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Page 1: Markov Semigroups - Dipartimento di Matematica …daipra/didattica/Bologna12/MarkovSem...Markov Semigroups Doctoral School, Academic Year 2011/12 Paolo Guiotto Contents 1 Introduction

Markov Semigroups

Doctoral School, Academic Year 2011/12

Paolo Guiotto

Contents

1 Introduction 1

2 Preliminaries: functional setting 2

3 Markov Processes 3

4 Feller processes 4

5 Strongly continuous semigroups on Banach spaces 8

6 Hille–Yosida theorem 13

7 Generators of Feller semigroups 17

8 Examples 208.2 Brownian motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

8.2.1 Case d = 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228.3.1 Case d > 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

8.11 Ising model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

9 Exponential Formula 28

1 Introduction

The main aim of these notes is to connect Markov processes to semigroups of linear operators on functionsspaces, an important connection that allows to a very useful and natural way to define Markov processesthrough their associated semigroup.

There’re lots of different definitions of Markov process in the literature. If this create a little bit ofconfusion at first sight, all of them are based of course on the same idea: a Markov process is someevolution phenomena whose future depends upon the past only by the present. Actually in most of theapplications we are interested in families of Markov processes living in some state space E characterized bya parameter x ∈ E which represents the starting point for the various processes of the family. Moreover,we could define the processes as usual stochastic processes (that is functions of time and of some randomparameter) or, and it is what we prefer here, through their laws, that is measures on the path space that,

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for a general Markov process, is the space DE [0,+∞[ of E−valued functions continuous from the rightand with limit from the left (so they may have jumps).

Like for ordinary dynamical systems an eventually non linear dynamics induces naturally a lineardynamics on observables, that is numerical functions defined on the state space E. We gain somethingin description (linearity) but we have to move to an infinite dimensional context (functions space). Apriori this is not better or worse, but for some questions may be better to treat with an eventuallyinfinite dimensional but linear setting. In many applications (e.g. Markov processes arising as diffusionsor interacting particle systems) this approach gives a very quick way to define the process itself definingjust a linear (generally unbounded) operator on observables: the so called infinitesimal generator.

2 Preliminaries: functional setting

Along this section we will provide the preliminaries about the main settings where we will work. In allwhat follows, (E, d) will play the role of state space and will be a locally compact metric space. We willcall

• B(E) the σ−algebra of Borel sets of E;

• B(E) the set of bounded and measurable real valued functions on E: in particular we recall thata function ϕ : E −→ R is called measurable if ϕ←(A) ∈ B(E) for any Borel set A.

• C0(E) the set of continuous real valued functions on E vanishing at infinity. By this we mean, inparticular that

∃x0 ∈ E, : ∀ε > 0, ∃R(ε) > 0, : |f(x)| 6 ε, ∀x ∈ E, : d(x, x0) > R(ε). (2.1)

Of course C0(E) ⊂ B(E). On these spaces it is defined a natural norm

‖ϕ‖∞ := supx∈E|ϕ(x)|, ϕ ∈ B(E).

It is a standard work to check that B(E) and C0(E) are Banach spaces with this norm. In general, afunction ϕ : E −→ R will be called observable. Moreover, if f ∈ C0(E) the sup–norm is actually a truemaximum as it is easily proved applying the Weierstrass theorem being E locally compact. Sometimesit will be useful to recall the

Theorem 2.1 (Riesz). The topological dual of C0(E) is the space of all bounded real valued measure onB(E). In particular

〈µ?, ϕ〉 =

∫E

ϕ(x) µ(dx).

Moreover the C0(E)? = closure〈δx : x ∈ E〉, where 〈δx, ϕ〉 = ϕ(x).

The natural space for trajectories of E−valued Markov processes is the space

DE [0,+∞[:= {γ : [0,+∞[−→ E, γ right continuous and with left limit} .

Frenchmen call this type of trajectories cadlag: continue a droite et avec limite a gauche. The space Eis called states space. We define also the classical coordinate mappings

πt : DE [0,+∞[−→ E, πt(γ) := γ(t), t > 0.

Moreover, we will define

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• F the smallest σ−algebra of DE [0,+∞[ such that all πt are measurable;

• Ft the smallest σ−algebra of DE [0,+∞[ such that all πs for s 6 t are measurable.

Clearly (Ft) is an increasing family of σ−algebras.

3 Markov Processes

Definition 3.1. Let (E, d) be a metric space. A family (Px)x∈E of probability measures on the path space(DE [0,+∞[,F ) is called Markov process if

i) Px (γ(0) = x) = 1, for any x ∈ E.

ii) (Markov property) Px (γ(t+ ]) ∈ F | Ft) = Pγ(t)(F ), for any F ∈ F and t > 0.

iii) the mapping x 7−→ Px(F ) is measurable for any F ∈ F .

Let (Px)x∈E be a Markov process. We denote by Ex the expectation w.r.t. Px, that is

Ex[Φ] =

∫DE [0,+∞[

Φ dPx, Φ ∈ L1(DE [0,+∞[,F ,Px).

The Markov property has a more flexible and general form by means of conditioned expectations:

Ex [Φ(γ(t+ ])) | Ft] = Eγ(t) [Φ] , ∀Φ ∈ L∞. (3.1)

We now introduce the fundamental object of our investigations: let’s define

S(t)ϕ(x) := Ex [ϕ(γ(t))] ≡∫DE [0,+∞[

ϕ(γ(t)) dPx(γ), ϕ ∈ B(E). (3.2)

We will see immediately that any S(t) is well defined for t > 0. The family (S(t))t>0 is called Markovsemigroup associated to the process (Px)x∈E . This is because of the following

Proposition 3.2. Let (Px)x∈E be a Markov process on E and (S(t)t>0 be the associated Markov semi-group. Then:

i) S(t) : B(E) −→ B(E) is a bounded linear operator for any t > 0 and ‖S(t)ϕ‖∞ 6 ‖ϕ‖∞ for anyϕ ∈ B(E), t > 0 (that is ‖S(t)‖ 6 1 for any t > 0).

ii) S(0) = I.

iii) S(t+ r) = S(t)S(r), for any t, r > 0.

iv) S(t)ϕ > 0 a.e. if ϕ > 0 a.e.: in particular, if ϕ 6 ψ a.e., then S(t)ϕ 6 S(t)ψ a.e..

v) S(t)1 = 1 a.e. (here 1 is the function constantly equal to 1).

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Proof — i) It is standard (about the measurability proceed by approximation: the statement is true for ϕ = χA,A ∈ B(E) by iii) of the definition of Markov process because

S(t)χA(x) = Px(γ(t) ∈ A) = Px (π←t (A)) ,

and F := π←t (A) ∈ F ; hence it holds for ϕ sum of χA, that is for simple functions; for general ϕ take firstf > 0 and approximate it by an increasing sequence of simple functions). Linearity follows by the linearity of theintegral. Clearly

|S(t)ϕ(x)| 6 Ex [|ϕ(γ(t))|] 6 ‖ϕ‖∞, ∀x ∈ E, =⇒ ‖S(t)ϕ‖∞ 6 ‖ϕ‖∞, ∀t > 0.

In other words S(t) ∈ L (B(E)) and ‖S(t)‖ 6 1.

ii) Evident.

iii) This involves the Markov property:

S(t+ r)ϕ(x) = Ex [ϕ(γ(t+ r))] = Ex [Ex [ϕ(γ(t+ r)) | Ft]](3.1)= Ex

[Eγ(t) [ϕ(γ(r))]

]= Ex [S(t)ϕ(γ(r))]

= S(r) [S(t)ϕ] (x).

iv), v) Evident.

4 Feller processes

To treat with bounded measurable observables is in general quite difficult because of their poor properties,so it’s better to restrict to continuous observables:

Definition 4.1 (Feller property). Let (S(t))t>0 be the Markov semigroup associated to a Markov process(Px)x∈E where (E, d) is locally compact. We say that the semigroup fulfills the Feller property if

S(t)f ∈ C0(E), ∀f ∈ C0(E), ∀t > 0.

This property turns out to give the strongly continuity of the semigroup:

Theorem 4.2 (strong continuity). Let (S(t))t>0 be the Markov semigroup associated to a Markov process(Px)x∈E where (E, d) is locally compact. If (S(t))t>0 fulfills the Feller property, it is then stronglycontinuous on C0(E), that is

S(·)ϕ ∈ C ([0,+∞[; C0(E)), ∀ϕ ∈ C0(E).

Proof — First we prove right weak-continuity, that is

limt→t0+

S(t)ϕ(x) = S(t0)ϕ(x), ∀x ∈ E, ∀ϕ ∈ Cb(E).

This follows immediately as an application of dominated convergence and because trajectories are right continuous.Indeed

limt→t0+

S(t)ϕ(x) = limt→t0+

∫DE [0,+∞[

ϕ(γ(t)) Px(dγ).

