Lecture 14. Stochastic completeness of the heat semigroup

In this Lecture, we will prove a first interesting consequence of the Bochner’s identity: We will prove that if, on a complete Riemannian manifold \mathbb{M}, the Ricci curvature is bounded from below, then the heat semigroup is stochastically complete, that is P_t 1=1. This result is due to S.T. Yau, and we will see this property is also equivalent to the uniqueness in L^\infty for solutions of the heat equation. The proof we give is due to D. Bakry.

Let \mathbb{M} be a complete Riemannian manifold and denote by L its Laplace-Beltrami operator. As usual, we denote by P_t the heat semigroup generated by L. Throughout the Lecture, we will assume that the Ricci curvature of \mathbb{M} is bounded from below by \rho \in \mathbb{R}. As seen in the previous Lecture, this is equivalent to the fact that for every f \in C^\infty(\mathbb{M}),
\Gamma_2(f,f) \ge \frac{1}{n} (Lf)^2 + \rho \Gamma(f,f).

We start with a technical lemma:

Lemma: If f \in L^2_\mu(\mathbb{M}), then for every t > 0, the functions \Gamma (P_t f), L\Gamma(P_t f), \Gamma(P_tf, LP_t f) and \Gamma_2 (P_t f) are in L^1_\mu(\mathbb{M}).

Proof: It is straightforward to see from the spectral theorem that \Gamma (P_t f) \in L^1_\mu(\mathbb{M}). Similarly, | \Gamma(P_tf, LP_t f) | \le \sqrt{\Gamma(P_tf) \Gamma(LP_tf) }   \in L^1_\mu(\mathbb{M}). Since, \Gamma_2 (P_t f) =\frac{1}{2} \left( L\Gamma(P_t f) -2  \Gamma(P_tf, LP_t f)\right), we are let with the problem of proving that \Gamma_2 (P_t f) \in  L^1_\mu(\mathbb{M}). If g \in C_0^\infty(\mathbb{M}), then an integration by parts easily yields \int_\mathbb{M} \Gamma_2 (g) d\mu=\int_\mathbb{M} (Lg)^2 d\mu. As a consequence,
\int_\mathbb{M} \Gamma_2 (g)-\rho \Gamma(g)  d\mu=\int_\mathbb{M} (Lg)^2 +\rho g Lg  d\mu,
and we obtain
\int_\mathbb{M} | \Gamma_2 (g)-\rho \Gamma(g) | d\mu \le\left(1 +\frac{1}{2}| \rho | \right) \int_\mathbb{M} (Lg)^2 d\mu +\frac{1}{2}| \rho | \int_\mathbb{M} g^2    d\mu.
Using a density argument, it is then easily proved that for g \in \mathcal{D}(L)\cap C^\infty(\mathbb{M}) we have
\int_\mathbb{M} | \Gamma_2 (g)-\rho \Gamma(g) | d\mu \le\left(1 +\frac{1}{2}| \rho | \right) \int_\mathbb{M} (Lg)^2 d\mu +\frac{1}{2}| \rho | \int_\mathbb{M} g^2    d\mu.
In particular, we deduce that if g \in \mathcal{D}(L)\cap C^\infty(\mathbb{M}), then \Gamma_2(g) \in L^1_\mu(\mathbb{M}) \square

We will also need the following fundamental parabolic comparison theorem that shall be extensively used throughout these lectures.

Proposition: Let T > 0. Let u,v: \mathbb{M}\times [0,T] \to \mathbb{R} be smooth functions such that:

  • For every t \in [0,T], u(\cdot,t) \in L^2(\mathbb{M}) and \int_0^T \| u(\cdot,t)\|_2 dt <\infty;
  • \int_0^T \| \sqrt{\Gamma(u) (\cdot,t)} \|_p dt <\infty for some 1 \le p \le \infty;
  • For every t \in [0,T], v(\cdot,t) \in L^q(\mathbb{M}) and \int_ 0^T \| v(\cdot,t ) \|_q dt <\infty for some 1 \le q \le \infty.

If the inequality
Lu+\frac{\partial u}{\partial t} \ge v,
holds on \mathbb{M}\times [0,T], then we have
P_T u(\cdot,T)(x) \ge u(x,0) +\int_0^T P_s v(\cdot,s)(x) ds.

