Throughout the section, we assume that condition is satisfied, but do not suppose that homogenization holds (even in probability): this condition is added only at the very end of the section.

Fix , let be as in and define the metric space

Θ:={θ∈C0,1(Rd):θ(0)=0and‖Dθ‖∞≤CR} with distance, for all ,d(θ1,θ2):=supx∈Rd|θ1(x)−θ2(x)|1+|x|2. It is immediate that Θ is a compact.

Next we enlarge the probability space to , which is endowed with the one-parameter group of transformations defined, for , by

τ˜x(ω,θ,s)=(τxω,θ(⋅+x)−θ(x),s); below, by an abuse of notation, we write .

Fix with , let be the solution to ((2.1)), define the map by

Φδ,p(ω)=(ω,vδ,p(⋅,ω)−vδ,p(0,ω),−δvδ,p(0,ω)), which is clearly measurable, and consider the push-forward measureμδ,p=Φδ,p♯P, which is a Borel probability measure on .

Note that, since the first marginal of is and Ω is a Polish space while is compact, the family of measures is tight.

Let *μ* be a limit, up to a subsequence , of the 's.

*For each* *, the transformation* *preserves the measure μ.*

Fix a continuous and bounded map . Since the map is continuous and bounded and converges weakly to *μ* , we have

∫Ω˜ξ(ω˜)τx♯μ(dω˜)=∫Ω˜ξ(τx(ω˜))μ(dω˜)=limn∫Ω˜ξ(τx(ω˜))μδn,p(dω˜). In view of the definition of and , we get∫Ω˜ξ(τx(ω˜))μδn,p(dω˜)=∫Ωξ(τxω,vδn,p(x+⋅,ω)−vδn,p(x,ω),−δnvδn,p(0,ω))dP(ω)=∫Ωξ(τxω,vδn,p(⋅,τxω)−vδn,p(0,τxω),−δnvδn,p(−x,τxω))dP(ω)=∫Ωξ(ω,vδn,p(⋅,ω)−vδn,p(0,ω),−δnvδn,p(−x,ω))dP(ω), the last line being a consequence of the stationarity of .

Using that is Lipschitz continuous uniformly in *δ* and *ξ* is continuous on the set , we find

∫Ω˜ξ(τx(ω˜))μδn,p(dω˜)=∫Ωξ(ω,vδn,p(⋅,ω)−vδn,p(0,ω),−δnvδn,p(0,ω)+O(δn))dP(ω)=∫Ω˜ξ(ω˜)dμδn,p(ω˜)+o(1). Letting , we finally get∫Ω˜ξ(ω˜)τx♯μ(dω˜)=∫Ω˜ξ(ω˜)dμ(ω˜), and, hence, the claim. □

The next lemma asserts that there exists some such that the restriction of *μ* to the last component is just a Dirac mass. If we know that homogenization holds, then is of course nothing but . Note that in what follows, abusing once again the notation, we denote by *μ* the restriction of *μ* to the first two components .

*There exits a constant* *such that, for any Borel measurable set* *,*

μ(E×[−Mp,Mp])=μ(E×{c‾}). *In particular, the sequence* *converges in probability to* *.*

Let large and , set and

Ek:={ω∈Ω:∃θ∈Θand∃s∈[tk,tk+1]such that(ω,θ,s)∈sppt(μ)}. Since the first marginal of *μ* is and , there exists such that .

It turns out that is translation invariant, that is, for each , . Indeed, if , there exists and such that and, hence, belongs to . Since *μ* is invariant under , so is its support. Hence and *ω* belongs to . The opposite implication follows in the same way.

The ergodicity of yields that , which means that *μ* is concentrated in some . Thus *μ* is also concentrated on . Letting implies that there exists such that *μ* is concentrated of the set .

It remains to check that converges in probability to . This is a consequence of the classical Porte-Manteau Theorem, since, for any ,

limsupn→∞P[|δnvδn,p(0,⋅)+c‾|≥ε]=limsupμδn,p[Ω×Θ×([−Mp,Mp]\(c‾−ε,c‾+ε))]≤μ[Ω×Θ×([−Mp,Mp]\(c‾−ε,c‾+ε))]=0. □

The next lemma is the first step in finding a corrector and possibly identifying and , when the latter exists.

*Let* *be defined by* Lemma 3.2*. For μ-a.e.* *, θ is a solution to*

(3.1)−tr(A(Dθ+p,x,ω)D2θ)+H(Dθ+p,x,ω)=c‾in Rd.

