combining this inequality with the above inequality we conclude

If now , then

and therefore

If instead , then for every we have . Choosing we find and therefore

For any we now choose such that to reach the desired conclusion