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Now: |ϕ(γ(t))| 6 ‖ϕ‖∞ which is Px−integrable, and γ(t) −→ γ(t0) as t −→ t0 because γ ∈ DE [0,+∞[. In asimilar way we have

∃ limt→t0−

S(t)ϕ(x), ∀x ∈ E, ∀ϕ ∈ C0(E), ∀t0 > 0.

So: fixing x ∈ E we have that the function t 7−→ S(t)ϕ(x) has limit from the left and from the right at any pointof [0,+∞[. It is a standard first year Analysis exercise to deduce that S(·)ϕ(x) has at most a countable numberof discontinuities, hence is measurable.

Now, define

Rλϕ(x) :=

∫ +∞

0

e−λtS(t)ϕ(x) dt, λ > 0, x ∈ E.

We will see later what is the meaning of this. The integral is well defined and convergent because

|e−λtS(t)ϕ(x)| = e−λt|S(t)ϕ(x)| 6 e−λt‖ϕ‖∞.

We say that Rλϕ ∈ C0(E) if ϕ ∈ C0(E). Indeed: everything follows immediately as application of dominatedconvergence because any S(t)ϕ ∈ C0(E) and because of the usual bound |S(t)ϕ(x)| 6 ‖ϕ‖∞.

We will show now strong continuity for ϕ of type Rλϕ, that is S(·)Rλϕ ∈ C ([0,+∞[; C0(E)). This willbe done in steps: the first is to prove strong right continuity, that is

S(·)RλϕC0(E)−→ S(t0)Rλϕ, as t −→ t0 + .

We start with the case t0 = 0. Notice that

S(t)Rλϕ(x) = S(t)

∫ +∞

0

e−λrS(r)ϕ(x) dr =

∫ +∞

0

e−λrS(r + t)ϕ(x) dr = eλt∫ +∞

t

e−λrS(r)ϕ(x) dr,

hence

S(t)Rλϕ(x)−Rλϕ(x) = (eλt − 1)

∫ +∞

t

e−λrS(r)ϕ(x) dr +

∫ t

0

e−λrS(r)ϕ(x) dr,

therefore

‖S(t)Rλϕ−Rλϕ‖∞ 6(eλt − 1

)∫ +∞

t

e−λr‖ϕ‖∞ dr +

∫ t

0

e−λr‖ϕ‖∞ dr 6eλt − 1

λ‖ϕ‖∞ + t‖ϕ‖∞ −→ 0,

as t −→ 0+. For generic t0 we have

‖S(t)Rλϕ− S(t0)Rλϕ‖∞ = ‖S(t0) (S(t− t0)Rλϕ−Rλϕ)‖∞ 6 ‖S(t− t0)Rλϕ−Rλϕ‖∞ −→ 0,

as t −→ t0+. Now we can prove the left continuity at t0: assuming now t < t0,

‖S(t)Rλϕ− S(t0)Rλϕ‖∞ = ‖S(t) (S(t0 − t)Rλϕ−Rλϕ)‖∞ 6 ‖S(t0 − t)Rλϕ−Rλϕ‖∞ −→ 0, t −→ t0 − .

We will now show that the set of Rλϕ (λ > 0) is dense in C0(E). To this aim take µ? ∈ C0(E)? such that

0 = 〈µ?, Rλϕ〉 =

∫E

Rλϕ(x) dµ(x), ∀ϕ ∈ C0(E).

The aim is to prove µ ≡ 0. Now notice that

λRλϕ(x) =

∫ +∞

0

e−λtS(t)ϕ(x) d(λt) =

∫ +∞

0

e−rS( rλ

)ϕ(x) dr.

Applying the right continuity in 0 of the semigroup and the dominated convergence it is easy to deduce that

λRλϕ(x) −→∫ +∞

0

e−rS(0)ϕ(x) dr = S(0)ϕ(x)

∫ +∞

0

e−r dr = ϕ(x).

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Moreover, always by previous formula

‖Rλϕ‖∞ 6∫ +∞

0

e−r‖ϕ‖∞ dr = ‖ϕ‖∞.

Therefore, applying the dominated convergence (µ is a finite measure) we have

0 =

∫E

λRλϕ dµ −→∫E

ϕ dµ, ∀ϕ ∈ C0(E).

But then 〈µ?, ϕ〉 = 0 for any ϕ ∈ C0(E), and this means µ = 0.

Conclusion: take ψ ∈⋃λ>0Rλ (C0(E)) such that ‖ϕ−ψ‖∞ 6 ε (such sequence exists by previous step). Each

S(·)ϕn ∈ C ([0,+∞[; C0(E)) by the first step. Therefore

‖S(t)ϕ− S(t0)ϕ‖∞ 6 ‖S(t)ϕ− S(t)ψ‖∞ + ‖S(t)ψ − S(t0)ψ‖∞ + ‖S(t0)ψ − S(t0)ϕ‖∞

6 2‖ϕ− ψ‖∞ + ‖S(t)ψ − S(t0)ψ‖∞ 6 2ε+ ‖S(t)ψ − S(t0)ψ‖∞,

thereforelim supt→t0

‖S(t)ϕ− S(t0)ϕ‖∞ 6 2ε+ lim supt→t0

‖S(t)ψ − S(t0)ψ‖∞ = 2ε,

and because ε is arbitrary the conclusion follows.

Because this will be the main subject of our investigations here we introduce the

Definition 4.3. A family of linear operators (S(t))t>0 ⊂ L (C0(E)), (E, d) locally compact metric space,is called a Feller semigroup if

i) S(0) = I;

ii) S(t+ r) = S(t)S(r) for any t, r > 0;

iii) S(·)ϕ ∈ C ([0,+∞[; C0(E));

iv) S(t)ϕ > 0 for any ϕ ∈ C0(E), ϕ > 0 and for all t > 0;

v) S(t)ϕ 6 1 for any ϕ ∈ C0(E), ϕ 6 1 and for all t > 0.

We say that a Feller semigroup is conservative if

∀(ϕn) ⊂ C0(E), : ϕn ↗ 1 on E, =⇒ S(t)ϕn ↗ 1, on E.

Remark 4.4. Notice in particular that by previous properties follows that ‖S(t)‖ 6 1. Indeed: if ‖ϕ‖∞ 6 1,in particular |ϕ(x)| 6 1, that is −1 6 ϕ(x) 6 1 for any x ∈ E. By iv) and v) we have

−1 6 ϕ 6 1, =⇒ −1 6 S(t)ϕ(x) 6 1, ∀x ∈ E, ⇐⇒ |S(t)ϕ(x)| 6 1, ∀x ∈ E, ⇐⇒ ‖S(t)ϕ‖∞ 6 1.

But this means exactly ‖S(t)‖ 6 1.

It is natural to ask if by a Feller semigroup it is possible to construct a Markov process. This is actuallytrue and the first step is the construction of a transition probability function:

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Proposition 4.5. Let (S(t))t>0 be a conservative Feller semigroup on C0(E), (E, d) locally compactmetric space. For any t > 0, x ∈ E there exists a probability measure

Pt(x, ·) : B(E) −→ [0, 1],

such that

S(t)ϕ(x) =

∫E

ϕ(y)Pt(x, dy), ∀ϕ ∈ C0(E).

Moreover:

i) x 7−→ Pt(x, F ) ∈ B(E) for any t > 0, F ∈ B(E).

ii) (Chapman–Kolomogorov equation) for any t, r > 0,

Pt+r(x, F ) =

∫E

Pt(x, dy)Pr(y, F ). (4.1)

Pt(x, dy) is called transition probability.

Proof — Fix t > 0 and x ∈ E and consider the functional ϕ 7−→ S(t)ϕ(x). It is clearly linear and continuous,so by Riesz representation theorem 2.1 there exists a finite Borel measure Pt(x, ·) on E such that

S(t)ϕ(x) =

∫E

ϕ(y) Pt(x, dy), ∀ϕ ∈ C0(E).

Moreover, because S(t)ϕ > 0 if ϕ > 0 we have Pt(x, ·) is a positive measure. Moreover, by conservativity andmonotone convergence,

Pt(x,E) =

∫E

Pt(x, dy) = limn

∫E

ϕn(y) Pt(x, dy) = limnS(t)ϕn(x) = 1, ∀x ∈ E, t > 0.

so Pt(x, dy) turns out to be a probability measure. By standard approximation methods i) follows. ii) follows bythe semigroup property: first notice that∫

E

ϕ(z)Pt+r(x, dz) = S(t+ r)ϕ(x) = S(t)S(r)ϕ(x) =

∫E

S(r)ϕ(y)Pt(x, dy)

=

∫E

(∫E

ϕ(z)Pr(y, dz)

)Pt(x, dy).

