Optimal stopping with irregular reward functions

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Stochastic Processes and their Applications 119 (2009) 3253–3284

www.elsevier.com/locate/spa

Optimal stopping with irregular reward functions

Damien Lamberton

^∗

Universit´e Paris-Est, Laboratoire d’Analyse et de Math´ematiques Appliqu´ees, Projet Mathfi, 5 Boulevard Descartes, 77 454 Marne-la-Vall´ee CEDEX 2, France

Received 25 April 2008; received in revised form 23 April 2009; accepted 12 May 2009 Available online 20 May 2009

Abstract

We consider optimal stopping problems with finite horizon for one-dimensional diffusions. We assume that the reward function is bounded and Borel-measurable, and we prove that the value function is continuous and can be characterized as the unique solution of a variational inequality in the sense of distributions.

c 2009 Elsevier B.V. All rights reserved.

MSC:60G40; 60J60; 60H10

Keywords:Optimal stopping; One-dimensional diffusions; Irregular reward functions

1. Introduction

In a recent paper with M. Zervos (see [10]), we studied optimal stopping problems with infinite horizon for one-dimensional diffusions. In particular, we proved that, under very general conditions, the value function is the unique solution (in the sense of distributions) of a stationary variational inequality. The purpose of the present paper is to examine optimal stopping problems with finite horizon and bounded Borel-measurable reward functions. We will prove that the value function is continuous and can be characterized as the unique solution (in the sense of distributions) of a suitable variational inequality.

The connection between optimal stopping and variational inequalities goes back to the work of Bensoussan and Lions [3] and Friedman [6]. This approach is very general and applies to multi- dimensional problems, but it requires uniform ellipticity of the diffusion and some regularity of

∗Tel.: +33 1 60 95 75 36; fax: +33 1 60 95 75 45.

E-mail address:damien.lamberton@univ-mlv.fr.

0304-4149/$ - see front matter c2009 Elsevier B.V. All rights reserved.

doi:10.1016/j.spa.2009.05.005

the reward function. Note that the techniques of viscosity solutions do not require ellipticity, but generally impose some continuity conditions on the reward function and the coefficients of the diffusion. For our results, we will not need any regularity assumption on the reward function, and we will deal with very general one-dimensional diffusions. On the other hand, our analysis will be limited to one-dimensional situations (cf. Remark 2.3).

The paper is organized as follows. In Section 2, we present our assumptions and the main results. In particular, we give the proper formulation of the variational inequality. In Section 3, we prove the continuity of the value function. In Section 4, we essentially relate the value function to the Snell envelope. Section 5 is devoted to the analytic interpretation of the supermartingale property. The proof that the value function satisfies the variational inequality is given in Section 6.

Uniqueness of the solution is proved in Section 7. In the last section, we have gathered a number of auxiliary results, which are classical in somewhat different contexts, but which require some justification under our assumptions. In particular, we derive regularity estimates for the semi- group of one-dimensional diffusions (see Theorem 8.11 and Corollary 8.13), which we have not found in the literature.

2. Assumptions and main results

We consider an open interval I = (α, β) (with −∞ ≤ α < β ≤ +∞ ) and a stochastic differential equation

d X

_t

= b(X

_t

) dt + σ (X

_t

) dW

_t

, X

0

= x ∈ I , (1)

where W is a standard one-dimensional Brownian motion, and b, σ : I →

R

are Borel- measurable functions satisfying the following condition.

A1. For all x ∈ (α, β), σ

²

(x ) > 0, and ∃ ε > 0,

R_x+ε x−ε

1+|b(y)|

σ²(y)

dy < ∞ .

Under assumption A1, we have existence and uniqueness in law of a weak solution of (1) up to a possible explosion time (cf. [9, Section 5.5C]). In fact, we will also assume that the diffusion is non-explosive. This assumption can be expressed in terms of the so-called scale function p(x ) and speed measure m (dx ), defined by

p(x) =

Z _x

c

exp

− 2

Z _y

c

b(z) σ

²

(z) dz

dy, for x ∈ I, (2)

m (dx ) = 2

σ

²

(x ) p

^′

(x) dx , (3)

where c is an arbitrary fixed element of I . The condition for no explosion can now be written as follows, according to Feller’s test (see Theorem 5.5.29 in Karatzas and Shreve [9]).