Fix and let be the set of such that *θ* such that, in the open ball ,

−tr(A(Dθ+p,x,ω)D2θ)+H(Dθ+p,x,ω)≥c‾−ε and−tr(A(Dθ+p,x,ω)D2θ)+H(Dθ+p,x,ω)≤c‾+ε. Recall that Lemma 3.2 gives that converge in probability to . Since solves ((2.1)) and is uniformly Lipschitz continuous, it follows that, as , .

Finally observing that is closed in , we infer, using again the Porte-Manteau Theorem, that .

As *R* and *ε* are arbitrary, we conclude that the set , for which the fact that the equation is satisfied in the viscosity sense is of full probability. □

Next we investigate some properties of *θ* .

*For any* *,* *.*

Since the map is continuous on and is stationary, we have

Eμ[θ(x)]=limEμδn,p[θ(x)]=limEP[vδn,p(x)−vδn,p(0)]=0. □

*For μ-a.e.* *and any direction* *, the (random) limit*

ρω˜(q):=limt→∞θ(tq)t *exists. Moreover,* *is invariant under* *for* *, that is,* ρτ˜x(ω˜)(q)=ρω˜(q)μ−a.e.

We first show that, for any , the limit

limt→+∞1t(∫Br(0)θ(tq+y)dy−∫Br(0)θ(y)dy) exists -a.s.

Since the uniform converge of uniformly Lipschitz continuous maps implies the -weak ⋆ convergence of their gradients, the map defined by

ξ((ω,θ)):=∫Br(0)Dθ(y)⋅qdy is continuous and bounded on .

Moreover,

1t(∫Br(0)θ(tq+y)dy−∫Br(0)θ(y)dy)=1t∫0t∫Br(0)Dθ(sq+y)⋅qdyds=1t∫0tξ(τ˜sq(ω˜))ds. It follows from the ergodic theorem that the above expression has, as and *μ* -a.s. a limit .

Choosing and letting , we also find that, as , has *μ* -a.s., a limit because *θ* is -Lipschitz continuous.

Fix and for which and are well defined; recall that this holds for *μ* -a.e. .

Then, in view of the Lipschitz continuity of *θ* , we have

ρτ˜x(ω˜)(q)=limt→+∞τ˜x(θ)(tq)t=limt→+∞1t(θ(x+tq)−θ(x))=ρω˜(q). □

*There exists a random vector* *such that, μ-a.s. and for any direction* *,*

limt→+∞θ(tv)t=rω˜⋅v.

Since *θ* is -Lipschitz continuous, it is enough to check that the map is linear on for *μ* -a.e. .

Let be a set of *μ* -full probability in Ω such that the limit in Lemma 3.5 exists for any .

Restricting further the set if necessary; we may also assume (see, for instance, the proof of Lemma 4.1 in [[1]]) that, for any and , there exists such that, for all with , all and ,

|θ(x+tq)−θ(x)t−ρω˜(q)|≤η(|x|+1). Fix , with , and , and let *T* be associated with as above. Then, for any , we haveθ(t(q1+q2))=θ(t(q1+q2))−θ(tq2)+θ(tq2). Thus|θ(t(q1+q2))t−ρω˜(q1)−ρω˜(q2)|≤|θ(t(q1+q2))−θ(tq2)t−ρω˜(q1)|+|θ(tq2)t−ρω˜(q2)|≤η(|q2|+t−1)+η Letting and yields the claim, since *η* and *M* are arbitrary. □

*Let* **r** *be defined as in* Lemma 3.6*. Then* *.*

Lemma 3.4 yields that, for any ,

0=limt→+∞Eμ[θ(tv)t]=Eμ[limt→+∞θ(tv)t]=Eμ[r⋅v]=Eμ[r]⋅v. □

As a straightforward consequence of the previous results, we have the existence of a corrector and, hence, homogenization for at least one vector .

*For μ-a.e.* *,* *exists for* *and is given by* *. Moreover,* *is a corrector for* *, in the sense that*

−tr(A(Dθ′+p′,x,ω)D2θ′)+H(Dθ′+p′,x,ω)=c‾in Rdwithlim|x|→+∞θ′(x)/|x|=0.

Another consequence of the above results is that homogenization holds if the law of under is a radially symmetric. By this we mean that, for any rotation matrix *R* , the law of is the same as the law of the pair given by

(A˜,H˜)(p,x,ω):=(RTAR,H)(Rp,Rx,ω). Note that this implies that has the same law as .

*Assume that,* *-a.s.,* *is* 0*-homogeneous in p, H satisfies, for all* *,*

(3.2)0≤H(λp,x,ω)≤λH(p,x,ω) *and suppose that the law of* *under* *is radially symmetric. Then homogenization holds in probability, that is, for any* *,* *in probability. Moreover, the map* *satisfies, for any* *,* 0≤c‾(s1)/s1≤c‾(s2)/s2.

