Isolated zeros for Brownian motion with variable drift

It is well known that standard one-dimensional Brownian motion B(t) has no isolated zeros almost surely. We show that for any alpha<1/2 there are alpha-H\"older continuous functions f for which the process B-f has isolated zeros with positive probability. We also prove that for any continuous function f, the zero set of B-f has Hausdorff dimension at least 1/2 with positive probability, and 1/2 is an upper bound if f is 1/2-H\"older continuous or of bounded variation.


Introduction
Let B be standard one-dimensional Brownian motion and f : I → R a continuous function defined on some interval I ⊂ R + . A standard result is that the zero set of B has no isolated points almost surely, see Theorem 2.28 in [15]. By the Cameron-Martin theorem (see Theorem 1.38 in [15] or Theorem 2.2 in Chapter 8 in [16]) the zero set of the process B − f has no isolated points almost surely if f is in the Cameron-Martin space D(I) (integrals of functions in L 2 (I)). We will prove that the same is true for any function f which is 1/2-Hölder continuous. Since all functions in D(I) are 1/2-Hölder continuous, this is a stronger statement than the one implied by the Cameron-Martin theorem. For any function g defined on some subset (or the whole) of R + denote by Z(g) the set of zeros of g in (0, ∞). We remove the origin from consideration since the origin is an isolated zero of the process B − f for any f growing fast enough in the neighborhood of the origin, say f (t) > t 1/3 . Proposition 1.1. For f : R + → R which is 1/2-Hölder continuous on compact intervals, the set Z(B − f ) has no isolated points almost surely.
The condition that f is 1/2-Hölder continuous is sharp in the following sense.   [1,2] (functions f γ,n from the construction in Section 3) for γ = 0.4 and n = 1, 2, 5. Approximations of the Cantor set (sets C γ,n from the construction in Section 3) are drawn in bold. Theorem 1.2 will follow directly from Proposition 3.1. An example of function f satisfying Theorem 1.2 is given in Section 3. For γ < 1/2, let C γ denote the middle (1 − 2γ)-Cantor set and let f γ be the corresponding Cantor function, shifted to an interval away from the origin; see Figure 1 and Section 3 for a precise definition. For γ < 1/4, the set Z(B − f γ ) has isolated zeros with positive probability, and all such zeros are contained in the Cantor set C γ . The proof of this claim consists of constructing a subset of C γ which contains zeros of B − f γ with positive probability, and in which any zero is isolated. En route we obtain the following result of independent interest. The case γ = 1/4 of the above theorem has already been resolved by Taylor and Watson (see Example 3 in [17]). Their interest in the graph of the restriction f γ | Cγ stemmed from the fact that, although the projection of this set on the vertical axis is an interval, the graph of Brownian motion does not intersect this set almost surely.
Part (ii) of Proposition 2.2 shows that isolated zeros of the process B − f can occur only where the function f increases or decreases very quickly. In the following theorem we bound the Hausdorff dimension of such sets. The Hausdorff dimension of a set A ⊂ R + will be denoted dim(A). It is a classical result that the zero set of Brownian motion has Hausdorff dimension 1/2 almost surely, see Theorem 4.24 in [15]. Of course, for any compact interval I not containing 0 and any continuous function f : I → R the event {Z(B − f ) = ∅} will have a non-zero probability, and it is easy to construct a function f : R + → R with the same property. However, we prove that adding a continuous drift can not decrease the Hausdorff dimension of the zero set almost surely. This is the content of the following theorem. Theorem 1.5. For continuous function f : R + → R, the set Z(B − f ) has Hausdorff dimension greater than or equal to 1/2 with positive probability.