A2. We have lim

_x↓α

l(x) = lim

_x↑β

l(x ) = ∞ , where l (x ) =

Z _x

c

[ p(x) − p(y)] m (dy), for x ∈ I .

Throughout the paper, assumptions A1 and A2 are in force. A weak solution of (1) is defined by a triple

(

Ω

,

F

, (

F_t

)

_t≥0

,

Px

), W , X

, where (

Ω

,

F

, (

F_t

)

_t≥0

,

Px

) is a filtered probability space with the filtration (

F_t

)

_t≥0

satisfying the usual conditions, W = (W

_t

)

_t≥0

is a standard (

F_t

)- Brownian motion and X is a continuous adapted process satisfying (1). Given such a weak solution, we denote by (

F⁰

t

)

_t≥0

the natural right-continuous filtration of X .

We now introduce an optimal stopping problem with a discounting rate (or interest rate) r . The function r : I →

R

is assumed to be non-negative, Borel-measurable and locally bounded on I . We denote by

T⁰

t

(resp.

T

¯

⁰

t

) the set of all stopping times with respect to the filtration (

F⁰

t

)

_t≥0

, with values in the interval [ 0, t ) (resp. [ 0, t ] ). Given a bounded Borel-measurable function f on I , we introduce the functions u

_f

and v

_f

defined on (0, +∞ ) × I as follows:

u

f

(t, x ) = sup

τ∈Tt⁰

Ex

h

e

^−Λτ

f ( X

τ

)

i

, (4)

v

_f

(t , x ) = sup

τ∈ ¯Tt⁰

Ex

h

e

^−Λτ

f (X

τ

)

i

, (5)

where

Λt

=

Z t 0

r ( X

_s

) ds.

Note that, due to the fact that we consider stopping times with respect to the natural filtration, the functions u

_f

and v

_f

depend only on the law of X , which is uniquely defined under assumptions A1 and A2. On the other hand, let

T_t

(resp.

T

¯

_t

) be the set of all stopping times with respect to the filtration (

F_t

)

_t≥0

, with values in the interval [ 0, t ) (resp. [ 0, t ] ). If we define by u ¯

_f

(resp. v ¯

_f

) the value function where

T⁰

t

(resp.

T

¯

⁰

t

) is replaced with

T_t

(resp.

T

¯

_t

), we have u ¯

_f

= u

_f

and v ¯

_f

= v

_f

(see Section 8, Remark 8.7).

We obviously have u

_f

≤ v

_f

. Our first observation is the following result, the proof of which is quite similar to the one given for infinite horizon problems (see [10], Lemma 7), and is therefore omitted.

Proposition 2.1. Let f : I →

R

be a bounded Borel-measurable function on I . Denote by f the ˆ upper semicontinuous envelope of f :

f ˆ (x ) = lim sup

y→x

f (y), x ∈ I.

Then we have u

_f

= u

_fˆ

.

Our next result concerns the joint continuity of the value function. The following theorem will be proved in Section 3.

Theorem 2.2. We have u

_f

= v

_f

and the function v

_f

is jointly continuous on (0, +∞ ) × I . Remark 2.3. The fact that we have a one-dimensional diffusion is essential for the continuity of the value function. Indeed, consider a two-dimensional Brownian motion (W

_t¹

, W

_t²

)

_t≥0

and let f be the indicator function of the singleton { 0 } . Since Brownian motion starting from x 6= 0 will never hit 0 with probability one, we clearly have (with similar notations as above) u

_f

= v

_f

= f , so that v

_f

is discontinuous. In fact, crucial to the continuity of the value function is the fact that the diffusion hits any given point close to the initial point with positive probability. Note that the regularization procedure that we develop in Section 5 also depends heavily on the one- dimensional setting.

In order to write the variational inequality satisfied by the value function, we need to introduce

the infinitesimal generator

L₀

of the diffusion. For a twice continuously differentiable function

u,

L₀

u is defined by

L₀

u(x) = σ

²

(x )

2 u

^′′

(x ) + b(x )u

^′

(x ), x ∈ I .

