Lecture 10. The distance associated to Laplace-Beltrami operators

In order to apply the diffusion semigroup theory developed in the first lectures and construct without ambiguity the semigroup associated to the Laplace-Beltrami operator L, we need to know if L is essentially self-adjoint. Interestingly, this property of essential self-adjointness is clodely related to a metric property of the underlying Riemannian structure: The completeness of the associated distance.

Given an absolutely continuous curve $\gamma: [0,T] \rightarrow \mathbb{R}^n$ , we define its Riemannian length by $L_g (\gamma)=\int_0^T \sqrt{g_{\gamma(s)} (\gamma'(s),\gamma'(s) )} ds.$ If $x,y \in \mathbb{R}^n$ , let us denote by $\mathcal{C}(x,y)$ the set of absolutely continuous curves $\gamma: [0,1] \rightarrow \mathbb{R}^n$ such that $\gamma(0)=x, \gamma(1)=y.$ The Riemannian distance between $x$ and $y$ is defined by $d(x,y)=\inf_{\gamma \in \mathcal{C}(x,y)} L_g(\gamma).$ By using reparametrization, we may define the Riemannian distance in a equivalent way by using the notion of sub-unit curve. Let $\gamma: [0,T] \to \mathbb{R}^n$ be an absolutely continuous curve. Since the vector fields $V_1,\cdots,V_n$ ‘s form a basis of $\mathbb{R}^n$ at each point, we may find continuous functions $\alpha_i:[0,T] \to \mathbb{R}^n$ such that $\gamma'(t)=\sum_{i=1}^n \alpha_i(t) V_i (\gamma(t)).$ The curve $\gamma$ is then said to be sub-unit if for almost every $t \in [0,T]$ , $\sum_{i=1}^n \alpha_i(t)^2 \le 1$ . By using reparametrization, it is easily seen that for $x,y \in \mathbb{R}^n$ , $d(x,y)=\inf \left\{ T \text{ such that there exists a sub-unit curve } \gamma:[0,T] \to \mathbb{R}^n, \gamma(0)=x, \gamma(1)=y \right\}$ .

Exercise: Let $\theta_1, \cdots , \theta_n \in \mathbb{R}$ . Show that for every $x \in \mathbb{R}^n$ , $d \left(x, e^{\sum_{i=1}^n \theta_i V_i } x \right) \le \sum_{i=1}^n \theta_i^2.$

An important fact is that $d$ hence defined is indeed a distance that induces the usual topology of $\mathbb{R}^n$ .

Definition: The function $d$ defined above is a distance that induces the usual topology of $\mathbb{R}^n$ .

Proof: Since any curve can be parametrized backwards and forwards, we have $d(x,y)=d(y,x)$ . The triangle inequality is easily proved by using juxtaposition of curves. Plainly $d(x,x)=0$ , so it remains to prove that if $x \neq y$ then $d(x,y) > 0$ . Let $x,y \in \mathbb{R}^n$ such that $x \neq y$ . Let us denote $R=\| x-y \|$ . The closed Euclidean ball $\bar{B}_e (x,R)$ is compact, therefore there exist two constants $\alpha, \beta > 0$ such that for every $z \in \bar{B}_e (x,R)$ and $u \in \mathbb{R}^n$ , $\alpha^2 \| u \|^2 \le g_z(u,u) \le \beta^2 \| u \|^2.$ Let now $\gamma:[0,1]\to \mathbb{R}^n$ be an absolutely continuous curve such that $\gamma(0)=x, \gamma(1)=y$ . Let $\tau =\inf \{ t, \| \gamma(t) \|=R \}.$
We have
$L_g(\gamma)$
$=\int_0^1 \sqrt{g_{\gamma(s)} (\gamma'(s),\gamma'(s) )} ds$
$\ge \int_0^\tau \sqrt{g_{\gamma(s)} (\gamma'(s),\gamma'(s) )} ds$
$\ge \alpha \int_0^\tau \| \gamma'(s) \| ds$
$\ge \alpha \int_0^\tau \| \gamma'(s) \| ds$
$\ge \alpha \| \gamma(\tau) - \gamma (0) \|$
$\ge \alpha \| x -y \|.$
As a consequence, we deduce that $d(x,y) \ge \alpha^2 \| x -y \| > 0.$ Therefore $d$ is indeed a distance. Moreover, it is shown as above that for every $z \in \mathbb{R}^n, R> 0$ , there are constants $C_1,C_2> 0$ such that for every $x,y \in \bar{B}_e (z,R)$ , $C_1 \| x-y \| \le d(x,y) \le C_2 \| x-y \|.$ This implies that $d$ induces the usual topology of $\mathbb{R}^n$ $\square$

