Lecture 2. Essentially self-adjoint diffusion operators

The goal of the next few lectures will be to introduce the semigroup generated by a diffusion operator. This semigroup will play pervasive role throughout these lectures and is the main tool associated to the curvature dimension inequalities. The construction of the semigroup is non trivial because diffusion operators are unbounded operators.

We consider a diffusion operator
L=\sum_{i,j=1}^n \sigma_{ij} (x) \frac{\partial^2}{ \partial x_i \partial x_j} +\sum_{i=1}^n b_i(x)\frac{\partial}{\partial x_i},
where b_i and \sigma_{ij} are continuous functions on \mathbb{R}^n and for every x \in \mathbb{R}^n, the matrix (\sigma_{ij}(x))_{1\le i,j\le n} is a symmetric and non negative matrix.

We will assume that L is symmetric with respect to a measure \mu which is equivalent to the Lebesgue measure, that is, for every smooth and compactly supported functions f,g : \mathbb{R}^n \rightarrow \mathbb{R} \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}), \int_{\mathbb{R}^n} g Lf d\mu= \int_{\mathbb{R}^n} fLg d\mu.

ExerciseShow that if L is symmetric with respect to \mu then, in the sense of distributions L'\mu=0, where L' is the adjoint of L in the distribution sense.

Exercise Show that if f : \mathbb{R}^n \rightarrow \mathbb{R} is a smooth function and if g \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}), then we still have the formula \int_{\mathbb{R}^n} f Lg d\mu =\int_{\mathbb{R}^n} gLf d\mu.

Exercise On \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}), let us consider the diffusion operator L=\Delta +\langle \nabla U, \nabla \cdot \rangle, where U: \mathbb{R}^n \rightarrow \mathbb{R} is a C^1 function. Show that L is symmetric with respect to the measure \mu (dx)=e^{U(x)} dx.

Exercise (Divergence form operator). On \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}), let us consider the operator Lf=\mathbf{div} (\sigma  \nabla f), where \mathbf{div} is the divergence operator defined on a C^1 function \phi: \mathbb{R}^n \rightarrow \mathbb{R}^n by
\mathbf{div} \text{ } \phi=\sum_{i=1}^n \frac{\partial \phi_i}{\partial x_i}
and where \sigma is a C^1 field of non negative and symmetric matrices. Show that L is a diffusion operator which is symmetric with respect to the Lebesgue measure.

For every smooth functions f,g: \mathbb{R}^n \rightarrow \mathbb{R}, let us define the so-called carre du champ, which is the symmetric first-order differential form defined by \Gamma (f,g) =\frac{1}{2} \left( L(fg)-fLg-gLf \right). A straightforward computation shows that \Gamma (f,g)=\sum_{i,j=1}^n  \sigma_{ij} (x) \frac{\partial f}{\partial x_i} \frac{\partial g}{\partial x_j}, so that for every smooth function f, \Gamma(f,f) \ge 0.

Exercise.

  • Show that if f,g :\mathbb{R}^n \rightarrow \mathbb{R} are C^1 functions and \phi_1,\phi_2: \mathbb{R} \rightarrow \mathbb{R} are also C^1 then,
    \Gamma (\phi_1 (f), \phi_2 (g))=\phi'_1 (f) \phi_2'(g) \Gamma(f,g).

  • Show that if f:\mathbb{R}^n \rightarrow \mathbb{R} is a C^2 function and \phi: \mathbb{R}\rightarrow \mathbb{R} is also C^2,
    L \phi (f)=\phi'(f) Lf+\phi''(f)  \Gamma(f,f).

The bilinear form we consider is given for f,g \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}) by \mathcal{E} (f,g)=\int_{\mathbb{R}^n} \Gamma (f,g) d\mu.
This is the energy functional (or Dirichlet form) associated to L. It is readily checked that \mathcal{E} is symmetric \mathcal{E} (f,g)=\mathcal{E} (g,f),
and non negative \mathcal{E} (f,f) \ge 0. It is easy to see that
\mathcal{E}(f,g)=-\int_{\mathbb{R}^n} fLg d\mu=-\int_{\mathbb{R}^n} gLf d\mu.

The operator L on the domain \mathcal{D}(L)= \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}) is a densely defined non positive symmetric operator on the Hilbert space \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}). However, in general, it is not self-adjoint, indeed we easily see that
\left\{ f \in  \mathcal{C}^\infty (\mathbb{R}^n,\mathbb{R}), \| f \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})} +\|Lf \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})} < \infty \right\} \subset \mathcal{D}(L^*).

A famous theorem of Von Neumann asserts that any non negative and symmetric operator may be extended into a self-adjoint operator. The following construction, due to Friedrich, provides a canonical non negative self-adjoint extension.

