Lecture 2. Semigroups on Hilbert spaces: The golden triangle

Let $(\mathcal{H},\langle \cdot, \cdot \rangle_\mathcal{H}$ be a Hilbert space and let $A$ be a densely defined operator on a domain $\mathcal{D}(A)$ . We have the following basic definitions.

The operator $A$ is said to be symmetric if for $f,g \in \mathcal{D}(A)$ ,
$\langle f , Ag \rangle_{\mathcal{H}}=\langle Af , g \rangle_{\mathcal{H}}.$
The operator $A$ is said to be non negative symmetric operator, if it is symmetric and if for $f \in \mathcal{D}(A)$ ,
$\langle f , Af \rangle_{\mathcal{H}} \ge 0.$
It is said to be non positive, if for $f \in \mathcal{D}(A)$ ,
$\langle f , Af \rangle_{\mathcal{H}} \le 0.$

The adjoint $A^*$ of $A$ is the operator defined on the domain $\mathcal{D}(A^*)=\{ f \in \mathcal{H}, \exists \text{ } c(f) \ge 0, \forall \text{ } g \in \mathcal{D}(A), | \langle f, Ag \rangle_{\mathcal{H}} | \le c(f) \| g \|_{\mathcal{H}} \}$

and given through the Riesz representation theorem by the formula
$\langle A^* f,g \rangle_{\mathcal{H}}=\langle f,Ag \rangle_{\mathcal{H}}$
where $g \in \mathcal{D}(A), f \in \mathcal{D}(A^*)$ . The operator $A$ is said to be self-adjoint if it is symmetric and if $\mathcal{D}(A^*)=\mathcal{D}(A)$ .

Let us observe that, in general, the adjoint $A^*$ is not necessarily densely defined, however it is readily checked that if $A$ is a symmetric operator then, from Cauchy-Schwarz inequality, $\mathcal{D}(A) \subset \mathcal{D}(A^*)$ . Thus, if $A$ is symmetric, then $A^*$ is densely defined.

We have the following first criterion for self-adjointness which may be useful.

Lemma: Let $A: \mathcal{D}(A) \subset \mathcal{H} \rightarrow \mathcal{H}$ be a densely defined operator. Consider the graph of $A$ :

$\mathbf{G}_A=\left\{ (v,Av), v \in \mathcal{D}(A) \right\} \subset \mathcal{H} \oplus \mathcal{H}$

and the complex structure

$\mathcal{J}: \mathcal{H} \oplus \mathcal{H} \rightarrow \mathcal{H} \oplus \mathcal{H} , (v,w) \rightarrow (-w,v).$

Then, the operator $A$ is self-adjoint if and only if

$\mathbf{G}_A^{\bot}=\mathcal{J} \left( \mathbf{G}_A \right).$

Proof: It is checked that for any densely defined operator $A$

$\mathbf{G}_{A^*}=\mathcal{J} \left( \mathbf{G}_A^{\bot} \right),$
and the conclusion follows from routine computations. $\square$

The following result is often useful.

Lemma: Let $A: \mathcal{D}(A) \subset \mathcal{H} \rightarrow \mathcal{H}$ be an injective densely defined self-adjoint operator. Let us denote by $\mathcal{R}(A)$ the range of $A$ . The inverse operator $A^{-1} : \mathcal{R}(A) \rightarrow \mathcal{H}$ is a densely defined self-adjoint operator.

Proof: First, let us observe that
$\mathcal{R}(A)^\bot =\mathbf{Ker} ( A^*)= \mathbf{Ker} ( A) =\{ 0 \}.$
Therefore $\mathcal{R}(A)$ is dense in $\mathcal{H}$ and $A^{-1}$ is densely defined. Now,
$\mathbf{G}_{A^{-1}}^\bot =\mathcal{J} \left( \mathbf{G}_{-A} \right)^\bot$
$=\mathcal{J} \left( \mathbf{G}_{-A}^\bot \right)$
$=\mathcal{J}\mathcal{J} \left( \mathbf{G}_{-A} \right)$
$= \mathcal{J} \left( \mathbf{G}_{A^{-1}} \right).$

$\square$

A major result in functional analysis is the spectral theorem.

Theorem: (Spectral theorem) Let $A$ be a non negative self-adjoint operator on a separable Hilbert space $\mathcal{H}$ . There is a measure space $(\Omega, \nu)$ , a unitary map $U: L^2(\Omega,\nu) \rightarrow \mathcal{H}$ and a non negative real valued measurable function $\lambda$ on $\Omega$ such that $U^{-1} A U f (x)=\lambda(x) f(x)$ , for $x \in \Omega$ , $Uf \in \mathcal{D}(A)$ . Moreover, given $f \in L^2(\Omega,\nu)$ , $Uf$ belongs to $\mathcal{D}(A)$ if only if $\int_{\Omega} \lambda^2 f^2 d\nu <+\infty$ .