As shown in the following proposition, the distance $d$ is intrinsically associated to the Laplace-Beltrami operator.

Proposition: For $x,y \in \mathbb{R}^n$ , we have $d(x,y) =\sup \{ | f(x)-f(y) |, f \in \mathcal{C}_c^\infty(\mathbb{R}^n,\mathbb{R}), \| \Gamma(f,f) \|_\infty \le 1 \}.$

Proof: Let $x,y \in \mathbb{R}^n$ . We denote $\delta (x,y)=\sup \{ | f(x)-f(y) |, f \in \mathcal{C}_c^\infty(\mathbb{R}^n,\mathbb{R}), \| \Gamma(f,f) \|_\infty \le 1 \}.$ Let $\gamma: [0,T] \to \mathbb{R}$ be a sub-unit curve such that $\gamma(0)=x, \gamma(T)=x.$ We can find $\alpha_1,\cdots,\alpha_n:[0,T] \to \mathbb{R}^n$ such that $\gamma'(t)=\sum_{i=1}^n V_i (\gamma(t)) \alpha_i (t),$ and $\sum_{i=1}^n \alpha_i^2 \le 1$ . Let now $f\in \mathcal{C}^\infty(\mathbb{R}^n,\mathbb{R}), \| \Gamma(f,f) \|_\infty \le 1$ . From the change of variable formula we have, $f(\gamma(T))= f(\gamma(0)) + \sum_{i=1}^n \int_0^T V_i f(\gamma(s)) \alpha_i(s) ds.$ Therefore, from Cauchy-Schwarz inequality, $\left| f(y)-f(x) \right| \le T.$ As a consequence $\delta (x,y) \le d(x,y).$

We now prove the converse inequality which is trickier. The idea is to use the function $f(y)=d(x,y)$ that satisfies $| f(x) -f(y)|=d(x,y)$ and $"\Gamma(f,f) =1"$ . However, giving a precise meaning to $\Gamma(f,f) =1$ is not so easy, because it turns out that $f$ is not differentiable at $x$ . It suggests to use an approximation of the identity to regularize $f$ and avoid the discussion of this differentiability issue. More precisely, fix $x_o ,y_o\in \mathbb{R}^n$ , and for $N \ge 1$ , consider the function $\Psi_N(y)= \eta \left( \int_{\mathbb{R}^n} \rho_N (t) d(x_o,y-t) dt \right),$ where $\rho \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ , $\rho \ge 0$ , $\int_{\mathbb{R}^n} \rho=1$ , $\rho_N(t)=N^n \rho(N t)$ and $\eta \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})$ , $\eta \ge 0$ , has the property that $\| \eta ' \|_\infty \le 1$ and $\eta (u)=u$ for $u \in [0,1+d(x_o,y_o)]$ . Since for any $\theta \in \mathbb{R}^{n}$ , $\| \theta \|=1$ , $t \ge 0$ $| d(x_o,y)-d\left(x_o, e^{t \sum_{i=1}^n \theta_i V_i} y\right)| \le d\left(y, e^{t \sum_{i=1}^n \theta_i V_i} y\right) \le t,$ it is easy to see that $\Gamma(\Psi_N,\Psi_N) \le 1+\frac{C}{N}$ , for some constant $> 0$ . Hence $\delta(x_o,y_o) \ge \lim \inf_{N \to +\infty} | \Psi_N(y_o) -\Psi_N(x_o) |=d(x_o,y_o)$ $\square$

The following theorem is known as the Hopf–Rinow theorem, it provides a necessary and sufficient condition for the completeness of the metric space $(\mathbb{R}^n,d)$ .