As the following example shows, upper bounds on the Hausdorff dimension of the zero set can not be obtained without additional assumptions on the drift f . Recall that fractional Brownian motion B (H) : R + → R with Hurst index 0 < H < 1 is a continuous, centered Gaussian process, such Taking the drift f to be an independent sample of fractional Brownian motion with Hurst index H, one gets that the Hausdorff dimension dim(Z(B − f )) is bounded from below by 1 − H, almost surely. This follows from the proof of the same fact for unperturbed fractional Brownian motion in Theorem 4 in Chapter 18 of [9] (see also Proposition 5.1 and Section 7.2 in [2]). Theorem 1.6. Let f : R + → R be either 1/2-Hölder continuous on compact intervals or of bounded variation on compact intervals. Then the Hausdorff dimension of Z(B − f ) is at most 1/2, almost surely.
Arguments from the proof of Theorem 1.5 and Theorem 1.6 imply the following corollary. Remark 1.8. For a function g : R + → R, define M g (t) = max 0≤s≤t g(s) and denote the set of its record times by Rec(g) = {t > 0 : M g (t) = g(t)}. For standard Brownian motion B, a result of Lévy says that the processes (M B (t) − B(t)) t and (|B(t)|) t have the same distribution (see e.g. Theorem 2.34 in [15]), which implies that sets Z(B) and Rec(B) have the same distribution (on the Borel sigma algebra of families of closed subsets of R + , generated by the Hausdorff metric). In general, for Brownian motion with drift there is no such correspondence. Actually, one can see that there are no isolated points in the set of record times of the process B − f almost surely. This is proven in part (ii) of Proposition 3.4.
1.1. Related results. Connection between Hölder continuity of drift f and path properties of B − f has already been observed. For d ≥ 2, a function f : R + → R d is called polar if, for d-dimensional Brownian motion B started at the origin and any point x ∈ R d , the probability that there is a t > 0 such that B(t) − f (t) = x is positive. In [7] Graversen constructed α-Hölder continuous functions which are polar for two dimensional Brownian motion, for α < 1/2. In [11] Le Gall showed that 1/2-Hölder continuous functions are not polar for two dimensional Brownian motion and that the same conclusion holds in higher dimensions when f satisfies a slightly stronger condition than 1/d-Hölder continuity. In a recent paper [1] it was shown that for any α < 1/d there are α-Hölder continuous functions f such that the image of B − f covers an open set almost surely.
We will briefly review some related results on intersections of Brownian trajectories with non-smooth paths. Let S α be the family of all functions f : [0, 1] → R such that sup 0≤t≤1 |f (t)| ≤ 1 and sup 0≤s,t≤1 |f (s) − f (t)|/|s − t| α ≤ 1. Define the local time of Brownian motion B on f by the formula It was proved in [3,4] that the supremum over f ∈ S α of L f 1 is finite for α > 5/6 and infinite for α < 1/2. The function (f, t) → L f t is continuous over S α × [0, 1] for α > 5/6. Suppose that g : R + → R is a continuous function and let X be Brownian motion reflected on g; see [5] for a precise definition. Let A g be the set of all t > 0 such that P(X(t) = g(t)) > 0. It was proved in [5] that for every continuous function g we have dim(A g ) ≤ 1/2 and for some continuous functions g we have dim(A g ) = 1/2.

Isolated zeros -general results
For an interval I we denote its length by |I| and say it is dyadic if it is of the form I = [k2 m , (k + 1)2 m ] for integers k > 0 and m. For intervals I and J, we will write I < J if J is located to the right of I.
Remark 2.1. We will repeatedly use the following simple observations.
(i) Suppose that F k , k ≥ 1, are events and for some p > 0 and all k we have P(F k ) ≥ p. Recall that lim sup k F k = n≥1 k≥n F k is the event that infinitely many F k 's occur. Then P(lim sup k F k ) ≥ p.
(ii) As an easy consequence of the Cauchy-Schwarz inequality we have P(Z > 0)E(Z 2 ) ≥ (EZ) 2 , for any nonegative variable Z. See Lemma 3.23 in [15]. Proposition 2.2. Let f : R + → R be a continuous function.