As should be expected, the variational inequality will involve the operator −

∂^∂t

+

L

, where the operator

L

is defined by

L

u(t , x ) =

L₀

u(t, x ) − r (x )u(t, x ), (t , x ) ∈ (0, +∞ ) × I .

In fact, in order to be able to apply the operator to possibly non-smooth functions, we will rather consider the operator

A

, where

A

= 2 σ

²

p

^′

− ∂

∂ t +

L

.

The following proposition is the key to the extension of

A

to irregular functions, in the sense of distributions.

Proposition 2.4. If u ∈ C

^1,2

((0, +∞ ) × I ), for any C

^∞

function

Φ

, with compact support in (0, +∞ ) ×

R

, we have

Z Z

A

u (t, x )

Φ

(t , x )dt dx =

Z Z

u(t , x ) ∂

Φ

∂ t +

LΦ

(t, x ) dt m (dx ).

Proof. It follows from integration by parts with respect to time and from the definition of the speed measure that

−

Z Z

2 σ

²

(x ) p

^′

(x )

∂ u

∂ t (t , x )

Φ

(t, x )dt dx =

Z Z

u(t , x ) ∂

Φ

∂ t (t, x ) dt m (dx ).

On the other hand, using the fact that the scale function p satisfies d

dx

1

p

^′

= 2b σ

²

1 p

^′

, we have

L₀

u = σ

²

p

^′

2

1 p

^′

∂

²

u

∂ x

²

+ 2b σ

²

1 p

^′

∂ u

∂ x

= σ

²

p

^′

2

∂

∂ x

1

p

^′

∂ u

∂ x

.

Hence, integrating by parts twice with respect to x,

Z Z

2

σ

²

(x) p

^′

(x )

L₀

u(t , x)

Φ

(t, x )dt dx =

Z Z

u(t, x )

L₀Φ

(t , x )dt m (dx ).

The result now follows easily. ⋄

In view of Proposition 2.4, it is natural, given a locally bounded measurable function u on (0, +∞ ) × I , to define the distribution

A

u by setting, for any smooth test function

Φ

,

h

A

u,

Φ

i =

Z Z

u(t, x ) ∂

Φ

∂ t +

LΦ

(t , x ) dt m(dx ).

Remark 2.5. We will also need the distribution

A

˜ u, defined, for u locally bounded on (0, T ) × I (where T is a fixed positive number) by

h ˜

A

u,

Φi = Z Z

u(t , x )

− ∂

Φ

∂ t +

LΦ

(t , x) dt m(dx ),

for

Φ

smooth with compact support in (0, T ) × I . Note that one can prove, as in Proposition 2.4 that if u ∈ C

^1,2

((0, T ) × I ),

A

˜ u =

_σ2²p^′

+

∂^∂t

+

L

u.

We can now state our main result. Recall that f ˆ denotes the upper semicontinuous envelope of f . Theorem 2.6. The value function v

_f

is the only continuous and bounded function on the open set (0, +∞ ) × I satisfying the following conditions.

1. On the set (0, +∞ ) × I , we have v ≥ f , and the distribution

A

v satisfies

A

v ≤ 0.

2. We have

A

v = 0 on the open set U := { (t , x ) ∈ (0, +∞ ) × I | v(t, x ) > f ˆ (x ) } . 3. For every x ∈ I , lim

_t↓0

v(t , x ) = ˆ f (x ).

3. Continuity of the value function

This section is devoted to the proof of Theorem 2.2. At the end of the section, we also include a proposition concerning the behaviour of the value function for small time (see Proposition 3.4).

The equality u

_f

= v

_f

is an easy consequence of the following proposition.

Proposition 3.1. Let τ be a stopping time with values in [ 0, t ] . We have

Ex

h

e

^−Λτ

f (X

τ

)

i

= lim

s→t,s<tEx

h

e

^−Λτ∧s

f (X

τ∧s

)

i

.

