aa r X i v : . [ m a t h . P R ] O c t A Probabilistic Characterization of g-HarmonicFunctions
Liang CaiDepartment of Mathematics, Beijing Institute of TechnologyBeijing 100081, People’s Republic of ChinaE-mail: [email protected] 1, 2018
Abstract
This paper gives a definition of g-harmonic functions and shows therelation between the g-harmonic functions and g-martingales. It’s directto construct such relation under smooth case, but for continuous casewe need the theory of viscosity solution. The results show that underthe nonlinear expectation mechanism, we also can get the similar relationbetween harmonic functions and martingales. Finally, we will give a resultabout the strict converse problem of mean value property of g-harmonicfunctions.
Key words or phrases : BSDE, g-martingale, g-harmonic function,nonlinear Feynman-Kac formula, viscosity solution.
Harmonic function (∆ u = 0) has a probabilistic interpretation as that if ∆ u = 0on R n , then u ( B xt ) is a martingale for any x ∈ R n (see for example [6]). Thisrelation between martingale and harmonic function connects probability withpotential analysis. It helps us to give probabilistic characterizations for har-monic function and more generalized X-harmonic function(see [6]). In 1997,Peng([9]) introduced the notions of g-expectation and conditional g-expectationvia a backward stochastic differential equation(BSDE) with a generator function g . Further, Peng([10]) introduced the notion of g-martingale which provided anheuristic tool to characterize some kind of harmonic function described by ellip-tic operator with a nonlinear term g by which we will define g-harmonic function.In this paper, with the help of nonlinear Feynman-Kac formula established fromBSDE(see for example [8]), we will give a probabilistic characterization of theg-harmonic function.Now we state our problem in detail. Let (Ω , F , P ) be a probability space en-dowed with the natural filtration {F t } t ≥ generated by an n-dimensional Brow-1ian motion { B xt } t ≥ . i.e. F t = σ { B s : s ≤ t } . Then we can define a g-martingale by an F t -adapted process { y t } t ≥ whichsatisfies the following BSDE for any 0 ≤ s ≤ t : y s = y t + Z ts g ( y r , z r ) dr − Z ts z r dB r . (1)Here g : R × R n −→ R , satisfies the condition:(H1). g ( y, ≡ ∃ C >
0, for any ( y , z ) , ( y , z ) ∈ R × R n we have | g ( y , z ) − g ( y , z ) | ≤ C ( | y − y | + | z − z | ) . And the equality(1) also can be formulated simply as[11]: E gs,t ( y t ) := y s . Then we can also get the definition of g-super(sub)martingale when E gs,t ( y t ) ≤ ( ≥ ) y s . This definition derives from the definition of g-expectation in the beginningpaper Peng[9]. When g ( y, z ) ≡ { X xt } t ≥ : dX xt = b ( X xt ) dt + σ ( X xt ) dB t , (2) X x = x ∈ R n , where b ( x ) : R n −→ R n , σ ( x ) : R n −→ R n × n satisfy the Lipschitz condition: ∃ C > | b ( x ) − b ( x ) | + | σ ( x ) − σ ( x ) | ≤ C | x − x | ∀ x , x ∈ R n , our problem is that: what kind of function u ( x ) : R n −→ R satisfies that u ( X xt )is a g-martingale for any x ∈ R n ?This problem also has its classical counterpart:First if { X xt } is just the Brownian motion { B xt } , then we have the resultthat when u ( x ) is harmonic on R n i.e.∆ u = Σ i ∂ u∂x i = 0 for any x ∈ R n , the process u ( B xt ) is a martingale for any x. And conversely if u ( x ) satisfies that u ( B xt ) is a martingale for any x, then u ( x ) must be harmonic on R n . The proof2ay have many editions, here we can give a sketch of one which may induce theextension to g-martingale case.If u ( x ) is harmonic on R n , then we use Itˆo formula to u ( B xt ) and get du ( B xt ) = X i ∂u∂x i ( B xt ) dB i,t + 12 X i ∂ u∂x i ( B xt ) dt = X i ∂u∂x i ( B xt ) dB i,t . Then we get u ( B xt ) is a martingale for any x ∈ R n . Conversely if u ( x ) iscontinuous on R n and for any x ∈ R n , u ( B xt ) is a martingale, then we have E [ u ( B xτ )] = u ( x ) for any stopping time τ . Particularly for any sphere S ( x, r ) = { y ∈ R n : | y − x | < r } , we have u ( x ) = E [ u ( B xτ S ( x,r ) )] = Z ∂S ( x,r ) u ( y ) dσ y where τ S ( x,r ) is the exit time of { B xt } from the sphere S ( x, r ), i.e. τ S ( x,r ) = inf { t > | B xt − x | ≥ r } , and σ y is the harmonic measure on the ∂S ( x, r ). Then from the familiar converseof the mean value property for harmonic function, we can get u ( x ) must beharmonic function.Further we can extend the Brownian motion { B xt } to the general diffusionprocess { X xt } :If u ( x ) ∈ C ( R n ) and satisfies X i b i ∂u∂x i ( x ) + 12 X i,j ( σσ τ ) i,j ∂ u∂x i ∂x j ( x ) = 0 , (3)then we have u ( X xt ) is a martingale for any x. The proof also uses the Itˆoformula. But conversely if u ( X xt ) is a martingale for any x, we can’t concludethat u ( x ) is smooth. Then with additional assumption u ( x ) ∈ C ( R n ) we canget that u ( x ) satisfies the PDE(3)(see [6]).Then naturally we will ask that what happens when we substitute the ex-pectation mechanism by the g-expectation mechanism. First we will define theinfinitesimal generator: Definition 1.
Let A Xg f ( x ) := lim t ↓ E g ,t [ f ( X xt )] − f ( x ) t , (4) then we call A Xg the infinitesimal generator of a diffusion process { X xt } underg-expectations. A Xg when f ∈ C ( R n ) by considering the following type ofquasilinear parabolic PDE: ( ∂u∂t − L u ( t, x ) − g ( u ( t, x ) , u x ( t, x ) σ ( x )) = 0 ,u (0 , x ) = f ( x ) . (5)Where L u ( t, x ) = X i b i ∂u∂x i ( t, x ) + 12 X i,j ( σσ τ ) i,j ∂ u∂x i ∂x j ( t, x ) . (6)When f ∈ C ( R n ), we assert that u ( t, x ) = E g ,t [ f ( X xt )] (7)is the solution of PDE(5). Then under the case t = 0, we get A Xg f ( x ) = L f ( x ) + g ( f ( x ) , f x ( x ) σ ( x )) . (8)Then we finish the preliminary and we can introduce our main results. Insection 2, we give a characterization of g-harmonic function under smooth case.In section 3, we characterize it under continuous case, where the differentialoperator is interpreted as viscosity solution. In section 4, we will investigate thestrict converse problem of mean value property of g-harmonic function evokedby its classical counterpart [7]. The equality (8) has the implication about the relation between the g-martingalesand the g-harmonic functions when f ∈ C ( R n ). In fact, the left side of (8) isrelated to a g-martingale and the right side is related to a harmonic PDE. Atfirst we will give the definition of g-harmonic functions: Definition 2.
Let f ∈ C ( R n ) . We call it a g-(super)harmonic function w.r.t. { X xt } if it satisfies A Xg f ( x )( ≤ ) = 0 for any x ∈ R n . (9)Then we suffice to construct the relation between the g-supermartingalesand the g-superharmonic functions. Theorem 1. If f ( x ) ∈ C ( R n ) , then the following assertions are equivalent:(1) f ( x ) is a g-superharmonic function.(2) { f ( X xt ) } is a g-supermartingale for any x ∈ R n . roof. (i) (1) ⇒ (2): f ∈ C ( R n ), by Itˆo formula, we can get f ( X xt ) is still an Itˆo diffusion process: f ( X xt ) = f ( X xs ) + Z ts L f ( X xr ) dr + Z ts f x ( X xr ) σ ( X xr ) dB r ≤ s ≤ t. and then we insert the term g ( f ( X xr ) , f x σ ( X rx )) and get: f ( X xs ) = f ( X xt ) − Z ts L f ( X xr ) dr − Z ts f x σ ( X xr ) dB r = f ( X xt ) + Z ts g ( f ( X xr ) , f x σ ( X rx )) dr − Z ts f x σ ( X xr ) dB r − Z ts {L f ( X xr ) + g ( f ( X xr ) , f x σ ( X rx )) } drf ( x ) is a g-superharmonic function, so: L f ( X xr ) + g ( f ( X xr ) , f x σ ( X xr )) = A Xg f ( X xr ) ≤ . And then according to the comparison theory of BSDE(cf.[10]), we can get { f ( X xt ) } is a g-supermartingale.(ii) (2) ⇒ (1):By the definition of the A Xg : A Xg f ( x ) = lim t ↓ E g ,t [ f ( X xt )] − f ( x ) t . { f ( X xt ) } is a g-supermartingale, so: E g ,t [ f ( X xt )] − f ( x ) ≤ , then A Xg f ( x ) ≤ . So we get f ( x ) is a g-superharmonic function. If we generalize the requirement of function f ( x ) to be only continuous on R n ,how we get a function f which satisfies that f ( X xt ) is a g-martingale for any x ∈ R n ? With the help of viscosity solution(cf. [3]) we can also refer to thequasi-linear second order PDEs. Here we need a lemma due to Peng[8]. Lemma 1.