This holds for any ϕ ∈ C0(E), hence by approximations, for any ϕ ∈ B(E), therefore also in the case ϕ = χF . Inthis case we obtain

Pt+r(x, F ) =

∫E

(∫E

χF (z)Pr(y, dz)

)Pt(x, dy) =

∫E

Pr(y, F ) Pt(x, dy),

which is just the Chapman–Kolmogorov equation.

Remark 4.6. If S(t) is not conservative, Pt(x,E) could be, possibly, strictly less than 1. This correspondto the case on which the underlying process we are trying to reconstruct leaving at time 0 from the state

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x die (or escape at infinity) at some time t > 0. If this is the case we could construct the one pointcompactification of E, E := E ∪ {∞} with the usual topology and set

Pt(x, F ) :=

Pt(x, F ), if x ∈ E, F ⊂ E,

1− Pt(x,E), if x ∈ E, F = {∞},

0, if x =∞, F ⊂ E,

1, if x =∞, F = {∞}.

It is easy to check that i) and ii) of previous Proposition are still true.

Transition probabilities are a good tool to construct Markov processes. The idea is to use the KolmogorovTheorem imposing the Markov property by the transition probabilities using the Chapman–Kolmogorovequation. Basically we want that

Px (γ ∈ DE [0,+∞[ : γ(0) = x, γ(t1) ∈ F1, . . . , γ(tn) ∈ Fn) =

=

∫F1

Pt1(x, dy1)

∫F2

Pt2−t1(y1, dy2) · · ·∫Fn

Ptn−tn−1(yn−1, dyn).

(4.2)

We are not interested here going in deep with this, see Revuz & Yor [4].

5 Strongly continuous semigroups on Banach spaces

In this section we will see some general facts that involves only the structure of continuous semigroup ona generic Banach space X.

Definition 5.1. A family (S(t))t>0 of bounded linear operators on a Banach space X is called stronglycontinuous semigroup if

i) S(0) = I;

ii) S(t+ r) = S(t)S(r), for all t, r > 0;

iii) t 7−→ S(t)ϕ ∈ C([0,+∞[;X) for all ϕ ∈ X.

If ‖S(t)‖ 6 1 for all t > 0 the semigroup is called contraction semigroup.

By what we have seen in the previous section we will be particularly interested in contraction semigroupsso we will limit the general discussion to this case even if all the theorem have extensions to the generalcase. It is natural to expect that, in a suitable sense, S(t) = etA for some A. Thinking to the caseA ∈ L (X),

A = limt→0+

etA − It

.

For this reason we will introduce the following

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Definition 5.2. Let (S(t))t>0 be a strongly continuous semigroup on a Banach space X. The operator

Aϕ := limh→0+

S(h)ϕ− ϕh

, x ∈ D(A) :=

{ϕ ∈ X : ∃ lim

h→0+

S(h)ϕ− ϕh

}.

is called infinitesimal generator of (S(t))t>0.

The reason why we call it infinitesimal generator will be clear by the Hille–Yosida theorem: we will firstcharacterize some properties an infinitesimal generator verifies; once we will have these properties wewill see that they are enough to construct a strongly continuous semigroup by an operator which verifiesthem. A first set of properties is given by the

Theorem 5.3. Let X be a Banach space, (S(t))t>0 a contraction semigroup and A its infinitesimalgenerator. Then

i) D(A) is dense in X;

ii) A is closed (i.e. G(A) := {(ϕ,Aϕ) ∈ X ×X : ϕ ∈ D(A)} is closed in the product topology ofX ×X).

Proof — i) Let ϕ ∈ X and define ψε := 1ε

∫ ε0S(r)ϕ dr. Clearly ψε −→ ϕ as ε↘ 0 (mean value theorem). Let’s

prove that ψε ∈ D(A) for all t > 0. We have

1

h(S(h)ψε − ψε) =

1

ε· 1

h

(S(h)

∫ ε

0

S(r)ϕ dr −∫ ε

0

S(r)ϕ dr

)=

1

εh

(∫ ε+h

h

S(r)ϕ dr −∫ ε

0

S(r)ϕ dr

)

=1

ε

(1

h

∫ ε+h

ε

S(r)ϕ dr − 1

h

∫ h

0

S(r)ϕ dr

)−→ 1

ε(S(ε)ϕ− ϕ),

always in force of the mean value theorem. This proves i).ii) We have to prove that

if (ϕn, Aϕn) −→ (ϕ,ψ), =⇒ (ϕ,ψ) ∈ G(A), i.e. ϕ ∈ D(A) and ψ = Aϕ.

In other words, we have to prove that

∃ limh→0+

S(h)ϕ− ϕh

= ψ.

NowS(h)ϕ− ϕ = lim

n(S(h)ϕn − ϕn) .

Because ϕn ∈ D(A) we know that t 7−→ S(t)ϕn is differentiable in t = 0 from the right. More:

Lemma 5.4. If ϕ ∈ D(A) then

∃ ddtS(t)ϕ = S(t)Aϕ = AS(t)ϕ, ∀t > 0.

Proof of the Lemma — We have to prove that t 7−→ S(t)ϕ is differentiable for all t > 0 and that S(t)ϕ ∈ D(A).We start with the first.First step: exists d+

dtS(t)ϕ. Indeed

d+

dtS(t)ϕ = lim

h→0+

S(t+ h)ϕ− S(t)

h= S(t) lim

h→0+

S(h)ϕ− ϕh

= S(t)Aϕ.

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Second step: exists d−

dtS(t)ϕ. Indeed

d−

dtS(t)ϕ = lim

h→0+

S(t− h)ϕ− S(t)

h= limh→0+

S(t− h)ϕ− S(h)ϕ

h.

Now: ∥∥∥∥S(t− h)ϕ− S(h)ϕ

h− S(t)Aϕ

∥∥∥∥ 6

∥∥∥∥S(t− h)

(ϕ− S(h)ϕ

h−Aϕ

)∥∥∥∥+ ‖S(t− h)Aϕ− S(t)Aϕ‖.

Clearly, by the strong continuity, the second term converges to 0. For the first, by the estimate ‖S(t)‖ 6 1 weobtain ∥∥∥∥S(t− h)

(ϕ− S(h)ϕ

h−Aϕ

)∥∥∥∥ 6

∥∥∥∥ϕ− S(h)ϕ

h−Aϕ

∥∥∥∥ −→ 0, h↘ 0.

By this the conclusion follows.Third step: S(t)ϕ ∈ D(A). Indeed

S(h)S(t)ϕ− S(t)ϕ

h= S(t)

S(h)ϕ− ϕh

−→ S(t)Aϕ, =⇒ AS(t)ϕ = S(t)Aϕ.

Coming back to the proof of the Theorem,

S(h)ϕn − ϕn =

∫ h

0

d

drS(r)ϕn dr =

∫ h

0

S(r)Aϕn dr −→∫ h

0

S(r)ψ dr.

To justify the last passage, notice that∥∥∥∥∫ h

0

S(r)Aϕn dr −∫ r

0

S(r)ψ dr

∥∥∥∥ 6∫ h

0

‖S(r)(Aϕn − ψ)‖ dr 6 h‖Aϕn − ψ‖ −→ 0.

Finally:

S(h)ϕ− ϕ =

∫ h

0

S(r)ψ dr.

Therefore,S(h)ϕ− ϕ

h=

1

h

∫ h

0

S(r)ψ dr −→ ψ, =⇒ ϕ ∈ D(A), e Aϕ = ψ.

The previous result gives the first indications about the properties of the infinitesimal generator. Ofparticular interest is the weak continuity property given by the closure of the operator A. This propertyis weaker than continuity and it is fulfilled by unbounded operators. For instance,

Example 5.5. Let X = C([0, 1]) be endowed with the sup–norm ‖ · ‖∞ and A given by

D(A) :={ϕ ∈ C1([0, 1]) : ϕ(0) = 0

}, Aϕ = ϕ′,

then A is closed.

Sol. — Indeed, if (ϕn) ⊂ D(A) is such that ϕn −→ ϕ in X with Aϕn = ϕ′n −→ ψ in X (that is, uniformly on[0, 1]), by a well known result ϕ ∈ C1([0, 1]) and ϕ′ = ψ, that is Aϕ = ψ. To finish just notice that ϕ(0) = 0,therefore ϕ ∈ D(A).

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11

Unfortunately, being A unbounded in general, it seems difficult to give a good meaning to the exponentialseries

S(t)ϕ ≡ etAϕ =

∞∑n=0

(tA)n

n!ϕ,

to define the semigroup by a given A. There’s however another possible formula to define the exponential,that is

etAϕ = limn→+∞

(I +

tA

n

)nϕ ≡ lim

n→+∞

(I− tA

n

)−nϕ. (5.1)

While the first limit seems bad because I + µA and its powers seems bad, the second is much moreinteresting, because we may expect that (I− µA)−1 is nicer than A.

Example 5.6. Compute (I− µA)−1 in the case of A defined in the previous example.