Proof: Let f,g \in C_0^\infty (\mathbb{M}), f,g \ge 0. We claim that we must have
\int_\mathbb{M} g P_T(fu(\cdot,T)) d\mu - \int_\mathbb{M} g f u(x,0) d\mu
\ge   -  \|\sqrt{\Gamma(f)}\|_\infty \int_0^T  \int_\mathbb{M}   (P_t g) \sqrt{\Gamma(u)}d\mu dt-  \| \sqrt{\Gamma(f)} \|_\infty \int_0^T \| \sqrt{\Gamma(P_t g) }\|_2 \| u(\cdot,t) \|_2   dt  +  \int_\mathbb{M} g \int_0^T  P_t( f v(\cdot,t)) d\mu dt

where for every 1\le p \le \infty and a measurable F, we have let ||F||_p = ||F||_{L^p(\mathbb{M})}. To establish this, we consider the function
\phi(t)=\int_\mathbb{M} g P_t (fu(\cdot,t)) d\mu.
Differentiating \phi we find
\phi'(t) =\int_\mathbb{M} g P_t \left(L( fu) + f\frac{\partial u}{\partial t} \right) d\mu
= \int_\mathbb{M} g P_t \left((L f) u+2 \Gamma (f,u) +f Lu + f\frac{\partial u}{\partial t} \right) d\mu
\ge \int_\mathbb{M} g P_t \left((L f) u+2 \Gamma (f,u)  \right) d\mu+\int_\mathbb{M} g P_t( f v) d\mu.
Since
\int_\mathbb{M} g P_t \left((L f) u\right) d\mu = \int_\mathbb{M}  (P_t g) (L f) u d\mu
= -\int_\mathbb{M}  \Gamma( f, u(P_t g)) d\mu
=-\left( \int_\mathbb{M}  P_t g \Gamma( f, u)+ u \Gamma(f,P_t g) d\mu\right),
we obtain
\phi'(t) \ge \int_\mathbb{M} P_t g \Gamma(f,u) d\mu - \int_\mathbb{M} u \Gamma(f,P_t g) d\mu + \int_\mathbb{M} g P_t(fv) d\mu.
Now, we can bound
\left| \int_\mathbb{M}  (P_t g) \Gamma( f, u) d\mu\right|   \le    \| \sqrt{\Gamma(f) } \|_\infty \int_\mathbb{M}  (P_t g)  \sqrt{\Gamma( u)} d\mu,
and for a.e. t\in [0,T] the integral in the right-hand side is finite. We have thus obtained
\phi'(t)  \ge -  \| \sqrt{\Gamma(f) } \|_\infty  \int_\mathbb{M}  (P_t g)  \sqrt{\Gamma(u)} d\mu- \int_\mathbb{M} u \Gamma(f , P_t g) d\mu+ \int_\mathbb{M} g P_t( f v(\cdot,t)) d\mu.
As a consequence, we find
\int_\mathbb{M} g P_T (fu(\cdot,T)) d\mu-\int_\mathbb{M} gfu(x,0)d\mu
\ge  -  \| \sqrt{\Gamma(f) } \|_\infty   \int_0^T  \int_\mathbb{M}   (P_t g) \sqrt{\Gamma(u)} d\mu dt -  \int_0^T \int_\mathbb{M}  u \Gamma\left(f,P_t g\right) d\mu dt + \int_0^T  \int_\mathbb{M} g  P_t(f v(\cdot,t)) d\mu dt
\ge   -  \| \sqrt{\Gamma(f) } \|_\infty   \int_0^T \int_\mathbb{M} (P_t g) \sqrt{\Gamma(u)} d\mu dt - \int_0^T \| u(\cdot,t) \|_2 \| \Gamma(f, P_t g) \|_2 dt  + \int_\mathbb{M} g \int_0^T  P_t( f v(\cdot,t)) dt d\mu
\ge     -  \| \sqrt{\Gamma(f) } \|_\infty   \int_0^T  \int_\mathbb{M} (P_t g) \sqrt{\Gamma(u)} d\mu dt -  \| \sqrt{\Gamma(f)} \|_\infty \int_0^T \| u(\cdot,t) \|_2  \| \sqrt{\Gamma(P_t g) }\|_2 dt + \int_\mathbb{M} g \int_0^T  P_t(f v(\cdot,t)) dt d\mu,
which proves what we claimed.
Let now h_k\in C^\infty_0(\mathbb{M}) be a sequence such that 0 \le h_k \le 1, \| \Gamma(h_k) \|_\infty \to 0 and h_k increases to 1.
Using h_k in place of f and letting k \to \infty, gives
\int_\mathbb{M} g P_T (u(\cdot,T)) d\mu - \int_\mathbb{M} g u(x,0)d\mu  \ge   \int_\mathbb{M} g \int_0^T  P_t(v(\cdot,t)) dt d\mu.
We observe that the assumption on v and Minkowski’s integral inequality guarantee that the function x\to \int_0^T  P_t(v(\cdot,t))(x) dt belongs to L^q(\mathbb{M}). We have in fact
\left(\int_\mathbb{M} \left|\int_0^T  P_t(v(\cdot,t)) dt\right|^q d\mu\right)^{\frac 1q}  \le \int_0^T \left| \int_\mathbb{M} \left|P_t(v(\cdot,t))\right|^q d\mu\right|^{\frac 1q} dt \le \int_0^T \left| \int_\mathbb{M} \left|v(\cdot,t)\right|^q d\mu\right|^{\frac 1q} dt
\le T^{\frac{1}{q'}}  \left(\int_0^T \int_\mathbb{M} \left|v(\cdot,t)\right|^q d\mu dt \right)^{\frac 1q} < \infty.