Proposition: The metric space $(\mathbb{R}^n,d)$ is complete (i.e. Cauchy sequences are convergent) if and only the compact sets are the closed and bounded sets.

Proof: It is clear that if closed and bounded sets are compact then the metric space $(\mathbb{R}^n,d)$ is complete; It comes from the fact that Cauchy sequence are convergent if and only if they have at least one cluster value. So, we need to prove that closed and bounded sets for the distance $d$ are compact provided that $(\mathbb{R}^n,d)$ is complete. To check this, it is enough to prove that closed balls are compact. Let $x \in \mathbb{R}^n$ . Observe that if the closed ball $\bar{B}(x, r)$ is compact for some $r> 0$ , then $\bar{B}(x, \rho)$ is closed for any $\rho < r$ . Define $R=\sup \{ r> 0, \bar{B}(x, r) \text{ is compact } \}.$ Since $d$ induces the usual topology of $\mathbb{R}^n$ , $R>0$ . Let us assume that $R < +\infty$ and let us show that it leads to a contradiction. We first show that $\bar{B}(x, R)$ is compact. Since $(\mathbb{R}^n,d)$ is assumed to be complete, it suffices to prove that $\bar{B}(x, R)$ is totally bounded: That is, for every $\varepsilon > 0$ there is a finite set $S_\varepsilon$ such that every point of $\bar{B}(x, R)$ belongs to a $\varepsilon$ -neighborhood of $S_\varepsilon$ .

So, let $\varepsilon > 0$ small enough. By definition of $R$ , the ball $\bar{B}(x, R-\varepsilon / 4)$ is compact; It is therefore totally bounded. We can find a finite set $S=\{ y_1,\cdots,y_N\}$ such that every point of $\bar{B}(x, R-\varepsilon / 4)$ lies in a $\varepsilon / 2$ -neighborhood of $S$ . Let now $y \in \bar{B}(x, R)$ . We claim that there exists $y' \in \bar{B}(x, R-\varepsilon / 4)$ such that $d(y,y') \le \varepsilon /2$ . If $y \in \bar{B}(x, R-\varepsilon / 4)$ , there is nothing to prove, we may therefore assume that $y \notin \bar{B}(x, R-\varepsilon / 4)$ . Consider then a sub-unit curve $\gamma: [0,R+\varepsilon / 4]$ such that $\gamma(0)=x$ , $\gamma(R+\varepsilon/4)=y$ . Let $\tau =\inf \{t, \gamma(t) \notin \bar{B}(x, R-\varepsilon / 4) \}.$ We have $\tau \ge R-\varepsilon / 4$ . On the other hand, $d(\gamma(\tau), \gamma(R+\varepsilon/4)) \le R+\varepsilon/4 -\tau.$ As a consequence, $d(\gamma(\tau),y) \le \varepsilon /2.$ In every case, there exists therefore $y' \in \bar{B}(x, R-\varepsilon / 4)$ such that $d(y,y') \le \varepsilon /2$ . We may then pick $y_k$ in $S$ such that $d(y_k,y') \le \varepsilon / 2$ . From the triangle inequality, we have $d(y,y_k) \le \varepsilon$ . So, at the end, it turns out that every point of $\bar{B}(x, R)$ lies in a $\varepsilon$ -neighborhood of $S$ . This shows that $\bar{B}(x, R)$ is totally bounded and therefore compact because $(\mathbb{R}^n,d)$ is assumed to be complete. Actually, the previous argument shows more, it shows that if every point of $\bar{B}(x, R)$ lies in a $\varepsilon /2$ -neighborhood of a finite $S$ , then every point of $\bar{B}(x, R+\varepsilon/4)$ will lie $\varepsilon$ -neighborhood of $S$ , so that the ball $\bar{B}(x, R+\varepsilon/4)$ is also compact. This contradicts the fact the definition of $R$ . Therefore every closed ball is a compact set, due to the arbitrariness of $x$ $\square$

Checking that the metric space $(\mathbb{R}^n,d)$ is complete is not always easy in concrete situations. From the Hopf-Rinow theorem, it suffices to prove that the closed balls are compact. The following proposition is therefore useful.

