# Estimates on the Green function of an elliptic second order differential operator.

Let $D$ be a linear differential elliptic operator of second order with infinitely smooth coefficients acting on real valued functions on a compact manifold $M$. Let us assume that $D$ has no free term, i.e. $D(1)=0$. Let us fix a smooth positive measure (density) $\mu$ on $M$. Does there exist a (integrable) Green function $G\colon M\times M\to \mathbb{R}$ with the following properties:

(1) $\int_M G(x,y) \cdot D\phi(y) d\mu(y) =\int_M\phi(y) d\mu(y) -\phi(x)$ for any function $\phi$ and $x\in M$ (this is the definition of Green function);

(2) $G$ is infinitely smooth outside of the diagonal;

(3) $G$ is bounded from below.

The last property can be asked in a stronger form:

(3') Does $G$ satisfy the asymptotic estimate near the diagonal: $$c|x-y|^{2-n}\leq G(x,y)\leq C|x-y|^{2-n}$$ where $c,C>0$ and $n=\dim M>2$. If $n=2$ there should be a logarithmic estimate.

I am pretty sure that this is true and should be well known. I would need a reference. The special case when $D$ is the Laplacian for a Riemannian metric on $M$ is contained explicitly in some textbooks I am familiar with.

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Semyon, (1) seems to assume that the kernel of $D$ consists only of constant functions but that's not necessarily the case. –  Deane Yang Jul 5 '11 at 9:25
To add to what Deane said: lower order terms sometimes matter when you try to apply maximum principle. Another condition is needed. –  Willie Wong Jul 5 '11 at 9:48
Thanks, Deane. You are right. In my situation $D$ has no free term. I have added this condition to the question. –  semyon alesker Jul 5 '11 at 9:56
Semyon, I learned how to prove (1) and (2) from books on pseudodifferential operators, probably the one by Chazarain and Piriou. In particular, (2) just follows from local elliptic regularity estimates that are quite easy to prove using pseudodifferential operators. But (3) and (3') are deeper and, as Willie says, probably require more assumptions that you have stated. My guess (and that's all it is) is that your conclusion holds if $Du = \partial_i(a^{ij}\partial_ju)$. But this is essentially the same case as the Laplacian for a Riemannian metric. –  Deane Yang Jul 5 '11 at 16:14
Deane, (1) and (2) are indeed standard. Regarding the operators of the form $Du=\partial_i(a^{ij}\partial_ju)$, they are not coordinate independent. Strictly speaking, even the Riemannian Laplacian does not have this form: $\partial_i$ should be replaced by covariant derivatives $\nabla_i$, and for this you need a metric. I would be surprised if there is no generalizations beyond Riemannian Laplacians. –  semyon alesker Jul 6 '11 at 7:34