Since this must hold for every non negative g \in C_0^\infty (\mathbb{M}), we conclude that
P_T(u(\cdot,T))(x) \ge u(x,0) +\int_0^T P_s (v(\cdot,s))(x) ds,
which completes the proof \square

We are in position to prove the first gradient bound for the semigroup P_t.

Proposition: If f is a smooth function in \mathcal{D}(L), then for every t \ge 0 and x \in \mathbb{M},
\sqrt{\Gamma(P_t f)}(x) \le e^{-\rho t} P_t \sqrt{\Gamma(f)} (x).

Proof: We fix T > 0 and consider the functional
\Phi(x,t)=e^{-\rho t} \sqrt{ \Gamma(P_{T-t} f)}(x).
We first assume that (x,t)\to \Gamma(P_t f)(x) > 0 on \mathbb{M} \times [0,T]. From the previous lemma, we have \Phi(t) \in L^2(\mathbb{M}). Moreover \Gamma(\Phi)(t)=e^{-2\rho t}\frac{\Gamma(\Gamma(P_{T-t} f))}{4 \Gamma(P_{T-t} f)}. So, we have \Gamma( \Phi)(t) \le e^{-2\rho t}( \Gamma_2(P_{T-t} f)-\rho\Gamma(P_tf)). Therefore, again from the previous proposition , we deduce that \Gamma( \Phi)(t) \in L^1(\mathbb{M}). Next, we easily compute that
\frac{\partial \Phi}{\partial t}+ L\Phi =e^{-\rho t} \left( \frac{\Gamma_2(P_{T-t} f)}{\sqrt{\Gamma(P_{T-t} f)}}-\frac{\Gamma(\Gamma(P_{T-t} f))}{4 \Gamma(P_{T-t} f)^{3/2} } -\rho \sqrt{\Gamma(P_{T-t} f)} \right).
Thus,
\frac{\partial \Phi}{\partial t}+ L\Phi \ge 0.
We can then use the parabolic comparison theorem to infer that
\sqrt{ \Gamma(P_{T} f)} \le e^{-\rho T} P_T \left(\sqrt{\Gamma (f)} \right).
If (x,t) \to \Gamma(P_t f)(x) vanishes on \mathbb{M} \times [0,T], we consider the functional
\Phi(t)=e^{-\rho t} g_\varepsilon (\Gamma(P_{T-t} f) ),
where, for 0 <  \varepsilon < 1,
g_\varepsilon (y)=\sqrt{ y+\varepsilon^2}-\varepsilon.
Since \Phi(t) \in L^2(\mathbb{M}), an argument similar to that above (details are let to the reader) shows that
g_\varepsilon (\Gamma(P_{T} f) )\le e^{-\rho T} P_T \left( g_\varepsilon( \Gamma (f)) \right).
Letting \varepsilon \to 0, we conclude that
\sqrt{ \Gamma(P_{T} f)} \le  e^{-\rho T} P_T \left(\sqrt{\Gamma (f)} \right) \square

We now prove the promised stochastic completeness result:

Theorem: For t \ge 0, one has P_t 1 =1.