Sol. —(I− µA)ϕ = ϕ− µϕ′ = ψ, ⇐⇒ ϕ = (I− µA)−1ψ.

We may think that given ψ we have to solve the differential equation

ϕ′ =1

µϕ− 1

µψ, ⇐⇒ ϕ(x) = e

∫ x0

1µdy

(∫ x

0

e−

∫ y0

1µdz 1

µψ(y) dy + C

)= e

(1

µ

∫ x

0

e− yµψ(y) dy + C

).

Now, imposing that ϕ ∈ D(A) we have

ϕ(0) = 0, ⇐⇒ C = 0, ⇐⇒ ϕ(x) ≡ (I− µA)−1 ψ(x) =exµ

µ

∫ x

0

e− yµψ(y) dy.

Writing

(I− µA)−1 =1

µ

(1

µI−A

)−1=:

1

µR 1µ,

we find a very familiar concept of Functional Analysis:

Definition 5.7. Let λ ∈ C. If Rλ := (λI−A)−1 ∈ L (X) we say that λ ∈ ρ(A) (resolvent set) and wecall Rλ resolvent operator. The set σ(A) := C\ρ(A) is called spectrum of A. Analytically the spectrumis divided in

• point spectrum, denoted by σp(A), that is the set of λ ∈ C such that λI − A is not injective (inother words: the elements of σp(A) are the eigenvalues);

• continuum spectrum, denoted by σc(A), that is the set of λ ∈ C such that λI−A is bijective butthe inverse is not continuous;

• residual spectrum, what it remains, that is σ(A)\{σp(A) ∪ σc(A)}.

Looking at the second limit in (5.1), we would expect that ρ(A) ⊃ [λ0,+∞[ for some λ0. To this aimwe need a relationship between the resolvent operator and the semigroup. This is formally easy: writingS(t) = etA and treating λI −A as a negative number we have∫ +∞

0

e−λrS(r)ϕ dr =

∫ +∞

0

e−r(λ−A)ϕ dr =

[e−r(λ−A)

λ−A

]r=+∞

r=0

ϕ = (λI−A)−1ϕ.

This formula turns out to be true. Precisely we have the

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12

Theorem 5.8. Let (S(t))t>0 be a contraction semigroup on a Banach space X. Then:

i) ρ(A) ⊃ {Reλ > 0} and

Rλϕ =

∫ +∞

0

e−λrS(r)ϕ dr, ∀λ ∈ C : Reλ > 0, ∀ϕ ∈ X; (5.2)

ii) The following estimate holds:

‖Rλ‖ 61

Reλ, ∀λ ∈ C : Reλ > 0. (5.3)

ΣHAL

Figure 1: The spectrum contained in the half plane Reλ < 0.

Proof — We first prove that the integral in (5.2) is well defined. This is easy because, recalling that ‖S(t)‖ 6 1for any t > 0, we have

‖e−rλS(r)ϕ‖ 6 e−rReλ‖ϕ‖ ∈ L1([0,+∞[), ∀λ ∈ C : Reλ > 0.

Therefore, being r 7−→ e−rλS(r)ϕ ∈ C([0,+∞[;X) the integral is well defined. By the same estimate we get

‖Rλϕ‖ 6∫ +∞

0

‖ϕ‖e−rReλ dr =1

Reλ‖ϕ‖, ∀ϕ ∈ X,

that is (5.3). It remain to prove that the integral operator is indeed the resolvent operator, that is:

a) Rλϕ ∈ D(A), ∀ϕ ∈ X, b) (λI −A)Rλϕ = ϕ, ∀ϕ ∈ X, c) Rλ(λI −A)ϕ = ϕ, ∀ϕ ∈ D(A).

We start from the first. Notice that

S(h)Rλϕ−Rλϕh

=1

h

(S(h)

∫ +∞

0

e−λrS(r)ϕ dr −∫ +∞

0

e−λrS(r)ϕ dr

)

=1

h

(∫ +∞

h

e−λ(r−h)S(r)ϕ dr −∫ +∞

0

e−λrS(r)ϕ dr

)

=eλh − 1

h

∫ +∞

h

e−λrS(r)ϕ dr +1

h

(∫ +∞

h

e−λrS(r)ϕ dr −∫ +∞

0

e−λrS(r)ϕ dr

)

−→ λ

∫ +∞

0

e−λrS(r)ϕ dr − ϕ = λRλϕ− ϕ.

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13

Therefore, Rλϕ ∈ D(A) and

ARλϕ = λRλϕ− ϕ, ⇐⇒ (λI −A)Rλϕ = ϕ, ∀ϕ ∈ X,

that is (λI −A)Rλ = I, that is b). To finish, we have to prove c). Let ϕ ∈ D(A). Notice that

Rλ(λI −A)ϕ = λ

∫ +∞

0

e−λrS(r)ϕ dr −∫ +∞

0

e−λrS(r)Aϕ dr.

Because ϕ ∈ D(A), by Lemma 5.4 we have that S(r)Aϕ = (S(r)ϕ)′. Therefore, integrating by parts,∫ +∞

0

e−λrS(r)Aϕ dr =

∫ +∞

0

e−λr (S(r)ϕ)′ dr =[e−λrS(r)ϕ

]r=+∞

r=0−∫ +∞

0

−λe−λrS(r)ϕ dr = −ϕ+ λRλϕ,

that is the conclusion.

6 Hille–Yosida theorem

In the previous section we have seen that:

Corollary 6.1. The infinitesimal generator A : D(A) ⊂ X −→ X of a contraction semigroup fulfills thefollowing properties:

i) A is densely defined and closed;

ii) ρ(A) ⊃ {λ ∈ C : Reλ > 0} and ‖Rλ‖ = ‖(λI−A)−1‖ 6 1

Reλ for any λ ∈ C such that Reλ > 0.

Actually we may notice that to be closed is redundant because the following general fact

Proposition 6.2. Let A : D(A) ⊂ X −→ X be a linear operator such that ρ(A) 6= ∅. Then A is closed.

Proof — Let (ϕn) ⊂ D(A) such thatϕn −→ ϕ, Aϕn −→ ψ.

We need to prove ϕ ∈ D(A) and Aϕ = ψ. Now let λ ∈ ρ(A) and consider

λϕn −Aϕn = (λI−A)ϕn, ϕn = (λI−A)−1 (λϕn −Aϕn) −→ (λI−A)−1 (λϕ− ψ)

because (λI−A)−1 ∈ L (X). In particular

ϕ = (λI−A)−1 (λϕ− ψ) ∈ D(A), and λϕ−Aϕ = λϕ− ψ, ⇐⇒ Aϕ = ψ.

In this section we will see that these conditions are sufficient to construct a unique contraction semigroupwhich generator is just A. The idea is to construct etA by approximations

etAϕ = limλ→+∞

etAλϕ,

where Aλ ∈ L (X) are suitable approximations of A. Such approximations, that will be called Yosidaregularizations, are extraordinarily intuitive because are given by

Aλ :=A

I− 1λA

”λ−→+∞−→ ” A.

Let us introduce formally these operators:

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14

Definition 6.3. Let A : D(A) ⊂ X −→ X be a densely defined linear operator such that

ρ(A) ⊃ {λ ∈ C : Reλ > 0} , ‖Rλ‖ = ‖(λI−A)−1‖ 6 1

Reλ, ∀λ ∈ C : Reλ > 0.

We call Yosida regularization of A the family (Aλ)λ>0 ⊂ L (X) defined as

Aλ := λARλ = λA(λI−A)−1, ∀λ > 0.

Remark 6.4. At first sight may be is not evident that Aλ ∈ L (X): indeed

(λI−A)Rλ = IX , =⇒ Aλ = λARλ = λ(λRλ − IX) ∈ L (X).

Morally Aλ −→ A as λ −→ +∞. Indeed we have the

Lemma 6.5. Let A : D(A) ⊂ X −→ X an operator fulfilling hypotheses of Definition 6.3 on X Banach.Then

limλ→+∞

Aλϕ = Aϕ, ∀ϕ ∈ D(A).

Proof — Because, as in the remark, ARλ = λRλ − IX we can write

Aλ = λARλ = λ2Rλ − λIX .

Therefore, if ϕ ∈ D(A) we have Aλϕ = λ(λRλϕ− ϕ). But Rλ(λI−A) = ID(A), so

Rλ(λϕ−Aϕ) = ϕ, ⇐⇒ λRλϕ− ϕ = RλAϕ, =⇒ Aλϕ = λRλAϕ.

Set ψ := Aϕ. If we prove thatlim

λ→+∞λRλψ = ψ, ∀ψ ∈ X, (6.1)

we are done. Assume first that ψ ∈ D(A): by the same identity as before,

λRλψ − ψ = RλAψ, =⇒ ‖λRλψ − ψ‖ = ‖RλAψ‖ 6 ‖Rλ‖‖Aψ‖ 61

λ‖Aψ‖ −→ 0, λ −→ +∞.

