## 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.