Proof: Let f,g \in  C^\infty_0(\mathbb M), we have
\int_{\mathbb{M}} (P_t f -f) g d\mu = \int_0^t \int_{\mathbb{M}}\left( \frac{\partial}{\partial s} P_s f \right) g d\mu ds= \int_0^t\int_{\mathbb{M}}\left(L P_s f \right) g d\mu ds=- \int_0^t \int_{\mathbb{M}}\Gamma ( P_s f , g) d\mu ds.
By means of the previous Proposition and Cauchy-Schwarz inequality, we
find
\left| \int_{\mathbb{M}} (P_t f -f) g d\mu \right| \le \left(\int_0^t e^{-\rho s} ds\right) \sqrt{ \| \Gamma (f) \|_\infty  } \int_{\mathbb{M}}\Gamma (g)^{\frac{1}{2}}d\mu.

We now apply the previous inequality with f = h_n, and then let n\to \infty.
Since by Beppo Levi’s monotone convergence theorem we have P_t h_n(x)\nearrow P_t 1(x) for every x\in \mathbb{M}, we see that the left-hand side converges to \int_{\mathbb{M}} (P_t 1 -1) g d\mu. We thus reach the conclusion
\int_{\mathbb{M}} (P_t 1 -1) g d\mu=0,\ \ \ g\in C^\infty_0(\mathbb{M}).
It follows that P_t 1 =1 \square

A consequence of the stochastic completeness is the uniqueness in L^\infty of solutions of the heat equation. More precisely, the following L^\infty parabolic comparison theorem holds.

Proposition: Let T > 0. Let u,v: \mathbb{M}\times [0,T] \to \mathbb{R} be smooth functions such that for every T > 0, \sup_{t \in [0,T]} \| u(\cdot,t)\|_\infty < \infty, \sup_{t \in [0,T]} \| v(\cdot,t)\|_\infty <\infty; If the inequality

Lu+\frac{\partial u}{\partial t} \ge v

holds on \mathbb{M}\times [0,T], then we have

P_T(u(\cdot,T))(x) \ge u(x,0) +\int_0^T P_s(v(\cdot,s))(x) ds.

Proof: Let (X^x_t)_{t \ge 0} be the diffusion Markov process with semigroup (P_t)_{t \ge 0} and started at x \in \mathbb{M}. From P_t1=1, we deduce that (X^x_t)_{t\ge 0} has an infinite lifetime. We have then for t \ge 0,
u\left( X^x_t, t \right)=u\left( x,0\right)+\int_0^t \left( Lu+\frac{\partial u}{\partial t}\right)(X^x_s,s) ds +M_t,
where (M_t)_{t \ge0 } is a local martingale. From the assumption one obtains
u\left( X^x_t, t \right) \ge u\left( x,0\right)+\int_0^t v(X^x_s,s) ds +M_t.
Let now (T_n)_{n \in \mathbb{N}} be an increasing sequence of stopping times such that almost surely T_n \to +\infty and (M_{t\wedge T_n})_{t \ge 0} is a martingale.
From the previous inequality, we find
\mathbb{E}\left( u\left( X^x_{t\wedge T_n}, t\wedge T_n \right) \right) \ge u\left( x,0\right)+\mathbb{E}\left( \int_0^{t\wedge T_n} v(X^x_s,s) ds\right).
By using the dominated convergence theorem, we conclude
\mathbb{E}\left( u\left( X^x_{t}, t \right) \right) \ge u\left( x,0\right)+\mathbb{E}\left( \int_0^{t} v(X^x_s,s) ds\right),
which yields the conclusion \square

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One Response to Lecture 14. Stochastic completeness of the heat semigroup

  1. Jiayong Wu says:

    I would like to point out that stochastically complete condition is not sufficient to imply an $L^1$-Liouville theorem for (positive) harmonic functions. An example was provided by Li-Schoen. But it is sufficient to imply an $L^1$-Liouville theorem for nonnegative superharmonic functions. This was confirmed by Grigor’yan.

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