In the general case that ψ ∈ X, by density of D(A) in X there exists ψε ∈ D(A) such that ‖ψ − ψε‖ 6 ε.Therefore,

‖λRλψ − ψ‖ 6 ‖λRλψ − λRλψε‖+ ‖λRλψε − ψε‖+ ‖ψε − ψ‖ 6 λ1

λε+ ‖λRλψε − ψε‖+ ε,

and by this the conclusion follows easily.

We are now ready for the main result:

Theorem 6.6 (Hille–Yosida). Let X be a Banach space, A : D(A) ⊂ X −→ X fulfilling hypotheses ofDefinition 6.3 that is:

i) A is densely defined;

ii) ρ(A) ⊃ {λ ∈ C : Reλ > 0} and

‖Rλ‖ = ‖(λI−A)−1‖ 6 1

Reλ, ∀λ ∈ C : Reλ > 0. (6.2)

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There exists then a unique contraction semigroup (S(t))t>0 which generator is A. This semigroup will bedenoted by (etA)t>0.

Proof — As announched, we want to define S(t)ϕ := limλ→+∞ etAλϕ.

First step : construction of the semigroup. Because Aλ ∈ L (X), it is well defined the uniformly continuousgroup etAλ . By the fundamental theorem of calculus,

etAλϕ− etAµϕ =

∫ 1

0

(etrAλet(1−r)Aµϕ

)′dr =

∫ 1

0

(etrAλtAλe

t(1−r)Aµϕ+ etrAλet(1−r)Aµ(−tAµ)ϕ)dr.

Clearly, AλAµ = AµAλ. Therefore

etAλϕ− etAµϕ = t

∫ 1

0

etrAλet(1−r)Aµ (Aλ −Aµ)ϕ dr,

henceforth ∥∥∥etAλϕ− etAµϕ∥∥∥ 6 t

∫ 1

0

∥∥∥etrAλ∥∥∥ ∥∥∥et(1−r)Aµ∥∥∥ dr ‖(Aλ −Aµ)ϕ‖ .

Let us give an estimate of ‖euAλ‖. Recall first that Aλ = λ2Rλ − λIX . Therefore (1)

euAλ = eu(λ2Rλ−λIX ) = euλ

2Rλe−uλIX = e−uλeuλ2Rλ ,

so ∥∥∥euAλ∥∥∥ = e−uλ∥∥∥euλ2Rλ

∥∥∥ 6 e−uλeuλ2‖Rλ‖ 6 e−uλeuλ = 1. (6.3)

We deduce by this that ∥∥∥etAλϕ− etAµϕ∥∥∥ 6 t ‖(Aλ −Aµ)ϕ‖ . (6.4)

Now, if ϕ ∈ D(A) by previous Lemma Aλϕ −→ Aϕ. With this in hand it is easy to deduce that the sequenceof functions (e·Aλϕ)λ>0 is uniformly Cauchy on every interval [0, T ], for all T > 0. Call S(·)ϕ the uniform limitfunction, defined on [0,+∞[. Passing to the limit into the (6.4) we have∥∥∥etAλϕ− S(t)ϕ

∥∥∥ 6 t ‖Aλϕ−Aϕ‖ , ∀ϕ ∈ D(A). (6.5)

In this way, for all t > 0, we have defined S(t)ϕ := limλ→+∞ etAλϕ for all ϕ ∈ D(A). Clearly

i) S(t) : D(A) ⊂ X −→ X is linear;

ii) S(0)ϕ = ϕ, ∀ϕ ∈ D(A);

iii) S(t+ r)ϕ = S(t)S(r)ϕ, ∀ϕ ∈ D(A);

iv) t 7−→ S(t)ϕ ∈ C([0,+∞[;X), ∀ϕ ∈ D(A).

Moreover‖S(t)ϕ‖ = lim

λ→+∞‖etAλϕ‖ 6 ‖ϕ‖.

Being X complete and D(A) dense, it is easy to conclude that S(t) is extendible to all X and ‖S(t)‖ 6 1. Ofcourse, ii) and iii) hold true to all X. It remain to prove that also iv) extends to all X. Fix ϕ ∈ X and letϕε ∈ D(A) such that ‖ϕ− ϕε‖ 6 ε. Then, if h > 0,

‖S(t+ h)ϕ− S(t)ϕ‖ 6 ‖S(t+ h)(ϕ− ϕε)‖+ ‖S(t+ h)ϕε − S(t)ϕε‖+ ‖S(t)ϕε − S(t)ϕ‖

6 2‖ϕε − ϕ‖+ ‖S(t+ h)ϕε − S(t)ϕε‖

6 2ε+ ‖S(t+ h)ϕε − S(t)ϕε‖.1Here we are using the property eA+B = eAeB . Of course in general it is false, but if A and B commutes, as in the

present case, it is true.

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Therefore,lim suph→0+

‖S(t+ h)ϕ− S(t)ϕ‖ 6 2ε.

But ε > 0 is free, so lim suph→0+ ‖S(t+h)ϕ−S(t)ϕ‖ = 0 and this says that S(·)ϕ is right continuous on [0,+∞[.For left continuity, if h > 0

‖S(t− h)ϕ− S(t)ϕ‖ = ‖S(t− h)[ϕ− S(h)ϕ]‖ 6 ‖ϕ− S(h)ϕ‖ −→ 0, h −→ 0+,

by right continuity.

Second step: A is the generator of (S(t))t>0. Let

B : D(B) ⊂ X −→ X, D(B) :=

{ϕ ∈ X : ∃ lim

h→0+

S(h)ϕ− ϕh

}, Bϕ := lim

h→0+

S(h)ϕ− ϕh

, ϕ ∈ D(B),

the infinitesimal generator of (S(t))t>0. We will prove that D(B) = D(A) and A = B.

We start by proving that D(A) ⊂ D(B) and Aϕ = Bϕ for all ϕ ∈ D(A). Let ϕ ∈ D(A). By definition of S(h) wehave

S(h)ϕ− ϕ = limλ→+∞

(ehAλϕ− ϕ

)= limλ→+∞

∫ h

0

(erAλϕ

)′dr = lim

λ→+∞

∫ h

0

erAλAλϕ dr.

It’s natural to show that

limλ→+∞

∫ h

0

erAλAλϕ dr =

∫ h

0

S(r)Aϕ dr. (6.6)

We have ∥∥∥∥∫ h

0

erAλAλϕ dr −∫ h

0

S(r)Aϕ dr

∥∥∥∥ 6∫ h

0

∥∥∥erAλAλϕ− S(r)Aϕ∥∥∥ dr

6∫ h

0

∥∥∥erAλ(Aλϕ−Aϕ)∥∥∥+

∥∥∥erAλAϕ− S(r)Aϕ∥∥∥ dr

6 h‖Aλϕ−Aϕ‖+

∫ h

0

∥∥∥erAλAϕ− S(r)Aϕ∥∥∥ dr

By Lemma 6.5 the first term goes to 0 as λ −→ +∞. For the second term notice first that erAλψ −→ S(r)ψuniformly on [0, h] for all ψ ∈ X. Indeed: by construction, this is true if ψ ∈ D(A). On the other hand, beingD(A) dense in X, if ψε ∈ D(A) is such that ‖ψ − ψε‖ 6 ε, by estimate (6.5)

‖erAλψ − S(r)ψ‖ 6 ‖erAλ(ψ − ψε)‖+ ‖erAλψε − S(r)ψε‖+ ‖S(r)(ψ − ψε)‖ 6 2ε+ r‖Aλψε −Aψε‖,

Therefore

‖e·Aλψ − S(·)ψ‖∞,[0,h] 6 2ε+ h‖Aλψε −Aψε‖, =⇒ lim supλ→+∞

‖e·Aλψ − S(·)ψ‖∞,[0,h] 6 2ε.

Being ε > 0 arbitrary the conclusion follows. This explain completely (6.6). As consequence,

S(h)ϕ− ϕ =

∫ h

0

S(r)Aϕ dr, ∀ϕ ∈ D(A). (6.7)

By mean value theorem,

limh→0+

S(h)ϕ− ϕh

= limh→0+

1

h

∫ h

0

S(r)Aϕ dr = S(0)Aϕ = Aϕ,

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and this means exactly that D(A) ⊂ D(B) and Bϕ = Aϕ for ϕ ∈ D(A).

The inverse inclusion is much more soft. Indeed: because B is the generator of a contraction semigroup, itfulfills the conditions i) and ii) of Theorem 5.8. In particular, 1 ∈ ρ(B), that is (I − B)−1 : X −→ D(B) isbounded and D(B) = (I − B)−1X. On the other hand, we have seen that D(A) ⊂ D(B) and B∣∣∣

D(A)

= A. In

particular, (I − B)D(A) = (I − A)D(A). By our assumption, 1 ∈ ρ(A), so again D(A) = (I − A)−1X, that isX = (I−A)D(A). Hence, (I−B)D(A) = X and this is possible iff D(B) ⊂ D(A). By this the conclusion followseasily.

