Under (S1), for any t≥0 and x∈R2 , the integral in (6) is well defined.

According to the definition of the integral with respect to M, the integral is well defined, provided that (7) ∬R2|G(x,y,t)|αdy<∞. Taking into account (S1), we have the estimate |G(x,y,t)|=12πa|∫0t−x−yaσ(y,τ)a2(t−τ)2−|y−x|2dτ|1|x−y|<at≤C∫0t−x−yadτa2(t−τ)2−|y−x|21|x−y|<at=Caln(atx−y+a2t2x−y2−1)1|x−y|<at. Therefore, ∬R2|G(x,y,t)|αdy≤C∬B(x,at)|ln(atx−y+a2t2x−y2−1)|αdy=y1=x1+atrcosϕy2=x2+atrsinϕ=C∫02πdϕ∫01r(ln|r−1+r−2−1|)αdr≤C∫01r|ln(2r−1)|αdr≤C∫01r1−εdr<∞, where ε is a small positive number. This proves the statement.  □

Define Ω˜ξ={ωξ∈Ωξ∣∃k≥1:|ξk(ωξ)−x|≤δ}. Since {ξk,k≥1} are iid with everywhere positive density, it is clear that Pξ(Ω˜ξ)=1 . Fix some ωξ∈Ω˜ξ and ωΓ∈ΩΓ . Then U(x,t) has a centered Gaussian distribution, so that by the 0–1 law for Gaussian measures Pg(supy∈B(x,δ)|U(y,t)|<+∞)∈0,1. Suppose by contradiction that Pg(supy∈B(x,δ)|U(y,t)|<+∞)=1. Then by Fernique’s theorem (8) Eg[supy∈B(x,δ)|U(y,t)|2]<∞. On the other hand, Eg[supy∈B(x,δ)|U(y,t)|2]≥supy∈B(x,δ)Eg[|U(y,t)|2]=Cα2supy∈B(x,δ)∑k≥1Γk−2/αφ(ξk)−2/α|G(y,ξk,t)|2≥Cα2Γk(ωξ)−2/αφ(ξk(ωξ))−2/αsupy∈B(x,δ)G(y,ξk(ωξ),t)2, where k(ωξ) is an integer such that |ξk(ωξ)−x|<δ (it exists since ωξ∈Ω˜ξ ).

Since σ is continuous, there exists ε>0 0$]]> such that, for all y∈R2 with |x−y|<ε , σ(y,t)≠0 . Without loss of generality, we can assume that ε≥δ . Taking into account that ξk(ωξ)∈B(x,δ) , we get Eg[supy∈B(x,δ)|U(y,t)|2]≥Cα2Γk(ωξ)−2/αφ(ξk(ωξ))−2/αG(ξk(ωξ),ξk(ωξ),t)2. However, since σ(ξk(ωξ),t)≠0 , the observation preceding the theorem yields G(ξk(ωξ),ξk(ωξ),t)2=+∞ , which contradicts (8).

Consequently, for all ωξ∈Ω˜ξ and ωΓ∈ΩΓ , Pg(supy∈B(x,δ)|U(y,t)|<+∞)=0, whence P(supy∈B(x,δ)|U(y,t)|<+∞)=∫ΩΓ∫ΩξPg(supy∈B(x,δ)|U(y,t)|<+∞)dPξ(ωξ)dPΓ(ωΓ)=0, as claimed.  □

In the second part of this theorem, the exceptional event of probability zero may depend on θ.

Write the LePage representation for the left-hand side of Eq. (4): L(θ)=Cα ∑k=1∞Γk−1/αK(x,ξk,t)gk,K(x,y,t)=φ(ξk)−1/α∫0∞∬R2G(x,y,t)(∂2∂t2θ(x,t)−a2Δθ(x,t))dxdt and its right-hand side R(θ)=Cα∫0∞ ∑k=1∞Γk−1/αφ(ξk)−1/αθ(ξk,t)σ(ξk,t)gkdt. The proof consists of two steps: showing the convergence of the LePage series and then proving that (4) holds for partial sums of the LePage series.