(i) Let A ⊂ R + be a set such that for any t ∈ A there is an α < 1/2 such that lim inf s→t Then, almost surely any point in Proof. However, this is impossible since, by Levy's modulus of continuity, almost surely, there exists an h 1 > 0 such that for all h ∈ (0, h 1 ) and all 0 ≤ t ≤ N we have |B(t + h) − B(t)| ≤ 3 h log(1/h), see e. g. Theorem 1.14 in [15]. If A is unbounded, apply the above reasoning to A N = A ∩ [0, N ] and let N go to infinity.
(ii) First define τ q = min {t ≥ q : B(t) = f (t)} and notice that any isolated zero of the process B − f must equal τ q , for some q ∈ Q. This is because, for any zero s ∈ Z(B − f ), not of the form τ q , and a sequence of rational numbers (q n ) converging to s from below, we have lim n τ qn = s. Therefore, it is enough to prove that for each q ∈ Q + , the event that τ q / ∈ A + f ∪ A − f and that τ q is isolated in the set Z(B − f ), has probability zero.
Fix a positive integer M and define sequences of functions Since f is continuous, it is easy to see that for each n, the functions s + n and s − n are measurable. Also define For all t ∈ A f (M ) it holds that s + n (t) > 0 and s − n (t) > 0, for all n. Since τ q is a stopping time, the process B q (t) = B(τ q + t) − B(τ q ), is, by the strong Markov property, a Brownian motion independent of the sigma algebra F τq .
Let F − denote the event that B q (s − n (τ q )) ≥ M s − n (τ q ) happens for infinitely many n's. Since the random variables s − n (τ q ) are measurable with respect to F τq , Blumenthal's 0-1 law implies that P(F − | F τq ) is equal to 0 or 1 on the event {τ q ∈ A f (M )}. On the event {τ q ∈ A f (M )}, for every n, we have Since the right hand side does not depend on n, by Remark 2.
happens for infinitely many n's then P(F + | F τq ) = 1 on the event {τ q ∈ A f (M )}. By the definition of the sequences (s − n (t)) and (s + n (t)), if F − ∪ F + holds then τ q is not an isolated zero from the right. Therefore, the probability that τ q ∈ A f (M ) and that τ q is an isolated point of Z(B − f ) is equal to zero. Taking the union over all rational q's and observing that ( Every point t for which the limits are equal to ∞ or −∞ is a strict local minimum or maximum. Since a function can have only countable many strict local extrema, almost surely all isolated points of Z(B − f ) are contained in the set In particular, all isolated points of Z(B −f ) are contained in the set of points of increase or decrease of f (recall that t is a point of increase of f if for some and points of decrease are defined analogously). Therefore, if f is a function with at most countable many points of increase or decrease, then Z(B − f ) has no isolated points almost surely. Examples of such functions include functions constructed by Loud in [12] which satisfy a certain local reverse Hölder property at each point (see also the construction in [13]). These functions are defined as g(t) = ∞ k=1 g k (t) where g k (t) = 2 −2Aαk g 0 (2 2Ak t), for 0 < α < 1, a positive integer A such that 2A(1−α) > 1, and a continuous function g 0 which has value 0 at even integers, value 1 at odd integers and is linear at all other points. To prove that these functions have at most countably many points of increase or decrease, we proceed analogously as in the proof of the lower bound in [12]. We will take t / ∈ Q and show that t is not a point of increase. First observe that ⌊2 2Am t⌋ is odd for infinitely many integers m. For such an integer m assume that g m (t) ≥ 2 −2Aαm−1 and denote t m = t + 2 −2A(m+1) . Then by construction g Using the fact that 2A(1 − α) > 1, it is easy to check that the right hand side above is negative, and since t m > t can be arbitrarily close to t the claim follows. If g m (t) < 2 −2Aαm−1 then define t m = t − 2 −2A(m+1) , which now satisfies g m (t m ) = g m (t)+2 −2A(αm+1) and proceed analogously to prove that g(t m ) > g(t).
Proof of Proposition 1.1. This is straightforward from part (ii) of Proposition 2.2.