Theorem:(Friedrichs extension) On the Hilbert space \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}), there exists a densely defined non positive self-adjoint extension of L.

Proof: The idea is to work with a Sobolev type norm associated to the energy form \mathcal{E}. On \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}), let us consider the following norm
\| f\|^2_{\mathcal{E}}=\| f \|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }+\mathcal{E}(f,f).
By completing \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}) with respect to this norm, we get a Hilbert space (\mathcal{H},\langle \cdot , \cdot \rangle_{\mathcal{E}}). Since for f \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}), \| f \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) } \le \| f\|_{\mathcal{E}}, the injection map \iota : ( \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}), \| \cdot \|_{\mathcal{E}}) \rightarrow (\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}), \| \cdot \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }) is continuous and it may therefore be extended into a continuous map \bar{\iota}: (\mathcal{H}, \| \cdot \|_{\mathcal{E}}) \rightarrow (\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}), \| \cdot \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }). Let us show that \bar{\iota} is injective so that \mathcal{H} may be identified with a subspace of \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) . So, let f \in \mathcal{H} such that \bar{\iota} (f)=0. We can find a sequence f_n \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}), such that \| f_n -f \|_{\mathcal{E}} \to 0 and \| f_n \|_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) } \to 0. We have
\| f \|_{\mathcal{E}}  =\lim_{m,n \to + \infty} \langle f_n, f_m \rangle_{\mathcal{E}}
=\lim_{m \to + \infty} \lim_{n \to + \infty}  \langle f_n,f_m \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }+\mathcal{E}(f_n,f_m)
=\lim_{m \to + \infty} \lim_{n \to + \infty}  \langle f_n,f_m \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }- \langle f_n,Lf_m \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }=0
thus f=0 and \bar{\iota} is injective. Let us now consider the map
B=\bar{\iota} \cdot \bar{\iota}^* : \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})  \rightarrow \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) .
It is well defined due to the fact that since \bar{\iota} is bounded, it is easily checked that \mathcal{D}(\bar{\iota}^*)= \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}).

Moreover, B is easily seen to be symmetric, and thus self-adjoint because its domain is equal to \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}). Also, it is readily checked that the injectivity of \bar{\iota} implies the injectivity of B. Therefore, we deduce that the inverse
A=B^{-1}: \mathcal{R} (\bar{\iota} \cdot \bar{\iota}^*) \subset\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) \rightarrow \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})
is a densely defined self-adjoint operator on \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}). Now, we observe that for f,g \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}),
\langle f,g \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}-\langle Lf,g \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}
=  \langle \bar{i}^{-1}(f),\bar{i}^{-1}(g) \rangle_\mathcal{E}
= \langle (\bar{i}^{-1})^* \bar{i}^{-1} f,g \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}
= \langle  (\bar{i} \bar{i}^*)^{-1} f,g \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}
Thus A and \mathbf{Id}-L coincide on \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}).
By defining, -\bar{L}=A-\mathbf{Id}, we get the required self-adjoint extension of -L \square

The operator \bar{L}, as constructed above, is called the Friedrichs extension of L.

Definition: If \bar{L} is the unique non positive self-adjoint extension of L, then the operator L is said to be essentially self-adjoint on \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}). In that case, there is no ambiguity and we shall denote \bar{L}=L.

We have the following first criterion for essential self-adjointness.

Lemma: If for some \lambda > 0, \mathbf{Ker} (-L^* +\lambda \mathbf{Id} )= \{ 0 \}, then the operator L is essentially self-adjoint on \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}).

Proof: We make the proof for \lambda=1 and let the reader adapt it for \lambda \neq 0. Let -\tilde{L} be a non negative self-adjoint extension of -L. We want to prove that actually, -\tilde{L}=-\bar{L}. The assumption \mathbf{Ker} (-L^* + \mathbf{Id} )= \{ 0 \} implies that \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}) is dense in \mathcal{D}(-L^*) for the norm
\| f \|^2_{\mathcal{E}}=\| f \|^2_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) } -\langle f , L^* f \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) }.
Since, -\tilde{L} is a non negative self-adjoint extension of -L, we have
\mathcal{D}(-\tilde{L}) \subset  \mathcal{D}(-L^*).
The space \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}) is therefore dense in \mathcal{D}(-\tilde{L}) for the norm \| \cdot \|_{\mathcal{E}}.