Let ≤ t ≤ T and u ( t, x ) = E g ,T − t [ f ( X xT − t )] . hen u ( t, x ) is the viscosity solution of the following PDE on (0 , T ) × R n : ( ∂u∂t + L u ( t, x ) + g ( u ( t, x ) , u x ( t, x ) σ ( x )) = 0 u ( T, x ) = f ( x ) (10) Here g ( y, z ) and f ( x ) satisfy:(H2). Let F ( u, p ) = g ( u, pσ ( x )) , then ∃ C > s.t. | F ( u, p ) | ≤ C (1 + | u | + | p | ); | D u F ( u, p ) | , | D p F ( u, p ) | ≤ C, and (H3). f ( x ) is a continuous function with a polynomial growth at infinity. Definition 3.
Let u ( t, x ) ∈ C ( R × R n ) . u ( t, x ) is said to be a viscosity super-solution (resp. sub-solution) of the following PDE(11): ∂u∂t + L u ( t, x ) + g ( u ( t, x ) , u x ( t, x ) σ ( x )) = 0 , (11) if for any ( t, x ) ∈ R × R n and ϕ ∈ C , ( R × R n ) such that ϕ ( t, x ) = u ( t, x ) and ( t, x ) is a maximum (resp. minimum) point of ϕ − u , ∂ϕ∂t ( t, x ) + L ϕ ( t, x ) + g ( ϕ ( t, x ) , ϕ x ( t, x ) σ ( x )) ≤ . ( resp. ∂ϕ∂t ( t, x ) + L ϕ ( t, x ) + g ( ϕ ( t, x ) , ϕ x ( t, x ) σ ( x )) ≥ . ) u ( t, x ) is said to be a viscosity solution of PDE(11) if it is both a viscosity super-and sub-solution of (11). We also consider the viscosity solution of the following type of quasilinearelliptic PDE(12): L u ( x ) + g ( u ( x ) , u x ( x ) σ ( x )) = 0 . (12)We can directly get an relation between the two solutions of (11) and (12): Lemma 2.
Let ˜ u ( t, x ) = u ( x ) for all ( t, x ) ∈ R × R n , then we have: ˜ u ( t, x ) is the viscosity super-(sub-)solution of PDE (11) ⇔ u ( x ) is the viscositysuper-(sub-)solution of PDE (12).Proof. We only prove the case of viscosity super-solution. The ”sub-” case isan immediate conclusion of the ”super-” case.(i). ” ⇒ ”:For any ( t , x ) ∈ R × R n , and a function ϕ ( x ) ∈ C ( R n ) which satisfies ϕ ( x ) ≤ u ( x ) , ϕ ( x ) = u ( x ), we define ˜ ϕ ( t, x ) = ϕ ( x ) for all ( t, x ) ∈ R × R n .Then ∂ ˜ ϕ∂t = 0 , ˜ ϕ ( t , x ) = ˜ u ( t , x ) , ˜ ϕ ( t, x ) ≤ ˜ u ( t, x ) , u ( t, x ) is the viscosity super-solution of PDE(11),we have ∂ ˜ ϕ∂t ( t , x ) + L ˜ ϕ ( t , x ) + g ( ˜ ϕ ( t , x ) , ˜ ϕ x ( t , x ) σ ( x )) ≤ , i.e. L ϕ ( x ) + g ( ϕ ( x ) , ϕ x ( x ) σ ( x )) ≤ . So u ( x ) is the viscosity super-solution of PDE (12).(ii). ” ⇐ ”:For any ( t , x ) ∈ R × R n , and a function ϕ ( t, x ) ∈ C ( R × R n ) whichsatisfies ϕ ( t, x ) ≤ ˜ u ( t, x ) , ϕ ( t , x ) = ˜ u ( t , x ), then ∂ϕ∂t ( t , x ) = 0 , (13)and due to the assumption that u ( x ) is the viscosity super-solution of PDE(12),we have L ϕ ( t , x ) + g ( ϕ ( t , x ) , ϕ x ( t , x ) σ ( x )) ≤ . Combined with (13), we get ∂ϕ∂t ( t , x ) + L ϕ ( t , x ) + g ( ϕ ( t , x ) , ϕ x ( t , x ) σ ( x )) ≤ . So ˜ u ( t, x ) is the viscosity super-solution of PDE (11).Then we can introduce our main result of this section: Theorem 2.