7 Generators of Feller semigroups

What we have seen in the previous two sections are general results that involves the first three propertiesin the Definition 4.3 of Feller semigroup. Now clearly the question is: which other property we need inorder that, in the specific case of the Banach space X = C0(E), a linear operator A be the generatorof e Feller semigroup? Knowing the general connection between a semigroup and its generator we haveimmediately the

Proposition 7.1. Let (etA)t>0 be a Feller semigroup on C0(E), (E, d) locally compact metric space.Then the positive maximum principle holds:

if ϕ ∈ D(A) has a maximum at x0 with ϕ(x0) > 0 then Aϕ(x0) 6 0.

Proof — Let x0 a positive maximum for ϕ. Notice first that by definition

Aϕ = limh→0+

S(h)ϕ(x0)− ϕ(x0)

h.

The limit is intended in C0(E), therefore in the uniform convergence. This means, in particular, that

Aϕ(x0) = limh→0+

S(h)ϕ(x0)− ϕ(x0)

h.

Now, setting ϕ+ := max{ϕ, 0} ∈ C0(E), we have

S(h)ϕ 6 S(h)ϕ+ 6 ‖S(h)ϕ+‖∞ 6 ‖ϕ+‖∞ = ϕ(x0).

Therefore, if h > 0,

S(h)ϕ(x0)− ϕ(x0)

h6ϕ(x0)− ϕ(x0)

h= 0, =⇒ Aϕ(x0) 6 0.

The maximum principle gives basically the estimate (6.2). Actually

Proposition 7.2. Let A : D(A) ⊂ C0(E) −→ C0(E). If A fulfills the positive maximum principle thenit is dissipative that is

‖λϕ−Aϕ‖ > λ‖ϕ‖, ∀ϕ ∈ D(A), ∀λ > 0. (7.1)

Proof — Because ϕ ∈ C0(E), by Weierstrass thm there exists x0 ∈ E such that

|ϕ(x0)| = maxx∈E|ϕ(x)| = ‖ϕ‖.

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We may assume ϕ(x0) > 0 (otherwise replace ϕ with −ϕ). Therefore

‖λϕ−Aϕ‖ = supx∈E|λϕ(x)−Aϕ(x)| > |λϕ(x0)−Aϕ(x0)|.

By the positive maximum principle, Aϕ(x0) 6 0, therefore λϕ(x0)−Aϕ(x0) > 0, hence

‖λϕ−Aϕ‖ > |λϕ(x0)−Aϕ(x0)| = λϕ(x0)−Aϕ(x0) > λϕ(x0) = λ‖ϕ‖.

Notice in particular that

Lemma 7.3. Let A : D(A) ⊂ X −→ X, X normed space. If A is dissipative then λI−A is injective. Ifit is also surjective then λ ∈ ρ(A) and ‖Rλ‖ 6 1

λ for any λ > 0.

Proof — Immediate.

Dissipativity gives an immediate simplification to check the ii) of Hille–Yosida thm.

Theorem 7.4 (Phillips). Let X be a Banach space, A : D(A) ⊂ X −→ X be a densely defined linearoperator. Suppose that

i) A is dissipative;

ii) R(λ0I −A) = X for some λ0 > 0 (that is: λ0I−A is surjective).

The A generates a strongly continuous semigroup of contractions. (2)

Proof — By Hille–Yosida Theorem, we have to prove that

ρ(A) ⊃]0,+∞[, and ‖Rλ‖ 61

λ, ∀λ > 0.

By the Lemma 7.3 it is clear that λ0 ∈ ρ(A). Therefore the unique thing to prove is that ρ(A) ⊃]0,+∞[, that is:every λ > 0 belongs to the resolvent set. The idea is to prove that ρ(A)∩]0,+∞[ (which is non empty by whatwe have seen just now) is open and closed: it is therefore connected, hence equal to all ]0,+∞[.

First: ρ(A) is open, therefore ρ(A)∩]0,+∞[ is open in ]0,+∞[. This is a general fact so we state itseparately:

Lemma 7.5. The resolvent set of any linear (eventually unbounded) operator is open in C and λ 7−→ Rλ ∈C(ρ(A);L(X)). In particular: if λ ∈ ρ(A) then B(λ, 1

2‖Rλ‖[⊂ ρ(A).

Proof — The argument is based on the following formal identity: fixed λ ∈ ρ(A) and µ ∈ C,

Rµ” = ”1

µ−A =1

µ− λ+ (λ−A)=

1

λ−A1

1 + µ−λλ−A

” = ”Rλ (I + (µ− λ)Rλ)−1 . (7.2)

It is easy to check that if the right hand side makes sense as bounded linear operator then it is exactly Rµ. Togive a meaning to the right hand side, we have to justify that B = I + (µ − λ)Rλ is invertible with continuousinverse. To this aim recall that

if ‖I −B‖ < 1, =⇒ B−1 =∞∑n=0

(I −B)n ∈ L(X).

2Actually, if the space X is reflexive, the Phillips thm. is an iff.

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In our case

‖I −B‖ = ‖(µ− λ)Rλ‖ = |µ− λ|‖Rλ‖ < 1, ⇐⇒ |µ− λ| < 1

‖Rλ‖.

This means that B(λ, 1‖Rλ‖

[⊂ ρ(A). The continuity of the resolvent map follows again by the (7.2). Indeed

Rµ = Rλ

∞∑n=0

((λ− µ)Rλ)n = Rλ+(λ−µ)Rλ

∞∑n=0

((λ− µ)Rλ)n , =⇒ ‖Rµ−Rλ‖ 6 |λ−µ|‖Rλ‖∞∑n=0

‖(λ−µ)Rλ‖n.

Now, if |λ− µ| 6 12‖Rλ‖

it follows that

‖Rµ −Rλ‖ 6 |λ− µ|‖Rλ‖∞∑n=0

1

2n= 2‖Rλ‖|λ− µ|,

and now the conclusion is evident.

Second: ρ(A)∩]0,+∞[ is closed. Let (λn) ⊂ ρ(A)∩]0,+∞[ such that λn −→ λ ∈]0,+∞[. We want to provethat λ ∈ ρ(A). By previous arguments, this is equivalent to show that R(λI − A) = H. So, fix ψ ∈ H. We lookfor ϕ ∈ D(A) such that

λϕ−Aϕ = ψ.

Let ϕn ∈ D(A) such thatλnϕn −Aϕn = ψ.

The idea is to pass to the limit in this equation. We have just to prove that (ϕn) is convergent first, in particularof Cauchy. To this aim notice that

λnϕn −Aϕn = ψ,

λmϕm −Aϕm = ψ,=⇒ λnϕn−λmϕm−A(ϕn−ϕm) = 0, ⇐⇒ λn(ϕn−ϕm)−A(ϕn−ϕm) = (λm−λn)ϕm

so, by dissipativity,

|λm − λn|‖ϕm‖ = ‖λn(ϕn − ϕm)−A(ϕn − ϕm)‖ > |λn|‖ϕn − ϕm‖.

Now: because λn −→ λ ∈]0,+∞[ we can say that |λn| > α for all n ∈ N, for some α > 0. Suppose we have provedthat (ϕn) is bounded, that is ‖ϕn‖ 6 K for all n: we are done because, in this case,

‖ϕn − ϕm‖ 6K

α|λn − λm| 6 ε, ∀n,m > N0(ε),

i.e. (ϕn) would be a Cauchy sequence. Boundedness of (ϕn) follows by the same argument due to the dissipativity:indeed

λn‖ϕn‖ 6 ‖λnϕn −Aϕn‖ = ‖ψ‖, =⇒ ‖ϕn‖ 6‖ψ‖α, ∀n ∈ N.

Summarizing: (ϕn) is convergent. Then, because Aϕn = λnϕn − ψ, also (Aϕn) is convergent. Now, you willrecall that if ρ(A) 6= ∅ then A is closed. So, being this the case in our context, we have that ϕn −→ ϕ ∈ D(A),Aϕn −→ Aϕ and passing to the limit in the equation we would get λϕ − Aϕ = ψ, that is the conclusion. Withthis the proof is finished.

Definition 7.6. Let A : D(A) ⊂ C0(E) −→ C0(E) be a linear operator, (E, d) be a locally compactmetric space. We say that A is a Markov generator if

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20

i) D(A) is dense in C0(E).

ii) A fulfills the positive maximum principle.

iii) R(λ0I−A) = C0(E) for some λ0 > 0.

In other words, combining the Hille–Yosida thm with the Phillips thm we have the

Corollary 7.7. A linear operator A generates a Markov semigroup on C0(E) iff A is a Markov generator.