Isolated zeros
For 0 < γ < 1/2, we will define the middle (1 − 2γ)-Cantor set and denote it by C γ . Take a closed interval I of length |I|. Define C γ,1 as the set consisting of two disjoint closed subintervals of I of length γ|I|, the left one (for which the left endpoint coincides with the left endpoint of I) and the right one (for which the right endpoint coincides with the right endpoint of I). Continue recursively, if J ∈ C γ,n , then include in the set C γ,n+1 its left and right closed subintervals of length γ n+1 |I|. Define the set C γ,n as the union of all the intervals from C γ,n . For any n, the family C γ,n is the set of all connected components of the set C γ,n . The Cantor set is a compact set defined as C γ = ∞ n=1 C γ,n . It is easy to show that dim(C γ ) = log 2/ log(1/γ). Now we recall the construction of the standard Cantor function. Define the function f γ,1 so that it has values 0 and 1 at the left and the right endpoint of the interval I, respectively, value 1/2 on I\C γ,1 and interpolate linearly on the intervals in C γ,1 . Recursively, construct the function f γ,n+1 so that for every interval J = [s, t] ∈ C γ,n , the function f γ,n+1 agrees with f γ,n at s and t, it has value (f γ,n (s) + f γ,n (t))/2 on J\C γ,n+1 and interpolate linearly on the intervals in C γ,n+1 . See Figure 1. It is easy to see that the sequence of functions (f γ,n ) converges uniformly on I. We define the Cantor function f γ as the limit f γ = lim n f γ,n . Note that for any n and all m ≤ n the functions f γ and f γ,n agree at the endpoints of intervals J ∈ C γ,m .
Another way to characterize the Cantor set C γ and the Cantor function f γ is by representing it as fixed points of certain transformations. Define linear bijections g γ : [0, γ] → [0, 1] and h γ : , and the corresponding Cantor function is the unique continuous function that satisfies It is not hard to see that f γ is log 2/ log(1/γ)-Hölder continuous. The value γ = 1/4 is the threshold at which the functions f γ become 1/2-Hölder continuous. For γ < 1/4 the function f = f γ will give an example for Theorem 1.2. This threshold is sharp, since by Proposition 1.1, for γ ≥ 1/4 there are no isolated points in the zero set Z(B − f γ ) almost surely. For simplicity, we will assume the initial interval I to be [1,2], but we note that the analysis works for all compact intervals. To satisfy the assumptions of Theorem 1.2, the function f γ should of course be extended to R + \ [1,2], by, say, value 0 on [0, 1) and value 1 on (2, ∞).
Proof of Theorem 1.3. For an interval I = [r, s] ∈ C γ.n , define Z n (I) as the event B(s) ∈ [f γ (r), f γ (s)], and the random variable Z γ,n = I∈Cγ,n 1(Z n (I)), where 1(Z n (I)) is the indicator function of the event Z n (I). Note that there is a constant c 1 > 0, such that for any 0 < γ < 1/2 we have If Z γ,n > 0 happens for infinitely many n's, then we can find a sequence of intervals Since s k ∈ C γ,n k , the sequence (s k ) will have a subsequence converging to some s ∈ C γ , which obviously satisfies B(s) = f γ (s). Therefore To estimate the probabilities P(Z γ,n > 0) from below we will use the inequality from Remark 2.1 (ii), for which we need to bound the second moment E(Z 2 γ,n ) from above. First we express it as Now fix n and intervals I = [s 1 , t 1 ] and J = [s 2 , t 2 ] from C γ,n , so that I < J, and denote a i = f γ (s i ) and b i = f γ (t i ), for i = 1, 2. Let B be the process which is, by the Markov property and Brownian scaling, again a Brownian motion, independent of F t 1 , and thus independent of the event Z n (I) ∈ F t 1 . The event Z n (J) happens when Case γ < 1/4: Fix intervals I 0 and J 0 in C γ,ℓ+1 , which are contained in a single interval in C γ,ℓ . Label the intervals from C γ,n contained in I 0 by I 1 , . . . , I 2 n−ℓ−1 , and those contained in J 0 by J 1 , . . . , J 2 n−ℓ−1 , so that I i+1 < I i and J j < J j+1 . Set I = I i and J = J j for some 1 ≤ i, j ≤ 2 n−ℓ−1 , and define a i , b i , B and J as before. Conditional on Z n (I), the left endpoint of the interval J is at least (a 2 − b 1 )(t 2 − t 1 ) −1/2 , and since which, by summing over 1 ≤ i, j ≤ 2 n−ℓ−1 gives Here we used the trivial bound for i = j = 1. The sum on the right hand side can be written as is a bounded function on R + , (5) implies that for any fixed I 0 and J 0 as above for some c 2 > 0. Therefore, summing the inequality in (6) over all I 0 and J 0 and ℓ = 0, . . . , n − 1, and using it together with (4) and (2), we have which is bounded since 2 √ γ < 1.