Proof. We have

Ex

h

e

^−Λτ∧s

f ( X

τ∧s

)

i

=

Ex

h

e

^−Λτ

f (X

τ

)1

_{{τ <s}}i

+

Ex

h

e

^−Λs

f ( X

_s

)1

_{τ≥s}

i

, (6)

and

Ex

h

e

^−Λs

f (X

_s

)1

_{τ≥s}

i

=

Ex

h

e

^−Λt

f ( X

_s

)1

_{τ=t}

i

+

Ex

h

f (X

_s

)

e

^−Λs

1

_{τ≥s}

− e

^−Λt

1

_{τ=t}

i

.

By dominated convergence,

s→

lim

t,s<tEx

h

e

^−Λτ

f (X

τ

)1

_{{τ <s}} i

=

Ex

h

e

^−Λτ

f (X

τ

)1

_{{τ <t}} i

,

and

s→

lim

t,s<tEx

h

f (X

_s

)

e

^−Λs

1

_{τ≥s}

− e

^−Λt

1

_{τ=t}

= 0

i

.

We now want to prove

s→

lim

t,s<tEx

| f ( X

_s

) − f (X

_t

) | = 0. (7)

This is clearly true if f is continuous. If f is arbitrary, we have

Ex

| f (X

_s

) − f (X

_t

) | ≤

Ex

| f (X

_s

) − ϕ(X

_s

) | +

Ex

| ϕ(X

_s

) − ϕ(X

_t

) |

+

Ex

| ϕ( X

t

) − f ( X

t

) | .

So, in order to prove (7), we need only prove that, given ε > 0, one can find a bounded continuous function ϕ such that

sup

t/2≤s≤t

Ex

| f ( X

_s

) − ϕ( X

_s

) | ≤ ε.

From Corollary 8.13 we know that, given x ∈

I

, there exists a constant C

_x

> 0 such that, for all t > 0 and h ≥ 0,

P

_t

h(x ) ≤ C

_x

1 + 1

√ t

k h k

L²(m)

. (8)

Now assume that f ∈ L

²

(m ) (the extension to f bounded on I is straightforward). Given ε > 0, one can find a continuous function ϕ with compact support such that k f − ϕ k

L²(m)

< ε, so that (using (8) with h = | f − ϕ | )

sup

t/2≤s≤t

Ex

| f ( X

s

) − ϕ( X

s

) | ≤ C

x

1 +

√ 2

√ t

!

ε,

which completes the proof of (7). ⋄

Remark 3.2. The proof of Proposition 3.1 relies on the convergence (in probability) of f (X

s

) to f (X

_t

), when s → t , for f bounded and Borel-measurable. As proved in [4], this is related to the relative weak compactness of the laws of the random variables X

_s

. The argument we give can be seen as a way of proving this property.

For the proof of the continuity of the value function, we will also need the following lemma.

Lemma 3.3. Let (

Ω

,

F

, (

F_t

)

_t≥0

,

Px

, W, X ) be a weak solution of (1). For y ∈ I , define τ

_y

= inf { t ≥ 0 | X

_t

= y } .

We have lim

_y→xPx

(τ

_y

< ∞ ) = 1 and, for all t > 0, lim

_y→x Px

(τ

_y

≥ t ) = 0. We also have lim

_y→x Ey

(e

^−Λτx

) = 1 and lim

_y→xPy

(τ

_x

≥ t ) = 0, for all t > 0.

Proof. It is well known that the function (x, y) 7→

Ex

(e

^−τy

) is jointly continuous on I × I (see for instance [8]). In particular, we have lim

_y→x Ex

(e

^−τy

) = 1. We have

Px

(τ

_y

≥ t ) =

Px

(1 − e

^−τy

≥ 1 − e

^−t

) ≤

Ex

(1 − e

^−τy

) 1 − e

^−t

.

Hence, lim

_y→x Px

(τ

_y

≥ t) = 0. A similar argument gives lim

_y→x Py

(τ

_x

≥ t ) = 0. Since

Px

(τ

_y

= ∞ ) ≤

Px

(τ

_y

≥ t), we have lim

_y→x Px

(τ

_y

< ∞ ) = 1.

