A one-dimensional diffusion hits points fast

A one-dimensional, continuous, regular, and strong Markov process $X$ with state space $E$ hits any point $z \in E$ fast with positive probability. To wit, if $\tau_z = \inf \{t \geq 0:X_{t} = z\}$, then $P_\xi({ \tau}_z<\varepsilon)>0$ for all $\xi \in E$ and $\varepsilon>0$.


Introduction
Consider a measurable function σ : R → R \ {0} such that 1/σ 2 is locally integrable. Then Engelbert and Schmidt (1981) guarantee the existence of a filtered probability space (Ω, F , F, P), equipped with a Brownian motion W = (W t ) t≥0 , and the existence of a stochastic process Z = (Z t ) t≥0 such that holds. Moreover, Z is strong Markov and continuous. Let now z ∈ R, ε > 0, and τ Z z denote the first hitting time of z by Z. Then we know that P(τ Z z < ∞) > 0. Mijatović (2014) and Karatzas and Ruf (2015) ask whether also P(τ Z z < ε) > 0 holds for all ε > 0. Only a partial answer is provided: If 1/σ 4 is locally integrable (everywhere, apart from countably many points), then the answer is affirmative.
This note answers the question affirmatively in a general setup. To this end, we fix an open interval E of R, and denote its closure by E. We then consider a one-dimensional Markov process X = (X t ) t≥0 with state space E on the filtered space (Ω, F , F), along with a family of probability measures (P ξ ) ξ∈E . We denote the death-time of X by ζ. We assume that X is strong Markov, regular, continuous on [0, ζ), and lim tրζ X t exists and satisfies lim tրζ X t / ∈ E on {ζ < ∞}. We set X ζ+s = lim tրζ X t ∈ E for all s ≥ 0 on {ζ < ∞}. If Y = (Y t ) t≥0 is a stochastic process and ρ a stopping time, We now define the stopping times Since X is regular, we have P ξ (τ z < ∞) > 0 for all ξ, z ∈ E. Throughout the note we shall fix a starting point ξ ∈ E and a target point z ∈ E. We are now able to state the main result of this note.
Remark 2. We now provide some warnings concerning Thereom 1.
• The continuity of X is clearly important in Theorem 1. For instance, the compensated Poisson process with state space E = R is strong Markov and regular, but the assertion of Theorem After we had completed this note, Umut Cetin pointed out to us that Theorem 1 could also be derived from the arguments in Appendix II of Kotani and Watanabe (1982). We feel, however, that the arguments of this note are different and more direct (and cuter :-)).

Proof of Theorem 1
Before proving Theorem 1, we provide some auxiliary results.
Then the first variation of v| [0,t] is bounded by 2t, for each t ≥ 0.
Proof. Note that v can increase by at most t on the interval [0, t]. This, in conjunction with the nonnnegativity of v, then yields that v can drop by at most t as well, and hence the bound of 2t.
Recall that we have fixed a strong Markov process X with state space E and a starting point ξ ∈ E for which the following results are formulated.
Proposition 4. Let v : E → R be a measurable function and assume that the Markov process X is also a continuous P ξ -local martingale. Then the function Since Proposition 4 is the core step of this note's argument we provide three different proofs.
Preparation for the proofs of Proposition 4. Clearly, v being constant on E implies that v(X · ) is of finite first variation; thus it suffices to argue the reverse direction. Hence, from now on, we will assume that v(X · ) is of finite first variation on compact subintervals of [0, ∞). Note that v(X · ) is of finite first variation variation on {ζ < ∞}. If P ξ (ζ = ∞) > 0 let (a n ) n∈N be a strictly decreasing sequence and (b n ) n∈N a strictly increasing sequence such that E = n∈N (a n , b n ) and ξ ∈ (a 1 , b 1 ). Moreover, let Then we have ζ n < ∞ and v(X ζn · ) is of finite first variation for each n ∈ N. Moreover, note that v is constant on E if and only if v is constant on (a n , b n ) for each n ∈ N. Thus, we shall assume, without loss of generality, that v(X · ) is of finite first variation.