Thus we have shown that, for a fixed γ < 1/4, the second moments E(Z 2 γ,n ) are bounded from above. Now the lower bound in the second inequality in (2) and the inequality from Remark 2.1 (ii) imply that P(Z γ,n > 0) ≥ E(Z γ,n ) 2 /E(Z 2 γ,n ) is bounded from below and the claim follows from (3) and Remark 2.1 (i).
Case γ > 1/4: Again pick I, J ∈ C γ,n such that [s 1 , t 1 ] = I < J = [s 2 , t 2 ] and define a i , b i , B and J as before. By ℓ denote the largest integer such that both I and J are contained in a single interval from C γ,ℓ . Assume that Z n (I) happens. Clearly the endpoints of the interval J satisfy The sequence ((2 √ γ) −ℓ ) is bounded, and therefore the interval J is contained in a compact interval, which does not depend on the choice of n, ℓ, I or J.
Using this and the fact that the length of J is bounded from above by (1 − 2γ) −1/2 2 −n γ −ℓ/2 and from below by c ′ 2 −n γ −ℓ/2 , we get that for some positive constants c 3 and c 4 we have Note that, since the sequence ((2 √ γ) −ℓ ) is bounded for γ = 1/4, estimates (7) also hold for γ = 1/4. Substituting (7) and the upper bounds from (2) into (4), and summing over all intervals I and J, we obtain Since 2 √ γ > 1, we have bounded E(Z 2 γ,n ) from above by a constant not depending on n, and the claim follows as in the case γ < 1/4.
Case γ = 1/4: Assume that Z(B − f γ ) ∩ C γ = ∅ and define τ as the first zero of B − f γ in the Cantor set C γ (τ exists since Z(B − f γ ) ∩ C γ is a closed set). For an interval I = [s, t] ∈ C γ,n assume that τ ∈ I. Since τ is a stopping time, and by Brownian scaling, the conditional probability P Z n (I) | F τ , τ ∈ I is equal to the probability that Brownian motion at time 1 is between we see that y 1 ≤ 0 and y 2 ≥ 0. Moreover, the assumption γ = 1/4 implies that t − τ ≤ 4 −n = (f γ (t) − f γ (s)) 2 which leads to y 2 − y 1 ≥ 1. Thus we can bound the probability for some K > 0, see also Figure 3. Therefore, P(Z γ,n > 0 | τ ∈ I) ≥ K −1 and, since the events {τ ∈ I} are disjoint for different I ∈ C γ,n , we have P(Z γ,n > 0 | Z(B − f ) ∩ C γ = ∅) ≥ Figure 3. If γ ≤ 1/4, conditional on the event that there is a zero of B − f γ in an interval I ∈ C γ,n (τ is the first such zero) the probability of the event Z n (I) (Brownian motion intersecting the right hand side of the rectangle) is bounded from below. K −1 for every n. Thus we obtain The arguments leading to (8) are true for all γ ≤ 1/4 and therefore (9) holds for γ ≤ 1/4. For I = [s, t] ∈ C γ,n and 0 ≤ ℓ < n, let I ℓ denote the interval from C γ,ℓ that contains I, and let I ℓ 1 , I ℓ 2 ∈ C γ,ℓ+1 be the left and right subintervals of I ℓ , respectively. Let s = 1.a 1 a 2 . . . a n denote the 4-ary expansion of the left endpoint of I (note that the 4-ary expansion of s contains only the digits 0 and 3 and has length at most n, here we add zeros at the end, if necessary, to make it of length n). It is easy to see that I ⊂ I ℓ 1 if a ℓ+1 = 