Let us estimate the terms in the series for L(θ) . Assume that suppθ⊂[0,R]×B(0,R) . Then, denoting ψ(x,t)=∂2∂t2θ(x,t)−a2Δθ(x,t) , we have (9) |K(x,ξk,t)|≤∫0R∬B(0,R)|φ(ξk)|−1/α|G(x,ξk,t)||ψ(x,t)|dxdt≤(infB(0,R)|φ(x)|)−1/α×supx∈R2,t≥0|ψ(x,t)|∫0R∬B(0,R)|G(x,ξk,t)|dxdt≤CR,θ.

Consider first the case α∈(0,1) . By the strong law of large numbers and well-known properties of Gaussian random variables there exists Ω0∈F,P(Ω0)=1 , such that, for all ω∈Ω0 and k≥1 , Γk≥C1(ω)k,gk≤C2(ω)logk+1, where C1,C2 are some positive random variables. Therefore, the kth term in the series for L(θ) is bounded by CαΓk−1/α|K(t,x,ξk)gk|≤C(ω)k−1/αlogk+1. Consequently, the series for L(θ) is convergent for all ω∈Ω0 and θ(x,t)∈D(R2×R+) . Similarly, we can show the convergence of R(θ) .

For α∈[1,2) , the argument is changed slightly. Specifically, we show the almost sure convergence of the series for L(θ) and R(θ) for all θ(x,t)∈D(R2×R+) . Indeed, for fixed ωξ∈Ωξ,ωΓ∈ΩΓ , in view of (9), we have Eg[L(θ)2]=Cα2∑k≥1Γk−2/αK(x,ξk,t)2≤C(ω)∑k≥1k−2/α almost surely. Therefore, by the Kolmogorov theorem the series for L(θ) converges Pg -almost surely for almost all ωξ∈Ωξ and ωΓ∈ΩΓ and, therefore, P -almost surely. The almost sure convergence of R(θ) is shown in a similar way.

Now we prove that Eq. (4) holds for partial sums of the LePage series; the argument does not depend on the value of α. The counterpart of Eq. (4) for the partial sums reads as Cα ∑k=1NΓk−1/αφ(ξk)−1/α×∫0∞∬R2∫0tσ(ξk,τ)a2(t−τ)2−y−ξk21y−ξk<a(t−τ)ψ(y,t)dτdydt=Cα ∑k=1NΓk−1/αφ(ξk)−1/α∫0∞θ(ξk,τ)σ(ξk,τ)dτ. It suffices to show the equality of the corresponding terms, that is, to prove that, for any x∈R2 , ∫0∞∫τ∞∬B(x,a(t−τ))σ(y,τ)ψ(y,t)a2(t−τ)2−x−y2dydtdτ=∫0∞σ(x,τ)θ(x,τ)dτ. This equality, in turn, would follow if we show that (10) ∫τ∞∬B(x,a(t−τ))ψ(y,t)a2(t−τ)2−x−y2dydt=θ(x,τ) for all τ≥0,y∈R2 .

As before, assume that suppθ⊂[0,R]×B(0,R) . Define θ˜(x,u)=θ(x,R−u),u≤R. Then ∂2∂t2θ(x,R−u)=∂2∂t2θ˜(x,u);θ˜(x,0)=0;∂∂tθ˜(x,0)=0;Δθ(x,R−u)=Δθ˜(x,u). Consider the following Cauchy problem: (∂2∂t2−a2Δ)Vx,t=∂2∂t2θ˜(x,t)−a2Δθ˜(x,t),V(x,0)=0,∂V(x,t)∂t|t=0=0, Clearly, the function θ˜ is a solution. On the other hand, by Poisson’s formula, for all x∈R2 and r≥0 , θ˜(x,r)=12πa∫0r∬B(x,a(r−s))∂2∂t2θ˜(y,s)−a2Δθ˜(y,s)a2(r−s)2−x−y2dyds. Changing the variables r→R−τ , s→R−t and noticing that ψ(t,x) vanishes for τ≥R , we get (10) for all t∈[0,R] . For τ≥R , the both sides of the equality are zero, whence the proof follows.  □

Assume (S1)–(S3). Then for any T>0 0$]]> and x∈R2 the function U(x,·) is Hölder continuous on [0,T] almost surely with any exponent less than γ∧12 . This implies the required statement.