Next, observe that the Dambis-Dumbins-Schwarz theorem yields the existence of a Brownian motion B = (B t ) t≥0 with B 0 = ξ, possibly on an extension of the probability space, such that X = B [X] ; see, for instance, Theorem V.1.7 in Revuz and Yor (1999).
The first proof relies on an application of the Itô-Meyer-Tanaka formula.
Proof I of Proposition 4. Proceeding as in Section 5 in Ç inlar et al. (1980) we observe that v is a so called semimartingale function for a Brownian motion killed when hitting the boundary of E and thus, v is locally the difference of two convex functions. More precisely, with (a n ) n∈N and (b n ) n∈N as above, v| [an,bn] is the difference of two convex functions. It then suffices to prove that D − v| [an,bn] = 0, where D − v| [an,bn] denotes its left derivative, for each n ∈ N. To this end, let Then the Itô-Meyer-Tanaka formula yields where A = (A t ) t≥0 is a process of finite first variation. Since v(B ρn · ) is of finite first variation we obtain ·∧ρn 0 (D − v| [an,bn] (B t )) 2 dt = 0, and thus D − v| [an,bn] = 0 for each n ∈ N, as desired.
We remark that Aboulaïch and Stricker (1983) provide a similar proof. The next proof has been suggested by Vilmos Prokaj, to whom we are very grateful. The proof requires the additional assumption that v is of finite first variation and uses local time of Brownian motion.
Proof II of Proposition 4. Let N (x, y) denote the number of upcrossings of [x, y] made by B ρ for all x, y ∈ R with x < y. Moreover, let L ρ (x) denote the local time of B ρ at x ∈ E, fix ε > 0, and pick some sufficiently small δ > 0, possibly depending on ω ∈ Ω, such that for all x ∈ E. Such a δ exists almost surely, thanks to the uniform convergence of Theorem 2 in Chacon et al. (1981). Next, define the sequence (σ k ) k∈N 0 of stopping times inductively by σ 0 = 0 and σ k+1 = ρ ∧ inf{t > σ k : |B t − B σ k | = δ}. Suppose that the first variation Ξ of the function v(B ρ · ) is finite almost surely, directly implying that v is continuous on E. Then we have Letting now δ tend to zero and using the continuity of L ρ , argued in Theorem VI.1.7 in Revuz and Yor (1999), note that where TV(v) denotes the variation of v, which is finite by assumption. Next, letting ε tend to zero, taking expectations, and using Tonelli yields Since each expectation is strictly positive, we obtain that the function v is constant on E.
The third proof follows a pathwise argument and relies less on the one-dimensional character of X. The proof requires the additional assumption that v(ξ) = 0, v is nonnegative, and there exists a P ξ -nullset N such that for all s, t ≥ 0 and ω ∈ Ω \ N we have the upper-Lipschitz condition v (X t+s (ω)) − v (X t (ω)) ≤ s. (1) Proof III of Proposition 4. Again, clearly v is continuous on E. Fix now some ω ∈ Ω such that the function f : [0, ζ(ω)) → R, t → v(X t (ω)) is of finite first variation, (1) holds, and X(ω) has no point of monotonicity (see Theorem 2.9.13 in Karatzas and Shreve (1991)). Then f is continuous and Theorem 3.23(b) in Folland (2013) yields that f has a derivative f ′ almost everywhere. Levy's decomposition theorem, Hahn's decomposition theorem, and Proposition 3.30 in Folland (2013) yield the existence of two nonnegative measures µ − and µ + , both singular with respect to each other and to Lebesgue measure, such that Suppose now that f ′ (t) > 0 for some t > 0. Then we must have f (t + h) − f (t) > 0 for all sufficiently small h ∈ R, but then t is a point of monotonicity of X(ω). This contradicts the choice of ω. Thus f ′ ≤ 0 and we get in the same way that f ′ = 0. Therefore, df = −dµ − + dµ + Since, on intervals, we have df ≤ dt thanks to the upper-Lipschitz condition we get µ + ≤ m+µ − , where m denotes the Lebesgue measure. Thanks to a monotone class argument we also get µ + (D) ≤ m(D) + µ − (D) for all D ∈ B, the Borel sigma algebra of [0, ∞). Thus, µ + is both absolutely continuous and singular with respect to m + µ − , and we get µ + = 0. Finally, since f ≥ 0 and f (0) = 0, we have µ − = 0, and so f is constant.
