Lecture 21. The Poincaré inequality on domains

Let (\mathbb{M},g) be a complete Riemannian manifold and \Omega \subset \mathbb{M} be a non empty bounded set. Let \mathcal{D}^\infty be the set of smooth functions f \in C^\infty(\bar{\Omega}) such that for every g \in C^\infty(\bar{\Omega}), \int_\Omega g Lf d\mu=-\int_\Omega \Gamma(f,g) d\mu.
It is easy to see that L is essentially self-adjoint on \mathcal{D}^\infty . Its Friedrichs extension, still denoted L, is called the Neumann Laplacian on \Omega and the semigroup it generates, the Neumann semigroup. If the boundary \partial \Omega is smooth, then it is known from the Green’s formula that
\int_\Omega g Lf d\mu=-\int_\Omega \Gamma(f,g) d\mu+\int_{\partial \Omega} g Nf d\mu,
where N is the normal unit vector. As a consequence, f \in \mathcal{D}^\infty if and only if Nf=0. However, we stress that no regularity assumption on the boundary \partial \Omega is needed to define the Neumann Laplacian and the Neumann semigroup.

Since \bar{\Omega} is compact, the Neumann semigroup is a compact operator and -L has a discrete spectrum 0 =\lambda_0 <\lambda_1 \le \cdots. We get then, the so-called Poincaré inequality on \Omega: For every f \in C^\infty(\bar{\Omega}),
\int_\Omega (f-f_\Omega)^2 d\mu \le \frac{1}{\lambda_1} \int_\Omega \Gamma(f) d\mu.
Our goal is in this lecture will be to try to understand how the constant \lambda_1 depends on the size of the set \Omega. A first step in that direction was made by Poincaré himself in the Euclidean case.

Theorem: If \Omega \subset \mathbb{R}^n is a bounded open convex set then for a smooth f : \bar{ \Omega} \to \mathbb{R} with \int_\Omega f(x) dx=0,
\frac{C_n}{\mathbf{diam}(\Omega)^2} \int_\Omega f^2 (x)dx \le  \int_\Omega \| \nabla f (x) \|^2 dx.
where C_n is a constant depending on n only.

Proof: The argument of Poincare is beautifully simple.
\frac{1}{\mu( \Omega )}  \int_\Omega f^2 (x)dx  =\frac{1}{2} \frac{1}{\mu( \Omega )}  \int_\Omega f^2 (x)dx+\frac{1}{2} \frac{1}{\mu( \Omega )}  \int_\Omega f^2 (y)dy
=\frac{1}{2} \frac{1}{\mu( \Omega )^2} \int_\Omega \int_\Omega (f(x)-f(y))^2 dx dy.
We now have
f(x)-f(y)=\int_0^1 (x-y)\cdot \nabla f(tx +(1-t)y) dt,
which implies
(f(x)-f(y))^2 \le \mathbf{diam}(\Omega)^2\int_0^1 \| \nabla f \|^2 (tx +(1-t)y) dt.
By a simple change of variables, we see that
\int_\Omega \int_\Omega  \| \nabla f \|^2 (tx +(1-t)y) dxdy =\frac{1}{t^n}  \int_\Omega \int_{t \Omega+(1-t)y}   \| \nabla f \|^2 (u) dudy
=\frac{1}{t^n}  \int_\Omega \int_{ \Omega}  \mathbf{1}_{t \Omega+(1-t)y} (u) \| \nabla f \|^2 (u) dudy.
Now, we compute
\int_{ \Omega}  \mathbf{1}_{t \Omega+(1-t)y} (u) dy=\mu \left( \Omega \cap \frac{1}{1-t} (u-t\Omega) \right)\le \min \left( 1, \frac{t^n}{(1-t)^n} \right) \mu(\Omega).
As a consequence we obtain
\frac{1}{\mu( \Omega )}  \int_\Omega f^2 (x)dx \le \frac{ \mathbf{diam}(\Omega)^2}{2\mu( \Omega )}\int_0^1  \min \left( 1, \frac{t^n}{(1-t)^n} \right)\frac{dt}{t^n}  \int_\Omega \| \nabla f (x) \|^2 dx
\square

It is known (Payne-Weinberger) that the optimal constant C_n is \pi^2.