0 and I ⊂ I ℓ 2 if a ℓ+1 = 3. Call an interval I balanced if the sequence a 1 , . . . , a n contains at least n/3 zeros and otherwise unbalanced. For a balanced interval I = [s, t] ∈ C γ,n let A I denote the event that I is the leftmost balanced interval for which Z n (I) happens, and, as before, let 1.a 1 . . . a n denote the 4-ary expansion of s. Assume that for some 0 ≤ ℓ < n we have a ℓ+1 = 0 and pick an interval J ∈ C γ,n such that J ⊂ I ℓ 2 . Since A I ∈ F t 1 we can use the same arguments that lead to the lower bound in (7) to conclude that the probability P(Z n (J) | A I ) ≥ c 5 2 −(n−ℓ) , for some constant c 5 > 0. Summing over all J ⊂ I ℓ 2 and over all ℓ such that a ℓ+1 = 0 gives E(Z γ,n | A I ) ≥ c 5 |{1 ≤ ℓ ≤ n : a ℓ = 0}| ≥ c 5 n/3. By (2) and since events A I are disjoint, we have P(Z n (I) for some balanced interval I) To estimate the probability that Z n (I) happens for some unbalanced interval I notice that the number of such intervals is bounded from above by e −c 6 n 2 n for some c 6 > 0. By (2) this gives P(Z n (I) for some unbalanced interval I) ≤ e −c 6 n .
To prove this, fix an interval J and take the largest integer n such that |J| ≤ 2 −n . Notice that J can be covered by no more than two consecutive binary intervals J 1 and J 2 of length 2 −n . Moreover, there are consecutive I 1 , I 2 ∈ C γ,n such that f γ (I i ) = J i for i = 1, 2, and C γ ∩ f −1 γ (J) ⊂ I 1 ∪ I 2 , see Figure 4. Now using the notation from the proof of Theorem 1.3 and the arguments that lead to (8) we obtain P(Z n (I i ) | Z(C γ ∩ I i )) ≥ K −1 which yields P(Z(C γ ∩ I i )) ≤ KP(Z n (I i )). (13) But by the first inequality in (2), the probability on the right hand side is bounded from above by 2 −n . Using this fact in (13) and summing the expression for i = 1, 2, we obtain (12).
By Theorem 1.3 the set Z(B−f γ )∩C γ is non-empty with some probability p > 0. Take an arbitrary γ < γ 1 < 1/4 and n 0 such that n≥n 0 (2 √ γ 1 ) n ≤ p/(2c 1 ). For n ≥ n 0 consider the interval J k,n = [k2 −n −γ n/2 1 /2, k2 −n +γ n/2 1 /2] and define the set M n 0 = n≥n 0 0≤k≤2 n J k,n . By (12) and the choice of n 0 , we have that P(Z(C γ ∩f −1 γ (M n 0 ))) ≤ p/2. Therefore, the probability that there is a zero of B(t) − f γ (t) in the set C γ ∩ Int(C γ,n 0 )\f −1 γ (M n 0 ) is at least p/2 (here Int(C γ,n 0 ) is the interior of the set C γ,n 0 ). Now the claim will be proven if we show that any such zero is isolated. Take t ∈ C γ ∩Int(C γ,n 0 )\f −1 γ (M n 0 ) and any s = t in the same connected component of Int(C γ,n 0 ). The largest integer ℓ such that both s and t are contained in the same interval of C γ,ℓ  is also equal to p, for any n ≥ 1. Therefore, in view of Remark 2.1 (i), the probability that there is an isolated point of Z(B − f ) in the interval [4 −n−1 , 4 −n ], for infinitely many n's is bounded from below by p. By Blumenthal's zero-one law this event has probability one, which proves the claim.