Definition: Let $A$ be a non negative self-adjoint operator on $\mathcal{H}$ . Let $g: \mathbb{R}_{\ge 0} \to \mathbb{R}$ be a Borel function. With the notations of the spectral theorem, one defines the operator $g(A)$ by the requirement

$U^{-1} g(A) U f (x)=g(\lambda(x)) f(x),$

with $\mathcal{D}(g(A))=\{ Uf, (g \circ\lambda) f \in L^2(\Omega,\nu) \}$ .

Exercise: Show that if $A$ is a non negative self-adjoint operator on $\mathcal{H}$ and $g$ is a bounded Borel function, then $g(A)$ is a bounded operator on $\mathcal{H}$ .

As in the previous lecture, we have the following definition:
Definition: A strongly continuous self-adjoint contraction semigroup is a family of self-adjoint operators $(P_t)_{ t \ge 0}: \mathcal{H} \to \mathcal{H}$ everywhere defined on $\mathcal{H}$ such that:

For $s,t \ge 0$ , $P_t \circ P_s=P_{s+t}$ (semigroup property);
For every $f \in \mathcal{H}$ , $\lim_{t \to 0} P_t f =f$ (strong continuity);
For every $f \in \mathcal{H}$ and $t \ge 0$ , $\| P_t f \| \le \| f \|$ (contraction property).

Definition: A closed symmetric non negative bilinear form on $\mathcal{H}$ is a densely defined non negative quadratic form $\mathcal{E}: \mathcal{F}:=\mathcal{D}(\mathcal{E}) \to \mathbb{R}$ such that $\mathcal{F}$ equipped with the norm $\| f \|_\mathcal{F}^2= \| f \|^2 +\mathcal{E}(f)$ is a Hilbert space. If $\mathcal{E}$ is a closed symmetric non negative bilinear form on $\mathcal{H}$ , one can define for $f,g \in \mathcal{F}$ , $\mathcal{E}(f,g)=\frac{1}{4} ( \mathcal{E} (f+g) -\mathcal{E} (f-g))$ .

One has the following theorems:

Theorem 1: Let $(P_t)_{t \ge 0}$ be a strongly continuous self-adjoint contraction semigroup on $\mathcal{H}$ . Then its generator $A$ is a densely defined non positive self-adjoint operator on $\mathcal{H}$ . Conversely, if $A$ is a densely defined non positive self-adjoint operator on $\mathcal{H}$ , then it is the generator a strongly continuous self-adjoint contraction semigroup on $\mathcal{H}$ .

Proof: Let $(P_t)_{t \ge 0}$ be a strongly continuous self-adjoint contraction semigroup on $\mathcal{H}$ with generator $A$ . As we proved in the previous lecture, one has for $\lambda >0$
$\int_0^{+\infty} e^{-\lambda t} P_t dt=(\lambda \mathbf{Id} -A)^{-1}.$
However, the operator $\int_0^{+\infty} e^{-\lambda t} P_t dt$ is seen to be self-adjoint, thus $(\lambda \mathbf{Id} -A)^{-1}$ is. From previous lemma, we deduce that $\lambda \mathbf{Id} -A$ is self-adjoint, from which we deduce that $A$ is self-adjoint (exercise !).

On the other hand, let $A$ be a densely defined non positive self-adjoint operator on $\mathcal{H}$ . From spectral theorem, there is a measure space $(\Omega, \nu)$ , a unitary map $U: L^2(\Omega,\nu) \rightarrow \mathcal{H}$ and a non negative real valued measurable function $\lambda$ on $\Omega$ such that
$U^{-1} A U f (x)=-\lambda(x) f(x),$
for $x \in \Omega$ , $Uf \in \mathcal{D}(A)$ . We define then $P_t:\mathcal{H} \to \mathcal{H}$ such that
$U^{-1} P_t U f (x)=e^{-t\lambda(x)} f(x),$
and let as an exercise the proof that $(P_t)_{t \ge 0}$ is a strongly continuous self-adjoint contraction semigroup on $\mathcal{H}$ with generator $A$ . $\square$

Theorem 2: Let $(P_t)_{t \ge 0}$ be a strongly continuous self-adjoint contraction semigroup on $\mathcal{H}$ . One can define a closed symmetric non negative bilinear form on $\mathcal{H}$ by

$\mathcal{E} (f)= \lim_{t \to 0} \left\langle \frac{\mathbf{Id}_\mathcal{H} -P_t}{t} f, f \right\rangle.$

The domain $\mathcal{F}$ of this form is the set of $f$ ‘s for which the limit exists.