At that point, we use some notations introduced in the proof of the Friedrichs extension theorem. Since \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}) is dense in \mathcal{D}(-\tilde{L}) for the norm \| \cdot \|_{\mathcal{E}}, we deduce that the equality
\langle f,g \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}-\langle \tilde{L}f,g \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}=  \langle \bar{i}^{-1}(f),\bar{i}^{-1}(g) \rangle_\mathcal{E} ,
which is obviously satisfied for f,g \in \mathcal{C}_c(\mathbb{R}^n,\mathbb{R}) actually also holds for f,g \in \mathcal{D}(\tilde{L}). From the definition of the Friedrichs extension, we deduce that \bar{L} and \tilde{L} coincide on \mathcal{D}(\tilde{L}). Finally, since these two operators are self adjoint we conclude \bar{L}=\tilde{L} \square

Given the fact that -L is given here with the domain \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}), the condition \mathbf{Ker} (-L^* +\lambda \mathbf{Id} )= \{ 0 \}, is equivalent to the fact that if f \in \mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R}) is a function that satisfies in the sense of distributions -Lf+\lambda f=0, then f=0.

As a corollary of the previous lemma, the following proposition provides a useful sufficient condition for essential self-adjointness that is easy to check for several diffusion operators (including Laplace-Beltrami operators on complete Riemannian manifolds).

Proposition: If the diffusion operator L is elliptic with smooth coefficients and if there exists an increasing sequence h_n\in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}), 0 \le h_n \le 1, 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 on \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}).

Proof: Let \lambda > 0. According to the previous lemma, it is enough to check that if L^* f=\lambda f with \lambda > 0, then f=0. As it was observed above, L^* f=\lambda f is equivalent to the fact that, in the sense of distributions, Lf =\lambda f. From the hypoellipticity of L, we deduce therefore that f is a smooth function. Now, for h \in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}),
\int_{\mathbb{R}^n} \Gamma( f, h^2f) d\mu  =-\langle f, L(h^2f)\rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}
=-\langle L^*f ,h^2f \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}
=-\lambda  \langle f,h^2f\rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}
=-\lambda \langle f^2,h^2 \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})} \le 0.
Since \Gamma( f, h^2f)=h^2 \Gamma (f,f)+2 fh \Gamma(f,h), we deduce that
\langle h^2, \Gamma (f,f) \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})}+2 \langle fh, \Gamma(f,h)\rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})} \le 0.
Therefore, by Cauchy-Schwarz inequality
\langle h^2, \Gamma (f,f) \rangle_{\mathbf{L}_{\mu}^2 (\mathbb{R}^n,\mathbb{R})} \le 4 \| f |_2^2 \| \Gamma (h,h) \|_\infty.
If we now use the sequence h_n and let n \to \infty, we obtain \Gamma(f,f)=0 and therefore f=0, as desired \square

The assumption on the existence of the sequence h_n will be met several times in this course. We will see later that, from a geometric point of view, it says that the intrinsic metric associated to L is complete, or in other words that the balls of the diffusion operator L are compact.

Exercise: Let L be an elliptic diffusion operator with smooth coefficients. We assume that L defined on \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}) is symmetric with respect to the measure \mu. Let \Omega \subset \mathbb{R}^n be a non empty set whose closure \bar{\Omega} is compact. Show that the operator L is essentially self-adjoint on
\{ u :\bar{\Omega} \to \mathbb{R},\text{ u smooth}, \text{ } u=0 \text{ on } \partial\Omega \}.

Exercise:Let
L=\Delta +\langle \nabla U, \nabla \cdot\rangle,
where U is a smooth function on \mathbb{R}^n. Show that with respect to the measure \mu(dx)=e^{U(x)} dx, the operator L is essentially self-adjoint on \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}).

Exercise: On \mathbb{R}^n, we consider the divergence form operator
Lf=\mathbf{div} (\sigma  \nabla f),
where \sigma is a smooth field of positive and symmetric matrices that satisfies
a \|x \|^2  \le \langle x , \sigma x \rangle  \le b \|x \|^2, \quad x \in \mathbb{R}^n,
for some constant 0 < a \le b. Show that with respect to the Lebesgue measure, the operator L is essentially self-adjoint on \mathcal{C}_c (\mathbb{R}^n,\mathbb{R})

Exercise: On \mathbb{R}^n, we consider the Schrodinger type operator H=L-V, where L is a diffusion operator and V:\mathbb{R}^n \rightarrow \mathbb{R} is a smooth function. We denote
\Gamma (f,g) =\frac{1}{2} \left( L(fg)-fLg-gLf \right).
Show that if there exists an increasing sequence h_n\in \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}), 0 \le h_n \le 1, such that h_n\nearrow 1 on
\mathbb{R}^n, and ||\Gamma (h_n,h_n)||_{\infty} \to 0, as n\to \infty and that if V is bounded from below then H is essentially self-adjoint on \mathcal{C}_c (\mathbb{R}^n,\mathbb{R}).

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