Note that the map is increasing as soon as it is positive. Moreover, one easily checks that, if, in addition, *H* is 1-homogeneous in *p* and coercive, then for some positive constant .

It follows from the assumed bounds and the stationarity, that there exists a set with such that, for any and , and exist and are deterministic. The radial symmetry assumption and as well as (Corollary 3.9) imply that and, in addition, for all ,

0≤c±(λs)≤λc±(s). Also note that the maps are nondecreasing. Indeed given , choosing and , we findc±(s1)/s1≤c±(s2)/s2≤c±(s2)/s1. It follows that is increasing as soon as it is positive.

On the other hand, for any , we can find a subsequence of converging to some measure as *δ* tends to 0. By a diagonal argument, we can assume that this is the same subsequence for any .

Let be associated with the limit measure as above. It follows from (**H**) that the map is uniformly continuous and thus can be continuously extended to . Moreover, the assumed radial symmetry yields that . Finally, note that, for all ,

(3.3)0≤c−(s)≤c‾(s)≤c+(s). Let . Then ((3.3)) implies on .

Fix with , and let , and **r** be associated with *p* . For -a.e. with , is a corrector for with associated ergodic constant . It follows that , and, hence,

c+(|p|)≥c‾(|p|)=c+(|p+r|)μp−a.s. Since is increasing on , this inequality implies that a.s. Then gives -a.s., so that, for -a.e. , *θ* is a corrector for *p* with associated ergodic constant . It follows that . In conclusion, for any , and thus, by continuity, for any , which, in turn, proves that homogenization holds. □

Another application of the previous results is the convergence in law of the random variable when *H* is convex in the gradient variable. The argument is a variant of [[15]]. Of course, the result is much weaker than the a.s. convergence is established in [[14]]; see also [[1], [2]]. The proof is, however, rather simple.

*Assume that,* *-a.e.,* *is convex in the p variable and that* *does not depend on p. Then, for any* *, homogenization holds in probability, that is there exists* *such that* *in probability.*

Let *μ* be a measure built as in the beginning of the section. It follows that there exists a random family of measures on Θ such that, for any continuous map , one has

∫Ω×Θϕ(ω,θ)dμ(ω,θ)=∫Ω[∫Θϕ(ω,θ)dμω(θ)]dP(ω). Set . Since and *μ* are invariant with respect to and respectively, for any bounded measurable map and any , we have∫Ωϕ(ω)(θˆ(x+z,ω)−θˆ(z))dP(ω)=∫Ω×Θϕ(ω)(θ(x+z)−θ(z))dμ(ω,θ)=∫Ω×Θϕ(τ−zω)θ(x)τ˜z♯dμ(ω,θ)=∫Ωϕ(τ−zω)θˆ(x,ω)dP(ω)=∫Ωϕ(ω)θˆ(x,τzω)dP(ω). This shows that has stationary increments. Moreover, in view of Lemma 3.4, has mean zero, and, hence, is stationary with average 0. In particular, is -a.s. strictly sublinear at infinity. Since, for *μ* -a.e. , *θ* is a solution to (Lemma 3.3) and *H* is convex in the gradient variable, is a subsolution to (Lemma 3.3) and, thus a subcorrector. Following [[15]], this implies thatliminfδ→0δvδ,p(0,ω)≥−c‾. In particular, for any sequence that tends to 0 such that and converge respectively to a measure and a constant , we have . Exchanging the roles of and leads to the equality . The conclusion now follows. □

We are now ready to prove the main result.

We assume that homogenization holds in probability and is an extreme point of the convex hull of the set .

Let *μ* be a measure built as in the beginning of the section and **r** be defined by Lemma 3.6.

Then *μ* -a.s., that is belongs to *S μ* -a.s. Indeed Lemma 3.3 gives and

−tr(A(Dθ+p,x,ω)D2θ)+H(Dθ+p,x,ω)=c‾in Rd, while, in view of Lemma 3.6, for all ,limt→+∞θ(tx)t=r⋅x. Thus is a corrector for , that is it satisfies−tr(A(Dθ˜+p+r,x,ω)D2θ˜)+H(Dθ˜+p+r,x,ω)=c‾in Rdandlim|x|→+∞θ˜(x)/|x|=0. It follows that *μ* -a.s.

Next we recall (Lemma 3.7) that . Since *μ* -a.s. and *p* is an extreme point of the convex hull of *S* , the equality implies that *μ* -a.s. Therefore *μ* -a.s., which, together with the fact that *θ* solves the corrector equation for *p* , implies that *θ* is a corrector for *p* itself. □