For a measure ν satisfying (1), denote by (6) ftν,ε(x)=ExLtν(εW)=∫0t∫R12πsε2e−(y−x)22sε2ν(dy)ds,t≥0,x∈R, the characteristic of the local time Lν(εW) considered as a W-functional of  εW ; see [3], Chapter 6.

The following statement is a version of Khas’minskii’s lemma; see [9], Section 1.2.

Using the Markov property, as a simple corollary, we obtain, for arbitrary t>0 0$]]>, (8) supx∈RExeLtν(εW)≤21+t/s=2e(log2)(t/s), where s>0 0$]]> is such that (7) holds. This inequality, combined with (6), leads to the following estimate.

In what follows, we will repeatedly decompose μ into sums of two components and analyze separately the exponential moments of the local times that correspond to these components. We will combine these estimates and obtain an estimate for Ltμ(εW) itself using the following simple inequality. Let μ=ν+κ and p,q>1 1$]]> be such that 1/p+1/q=1 . Then Ltμ(εW)=Ltν(εW)+Ltκ(εW)=(1/p)Ltpν(εW)+(1/q)Ltqκ(εW), and therefore by the Hölder inequality we get (10) EeLtμ(εW)≤(EeLtpν(εW))1/p(EeLtqκ(εW))1/q. We will also use another version of this upper bound, which has the form (11) EeLtμ(εW)1A≤(EeLtpμ(εW))1/p(P(A))1/q,A∈F.

We denote Δ=supx∈Rμ({x}). We will prove Theorem 1 in several steps, in each of them extending the class of measures μ for which the required statement holds.

If μ=aδz is a weighted δ-measure at the point z, then we have Ltμ(εW)=aε−1Lt(z)(W), where Lt(z)(W)=limη→012η∫0t1|Ws−z|≤ηds is the local time of a Wiener process at the point z. The distribution of Lt(z)(W) is well known; see, e.g., [5], Chapter 2.2 and expression (6) in Chapter 2.1. Hence, the required statement in the particular case μ=aδz is straightforward, and we have the following: (12) limε→0ε2supxlogExeaε−1Lt(z)(W)=ta22. Note that in this formula the supremum is attained at the point x=z .

In this section, we will extend this result to the case where μ is a finite mixture of δ-measures, that is, μ= ∑j=1kajδzj. Let j∗ be the number of the maximal value in {aj} , that is, Δ=aj∗ . Then Ltμ(εW)≥Δε−1Lt(zj∗)(W) , and it follows directly from (12) that (13) lim infε→0ε2supx∈RlogExeLtμ(εW)≥tΔ22. In what follows, we prove the corresponding upper bound (14) lim supε→0ε2supx∈RlogExeLtμ(εW)≤tΔ22, which, combined with this lower bound, proves (3).

Observe that, for γ>0 0$]]> small enough, N(μ,γ)=Δ. Then by Lemma 2, for any λ≥1 , (15) lim supε→0ε2supx∈RlogExeλLtμ(εW)≤c1λ2tΔ2 with c1=(4log2)c0=8log2π(1+2 ∑k=1∞e−(2k−1)22)2. In particular, taking λ=1 , we obtain an upper bound of the form (14), but with a worse constant c1 instead of required 1/2 . We will improve this bound by using the large deviations estimates for εW , the Markov property, and the “individual” identities (12).

Denote μj=ajδzj,j=1,…,k . Then Ltμ(εW)= ∑j=1kLtμj(εW). Fix some family of neighborhoods Oj of zj,j=1,…,k , such that the minimal distance between them equals ρ>0 0$]]>, and denote Oj=R∖⋃i≠jOi. For some N≥1 whose particular value will be specified later, consider the partition tn=t(n/N) , n=0,…,N , of the segment [0,t] and denote Bn,j={f∈C(0,t):fs∈Oj,s∈[tn−1,tn]},j∈{1,…,k},n∈{1,…,N}, Cj1,…,jN= ⋂n=1NBn,jn,j1,…,jN∈{1,…,k}. Observe that if the process εW does not visit Oj on the time segment [u,v] , then Lμj(εW) on this segment stays constant. This means that, on the set {εW∈Cj1,…,jN} , we have Ltμ(εW)= ∑n=1N(Ltnμjn(εW)−Ltn−1μjn(εW)). Because Lμj(εW) is a time-homogeneous additive functional of the Markov process εW , we have Ex[eLtnμjn(εW)−Ltn−1μjn(εW)|Ftn−1]=EyeLt/Nμjn(εW)|y=εWtn−1. Then by (12), for any j1,…,jN∈{1,…,k} , lim supε→0ε2supx∈RlogExeLtμ(εW)1εW∈Cj1,…,jN≤t2N ∑n=1N(ajn)2≤tΔ22. Because we have a fixed number of sets Cj1,…,jN , this immediately yields (16) lim supε→0ε2supx∈RlogExeLtμ(εW)1εW∈C≤tΔ22 with C=⋃j1,…,jN∈{1,…,k}Cj1,…,jN. Hence, to get the required upper bound (14), it suffices to prove an analogue of (16) with the set C replaced by its complement D=C(0,t)∖C . Using (11) with p=2 , A={εW∈D} , and (15) with λ=2 , we get lim supε→0ε2supx∈RlogExeLtμ(εW)1εW∈D≤2c1tΔ2+12lim supε→0ε2supx∈RlogPx(εW∈D).