On the other hand, we have

Ey

(e

^−Λτx

) ≥

Ey

(e

^{−τx−Λτx}

) =

Ey

(e

^{−R0τx(1+r(Xs))ds}

) and we also have that (x, y) 7→

Ex

(e

^{−R0τy(1+r(Xs))ds}

) is jointly continuous on I × I (see for instance [8]), so that lim

_y→xEy

(e

^{−R0τx(1+r(Xs))ds}

) = 1. Hence, lim

_y→x Ey

(e

^−Λτx

) = 1. ⋄

Proof of Theorem 2.2. As mentioned above, the equality u

_f

= v

_f

follows easily from Propo- sition 3.1. On the other hand, since u

_f

= u

_fˆ

= v

_fˆ

, we also have v

_f

= v

_fˆ

. It is known that if f is upper semicontinuous, so is v

_f

(see [2], Proposition 17 or [5]). It remains to prove that v

_f

is lower semicontinuous.

Fix (t, x ) ∈ (0, +∞ ) × I and τ ∈ ¯

T_t⁰

. Since τ is a stopping time of (

F_t⁰

), we have { τ = 0 }

∈

F⁰

0

, and we deduce from the zero–one law (cf. Remark 8.4) that

Px

(τ = 0) ∈ { 0, 1 } .

Suppose

Px

(τ = 0) = 1. We then have

Ex

e

^−Λτ

f (X

τ

)

= f (x ). On the other hand, we have, for any (s, y) ∈ (0, +∞ ) × I , with M = sup

_y∈I

| f (y) | ,

v

_f

(s, y) ≥

Ey

e

^−Λτx∧s

f ( X

τ_x∧s

)

≥

Ey

e

^−Λτx∧s

f (X

τ_x∧s

)1

_{τ_x<s}

− M

Py

(τ

_x

≥ s)

= f (x )

Ey

(e

^−Λτx

1

_{τ_x<s}

) − M

Py

(τ

_x

≥ s ).

Hence v

_f

(s, y) ≥ f (x )

Ey

(e

^−Λτx

) − 2 M

Py

(τ

_x

≥ s). Using Lemma 3.3, we have

lim

_y→xEy

(e

^−Λτx

) = 1 and lim

_y→x Py

(τ

_x

≥ t /2) = 0. Hence f (x ) ≤ lim inf

_{(s,y)→(t,x)}

v

_f

(s, y).

We now assume that

Px

(τ = 0) = 0. From Proposition 3.1, we know that, given ε > 0, there exists δ ∈ (0, t) such that, for all s ∈ [ t − δ, t ] ,

Ex

(e

^−Λτ∧s

f (X

τ∧s

)) −

Ex

(e

^−Λτ

f ( X

τ

))

≤ ε. (9)

Obviously, this inequality is also true for s ≥ t . Now, we have, for all (s, y) ∈ (0, +∞ ) × I ,

Ex

e

^−Λτ∧s

f (X

τ∧s

)

=

Ex

e

^−Λτ∧s

f ( X

τ∧s

)1

_{τ_y≤τ∧s}

+

Ex

e

^−Λτ∧s

f (X

τ∧s

)1

_{τ_y>τ∧s}

.

We have

Ex

e

^−Λτ∧s

f (X

τ∧s

)1

_{τ_y>τ∧s}

≤ M

Px

(τ

_y

> τ ∧ s).

On the other hand, with the notation τ

^s

for τ ∧ s , we have

Ex

e

^−Λτs

f (X

τ^s

)1

_{τ_y≤τ^s}

=

Ex

e

^−Λτy

1

_{τ_y≤τ^s}

e

⁻

R(τs−τy)+

0 r(Xθ)dθ

f ( X

τ_y+(τ^s−τ_y)₊

)

≤

Ex

1

_{τ_y<∞}

e

^−R(τ

s−τy)+

0 r(Xθ)dθ

f (X

τ_y+(τ^s−τ_y)₊

)

=

Px

(τ

_y

< ∞ )

Ex

e

^−R

(τs−τy)+

0 r(X_θ)dθ

f (X

τ_y+(τ^s−τ_y)₊

) | τ

_y

< ∞

.