Proposition: Suppose that the vector fields $V_1,\cdots,V_n$ ‘s have globally Lipschitz coefficients . Then the closed ball $\bar B(x,R)$ is compact for every $x \in \mathbb{R}^n$ and $R > 0$ . As a consequence the metric space $(\mathbb{R}^n,d)$ is complete.
Proof: By the hypothesis on the $V_j$ ‘s there exists a constant $M > 0$ such
that $\|V(x)\|=\left(\sum_{j=1}^n \|V_j(x)\|^2\right)^\frac{1}{2}\leq M(1+\|x\|)$ for any $x\in\mathbb{R}^n$ . Fix $x_o,y\in\mathbb{R}^n$ and let $\gamma:[0,T]\to\mathbb{R}^n$ , be a sub-unit curve such that $\gamma(0)=x_o, \gamma(T)=y.$ Letting $y(t)=\|\gamma(t)\|^2$ we obtain $y'(t)=2\leq 2 \|\gamma(t)\|\|\gamma'(t)\| \leq 2\|\gamma(t)\|\|V(\gamma(t))\|.$ We infer that $y'(t)\leq C\left(\sqrt{y(t)}+y(t) \right)$ , for some $C>0$ depending only on $M$ . Integrating the latter inequality one has $\|\gamma(t)\| \leq (A+\| x_o \|) e^{\frac{C}{2}T},\quad t\in [0,T]$ . for some constant $A >0$ . The previous estimate shows in particular, that $B(x_o,R)\subset B_e(0, (A+\| x_o \|) e^{CR}).$ We conclude that $\bar B(x_o,R)$ is Euclidean compact. Since the metric and the Euclidean topology coincide, it is also $d$ -compact $\square$

Completeness of the metric space $(\mathbb{R}^n,d)$ is related to the essential self-adjointness of the Laplace-Beltrami operator.

Theorem: If the metric space $(\mathbb{R}^n,d)$ is complete, then the Laplace-Beltrami operator $L$ is essentially self-adjoint.

Proof: We know that if there exists an increasing sequence $h_n\in C_c(\mathbb{R}^n,\mathbb{R})$ such that $h_n\nearrow 1$ on $\mathbb{R}^n$ , and $||\Gamma(h_n,h_n)||_{\infty} \to 0$ , as $n\to \infty$ , then the operator $L$ is essentially self-adjoint. We are therefore reduced to prove the existence of such a sequence. Let us fix a base point $x_0\in\mathbb{R}^n$ . We can find an exhaustion function $\rho\in C^\infty(\mathbb{R}^n,\mathbb{R})$ such that $|\rho - d(x_0,\cdot)| \le L,\ \ \ \ \ \ |\Gamma(\rho,\rho)|\le L \ \ \text{on } \mathbb{R}^n.$ By the completeness of $(\mathbb{R}^n,d)$ and the Hopf-Rinow theorem, the level sets $\Omega_s = \{x\in \mathbb{M} \mid \rho(x) < s\}$ are relatively compact and, furthermore, $\Omega_s \nearrow \mathbb{R}^n$ as $s\to \infty$ . We now pick an increasing sequence of functions $\phi_n\in C^\infty([0,\infty))$ such that $\phi_n\equiv 1$ on $[0,n]$ , $\phi_n \equiv 0$ outside $[0,2n]$ , and $|\phi_n'|\le \frac{2}{n}$ . If we set $h_n(x) = \phi_n(\rho(x))$ , then we have $h_n\in C_c(\mathbb{R}^n,\mathbb{R})$ , $h_n\nearrow1$ on $\mathbb{R}^n$ as $n\to \infty$ , and $||\Gamma(h_n,h_n)||_{\infty} \le \frac{2L}{n}$ $\square$