We have the following two consequences:(i). For any f ( x ) ∈ C ( R n ) , and g ( y, z ) satisfying (H1), if ∀ x ∈ R n , f ( X xt ) is ag-supermartingale, then f ( x ) is a viscosity super-solution of PDE (12).(ii). For any f ( x ) satisfying (H3), and g ( y, z ) satisfying (H1) (H2), let f ( x ) is a viscosity super-solution of PDE (12), then { f ( X xt ) } is a g-supermartingalefor all x ∈ R n . Actually, the consequence (ii) is the answer of our main problem and theconsequence (i) is the converse of it. But (i) is easier to be proved, so we aregoing to prove (i) at first:
Proof. (i).For any x ∈ R n , let ϕ ∈ C ( R n ), ϕ ( x ) = f ( x ) where x is a maximum pointof ϕ − f . It means ∀ ˜ x ∈ R n , we have ϕ (˜ x ) ≤ f (˜ x ). Then from (8), we get L ϕ ( x ) + g ( ϕ ( x ) , ϕ x ( x ) σ ( x )) = A Xg ϕ ( x )= lim t ↓ E gt [ ϕ ( X xt )] − ϕ ( x ) t = lim t ↓ E gt [ ϕ ( X xt )] − f ( x ) t . E gt [ ϕ ( X xt )] ≤ E gt [ f ( X xt )] , and with the assumption { f ( X xt ) } is a g-supermartingale, we can get: E gt [ ϕ ( X xt )] − f ( x ) ≤ E gt [ f ( X xt )] − f ( x ) ≤ . Then A Xg ϕ ( x ) = lim t ↓ E gt [ ϕ ( X xt )] − f ( x ) t ≤ . i.e. L ϕ ( x ) + g ( ϕ ( x ) , ϕ x ( x ) σ ( x )) ≤ . By definition, it means f ( x ) is a viscosity super-solution of PDE (12).(ii).We want to prove { f ( X xt ) } is a g-supermartingale for any x ∈ R n . It meansthat we need to prove ∀ x ∈ R n and ∀ ≤ s ≤ t , we have E gs,t [ f ( X xt )] ≤ f ( X xs ) . Under the assumption, in fact b ( x ) , σ ( x ) and g ( y, z ) are all independent of timet, so we can get the Markovian property of E gs,t , i.e. E gs,t [ f ( X xt )] = E gt − s [ f ( X yt − s )] | y = X xs . Then we get an equivalence relation: { f ( X xt ) } is a g-(super)martingale for any x ∈ R n ⇔E gt [ f ( X xt )] = ( ≤ ) f ( x ) for any t ≥ x ∈ R n . (14)So we suffice to prove the latter assertion.For any T ≥
0, the assumption f ( x ) is a viscosity super-solution of PDE(12)implies that ˜ f ( t, x ) := f ( x ) is a viscosity super-solution to the following PDE: ( ∂u∂t + L u ( t, x ) + g ( u ( t, x ) , u x ( t, x ) σ ( x )) = 0 ,u ( T, x ) = f ( x ) , (15)according to lemma 2. And with the help of lemma 1, u ( t, x ) = E g ,T − t [ f ( X xT − t )]is actually the viscosity solution of PDE (15). Moreover by the maximum prin-ciple of the viscosity solution( see [1]), we can get: u ( t, x ) ≤ ˜ f ( t, x ) for any 0 ≤ t ≤ T. Especially, we have u (0 , x ) ≤ ˜ f (0 , x ) , i.e. E gT [ f ( X xT )] ≤ f ( x ) . orollary 1. (i). For any f ( x ) ∈ C ( R n ) , and g ( y, z ) satisfying (H1), if ∀ x ∈ R n , f ( X xt ) is a g-martingale, then f ( x ) is a viscosity solution of PDE(12).(ii). For any f ( x ) satisfying (H3), and g ( y, z ) satisfying (H1) (H2), let f ( x ) isa viscosity solution of PDE(12), then { f ( X xt ) } is a g-martingale for all x ∈ R n . It is an immediate consequence from the theorem 2.