Note that, conditionally on { τ

_y

< ∞} ,

(

Ω

,

F

, (

F_τ

y+θ

)

_θ≥0

,

P

), (W

τ_y+θ

− W

τ_y

), ( X

τ_y+θ

) is a weak solution of the stochastic differential equation with starting point y, and (τ

^s

− τ

_y

)

₊

is an (

F_τ

y+θ

)

_θ≥0

-stopping time. Hence (using Remark 8.7)

Ex

e

^−Λτs

f (X

τ^s

)1

_{τ_y≤τ^s}

≤

Px

(τ

_y

< ∞ )v

_f

(s, y).

Therefore, we have

Ex

e

^−Λτ∧s

f (X

τ∧s

)

≤

Px

(τ

_y

< ∞ )v

_f

(s, y) + M

Px

(τ

_y

> τ ∧ s).

Now, take s ∈ [ t − δ, +∞ ). We have, using (9),

Ex

e

^−Λτ

f (X

τ

)

− ε ≤

Ex

e

^−Λτ∧s

f (X

τ∧s

)

≤

Px

(τ

_y

< ∞ )v

_f

(s, y) + M

Px

(τ

_y

> τ ∧ s).

It follows from Lemma 3.3 that lim

_y→x Px

(τ

_y

< ∞ ) = 1 and lim

_(s,y)→(t,x)Px

(τ

_y

> τ ∧ s) = 0.

Hence

Ex

e

^−Λτ

f (X

τ

)

≤ lim inf

(s,y)→(t,x)

v

_f

(s, y).

We conclude that v

_f

is lower semicontinuous. ⋄

The following Proposition clarifies the asymptotic behaviour of the value function as time goes to zero.

Proposition 3.4. We have lim

_t↓0

v

_f

(t , x ) = ˆ f (x ).

Recall that f ˆ is the upper semicontinuous envelope of f . In view of Proposition 3.4, it is natural to extend the definition of v

_f

(t , x) at t = 0 by setting v

_f

(0, x ) = ˆ f (x ).

Proof of Proposition 3.4. Note that t 7→ v

_f

(t , x ) is clearly non-decreasing, so that the limit exists. Since v

_f

= v

_fˆ

, we have v

_f

(t, x ) ≥ ˆ f (x ), so that lim

_t→0

v

_f

(t, x ) ≥ ˆ f (x ). Now, let τ ∈ ¯

T⁰

t

. For any ε > 0 such that (x − ε, x + ε) ⊂ I , we have

Ex

e

^−Λτ

f (X

τ

)

≤ sup

y∈(x−ε,x+ε)

f (y) + sup

I

f

Px

(τ

_x−ε

∧ τ

_x+ε

≤ t ).

Hence

v

_f

(t , x ) ≤ sup

y∈(x−ε,x+ε)

f (y ) + sup

I

f (

Px

(τ

_x−ε

≤ t ) +

Px

(τ

_x+ε

≤ t )) .

Observe that lim

_t→0Px

(τ

_x±ε

≤ t ). Therefore lim

_t→0

v

_f

(t , x ) ≤ sup

_{y∈(x−ε,x+ε)}

f ( y), and by making ε go to 0, we get lim

_t↓0

v

_f

(t , x ) ≤ ˆ f (x ). ⋄

4. The value function along the paths

The main result of this section is the following.

Theorem 4.1. Fix a positive number T and let f : I →

R

be a bounded, non-negative and upper semicontinuous function. For any weak solution (

Ω

,

F

, (

F_t

)

_t≥0

,

Px

, W , X ) of (1), the process V defined by

V

_t

= e

^−Λt

v

_f

(T − t , X

_t

), 0 ≤ t ≤ T ,

is a supermartingale. Moreover, if τ ˆ = inf { t ≥ 0 | v

_f

(T − t , X

_t

) = f (X

_t

) } , the process (V

_{t∧ ˆτ}

)

_0≤t≤T

is a martingale.

This result is not surprising. It appears in various forms in the literature (see for instance [5]).

However, since, under our assumptions, it does not seem to follow directly from known results, we will give a complete proof.