Take some h>0 0$]]> and t∈[0,T−h] . For fixed ωξ∈Ωξ and ωΓ∈ΩΓ , we have the estimate Eg[(U1(x,t)−U1(x,t+h))2]=Cα2∑k≥1Γk−2/αφ(ξk)−2/α|G(x,ξk,t2)−G(x,ξk,t2)|2≤g(h), where g(h)=Cα2∑k≥1Γk−2/αφ(ξk)−2/αsupt1,t2∈[0,T]0<t1−t2<h|G(x,ξk,t1)−G(x,ξk,t2)|2. Let t1,t2∈[0,T] be such that 0<t2−t1<h . Write |G(x,y,t1)−G(x,y,t2)|2=|∫0t1−x−yaσ(y,τ)a2(t1−τ)2−|x−y|2dτ1x−y<at1−∫0t2−x−yaσ(y,τ)a2(t2−τ)2−|x−y|2dτ1x−y<at2|2. Consider first the case |x−y|<at1 . Changes of variable τ=t1−s and τ=t2−s in the first and second integrals, respectively, give |∫x−yat1σ(y,t1−s)dsa2s2−|x−y|2−∫x−yat2σ(y,t2−s)dsa2s2−|x−y|2|≤|∫x−yat1σ(y,t1−s)−σ(y,t2−s)a2s2−|x−y|2ds|+|∫t1t2σ(y,t1−s)dsa2s2−|x−y|2|=:I1+I2. Taking into account (S3), we have I1≤∫x−yat1σ(y,t1−s)−σ(y,t2−s)a2s2−|x−y|2ds≤t1−t2γ∫x−yat1dsa2s2−x−y2≤Ct1−t2γ∫x−yat1ds(as−x−y)(as+x−y)≤Ct1−t2γat1−x−yx−y≤Ct1−t2γx−y;I2≤C∫t1t2dsa2s2−x−y2≤C∫t1t2dsas−x−yas+x−y≤2Cax−y(at2−x−y−at1−x−y)≤Ct2−t1x−y. Now let at1<|x−y|<at2 . In this case, |G(t1,x,y)−G(t2,x,y)|=|∫0t2−x−yaσ(y,τ)a2(t2−τ)2−|x−y|2dτ1x−y<at2|≤C∫0t2−x−ya1x−y<aTdτa2(t2−τ)2−x−y2≤C∫0t2−x−ya1x−y<aTdτ(a(t2−τ)−x−y)(a(t2−τ)+x−y)≤Cx−y∫0t2−x−ya1x−y<aTdτ(a(t2−τ)−x−y)≤Cat2−x−yx−y1x−y<aT≤Chx−y1x−y<aT. Combining the estimates, we get g(h)≤Ch2β∑k≥1Γk−2/αφ(ξk)−2/α1x−ξk1x−ξk<aT, where β=γ∧12 . Taking the expectation w.r.t. ξ, we obtain Eξ[g(h)]≤Ch2β∑k≥1Γk−2/αEξ[φ(ξk)−2/αx−ξk1x−ξk<aT]≤Ch2β∑k≥1Γk−2/αinfB(x,aT)φ(y)−2/α∬B(x,aT)1x−ydy≤C(ωΓ)h2β. Hence, for any η>0 0$]]>, Eξ[ ∑n=1∞2nβn1+ηg(2−n)]≤C(ωΓ) ∑n=1∞n−1−η<∞. Therefore, ∑n=1∞2nβg(2−n)n1+η<∞ Pξ⊗PΓ -almost surely. In particular, 2nβg(2−n)n1+η→0 , n→∞ , Pξ⊗PΓ -almost surely. Since the function g is nondecreasing and the function f(x)=x−2β|lnx|1+η satisfies f(2x)≤Cf(x) for x small enough, the latter convergence implies g(h)h2βlnh1+η→0,h→0, Pξ⊗PΓ -almost surely. Therefore, the inequality Eg[(U(x,t1)−U(x,t2))2]≤(t2−t1)2β|ln(t2−t1)|1+η, holds Pξ⊗PΓ -almost surely for all t1,t2∈[0,T] close enough and such that t1<t2 . Since U has a centered Gaussian distribution for fixed ωξ∈Ωξ and ωΓ∈ΩΓ , the last observation yields that |U(x,t1)−U(x,t2)|≤C(ω)t1−t2β|lnt2−t1+1|1+η/2 P -almost surely for all t1,t2∈[0,T] close enough.  □