## Lecture 15. Convergence of the semigroup, Poincare and log-Sobolev

Let $\mathbb{M}$ be a complete $n$-dimensional 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 $\rho > 0$. We recall that this is equivalent to the fact that for every $f \in C^\infty(\mathbb{M})$,
$\Gamma_2(f,f) \ge \frac{1}{n} (Lf)^2 + \rho \Gamma(f,f).$
Readers knowing Riemannian geometry know that from Bonnet-Myers theorem, the manifold needs to be compact and we therefore expect the semigroup to converge to equilibrium. However for several lectures, our goal will be to not use the Bonnet-Myers theorem, because eventually we shall provide a proof of this fact using semigroup theory. Thus the results in this Lecture will not use the compactness of $\mathbb{M}$.

Lemma: The Riemannian measure $\mu$ is finite, i.e. $\mu(\mathbb{M}) < +\infty$ and for every $f \in L_\mu^2(\mathbb{M})$, the following convergence holds pointwise and in $L^2_\mu(\mathbb{M})$,
$P_t f \to_{t \to +\infty} \frac{1}{\mu(\mathbb{M})} \int_\mathbb{M} f d\mu.$

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 Cauchy-Schwarz inequality, we
find
$\left| \int_{\mathbb{M}} (P_t f -f) g d\mu \right| \le \left(\int_0^t e^{-2\rho s} ds\right) \sqrt{ \| \Gamma (f) \|_\infty } \int_{\mathbb{M}}\Gamma (g)^{\frac{1}{2}}d\mu.$
Now it is seen from spectral theorem that in $L^2(\mathbb{M})$ we have a convergence $P_t f \to P_\infty f$, where $P_\infty f$ belongs to the domain of $L$. Moreover $LP_\infty f=0$. By ellipticity of $L$ we deduce that $P_\infty f$ is a smooth function. Since $LP_\infty f=0$, we have $\Gamma(P_\infty f)=0$ and therefore $P_\infty f$ is constant.

Let us now assume that $\mu(\mathbb{M})=+\infty$. This implies in particular that $P_\infty f =0$ because no constant besides $0$ is in $L^2(\mathbb{M})$. Using then the previous inequality and letting $t \to +\infty$, we infer
$\left| \int_{\mathbb{M}} f g d\mu \right| \le \left(\int_0^{+\infty} e^{-2\rho s} ds\right) \sqrt{ \| \Gamma (f) \|_\infty } \int_{\mathbb{M}}\Gamma (g)^{\frac{1}{2}}d\mu.$
Let us assume $g \ge 0$, $g \ne 0$ and take for $f$ the usual localizing sequence $h_n$. Letting $n \to \infty$, we deduce $\int_\mathbb{M} g d\mu \le 0,$ which is clearly absurd. As a consequence $\mu (\mathbb{M}) < +\infty$.

The invariance of $\mu$ implies then $\int_\mathbb{M} P_\infty f d\mu =\int_\mathbb{M} f d\mu$,
and thus $P_\infty f =\frac{1}{\mu(\mathbb{M})} \int_\mathbb{M} f d\mu$. Finally, using the Cauchy-Schwarz inequality, we find that for $x \in \mathbb{M}$, $f \in L^2(\mathbb{M})$, $s,t,\tau \ge 0$,
$| P_{t+\tau} f (x)-P_{s+\tau} f (x) | = | P_\tau (P_t f -P_s f) (x) |$
$=\left| \int_\mathbb{M} p(\tau, x, y) (P_t f -P_s f) (y) \mu(dy) \right|$
$\le \int_\mathbb{M} p(\tau, x, y)^2 \mu(dy) \| P_t f -P_s f\|^2_2$
$\le p(2\tau,x,x) \| P_t f -P_s f\|^2_2.$

Thus, we also have $P_t f (x) \to_{t \to +\infty} \frac{1}{\mu(\mathbb{M})} \int_{\mathbb{M}} f d\mu$ $\square$