The following proposition shows that, with positive probability, the set Z(B − f ) can have only isolated points, or even only one point. Proof. Take f γ to be the Cantor function with γ < 1/4 defined on the interval [1,2]. Since Z(B − f γ ) has isolated points with positive probability, there are two rational numbers q 1 < q 2 such that, with positive probability, there will be only one zero of B − f γ in the interval (q 1 , q 2 ). Denote this event by D and on this event the unique zero by τ . Note that on the event D ∩ {(B(q 1 ) − f γ (q 1 ))(B(q 2 ) − f γ (q 2 )) > 0} the unique zero τ is necessarily a local extremum of the process B − f γ and by part (i) of the upcoming Proposition 3.4 this event has probability zero. Furthermore, on the event D ∩ {B(q 1 ) < f γ (q 1 )} ∩ {B(q 2 ) > f γ (q 2 )} the unique zero τ is necessarily a local point of increase of the Brownian motion. By a result of Dvoretzky, Erdös and Kakutani in [6], almost surely, Brownian motion has no points of increase and thus P(D, B(q 1 ) < f γ (q 1 ), B(q 2 ) > f γ (q 2 )) = 0, see also Theorem 5.14 in [15]. Next define S = {y > f γ (q 1 ) : P(D | B(q 1 ) = y) > 0}, and notice that by the Markov property and the discussion above P(B(q 1 ) ∈ S) > 0. This implies that the set S is of positive Lebesgue measure and so is S 1 = S ∩ (f γ (q 1 ) + ǫ, ∞), for ǫ small enough. Now the claim will follow if we prove that there is a modification f of the function f γ on the intervals (0, q 1 ) and (q 2 , ∞), such that 1) with positive probability there are no zeros of B − f in (0, q 1 ) and B(q 1 ) ∈ S 1 , 2) for any y < f γ (q 2 ) − ǫ, conditional on B(q 2 ) = y, with positive probability there are no zeros of B − f in (q 2 , ∞).
For 1) define f to be linear on [0, q 1 ] with f (0) = −ǫ and f (q 1 ) = f γ (q 1 ), if we do not require f (0) = 0. To prove that the probability that both While this is intuitively obvious it can be proven by using the reflection principle at the first time the Brownian motion hits the level −ǫ to conclude that the probability that both min 0<t<q 1 B(t) ≤ −ǫ and B(q 1 ) ∈ S 2 happen is equal to the probability that B(q 1 ) ∈ S 3 , where S 3 is obtained by reflecting the set S 2 around −ǫ. Since S 2 ⊂ R + we have P(B(q 1 ) ∈ S 3 ) < P(B(q 1 ) ∈ S 2 ) and therefore P min Now the probability that there is a unique zero in the interval (0, q 2 ) is equal to some p > 0. If we do require f (0) = 0, redefine f on the interval (0, δ) as f (t) = −ct 1/3 . Here c > 0 is chosen large enough so that for δ > 0, for which f is continuous, we have P(Z(B − f ) ∩ (0, δ)) < p.
To satisfy the condition 2), replace f γ on (q 2 , ∞) by a linear function of slope 1, that is f (q 2 + t) − f γ (q 2 ) = t. To prove that this f satisfies the required condition on (q 2 , ∞) it is enough to prove that, for standard Brownian motion B, with positive probability there are no zeros of the process B(t) − t − ǫ. This probability is equal to 1 − e −2ǫ by (5.13), Sect. 3.5 in [10]. See Figure 6.  Proof. (i) Take an interval [q 1 , q 2 ] with q 1 > 0 and let M be the maximum of the process X on this interval. Then, since the process X has independent increments, X(q 1 ) and M − X(q 1 ) are independent. Since X(q 1 ) has a continuous distribution, so has M = (M − X(q 1 )) + X(q 1 ), and therefore P(M = x) = 0 for any x ∈ R. Taking x = 0 and a union over all rational q 1 < q 2 proves the claim for local maxima. Similarly the statement holds for local minima. See Figure 7.