8 Examples

To check that an operator A is a Markov generator is not, in general, an easy business. In particular, isthe third conditions that usually is a little bit difficult to check and often it is useful to have a sort ofmild version of it. The problem is that we have in particular to solve the equation

λϕ−Aϕ = ψ,

for a given ψ ∈ C0(E). This is not generally easy. Let’s see a first example.

Example 8.1. The operator

A : D(A) ⊂ C0([0, 1]) −→ C0([0, 1]), Aϕ := ϕ′′, D(A) :={ϕ ∈ C 2([0, 1]) ∩ C0([0, 1]) : ϕ′′ ∈ C0([0, 1])

},

is a Markov generator.

Sol. — Clearly the first two properties are true. Let’s see the third. Take ψ ∈ C0([0, 1]) and consider theequation

λϕ(x)− ϕ′′(x) = ψ(x), x ∈ [0, 1].

As well known by ODEs theory, the general solution of previous equation is

ϕ(x) = c1w1(x) + c2w2(x) + U(x),

where (w1, w2) is a fundamental system for the homogeneous equation λϕ− ϕ′′ = 0. If λ > 0 (as in our case) we

have w1,2(x) = e±√λx. By the variation of constants formula

U(x) =

(− 1

2√λ

∫ x

0

ψ(y)e√λy dy

)e−√λx +

(1

2√λ

∫ x

0

ψ(y)e−√λy dy

)e√λx

=1√λ

∫ x

0

ψ(y)e√λ(x−y) − e−

√λ(x−y)

2dy =

1√λ

∫ x

0

ψ(y) sinh(√

λ(x− y))dy.

Therefore

ϕ(x) = c1e−√λx + c2e

√λx +

1√λ

∫ x

0

ψ(y) sinh(√

λ(x− y))dy.

Here there’s a first problem we meet: if ψ ∈ C ([0, 1]) it is not evident that ϕ ∈ C 2([0, 1]). Indeed, it is easy tocheck that ϕ ∈ C 1([0, 1]) and

ϕ′(x) = −c1√λe−

√λx + c2

√λe√λx +

1√λψ(x) sinh(

√λ · 0) +

∫ x

0

ψ(y) cosh(√λ(x− y)) dy

= −c1√λe−

√λx + c2

√λe√λx +

∫ x

0

ψ(y) cosh(√λ(x− y)) dy.

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21

Repeating the procedure we see that ϕ ∈ C 2([0, 1]). Let’s impose that ϕ,ϕ′′ ∈ C0([0, 1]). This meansϕ(0) = ϕ(1) = 0,

ϕ′′(0) = ϕ′′(1) = 0.

Notice that by the equation we haveϕ′′(x) = λϕ(x)− ψ(x),

therefore once we know ϕ ∈ C0([0, 1]) we get ϕ′′ ∈ C0([0, 1]) because ψ ∈ C0([0, 1]). So the previous conditionsreduce only to the first: ϕ(0) = ϕ(1) = 0, that is

c1 + c2 = 0,

c1e−√λ + c2e

√λ +

1√λ

∫ 1

0

ψ(y) sinh(√

λ(1− y))dy = 0.

This is a 2× 2 system in c1, c2 with determinant e√λ − e−

√λ = 2 sinh

√λ 6= 0 if λ 6= 0. It is clear that it admits

a unique solution (c1, c2). This means that there exists a unique ϕ ∈ D(A).

As you see most of the difficulties are due to the check that λI − A is surjective because this involvesthe solution of a more or less complicate equation. Other problems are that in general it is not easy todescribe a generator by giving his exact domain. We will meet this problem in the next examples.

Moreover, when we define processes starting by their infinitesimal generators, we need to have a clearunderstanding of the meaning of the generator for the underlying processes. To this aim notice that wecould write,

S(h)ϕ− ϕ = hAϕ+ o(h), ⇐⇒ Ex [ϕ(γ(h))− ϕ(γ(0))] = hAϕ(x) + o(h),

or, more in general becauseS(t+ h)ϕ− S(t)ϕ = hAϕ+ o(h),

we could writeEx [ϕ(γ(t+ h))− ϕ(γ(t)) | Ft] = hAϕ(γ(t)) + o(h). (8.1)

Therefore, we could think to Aϕ as the rate of infinitesimal variation of the observable ϕ along theinfinitesimal displacement of state from γ(t) to γ(t+ h).

8.2 Brownian motion

As well known the case of BM on the state space E := Rd corresponds to the case

Aϕ :=1

2∆ϕ.

Actually, there’re some technical aspects that complicate the discussion, mainly due to the problem tosolve the equation

λϕ−Aϕ = ψ, ⇐⇒ λϕ− 1

2∆ϕ = ψ.

We discuss separately as d = 1 and d > 2 this equation.

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22

8.2.1 Case d = 1

Let’s start by solving the equation

λϕ− 1

2ϕ′′ = ψ, ⇐⇒ 2λϕ− ϕ′′ = 2ψ, ψ ∈ C0(R).

Because the equation is basically the same of Example 8.1 so we have, for general solution

ϕ(x) = c1e−√2λx + c2e

√2λx +

2√2λ

∫ x

0

ψ(y) sinh(√

2λ(x− y))dy.

By the same argument of Example 8.1 we deduce that ϕ ∈ C 2(R). Let’s see if ϕ ∈ C0(R), that is when ϕ(±∞) = 0.To this aim rewrite the solution in the form,

ϕ(x) =

(c1 +

1√2λ

∫ x

0

ψ(y)e−√2λy dy

)e√2λx +

(c2 −

1√2λ

∫ x

0

ψ(y)e√2λy dy

)e−√2λx.

Notice that we have a unique possibility in such a way that ϕ(+∞) = 0, that is

c1 +1√2λ

∫ +∞

0

ψ(y)e−√2λy dy = 0. (8.2)

Indeed: first notice thatc2e−√2λx −→ 0, x −→ +∞,

and

e−√2λx

∫ x

0

ψ(y)e√2λy dy =

∫ x0ψ(y)e

√2λy dy

e√2λx

(H)=

ψ(x)e√2λx

√2λe√2λx

=1√2λψ(x) −→ 0, x −→ +∞,

because ψ ∈ C0(R). Moreover, the integral in (8.2) is convergent because |ψ(y)e−√2λy| 6 ‖ψ‖∞e−

√2λy. Therefore

if the (8.2) is not 0, by previous considerations, we would have(c1 +

1√2λ

∫ x

0

ψ(y)e−√2λy dy

)e√2λx −→ ±∞.

So the unique possibility is that (8.2) holds true. In that case applying again the Hopital rule we would have

c1 + 1√2λ

∫ x0ψ(y)e−

√2λy dy

e−√2λx

(H)=

1√2λψ(x)e−

√2λx

−√

2λe−√2λx

= − 1

2λψ(x) −→ 0, x −→ +∞,

again because ψ ∈ C0(R). The moral is: the unique possible choice for c1 such that ϕ(+∞) = 0 is given by (8.2).Similarly, at −∞ we will have

c2 −1√2λ

∫ −∞0

ψ(y)e√2λy dy = 0. (8.3)

This means that there’s a unique ϕ ∈ C 20 (R) such that λϕ−Aϕ = ψ. We can summarize the discussion with the

statement

Theorem 8.3. The operator

A :=1

2

d2

dx2, D(A) :=

{ϕ ∈ C 2(R) ∩ C0(R) : ϕ′′ ∈ C0(R)

}is a Markov generator.

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23

8.3.1 Case d > 2

This case is sensibly different from the previous one because now the equation λϕ−Aϕ = ψ is a a PDE

λϕ− 1

2∆ϕ = ψ. (8.4)

To solve this equation we invoke the Fourier transform. To this aim recall the Schwarz space

S (Rd) :=

{ϕ ∈ C∞(Rd) : sup

x∈Rd(1 + |x|)n|∂αϕ(x)| < +∞, ∀n ∈ N, α ∈ Nd

},

where ∂α stands for the usual multi-index notation for derivatives. It is well known that the Fourier transformis a bijection on S (Rd). To solve (8.4) in the Schwarz space is a standard application of the Fourier transform.Indeed: we take ψ ∈ S (Rd) and we look for a solution ϕ ∈ S (Rd). Applying the Fourier transform we have

λϕ− 1

2∆ϕ = ψ, ⇐⇒ λϕ+

4π2|ξ|2

2ϕ = ψ, ⇐⇒ (λ+ 2π2|ξ|2)ϕ = ψ, ⇐⇒ ϕ =

1

λ+ 2π2|ξ|2 ψ.

Now it is easy to check that if ψ ∈ S (Rd) then 1λ+2π2|]|2 ψ ∈ S (Rd): therefore, inverting the Fourier transform,

there exists a unique ϕ ∈ S (Rd) such that the previous equation holds. So if we would define

A :=1

2∆, D(A) := S (Rd),

we would haveR(λI−A) ⊃ S (Rd) =: Y.