The proof of Theorem 4.1 will be based on the following lemma.

Lemma 4.2. Let f : I →

R

be bounded, non-negative and Borel-measurable. For all T > 0 and for all τ ∈ ¯

T⁰

T

, we have

∀ x ∈ I ,

Ex

e

^−Λτ

v

_f

(T − τ, X

τ

)

≤ v

_f

(T , x ).

Proof. For the proof of this lemma, we will need the strong Markov property. So, we will work with the canonical realization of the process X . More precisely, denote by (

Ω

,

F⁰

) the canonical space, where

Ω

is the set of all continuous functions on

R₊

, with values in I , and

F⁰

is the σ -algebra generated by the finite-dimensional cylinder sets. We endow this space with the right- continuous natural filtration (

F_t⁰

) of the coordinate mapping process X defined by X

_t

(ω) = ω(t ), for t ≥ 0 and ω ∈

Ω

. This space supports the family of shift operators (θ

_t

, t ≥ 0), defined by θ

_t

(ω) = ω(t + · ), for t ≥ 0 and ω ∈

Ω

. Given an initial condition x ∈ I , we denote by

Px

the (unique) law of a weak solution of (1). The fact that we have the strong Markov property for the family of probability measures (

Px

, x ∈ I ) on the canonical space follows from the Markov property and the fact that the semi-group preserves continuity (the weak Markov property and the fact that the semi-group preserves continuity are proved in Section 8, and the strong Markov property can be deduced by classical arguments, see [11], chapter III, Section 3).

For (t , x ) ∈ [ 0, T ]× I , set U (t , x ) = v

_f

(T − t , x ). Note that, since f is upper semicontinuous, U (T , x ) = f (x ). Given a finite subset F of the interval I , we denote by

T ^F

T

the set of all stopping times in

T

¯

⁰

T

, such that, on the set { τ < T } , X

τ

takes its values in F . We will first prove that

∀ τ ∈

T ^F

T

,

Ex

e

^−Λτ

U (τ, X

τ

)

≤ U (0, x ). (10)

Suppose F = { a

₁

, . . . , a

_n

} , with a

₁

< · · · < a

_n

, and let τ ∈

T ^F

T

. Note that, since X

_T

has a density (this is an immediate consequence of Corollary 8.13), we have

Px

(X

T

∈ F ) = 0, so that, with probability 1,

e

^−Λτ

U (τ, X

τ

) = e

^−ΛT

U (T , X

_T

)1

_{τ=T}

+

Xn

i=1

e

^−Λτ

U (τ, a

_i

)1

_{Xτ=ai}

.

Let ρ = (ρ

₀

= 0 < ρ

₁

< · · · < ρ

_m−1

< ρ

_m

= T ) be a subdivision of the interval [ 0, T ] . For i ∈ { 1, . . . , n } and j ∈ { 1, . . . , m } , define

A

_{i j}

= { ρ

_j−1

≤ τ < ρ

_j

} ∩ { X

τ

= a

_i

} . We have (using U (T , · ) = f )

e

^−Λτ

U (τ, X

τ

) = e

^−ΛT

f (X

_T

)1

_{τ=T}

+

n

X

i=1 m

X

j=1

e

^−Λτ

U (τ, a

_i

)1

_{Ai j}

=

Xn i=1

m−1

X

j=1

e

^−Λτ

U (ρ

_j

, a

_i

)1

_{Ai j}

+ e

^−Λτ

U (τ, X

τ

)1

_{ρ_m−1≤τ <T}

+ e

^−ΛT

f ( X

_T

)1

_{τ=T}

+

n

X

i=1 m−1

X

j=1

e

^−Λτ

U (τ, a

_i

) − U (ρ

_j

, a

_i

)

1

_{Ai j}

. (11)

We have

U (ρ

_j

, a

_i

) = u

_f

(T − ρ

_j

, a

_i

) = sup

τ∈T_T⁰_−ρ

j

Eai

e

^−Λτ

f (X

τ

)

.