In this Lecture, we extend the above inequality to the case of Riemannian manifolds with non negative Ricci curvature. The key point is a lower bound on the Neumann heat kernel of \Omega. From now on we assume that \mathbf{Ric} \ge 0 and consider an open set in \mathbb{M} that has a smooth and convex boundary in the sense the second fundamental form of \partial \Omega is non negative. Due to the convexity of the boundary, all the results we obtained so far may be extended to the Neumann semigroup. In particular, we have the following lower bound on the Neumann heat kernel:

Theorem:
Let p^N (x,y,t) be the Neumann heat kernel of \Omega. There exists a constant C depending only on the dimension of \mathbb{M} such that for every t > 0, x,y \in \mathbb{M},
p^N (x,y,t) \ge \frac{C}{\mu( B(x,\sqrt{t}))} \exp \left(-\frac{d(x,y)^2}{3t}\right).

As we shall see, this directly implies the following Poincare inequality:

Theorem: For a smooth f : \bar{ \Omega} \to \mathbb{R} with \int_\Omega f  d\mu=0,
\frac{C_n}{\mathbf{diam}(\Omega)^2} \int_\Omega f^2 d\mu \le  \int_\Omega \Gamma(f) d\mu.
where C is a constant depending on the dimension of \mathbb{M} only.

Proof: We denote by R the diameter of \Omega. From the previous lower bound on the Neumann kernel of \Omega, we have
p^N (x,y,R^2) \ge \frac{C}{\mu( \Omega )},
where C only depends on n. Denote now by P_t^N the Neumann semigroup. We have for f \in \mathcal{D}^\infty
P^N_{R^2} (f^2)-(P^N_{R^2} f)^2 =\int_0^{R^2} \frac{d}{dt}  P^N_{t} ((P^N_{R^2 -t} f)^2) dt.
By integrating over \Omega, we find then,
\int_\Omega P^N_{R^2} (f^2)-(P^N_{R^2} f)^2 d\mu   =-\int_0^{R^2} \int_\Omega  \frac{d}{dt}  (P^N_{t} f)^2 d\mu dt
=2 \int_0^{R^2} \int_\Omega  \Gamma( P^N_{t} f, P^N_{t} f ) d\mu dt
\le 2 R^2 \int_\Omega \Gamma(f) d\mu.
But on the other hand, we have
P^N_{R^2} (f^2)(x)-(P^N_{R^2} f)^2(x) =P^N_{R^2}\left[ \left( f -(P^N_{R^2} f)(x) \right)^2 \right](x)
\ge \frac{C}{\mu(\Omega)} \int_\Omega (f(y)- (P^N_{R^2} f)(x) )^2 d\mu(y)
which gives
\int_\Omega P^N_{R^2} (f^2)-(P^N_{R^2} f)^2 d\mu \ge C \int_\Omega \left( f(x) -\frac{1}{\mu(\Omega)}  \int_\Omega f d\mu \right)^2 d\mu(x)
The proof is complete \square

In applications, it is often interesting to have a scale invariant Poincare inequality on balls. If the manifold \mathbb{M} has conjugate points, the geodesic spheres may not be convex and thus the previous argument does not work. However the following result still holds true:

Theorem: There exists a constant C_n > 0 depending only on the dimension of \mathbb{M} such that for every r > 0 and every smooth f :B(x,r) \to \mathbb{R} with \int_{B(x,r)} f  d\mu=0,
\frac{C_n}{r^2} \int_{B(x,r)}  f^2 d\mu \le  \int_{B(x,r)} \Gamma(f) d\mu.

We only sketch the argument. By using the global lower bound
p (x,y,t) \ge \frac{C}{\mu( B(x,\sqrt{t}))} \exp \left(-\frac{d(x,y)^2}{3t}\right),
for the heat kernel, it is possible to prove a lower bound for the Neuman heat kernel on the ball B(x_0,r): For x,y \in B(x_0.r/2),
p^N (x,y,r^2) \ge \frac{C}{\mu( B(x_0,r))},
Arguing as before, we get
\frac{C_n}{r^2} \int_{B(x_0,{r/2})}  f^2 d\mu \le  \int_{B(x_0,r)} \Gamma(f) d\mu.
and show then that the integral on the left hand side can be taken on B(x_0,{r}) by using a Whitney’s type covering argument.

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