Lecture 2. Essentially self-adjoint diffusion operators

Posted on August 31, 2016 by Fabrice Baudoin

The goal of the next few lectures will be to introduce the semigroup generated by a diffusion operator. 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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