Fix ε > 0 and denote by τ

_{i j}

a stopping time in

T⁰

T−ρ_j

, such that U (ρ

_j

, a

_i

) ≤

Eai

e

^{−Λτi j}

f (X

τ_{i j}

)

+ ε.

Let

˜

τ = T 1

_{ρ_m−1≤τ≤T}

+ 1

_{0≤τ <ρ_m−1}

τ +

n

X

i=1 m−1

X

j=1

1

_{Ai j}

τ

_{i j}

◦ θ

_τ

!

.

This clearly defines a stopping time with values in [ 0, T ] . Therefore U (0, x ) ≥

Ex

e

^−Λτ˜

f (X

τ_˜

)

=

Ex

e

^−ΛT

f (X

_T

)1

_{ρ_m−1≤τ≤T}

+

n

X

i=1 m−1

X

j=1

Ex

1

_{Ai j}

e

^{−Λτ+τi j◦θτ}

f ( X

τ+τ_{i j}◦θ_τ

) .

Using the strong Markov property and A

_{i j}

∈

F⁰

τ

, we have

Ex

1

_{Ai j}

e

^{−Λτ+τi j◦θτ}

f (X

τ+τ_{i j}◦θ_τ

)

=

Ex

h

1

_{Ai j}E

e

^{−Λτ+τi j◦θτ}

f ( X

τ+τ_{i j}◦θ_τ

) |

F⁰

τ

i

=

Ex

h

1

_{Ai j}

e

^−ΛτEai

e

^{−Λτi j}

f ( X

τ_{i j}

)

i

≥

Ex

1

_{Ai j}

e

^−Λτ

U (ρ

_j

, a

_i

) − ε .

Hence, using r ≥ 0, U (0, x ) ≥

Ex

e

^−ΛT

f (X

_T

)1

_{ρ_m−1≤τ≤T}

+

Ex

n

X

i=1 m−1

X

j=1

1

_{Ai j}

e

^−Λτ

U (ρ

_j

, a

_i

) − ε

!

≥

Ex

e

^−ΛT

f (X

_T

)1

_{ρ_m−1≤τ≤T}

+

Ex

n

X

i=1 m−1

X

j=1

1

_{Ai j}

e

^−Λτ

U (ρ

_j

, a

_i

)

!

− ε.

It follows from (11) that

n

X

i=1 m−1

X

j=1

1

_{Ai j}

e

^−Λτ

U (ρ

_j

, a

i

) = e

^−Λτ

U (τ, X

τ

) − e

^−Λτ

U (τ, X

τ

)1

_{ρ_m−1≤τ≤T}

−

Xn

i=1 m−1

X

j=1

e

^−Λτ

U (τ, a

_i

) − U (ρ

_j

, a

_i

) 1

_{Ai j}

.

Hence

U (0, x ) ≥

Ex

e

^−Λτ

U (τ, X

τ

)

+

Ex

h

e

^−ΛT

f (X

_T

) − e

^−Λτ

U (τ, X

τ

)

1

_{ρ_m−1≤τ <T}

i

−

Ex n

X

i=1 m−1

X

j=1

e

^−Λτ

U (τ, a

i

) − U (ρ

_j

, a

i

) 1

_{Ai j}

!

− ε.

Let | ρ | = max

_1≤j≤m

| ρ

_j

− ρ

_j−1

| . By passing to the limit as | ρ | → 0, we get, using the continuity of U on [ 0, T ) × I , U (0, x ) ≥

Ex

e

^−Λτ

U (τ, X

τ

)

− ε, and, since ε is arbitrary, (10) is proved.

Now, suppose τ ∈ ¯

T⁰

T

, and denote by (a

n

)

_n≥1

a dense sequence of elements in I . Set F

_n

= { a

1

, . . . , a

_n

} and

τ

_n

= inf { t ≥ τ | X

_t

∈ F

_n

} ∧ T .

We have τ

_n

∈

T ^Fn

, so that, according to (10),

Ex

e

^−Λτn

U (τ

_n

, X

τ_n

)

≤ U (0, x ). On the other

hand, the sequence (τ

_n

)

_n≥1

is non-increasing and lim

_n→∞

τ

_n

= τ . Indeed, if we denote the limit