Proof: Let $A$ be the generator of the semigroup $(P_t)_{t \ge 0}$ . We use spectral theorem to represent $A$ as
$U^{-1} A U g (x)=-\lambda(x) g(x),$
so that
$U^{-1} P_t U g (x)=e^{-t\lambda(x)} g(x).$
We then note that for every $g \in L^2(\Omega,\nu)$ ,
$\left\langle \frac{\mathbf{Id}_\mathcal{H} -P_t}{t} Ug, Ug \right\rangle=\int_\Omega \frac{1-e^{-t\lambda(x)} }{t} g(x)^2 d\nu(x).$
This proves that for every $f \in \mathcal{H}$ , the map $t \to \left\langle \frac{\mathbf{Id}_\mathcal{H} -P_t}{t} f, f \right\rangle$ is non increasing. Therefore, the limit $\lim_{t \to 0} \left\langle \frac{\mathbf{Id}_\mathcal{H} -P_t}{t} f, f \right\rangle$ exists if and only if $\int_\Omega (U^{-1}f )^2 (x) \lambda (x) d\nu(x)<+\infty$ , which is equivalent to the fact that $f \in \mathcal{D}( (-A)^{1/2})$ . In which case we have
$\lim_{t \to 0} \left\langle \frac{\mathbf{Id}_\mathcal{H} -P_t}{t} f, f \right\rangle =\| (-A)^{1/2} f \|^2.$
Since $(-A)^{1/2}$ is a densely defined self-adjoint operator, the quadratic form
$\mathcal{E}(f):=\| (-A)^{1/2} f \|^2$
is closed and densely defined on $\mathcal{F}:=\mathcal{D}((-A)^{1/2})$ .

Theorem 3: If $\mathcal{E}$ is a closed symmetric non negative bilinear form on $\mathcal{H}$ . There exists a unique densely defined non positive self-adjoint operator $A$ on $\mathcal{H}$ defined by

$\mathcal{D}(A)=\left \{ f \in \mathcal{F}, \exists g \in \mathcal{H}, \forall h \in \mathcal{F}, \mathcal{E}(f,h)=-\left\langle h, g \right\rangle \right\}$

$Af=g$

The operator $A$ is called the generator of $\mathcal{E}$ . Conversely, if $A$ is a densely defined non positive self-adjoint operator on $\mathcal{H}$ , one can define a closed symmetric non negative bilinear form $\mathcal{E}$ on $\mathcal{H}$ by

$\mathcal{F}=\mathcal{D}( (-A)^{1/2}), \quad \mathcal{E} (f)= \| A^{1/2} f \|^2.$

Proof: Let $\mathcal{E}$ be a closed symmetric non negative bilinear form on $\mathcal{H}$ . As usual, we denote by $\mathcal{F}$ the domain of $\mathcal{E}$ . We note that for $\lambda >0$ , $\mathcal{F}$ equipped with the norm $(\| f \|^2 +\lambda \mathcal{E} (f) )^{1/2}$ is a Hilbert space because $\mathcal E$ is closed. From the Riesz representation theorem, there exists then a linear operator $\mathbf{R}_\lambda :\mathcal H \to \mathcal F$ such that for every $f \in \mathcal{H},g \in \mathcal{F}$
$\langle f, g \rangle=\lambda \langle \mathbf{R}_\lambda f , g \rangle + \mathcal{E} (\mathbf{R}_\lambda f,g) .$
From the definition, the following properties are then easily checked:

1) $\| \mathbf{R}_\lambda f \| \le \frac{1}{\lambda} \| f \|$
2) For every $f,g \in \mathcal H$ , $\langle \mathbf{R}_\lambda f , g\rangle=\langle f , \mathbf{R}_\lambda g \rangle$ ;

3) $\mathbf{R}_{\lambda_1}-\mathbf{R}_{\lambda_2}+(\lambda_1-\lambda_2)\mathbf{R}_{\lambda_1}\mathbf{R}_{\lambda_2}=0$ ;
4) For every $f \in \mathcal{H}$ , $\lim_{\lambda \to +\infty} \| \lambda \mathbf{R}_\lambda f -f \|=0$ .

