Lecture 8. Positivity preserving property of diffusion semigroups

In the previous lectures, we have proved that if L is a diffusion operator that is essentially self-adjoint then, by using the spectral theorem, we can define a self-adjoint strongly continuous contraction semigroup with generator L and this semigroup is unique. A remarkable property of the semigroup is that it preserves the positivity of functions.

More precisely, we are going to prove that if L is am essentially self-adjoint operator with respect to a measure \mu then, by denoting (P_t)_{t \ge 0} the semigroup generated by L: If f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) satisfies 0 \le f \le 1, then 0 \le \mathbf{P}_t f \le 1, t \ge 0. This property is called the submarkov property of the semigroup (\mathbf{P}_t)_{t \ge 0}. The terminology stems from the connection with probability theory where (\mathbf{P}_t)_{t \ge 0} is interpreted as the transition semigroup of a sub-Markov process.

As a first step, we prove the positivity preserving property, which is a consequence of the following functional inequality satisfied by diffusion operators:

Lemma: (Kato’s inequality for diffusion operators). Let L be a diffusion operator on \mathbb{R}^n which is symmetric with respect to a measure \mu. Let u \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}). Define \mathbf{sgn} \text{ }u(x)=0 \quad \text{ if } u(x)=0 and \mathbf{sgn} \text{ }u (x)=\frac{u(x)}{|u(x)|} \quad \text{ if } u(x)\neq 0. In the sense of distributions, we have the following inequality L |u | \ge ( \mathbf{sgn} \text{ }u ) Lu. 

Proof: If \phi is a smooth and convex function and if u is assumed to be smooth, it is readily checked that L \phi(u)=\phi'(u) Lu +\phi''(u) \Gamma(u,u) \ge \phi'(u)Lu. By choosing for \phi the function \phi_\varepsilon(x)=\sqrt{x^2 +\varepsilon^2}, \quad \varepsilon > 0, we deduce that for every smooth function u \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}), L\phi_\varepsilon (u) \ge \frac{u}{\sqrt{x^2 +\varepsilon^2}} Lu. As a consequence this inequality holds in the sense of distributions, that is for every f \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}), f \ge 0, \int_{\mathbb{R}^n} f L\phi_\varepsilon (u) d\mu \ge \int_{\mathbb{R}^n} f \frac{u}{\sqrt{u^2 +\varepsilon^2}} Lu d\mu. Letting \varepsilon \to 0 gives the expected result \square

We are now in position to state and prove the positivity preserving theorem.

Proposition: Let L be an essentially self-adjoint diffusion operator on \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}). If f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) is almost surely nonnegative (f \ge 0), then we have for every t \ge 0, \mathbf{P}_t f \ge 0 almost surely.

Proof: The main idea is to prove that for \lambda > 0, the resolvent operator (\lambda \mathbf{Id}-L)^{-1} which is well defined due to essential self adjointness preserves the positivity of function. Then, we may conclude by the fact that, as it is seen from spectral theorem, \mathbf{P}_t=\lim_{n \to +\infty} \left( \mathbf{Id} -\frac{t}{n} L\right)^{-n}.

We first extend Kato’s inequality to a larger class of functions.