For classical harmonic function, many generalized results of the converse prob-lem of mean value property have been investigated (cf. [5][7]). In [7], Øksendaland Stroock give a technique to solve a strict converse of the mean value prop-erty for harmonic functions. Now we will generalize it to the case of g-harmonicfunction. Here the strictness means that for each x ∈ R n we don’t need justifythat for any stopping time τ whether E g ,τ ( f ( X xτ )) equals f ( x ). We only needto justify one appropriate stopping time of each x.In the sequel we put ∆( x, r ) = { y ∈ R n ; | y − x | < r } for any x ∈ R n and r >
0. Let τ U = inf { t > X xt ∈ U c } for any open set U. And we suppose theoperator (6) is elliptic on R n . Theorem 3. f ( x ) is a local bounded continuous function on R n . If for any x ∈ R n , there exists a radius r ( x ) , the mean value property holds: E g ,τ x [ f ( X xτ x )] = f ( x ) here τ x = τ ∆( x,r ( x )) . (16) And r ( x ) is a measurable function of x and satisfies that for each x, there existsa bounded open set U x , x ∈ U x and moreover r ( y ) , y ∈ U x should satisfy thefollowing two conditions: ≤ r ( y ) ≤ dist ( y, ∂U x ) , (17) and inf { r ( y ); y ∈ K } > for all closed subsets K of U x with dist ( K, ∂U x ) > . Then we can get:(i). For each y ∈ U x the mean value property holds on the boundary: E g ,τ y [ f ( X yτ y )] = f ( y ) , here τ y = inf { t > X yt ∈ U cx } . and furthormore:(ii). f ( x ) is the viscosity solution of PDE(12).Proof. ( i ) ⇒ ( ii ) is also based on the nonlinear Feynman-Kac formula for ellipticPDE(cf.[8]). So we sufficiently prove the first conclusion.For each y ∈ U x , we define a sequence of stopping times τ k for { X yt } byinduction as follows: τ ≡ τ k = inf { t ≥ τ k − ; | X yt − X yτ k − | ≥ r ( X yτ k − ) } ; k ≥ .
9y the mean property(16), and the strong markovian property we can get E g ,τ k [ f ( X yτ k )] = E g ,τ k − [ E gτ k − ,τ k [ f ( X yτ k )]]= E g ,τ k − [ E g ,τ k − τ k − [ f ( X X yτk − τ k − τ k − )]]= E g ,τ k − [ f ( X yτ k − )] , then by induction we get E g ,τ k [ f ( X yτ k )] = f ( y ) . In the following we will prove τ k → τ y a.e. when k → ∞ . Obviously τ k ≥ τ k − , so there exists a stopping time τ s.t. τ k ↑ τ . If τ = τ y , then there exists ǫ > X yτ k , ∂U x ) ≥ ǫ for any k. Let r k = r ( X yτ k ), according to the condition(18), we get there exists r > r k ≥ r for any k. It means dist( X yτ k , X yτ k − ) ≥ r. And since X yt is continuous, then τ k → ∞ , which implies τ y = ∞ . So P ( τ k don’t converge to τ y ) ≤ P ( τ y = ∞ ) . But for (6) is elliptic and U x is bounded, we have P ( τ y < ∞ ) = 1. So P ( τ k converge to τ y ) = 1 . Then we get f ( y ) = E g ,τ k [ f ( X yτ k )]= lim k →∞ E g ,τ k [ f ( X yτ k )]= E g ,τ y [ f ( X yτ y )]So we have finished the proof. Acknowledgement.
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