Proposition: The following Poincare inequality is satisfied: For $f \in \mathcal{D}(L)$, $\frac{1}{\mu(\mathbb{M})}\int_{\mathbb{M}} f^2 d\mu \le \left( \frac{1}{\mu(\mathbb{M})} \int_{\mathbb{M}} f d\mu \right)^2 +\frac{n-1}{n\rho} \frac{1}{\mu(\mathbb{M})} \int_{\mathbb{M}} \Gamma(f,f) d\mu.$

Let $f \in C_0^\infty(\mathbb{M})$. We have by assumption $\Gamma_2(f,f) \ge \frac{1}{n} (Lf)^2 + \rho \Gamma (f,f).$ Therefore, by integrating the latter inequality we obtain
$\int_{\mathbb{M}} \Gamma_2(f,f) d\mu \ge \frac{1}{n} (Lf)^2+ \rho \int_{\mathbb{M}} \Gamma (f,f) d\mu.$
But we have
$\int_{\mathbb{M}} \Gamma_2(f,f) d\mu=-\int_{\mathbb{M}} \Gamma(f,Lf) d\mu=\int_{\mathbb{M}} (Lf)^2 d\mu.$
Therefore we obtain
$\int_{\mathbb{M}} (Lf)^2 d\mu \ge \rho \int_{\mathbb{M}} \Gamma (f,f) d\mu=-\frac{n\rho}{n-1} \int_{\mathbb{M}} fLf d\mu.$
By density, this last inequality is seen to hold for every function $f\in \mathcal{D}(L)$. It means that the $L^2$ spectrum of $-L$ lies in $\{0\} \cup \left[\frac{n\rho}{n-1} ,+\infty\right)$. Since from the previous proof the projection of $f$ onto the $0$-eigenspace is given by $\frac{1}{\mu(\mathbb{M})} \int_{\mathbb{M}} f d\mu$, we deduce that
$\int_\mathbb{M} \left( f-\frac{1}{\mu(\mathbb{M})} \int_\mathbb{M} f d\mu\right)^2d\mu \le \frac{n-1}{n\rho} \int_{\mathbb{M}} \Gamma(f,f) d\mu$
which is exactly the inequality we wanted to prove $\square$

As observed in the proof, the Poincare inequality
$\int_{\mathbb{M}} f^2 d\mu \le \left( \int_{\mathbb{M}} f d\mu \right)^2 +\frac{n-1}{n\rho} \int_{\mathbb{M}} \Gamma(f,f) d\mu.$
is equivalent to the fact that the $L^2$ spectrum of $-L$ lies in $\{0\} \cup \left[\frac{n\rho}{n-1} ,+\infty\right)$, or in other words that $-L$ has a spectral gap of size at least $\frac{n\rho}{n-1}$. This is Lichnerowicz estimate. It is sharp, because on the $n$-dimensional sphere it is known that $\rho=n-1$ and that the first non zero eigenvalue is exactly equal to $n$.

As a basic consequence of the spectral theorem and of the above spectral gap estimate, we also get the rate convergence to equilibrium in $L^2_\mu(\mathbb{M})$ for $P_t$.

Proposition: Let $f \in L^2_\mu(\mathbb{M})$, then for $t \ge 0$,
$\left\| P_tf -\frac{1}{\mu(\mathbb{M})}\int_\mathbb{M} f d\mu \right\|^2_2 \le e^{-\frac{2n\rho}{n-1} t} \left\| f -\frac{1}{\mu(\mathbb{M})}\int_\mathbb{M} f d\mu \right\|^2_2 .$

Exercise: By using the Riesz-Thorin interpolation theorem, show that for $p \in [2,+\infty)$, and $f \in L^p_\mu(\mathbb{M})$,
$\left\| P_tf -\frac{1}{\mu(\mathbb{M})}\int_\mathbb{M} f d\mu \right\|^p_p \le 2^{p-2} e^{-\frac{2n\rho}{n-1} t} \left\| f -\frac{1}{\mu(\mathbb{M})}\int_\mathbb{M} f d\mu \right\|^p_p.$
By using duality, prove a corresponding statement when $p \in (1,2]$.

As we have just seen, the convergence in $L^2$ of $P_t$ is connected and actually equivalent to the Poincare inequality.