By the LDP for the Wiener process ([4], Chapter 3, §2), lim supε→0ε2supx∈RlogPx(εW∈D)=−inff∈closure(D)I(f), where I(f)=(1/2)∫0t(fs′)2ds,fis absolutely continuous on[0,t];+∞otherwise. For any trajectory f∈D , there exists n such that f visits at least two sets Oj on the time segment [tn−1,tn] . Therefore, any trajectory f∈closure(D) exhibits an oscillation ≥ρ on this time segment. On the other hand, for an absolutely continuous f, |fu−fv|=|∫uvfs′ds|≤|u−v|1/2(∫0t(fs′)2ds)1/2. This means that, for any f∈closure(D) , I(f)≥ρ2N2t, which yields lim supε→0ε2supx∈RlogExeLtμ(εW)1εW∈D≤2c1tΔ2−ρ2N2t. If in this construction, N was chosen such that N≥(4c1−1)ρ−2t2Δ2, then the latter inequality guarantees the analogue of (16) with D instead of C. This completes the proof of (14).

Exactly the same argument as that used in Section 2.2 provides the lower bound (13). In this section, we prove the upper bound (14) for a finite measure μ and thus complete the proof of the first assertion of the theorem. For finite μ and any χ>0 0$]]>, we can find γ>0 0$]]> and decompose μ=μ0+ν in such a way that μ0 is a finite mixture of δ-measures and N(ν,γ)<χ . Let p,q>1 1$]]> be such that 1/p+1/q=1 . The measure pμ0 has the maximal weight of an atom equal to pΔ . Since we have already proved the required statement for finite mixtures of δ-measures, we have (17) lim supε→0ε2supx∈Rlog(ExeLtpμ0(εW))1/p≤t2pΔ2. On the other hand, we have N(ν,γ)<χ and then by Lemma 2 lim supε→0ε2supx∈Rlog(ExeLtqν(εW))1/q≤c1qtχ2. Hence, by (10), lim supε→0ε2supx∈RlogExeLtμ(εW)≤t2pΔ2+c1qtχ2. Now we can finalize the argument. Fix Δ1>Δ:=maxx∈Rμ({x})2 \varDelta :=\max _{x\in \mathbb{R}}\mu {(\{x\})}^{2}$]]> and choose p,q>1 1$]]> such that 1/p+1/q=1 and pΔ2<Δ12 . Then there exists χ>0 0$]]> small enough such that pΔ2+2c1qtχ2<Δ12. Taking the decomposition μ=μ0+ν that corresponds to this value of χ and applying the previous calculations, we obtain an analogue of the upper bound (14) with Δ replaced by Δ1 . Since Δ1>Δ \varDelta $]]> is arbitrary, the same inequality holds for Δ.

In this section, we prove the second assertion of the theorem. As before, the lower bound can be obtained directly from the case μ=aδz , and hence we concentrate ourselves on the proof of the upper bound (18) lim supε→0ε2logExeLtμ(εW)≤tΔ22,x∈R. We will use an argument similar to that from the previous section and decompose μ into a sum μ=μ0+ν with finite μ0 and ν, which is negligible in a sense. However, such a decomposition relies on the initial value x, and this is the reason why we obtain an individual upper bound (18) instead of the uniform one (14).

Namely, for a given x, we define μ0,ν by restricting μ to [x−R,x+R] and its complement, respectively. Without loss of generality, we assume that for each R, the corresponding ν is nonzero. Since we have already proved the required statement for finite measures, we get (17).

Next, denote M=supx∈Rμ([x−1,x+1]) and observe that N(ν,1)≤M . Then by Lemma 2 with γ=1 and the strong Markov property, for any stopping time τ, the exponential moment of Ltqν(εW) conditioned by Fτ is dominated by 2ec1M2tq2ε−2 . This holds for ε≤εq,1x,R , where we put the indices x,R in order to emphasize that this constant depends on ν, which, in turn, depends on x,R . Since we have assumed that, for any x,R , the respective ν is nonzero, the constants εq,1x,R are strictly positive.

Now we take by τ the first time moment when |εWτ−x|=R . Observe that Ltν(εW) equals 0 on the set {τ>t} t\}$]]> and it is well known that Px(τ<t)≤4Px(εWt>R)≤Ce−tR2ε−2/2. R)\le C{e}^{-t{R}^{2}{\varepsilon }^{-2}/2}.\]]]> Summarizing the previous statements, we get ExeLtqν(εW)≤1+2Cetε−2(c1M2q2−R2/2),ε≤ελ,1x,R, which implies (19) lim supε→0ε2log(ExeLtqν(εW))1/q≤t(c1M2q−R2/(2q))+, where we denote a+=max(a,0) . By (10) inequalities (17) and (19) yield lim supε→0ε2logExeLtμ0(εW)≤t2pΔ2+t(c1M2q−R2/(2q))+.

Now we finalize the argument in the same way as we did in the previous section. Fix Δ1>Δ \varDelta $]]> and take p>1 1$]]> such that pΔ2≤Δ12 . Then take R large enough so that, for the corresponding q, c1M2q−R2/(2q)≤0. Under such a choice, the calculations made before yield (18) with Δ replaced by Δ1 . Since Δ1>Δ \varDelta $]]> is arbitrary, the same inequality holds for Δ.