Lemma 5. Assume that the Markov process X is also a continuous P ξ -local martingale. Then the quadratic variation process [X] is P ξ -almost surely strictly increasing on [0, ζ).
Proof. Proposition III.3.13 and the discussion proceeding it in Revuz and Yor (1999) yield that X cannot be constant on an interval. Proposition IV.1.13 in Revuz and Yor (1999) then yields the statement.
Before stating the next lemma we introduce some notation. Assume that E is of the form E = (a, b) for a, b ∈ R with a < b. For each x ∈ E we now define the deterministic function Note that u x is nonincreasing before x and nondecreasing after x; thus, in particular, the limits in (2) always exist, for each x ∈ E. Moreover, u x is nonnegative, of finite first variation, and satisfies u x (x) = 0, for each x ∈ E. Observe that an equivalent formulation of Theorem 1 is the statement that u ξ is constant.
Lemma 6. Assume that the Markov process X is also a continuous P ξ -local martingale. The function u ξ , given in (2), satisfies the following two claims.
Proof. To start, for all x, w ∈ E, we have the triangle inequality Indeed, this is clear if either one of the two summands equals one. To see the distributional property of (3) otherwise, fix x, w, y ∈ E and assume for the moment that the underlying probability space is the canonical one; see Section I.3 in Revuz and Yor (1999). Then, for each path ω we have the weak inequality τ y (ω) ≤ τ w (ω) + τ y (θ τw(ω) (ω)), where θ denotes the shift operator; that is θ t (ω)(·) = ω(t + ·) for all t ≥ 0, see also the discussion on page 104 in Revuz and Yor (1999). Fix now ε > 0 and t 1 = u x (w) + ε/2 and t 2 = u w (y) + ε/2. Then we have where the equality follows the strong Markov property of X and the last inequality follows from the definition of t 1 and t 2 . This yields directly that u x (y) ≤ t 1 + t 2 = u x (w) + u w (y) + ε. Letting ε tend to zero then gives (3). Claim (i): First, for any w ∈ E, the continuity of u w at w follows from the fact that X is not constant on any interval (see the proof of Lemma 5), in conjunction with the strong Markov property. Let us now study the continuity of u ξ at some y ∈ E. Without loss of generality, we may assume that y > ξ. The right-continuity then follows from (3) and the continuity of u y at y. For the left-continuity of u ξ at y, Section 3.3 in Itô and McKean (1965) or Lemma 4.1, in particular (4.5), in Karatzas and Ruf (2015) also hold for the case of the regular, strong Markov process X, thanks to Lemma 5. Thus, for each ε > 0 there exists w ∈ (ξ, y) such that P w (τ y ≤ ε) > 0. The left-continuity of u ξ at y then follows by another application of (3).
Claim (ii): Assume first that there exists some t ≥ 0 such that P w (u w (X t ) > t) > 0 for some w ∈ E. This then implies that there exists some y ∈ E such that u w (y) > t and P w (X t > y) > 0 if y > ξ and P w (X t < y) > 0 if y < ξ, respectively. This, in conjunction with the continuity of X, however, contradicts the definition of u w in (2). We therefore have P w (u w (X t ) ≤ t) = 1 for all t ≥ 0 and w ∈ E.
We now note that (3) and (4) imply (5). The claim then follows from the continuity of u ξ and X.
We are now ready to prove this note's main result.
Proof of Theorem 1. Thanks to Propositions VII.3.2, VII.3.4, and VII.3.5 in Revuz and Yor (1999) we may assume, without loss of generality, that X is in natural scale and thus a P ξ -local martingale. Next, we recall the function u ξ , given in (2). Now Lemma 3, in conjunction with Lemma 6(ii), yields that the function [0, ∞) ∋ t → u ξ (X t ) has finite first variation on compact subintervals of [0, ∞), P ξ -almost surely. Proposition 4 now implies that u ξ is constant. This yields that u ξ (z) = u ξ (ξ) = 0, and thus, the assertion of the theorem follows.