With respect to or space X := C0(Rd) we have that S (Rd) is dense in X. So we have that 12∆ fulfills the first

two conditions of Markov generator and the third is a little bit weaker being

R(λI−A) is dense in C0(Rd),

for any λ > 0. In this case it seems a little bit impossible that when we start with ψ ∈ C0(Rd) we get ϕ ∈ S (Rd).In other words: the domain of A seems too small to represent the natural domain for A (and indeed: A involves

only second order derivatives whereas in S (Rd) there’re very regular functions). On the other hand it is not at

all easy to solve the equation directly in C 2(Rd). So, how can we solve this impasse? The point is that there’s,

in a suitable sense, a unique possible extension of A which is a Markov generator. Let’s first treat

this question in general.

We start introducing some definitions in order to make clear what does it means ”extension” here. It’snot natural to talk about continuous extension because our operators wont be, in general, continuous.The right concept turns out to be that on of closed extension.

Definition 8.4. Let A : D(A) ⊂ X −→ X be a linear operator. We say that A is closable if G(A) isthe graph of some linear (clearly closed) operator. We will denote by A such operator. In particular:

G(A) = G(A).

Lemma 8.5. Let A : D(A) ⊂ X −→ X be a densely defined dissipative linear operator on X Banachspace. Then

i) A is closable.

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ii) A is dissipative.

iii)R(λI−A) = R

(λI−A

), ∀λ > 0. (8.5)

Proof — i) A is closable. Take G(A). It is enough to check that if (ϕ,ψ1), (ϕ,ψ2) ∈ G(A) then ψ1 = ψ2. Inthis case G(A) will be the graph of a well defined closed linear operator. By definition

∃(ϕn, Aϕn) ⊂ G(A), (ϕn, Aϕn) −→ (ϕ,ψ1), ∃(ϕn, Aϕn) ⊂ G(A), (ϕn, Aϕn) −→ (ϕ,ψ2).

By linearity taking differences and calling φn := ϕn − ϕn we have

(φn, Aφn) −→ (0, ψ), where ψ := ψ1 − ψ2,

so we are reduced to prove that ψ = 0. Because D(A) is dense, take (ψn) ⊂ D(A) such that ψn −→ ψ. Noticedthat

λφn −Aφn −→ −ψ,we have

‖(λI−A)ψm − λψ‖ = limn ‖(λI−A)ψm + (λI−A)(λφn)‖ = limn ‖(λI−A)(ψm + λφn)‖

> limn λ ‖ψm + λφn‖ = λ‖ψm‖.

Dividing by λ and letting it to +∞ we have

‖ψm‖ 6∥∥∥∥ψm − 1

λAψm − ψ

∥∥∥∥ −→ ‖ψm − ψ‖.Finally, letting m −→ +∞ we deduce ‖ψ‖ 6 0.

ii) A is dissipative. This is straightforward: if ϕ ∈ D(A) there exists (ϕn) ⊂ D(A) such that (ϕn, Aϕn) −→(ϕ,Aϕ). By dissipativity of A we have

‖λϕn −Aϕn‖ > λ‖ϕn‖, =⇒ ‖λϕ−Aϕ‖ > λ‖ϕ‖.

iii) We prove now the (8.5). It is clear that being G(A) ⊂ G(A) we have

R(λI−A) ⊂ R(λI−A), =⇒ R(λI−A) ⊂ R(λI−A).

Let’s see that R(λI−A) is closed: by this ⊂ in (8.5) will follows. Now, the conclusion follows because A is closed.Indeed: if

ψ ∈ R(λI−A), =⇒ ψ = limn

(λϕn −Aϕn

)=: lim

nψn, (ϕn) ⊂ D(A).

Notice that, being A dissipative,

‖ψn − ψm‖ =∥∥λ(ϕn − ϕm)−A(ϕn − ϕm)

∥∥ > λ‖ϕn − ϕm‖.

We deduce that (ϕn) is Cauchy, hence converges to some ϕ. But then Aϕn = λϕn−ψn −→ λϕ−ψ, and becauseA is closed it follows ϕ ∈ D(A) and Aϕ = λϕ− ψ, that is ψ = (λI−A)ϕ, so ψ ∈ R(λI−A) as advertised.

To prove the other inclusion ⊃ take ψ ∈ R(λI − A): ψ = λϕ − Aϕ for some ϕ ∈ D(A). Now: there exists(ϕn) ⊂ D(A) such that (ϕn, Aϕn) −→ (ϕ,Aϕ). But then

λϕn −Aϕn −→ λϕ−Aϕ = ψ, =⇒ ψ ∈ R(λI−A).

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In particular we deduce the following

Theorem 8.6. Let A : D(A) ⊂ X −→ X be a densely defined, dissipative linear operator such that

R (λ0I−A) is dense in X for some λ0 > 0.

Then A is closable and A generates a strongly continuous semigroup of contractions on X.

Proof — By the Lemma A is well defined and dissipative (and of course densely defined). By (8.5) and ourassumption

R(λ0I−A) = X,

so the conclusion follows by the Phillips Thm.

Let’s now particularize the discussion to the case of Markov processes. It is useful to introduce thefollowing

Definition 8.7. Let A : D(A) ⊂ C0(E) −→ C0(E) be a linear operator, (E, d) be a locally compactmetric space. We say that A is a Markov pre–generator if

i) D(A) is dense in C0(E).

ii) A fulfills the positive maximum principle.

iii) R(λ0I−A) = C0(E) for some λ0 > 0.

By previous Thm. it follows that if A is a Markov pre–generator, then A is closable and A generatesa strongly continuous semigroup of contractions. Is it a Markov semigroup? We need to check that Afulfills the positive maximum principle:

Proposition 8.8. If A is a Markov pre–generator then A is a Markov–generator.

Proof — Exercise.

Therefore, applying this result to our case we could say what follows: A = 12∆ defined on D(A) = S (Rd)

is closable and A generates a Markov semigroup. Of course A coincide with A on S (Rd).Now the further question is: is it possible that A is the restriction of some other operator B 6= A

generating a strongly continuous semigroup of contractions? In other words: does A identify uniquelythe Markov semigroup? To this aim we introduce the following concept:

Definition 8.9. Let A : D(A) ⊂ X −→ X be a linear operator. A set D ⊂ D(A) is called a core for Aif A coincides with the closure of its restriction to D, that is if

G(A|D ) = G(A).

HereG(A|D ) := {(ϕ,Aϕ) : ϕ ∈ D} ,

and of course the closure is intended to be in the space X ×X.

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It is clear that if A and B are two densely defined closed linear operators such that D is a core for bothand they coincide on D, then A = B. So if D = S (Rd) is a core for A, any other generator B having Das core and coinciding with 1

2∆ on D will generate the same semigroup of A. How to check if D is a corethen? The following proposition is sometimes useful:

Proposition 8.10. Let A be the generator of a strongly continuous semigroup of contractions on aBanach space X. A set D ⊂ X is a core for A iff

i) D is dense in X;

ii) R(λ0I−A|D ) = X for some λ0 > 0.

Proof — =⇒ Assume D is a core. Because A is densely defined (as generator) fixed ϕ ∈ X there exists ϕε ∈ D(A)such that ‖ϕ− ϕε‖ 6 ε. But G(A) = G(A|D ): in particular, taking (ϕε, Aϕε) there exists (ϕε, AID ϕε) ∈ G(AID )close to it. This means that ‖ϕε − ϕε‖ 6 ε, therefore ‖ϕ − ϕε‖ 6 2ε. This proves i). About ii) fix ψ ∈ X andconsider ϕ ∈ D(A) such that

λ0ϕ−Aϕ = ψ.

Now by assumption there exists (ϕn) ⊂ D such that (ϕn, Aϕn) −→ (ϕ,Aϕ). Clearly λ0ϕn − Aϕn −→ ψ, that isψ ∈ R(λ0I−A|D ).

⇐= Take (ϕ,Aϕ) ∈ G(A) and consider ψ := λ0ϕ − Aϕ. By ii) there exists (ϕn) ∈ D such that ψn := λ0ϕn −Aϕn −→ ψ. Being ϕn = Rλ0ψn we deduce that ϕn −→ ϕ := Rλ0ψ. Therefore Aϕn −→ λ0ϕ−ψ = Aϕ. But then(ϕn, Aϕn) −→ (ϕ,Aϕ), that is G(A) ⊂ G(A|D ). The other inclusion is evident.

In particular, in our context, D = S (Rd) is a core for the generator of the Brownian motion.It is in general not easy to characterize the domains of generators of strongly continuous semigroups.

In the present case it is possible to show that C 20 (Rd) is contained into the domain, but this is not,

however, the full domain. Actually it is possible to show that the domain of the generator is the subsetof C0(Rd) with ∆ϕ ∈ C0(Rd) in distributional sense. We don’t enter in these details.


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