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 (R^d) :=

(

ϕ ∈C^∞(R^d) : sup

x∈R^d

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

,

where ∂^α stands for the usual multi-index notation for derivatives. It is well known that the Fourier transform is a bijection onS (R^d). To solve (8.4) in the Schwarz space is a standard application of the Fourier transform.

Indeed: we take ψ ∈S (R^d) and we look for a solution ϕ ∈S (R^d). Applying the Fourier transform we have λϕ −1

2∆ϕ = ψ, ⇐⇒ λϕ +b 4π²|ξ|²

2 ϕ = bb ψ, ⇐⇒ (λ + 2π²|ξ|²)ϕ = bb ψ, ⇐⇒ ϕ =b 1 λ + 2π²|ξ|²ψ.b Now it is easy to check that if ψ ∈S (R^d) then _λ+2π¹_2|]|2ψ ∈b S (R^d): therefore, inverting the Fourier transform, there exists a unique ϕ ∈S (R^d) such that the previous equation holds. So if we would define

A :=1

2∆, D(A) :=S (R^d), we would have

R(λI − A) ⊃ S (R^d) =: Y.

With respect to or space X :=C0(R^d) we have thatS (R^d) 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(R^d),

for any λ > 0. In this case it seems a little bit impossible that when we start with ψ ∈C0(R^d) we get ϕ ∈S (R^d).

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 inS (R^d) there’re very regular functions). On the other hand it is not at all easy to solve the equation directly inC²(R^d). 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’s not 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) is the 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 Banach space. Then

i) A is closable.

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. In this case G(A) will be the graph of a well defined closed linear operator. By definition

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

By linearity taking differences and calling φn:= ϕn−ϕen 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. Noticed that

λφn− Aφn−→ −ψ, we have

k(λI − A)ψ^m− λψk = limnk(λI − A)ψ^m+ (λI − A)(λφn)k = limnk(λI − A)(ψ^m+ λφn)k

> limnλ kψm+ λφnk = λkψmk.

Dividing by λ and letting it to +∞ we have kψmk 6

ψm− 1

λAψm− ψ

−→ kψm− ψk.

Finally, letting m −→ +∞ we deduce kψk 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

kλϕn− Aϕnk > λkϕⁿk, =⇒ kλϕ − Aϕk > λkϕk.

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 thatR(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 = lim

n λϕn− Aϕn
=: lim

n ψn, (ϕn) ⊂ D(A).

Notice that, being A dissipative,

kψn− ψmk =

λ(ϕn− ϕm) − A(ϕn− ϕm)

> λkϕn− ϕmk.

We deduce that (ϕn) is Cauchy, hence converges to some ϕ. But then Aϕn= λϕn− ψn−→ λϕ − ψ, and because A 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).

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 our assumption

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 the following

Definition 8.7. Let A : D(A) ⊂ C0(E) −→ C0(E) be a linear operator, (E, d) be a locally compact metric 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 generates a strongly continuous semigroup of contractions. Is it a Markov semigroup? We need to check that A fulfills 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 (R^d) is closable and A generates a Markov semigroup. Of course A coincide with A onS (R^d).

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 uniquely the 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 A if A coincides with the closure of its restriction to D, that is if

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

Here

G(A_|D) := {(ϕ, Aϕ) : ϕ ∈ D} , and of course the closure is intended to be in the space X × X.

It is clear that if A and B are two densely defined closed linear operators such that D is a core for both and they coincide on D, then A = B. So if D =S (R^d) is a core for A, any other generator B having D as core and coinciding with ¹₂∆ on D will generate the same semigroup of A. How to check if D is a core then? The following proposition is sometimes useful:

Proposition 8.10. Let A be the generator of a strongly continuous semigroup of contractions on a Banach 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 kϕ − ϕεk 6 ε. But G(A) = G(A^|D): in particular, taking (ϕε, Aϕε) there exists (ϕeε, AI_Dϕeε) ∈ G(AI_D) close to it. This means that kϕε−ϕeεk 6 ε, therefore kϕ −ϕeεk 6 2ε. This proves i). About ii) fix ψ ∈ X and consider 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 is ψ ∈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= Rλ₀ψnwe 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 (R^d) 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 C0²(R^d) 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 subset ofC0(R^d) with ∆ϕ ∈C0(R^d) in distributional sense. We don’t enter in these details.