We then claim that $\mathbf{R}_{\lambda}$ is invertible. Indeed, if $\mathbf{R}_{\lambda} f=0$ , then for $\alpha > \lambda$ , one has from 3, $\mathbf{R}_{\alpha} f=0$ . Therefore $f =\lim_{\alpha \to +\infty } \mathbf{R}_{\alpha} f=0$ . Denote then
$Af=\lambda f -\mathbf{R}_{\lambda}^{-1} f,$
and $\mathcal{D}(A)$ is the range of $\mathbf{R}_{\lambda}$ . It is straightforward to check that $A$ does not depend on $\lambda$ . The operator $A$ is a densely defined self-adjoint operator that satisfies the properties stated in the theorem (Exercise !). $\square$

As a conclusion one has bijections between the set of non positive self-adjoint operators, the set of closed symmetric non negative bilinear form and the set of strongly continuous self-adjoint contraction semigroups. This is the golden triangle of the theory of heat semigroups on Hilbert spaces !

Let $A: \mathcal{D}(A) \to \mathcal{H}$ be a densely defined operator. A densely defined operator $\bar{A}$ is called an extension of $A$ if $\mathcal{D}(A) \subset \mathcal{D}(\bar A)$ and for every $f \in \mathcal{D}(A)$ , $\bar A f=Af$ .

Theorem: (Friedrichs extension) Let $A$ be a densely defined non positive symmetric operator on $\mathcal{H}$ . There exists at least one self-adjoint extension of $A$ .

Proof: On $\mathcal{D} (A)$ , let us consider the following norm
$\| f\|^2_{A}=\| f \|^2-\langle Af ,f \rangle.$
By completing $\mathcal{D} (A)$ with respect to this norm, we get an abstract Hilbert space $(\mathcal{H}_A,\langle \cdot , \cdot \rangle_{A})$ . Since for $f \in \mathcal{D} (A)$ , $\| f \| \le \| f\|_{A}$ , the injection map $\iota : ( \mathcal{D} (A), \| \cdot \|_{A}) \rightarrow (\mathcal{H}, \| \cdot \|)$ is continuous and it may therefore be extended into a continuous map $\bar{\iota}: (\mathcal{H}_A, \| \cdot \|_{A}) \rightarrow (\mathcal{H}, \| \cdot \|)$ . Let us show that $\bar{\iota}$ is injective so that $\mathcal{H}_A$ may be identified with a subspace of $\mathcal{H}$ . So, let $f \in \mathcal{H}_A$ such that $\bar{\iota} (f)=0$ . We can find a sequence $f_n \in \mathcal{D} (A)$ , such that $\| f_n -f \|_{A} \to 0$ and $\| f_n \| \to 0$ . We have then

$\| f \|_{A} =\lim_{m,n \to + \infty} \langle f_n, f_m \rangle_{A} =\lim_{m \to + \infty} \lim_{n \to + \infty} \langle f_n,f_m \rangle-\langle Af_n,f_m\rangle =0,$
thus $f=0$ and $\bar{\iota}$ is injective. Therefore, $\mathcal{H}_A$ may be identified with a subspace of $\mathcal{H}$ . Since $\mathcal{D}(A) \subset \mathcal{H}_A$ , one has that $\mathcal{H}_A$ is dense in $\mathcal{H}$ . We consider now the quadratic form on $\mathcal{H}$ defined by
$\mathcal{E}(f)=\| f\|^2_{A}-\| f \|^2, \quad f \in \mathcal{H}_A$
It is closed because $(\mathcal{H}_A,\langle \cdot , \cdot \rangle_{A})$ is a Hilbert space. The generator of this quadratic form is then a self-adjoint extension of $A$ .
$\square$
In general self-adjoint extensions of a given symmetric operator are not unique. The operator constructed in the proof above is called the Friedrichs extension of $A$ . It is the minimal self-adjoint extension of $A$ .
Definition: Let $A$ be a densely defined non positive symmetric operator on $\mathcal{H}$ . We say that $A$ is essentially self-adjoint if it admits a unique self-adjoint extension.

We have the following criterion for essential self-adjointness.

Exercise: Let $A$ be a densely defined non positive symmetric operator on $\mathcal{H}$ . If for some $\lambda >0$ ,
$\mathbf{Ker} (-A^* +\lambda \mathbf{Id} )= \{ 0 \},$

then the operator $A$ is essentially self-adjoint.

Lecture 2. Semigroups on Hilbert spaces: The golden triangle

Leave a comment Cancel reply

Categories

Blog Stats

Follow Blog via Email

Blogroll

Recent Posts

Meta

Lecture 2. Semigroups on Hilbert spaces: The golden triangle

Share this:

Related

Leave a comment Cancel reply

Categories

Blog Stats

Follow Blog via Email

Blogroll

Recent Posts

Meta