Let \lambda > 0. We consider on \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}) the norm \| f \|^2_{\lambda} =\| f \|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }+\lambda\mathcal{E}(f,f) =\| f \|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }+\lambda \int_{\mathbb{R}^n} \Gamma (f,f) d\mu and denote by \mathcal{H}_\lambda the completion of \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}). Our goal will be to show that the Kato’s inequality is also satisfied for u \in \mathcal{H}_\lambda. As in the proof of Kato’s inequality, we first consider smooth approximations of the absolute value. For \varepsilon > 0 we introduce the function \phi_\varepsilon(x)=\sqrt{x^2 +\varepsilon^2}-\varepsilon, \quad \varepsilon > 0. It is easily seen that for u \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}), \phi_\varepsilon(u) \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}). Let now u_n \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}) be a Cauchy sequence for the norm \| \cdot \|_{\lambda}. We claim that the sequence \phi_\varepsilon(u_n) is also a Cauchy sequence. Indeed, since | \phi_\varepsilon(x)-\phi_\varepsilon(y) | \le |x-y|, we have, on one hand \| \phi_\varepsilon(u_n)-\phi_\varepsilon(u_m) \|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) } \le \| \phi_\varepsilon(u_n)-\phi_\varepsilon(u_m) \|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}. Now, keeping in mind that \mathcal{E} is a nonnegative bilinear form and thus satisfies Cauchy-Schwarz-inequality, we have on the other hand
\sqrt{\mathcal{E}(\phi_\varepsilon(u_n)-\phi_\varepsilon(u_m),\phi_\varepsilon(u_n)-\phi_\varepsilon(u_m)) }
\le \|\phi'_\varepsilon(u_n)\|_\infty \sqrt{\mathcal{E}(u_n-u_m,u_n-u_m) }+\sqrt{\| \Gamma(u_m,u_m)\|_\infty } \|\phi'_\varepsilon(u_n)-\phi'_\varepsilon(u_m) \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}
\le \sqrt{\mathcal{E}(u_n-u_m,u_n-u_m) }+\frac{1}{\varepsilon} \sqrt{\| \Gamma(u_m,u_m)\|_\infty } \|u_n-u_m \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}.
As a consequence, \phi_\varepsilon(u_n) is a Cauchy sequence and thus converges toward an element of \mathcal{H}_\lambda. If u denotes the limit of u_n in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}), the limit in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) of \phi_\varepsilon(u_n) is \phi_\varepsilon(u). As a conclusion, if u \in \mathcal{H}_\lambda then \phi_{\varepsilon}(u) \in \mathcal{H}_\lambda.

From the proof of Kato’s inequality, if u \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}) then for every f \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}), f \ge 0, \int_{\mathbb{R}^n} f L\phi_\varepsilon (u) d\mu \ge \int_{\mathbb{R}^n} f \frac{u}{\sqrt{u^2 +\varepsilon^2}} Lu d\mu. This may be rewritten as - \int_{\mathbb{R}^n} f \frac{u} {\sqrt{u^2 +\varepsilon^2}} Lu d\mu \ge \mathcal{E} (f,\phi_\varepsilon (u)).
Let f \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}) and u \in \mathcal{D}(L)\subset \mathcal{H}_\lambda. We consider a sequence u_n \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}) such that u_n \to u for the norm \| \cdot \|_\lambda. In particular u_n \to u for the norm \| \cdot \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}, so that by passing to a subsequence we can suppose that u_n(x) \to u(x) pointwise almost surely. Applying the inequality to u_n and letting n \to +\infty leads to the conclusion that the inequality also holds for u \in \mathcal{D}(L) and f \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}).

Finally, by using the same type of arguments as above, it is shown for u \in \mathcal{H}_\lambda, when \varepsilon \to 0, \varepsilon (u) tends to |u| in the norm \| \cdot \|_\lambda. Thus, if u \in \mathcal{H}_\lambda, |u| \in \mathcal{H}_\lambda.

As a consequence of all this, if u \in \mathcal{D}(L), |u| \in \mathcal{H}_\lambda, and moreover for f \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}), - \int_{\mathbb{R}^n} f (\mathbf{sgn} \text{ }u )Lu d\mu \ge \mathcal{E} (f,|u|).
And this last inequality is easily extended to f \in \mathcal{H}_\lambda by density of \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}) in \mathcal{H}_\lambda for the norm \| \cdot \|_\lambda. In particular when f=|u| we get that for u \in \mathcal{D}(L), \mathcal{E}(|u|, |u|) \le \mathcal{E}(u, u). Again by density, this inequality can be extended to every u \in \mathcal{H}_\lambda. Since L is essentially self-adjoint we can consider the bounded operator \mathbf{R}_\lambda=( \mathbf{Id}-\lambda L)^{-1} that goes from \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) to \mathcal{D}(L)\subset \mathcal{H}_\lambda. For f \in \mathcal{H}_\lambda and g \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) with g \ge 0, we have
\langle | f | , \mathbf{R}_\lambda g \rangle_\lambda = \langle | f | , \mathbf{R}_\lambda g \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}-\lambda \langle |f| , L\mathbf{R}_\lambda g \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}
=\langle |f|, (\mathbf{Id}-\lambda L) \mathbf{R}_\lambda g \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}
=\langle |f|, g \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}
\ge | \langle f, g \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}|
\ge |\langle f , \mathbf{R}_\lambda g \rangle_\lambda|.
Moreover, from the previous inequality, for f \in \mathcal{H}_\lambda,
\|\text{ } | f|\text{ } \|_\lambda^2 =\| \text{ }| f |\text{ } \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}^2+\lambda \mathcal{E}(|f|,|f|)
\ge \| f \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}^2+\lambda \mathcal{E}(f,f) \ge \| f \|_\lambda^2.
By taking f= \mathbf{R}_\lambda g in the two above sets of inequalities, we draw the conclusion
|\langle \mathbf{R}_\lambda g , \mathbf{R}_\lambda g \rangle_\lambda| \le \langle | \mathbf{R}_\lambda g | , \mathbf{R}_\lambda g \rangle_\lambda \le \|\text{ } | \mathbf{R}_\lambda g|\text{ } \|_\lambda \|\mathbf{R}_\lambda g\|_\lambda\le |\langle \mathbf{R}_\lambda g , \mathbf{R}_\lambda g \rangle_\lambda|.
The above inequalities are therefore equalities which implies \mathbf{R}_\lambda g = | \mathbf{R}_\lambda g|. As a conclusion if g \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) is \ge 0, then for every \lambda > 0, ( \mathbf{Id}-\lambda L)^{-1} g \ge 0. Thanks to spectral theorem, in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}), \mathbf{P}_t g=\lim_{n \to +\infty} \left( \mathbf{Id} -\frac{t}{n} L\right)^{-n}g. By passing to a subsequence that converges pointwise almost surely, we deduce that \mathbf{P}_t g \ge 0 almost surely \square