We now turn to the so-called log-Sobolev inequality which is connected to the convergence in entropy for $P_t$. This inequality is much stronger (and more useful) than the Poincare inequality. To simplify a little the expressions, we assume in the sequel that $\mu(\mathbb{M})=1$ (Otherwise, just replace $\mu$ by $\frac{\mu}{\mu(\mathbb{M})}$ in the following results).

Proposition: For $f \in \mathcal{D}(L)$, $f \ge 0$,
$\int_{\mathbb{M}} f^2 \ln f^2 d\mu \le \int_{\mathbb{M}} f^2 d\mu \ln \left( \int_{\mathbb{M}} f^2 d\mu \right) +\frac{2}{ \rho} \int_{\mathbb{M}} \Gamma(f,f) d\mu.$

Proof: By considering $\sqrt{f}$ instead of $f$, it is enough to show that if $f$ is positive,
$\int_{\mathbb{M}} f \ln f d\mu \le \int_{\mathbb{M}} f d\mu \ln \left( \int_{\mathbb{M}} f d\mu \right) +\frac{1}{2 \rho} \int_{\mathbb{M}} \frac{\Gamma(f,f)}{f} d\mu.$
We now have
$\int_{\mathbb{M}} f \ln f d\mu - \int_{\mathbb{M}} f d\mu \ln \left( \int_{\mathbb{M}} f d\mu \right)= - \int_0^{+\infty} \frac{d}{dt} \int_{\mathbb{M}} P_t f \ln P_t f d\mu dt$
$=- \int_0^{+\infty} \int_{\mathbb{M}} L P_t f \ln P_t f d\mu dt$
$= \int_0^{+\infty} \int_{\mathbb{M}} \Gamma (P_t f, \ln P_t f) d\mu dt$
$=\int_0^{+\infty} \int_{\mathbb{M}} \frac{\Gamma (P_t f, P_t f) }{P_t f} d\mu dt$
Now, we know that
$\Gamma (P_t f, P_t f) \le e^{-2\rho t} \left( P_t \sqrt{\Gamma(f,f)}\right)^2.$ And, from Cauchy-Schwarz inequality, $(P_t \sqrt{\Gamma(f,f)})^2 \le P_t \frac{\Gamma(f,f)}{f} P_t f.$ Therefore,
$\int_{\mathbb{M}} f \ln f d\mu - \int_{\mathbb{M}} f d\mu \ln \left( \int_{\mathbb{M}} f d\mu \right) \le \int_0^{+\infty} e^{-2\rho t} dt \int_{\mathbb{M}} \frac{\Gamma(f,f)}{f} d\mu,$
which is the inequality we claimed $\square$

We finally prove the entropic convergence of $P_t$.

Theorem: Let $f \in L^2_\mu(\mathbb{M})$, $f \ge 0$. For $t \ge 0$,
$\int_{\mathbb{M}} P_t f \ln P_t f d\mu - \int_{\mathbb{M}} P_t f d\mu \ln \left( \int_{\mathbb{M}} P_t f d\mu\right) \le e^{-2\rho t} \left( \int_{\mathbb{M}} f \ln f d\mu- \int_{\mathbb{M}} f d\mu \ln \left( \int_{\mathbb{M}} f d\mu\right)\right).$

Proof: Let us assume $\int_{\mathbb{M}} f d\mu=1$, otherwise we use the following argument with $\frac{f}{\int_\mathbb{R}^n f d\mu}$ and consider the functional
$\Phi(t)= \int_{\mathbb{M}} P_t f \ln P_t f d\mu,$
which by differentiation gives
$\Phi'(t)= \int_{\mathbb{M}} LP_t f \ln P_t f d\mu=-\int_{\mathbb{M}} \frac{ \Gamma(P_t f)}{P_t f} d\mu.$
Using now the log-Sobolev inequality, we obtain
$\Phi'(t)\le -2\rho \Phi(t).$
The Gronwall’s differential inequality implies then:
$\Phi(t) \le e^{-2\rho t} \Phi(0),$
that is
$\int_{\mathbb{M}} P_t f \ln P_t f d\mu \le e^{-2\rho t} \int_{\mathbb{M}} f \ln f d\mu$
$\square$

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