Exercise: Let L be an elliptic diffusion operator with smooth coefficients that is essentially self-adjoint. Denote by p(t,x,y) the heat kernel of \mathbf{P}_t. Show that p(t,x,y) \ge 0. (Remark: It actually possible to prove that p(t,x,y) > 0). 

Besides the positivity preserving property, the semigroup is a contraction on \mathbf{L}_{\mu}^\infty (\mathbb{R}^n,\mathbb{R}). More precisely,

Proposition: Let L be an essentially self adjoint diffusion operator on \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}). If f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) \cap \mathbf{L}_{\mu}^\infty (\mathbb{R}^n,\mathbb{R}), then \mathbf{P}_t f \in \mathbf{L}_{\mu}^\infty (\mathbb{R}^n,\mathbb{R}) and \| \mathbf{P}_t f \|_\infty \le \| f \|_\infty. 

Proof: The proof is close and relies on the same ideas as the proof of the positivity preserving property. So, we only list below the main steps and let the reader fills the details.

As before, for \lambda > 0, we consider on \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}) the norm \| f \|^2_{\lambda} =\| f \|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }+\lambda\mathcal{E}(f,f) and denote by \mathcal{H}_\lambda the completion of \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}).

The first step is to show that if 0 \le f \in \mathcal{H}_\lambda, then 1 \wedge f (minimum between 1 and f) also belongs to \mathcal{H}_\lambda and moreover \mathcal{E}( 1 \wedge f, 1 \wedge f) \le \mathcal{E}(f,f).

Let f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) satisfy 0 \le f \le 1 and put g=\mathbf{R}_\lambda f=( \mathbf{Id}-\lambda L)^{-1} f \in \mathcal{H}_\lambda and h=1 \wedge g. According to the first step, h \in \mathcal{H}_\lambda and \mathcal{E}( h, h ) \le \mathcal{E}(g,g). Now, we observe that:
\| g-h \|_\lambda^2 =\| g \|_\lambda^2 -2 \langle g,h \rangle_\lambda +\| h \|_\lambda^2
=\langle \mathbf{R}_\lambda f,f \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }-2 \langle f,h \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) } +\| h\|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) } +\lambda \mathcal{E}(h,h)
= \langle \mathbf{R}_\lambda f,f \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }-\| f\|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }+\| f-h\|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }+\lambda \mathcal{E}(h,h)
\le \langle \mathbf{R}_\lambda f,f \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }-\| f\|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }+\| f-g\|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }+\lambda \mathcal{E}(g,g)=0
As a consequence g=h, that is 0\le g \le 1.

The previous step shows that if f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) satisfies 0 \le f \le 1 then for every \lambda > 0, 0 \le ( \mathbf{Id}-\lambda L)^{-1} f \le 1. As in the previous proposition, we deduce that 0 \le \mathbf{P}_t f \le 1 almost surely \square

This entry was posted in Uncategorized. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s