(ii) For any continuous function g, any record time s ∈ Rec(g) is a maximum of g on the interval [s − ǫ, s], for every ǫ > 0. Let s > 0 be an isolated point in Rec(g). Then s is a local maximum, because otherwise we would have record times to the right of s, arbitrarily close to s. Since g is continuous and there are no record times in the interval (s − ǫ, s) for some ǫ > 0, there has to be an r ∈ Rec(g) ∪ {0}, which is also a local maximum and such that r < s and g(r) = g(s). Applying these observations to g(t) = X(t) = B(t) − f (t), we see that, in order to prove the claim, it is enough to show that the process X = B − f does not have two equal local maxima almost surely. See Figure 8. This is well known for standard Brownian motion and can be proven in the same way for the process X. Namely, for two intervals [q 1 , r 1 ] and [q 2 , r 2 ], with r 1 < q 2 , define the random variables Y 1 = X(r 1 ) − max q 1 ≤t≤r 1 X(t) and Y 3 = max q 2 ≤t≤r 2 X(t) − X(q 2 ), and let Y 2 = X(q 2 ) − X(r 1 ). Clearly these three random variables are independent. Since Y 2 is a continuous random variable, so is Y 1 + Y 2 + Y 3 and P(Y 1 + Y 2 + Y 3 = 0) = 0. Therefore, almost surely the maxima on [q 1 , r 1 ] Figure 8. If X(t) = B(t) − f (t) has an isolated record time at time s then there exists a record time r < s for which X(r) = X(s). In this case both record times r and s have to be local maxima. This is an event of probability zero. and [q 2 , r 2 ] are different. Taking the union over all possible rational q 1 , r 1 , q 2 and r 2 as above proves the claim.

On Hausdorff dimension of zero sets
In this Section we prove Theorems 1.4, 1.5 and 1.6. Recall that an interval I is called dyadic if it is of the form I = [k2 m , (k + 1)2 m ] for some integers k > 0 and m. For a dyadic interval I, the set of its subintervals which are dyadic and of length 2 −n |I|, will be denoted by G n (I). For intervals I and J we will write I < J if J is located to the right of I.
Proof of Theorem 1.4. Recall the definition of sets A + f and A − f from part (ii) of Proposition 2.2. It is enough to prove that both of these sets have Hausdorff dimension at most 1/2. This follows from the case α = 1/2 of part (i) of the following lemma, which estimates the Hausdorff dimension of larger sets.
have Hausdorff dimension at most α. (ii) Assume f : R + → R has bounded variation on compact sets. For any 0 < α < 1 the set has Hausdorff dimension at most α.
Note that for any positive integer k, and any k points t 1 < t 2 < · · · < t k from B + f,n such that t k ≤ t 1 + 2 −n , we have Therefore, for any dyadic interval I ⊂ R + of length 2 −n there is a constant c 1 such that, for any k points t 1 < · · · < t k from I ∩ B + f,n , we have Now for any positive integer m and any dyadic interval J ∈ G m (I) such that ∅} is a cover of the set B + f,n ∩ I with intervals of diameter at most 2 −n−m . From (14) it is easy to see that Since m was arbitrary, by the definition of Hausdorff dimension we obtain dim(B + f,n ∩ I) ≤ α, for any dyadic interval I of length 2 −n . Therefore dim(B + f,n ) ≤ α and taking the union over all n gives dim(B + f ) ≤ α. (ii) It is enough to prove that for any c > 0 the set has Hausdorff dimension bounded by α from above. Representing B f as the union B f = ∞ n=1 B f (1/n) will prove the claim. Fix a dyadic interval I and an ǫ > 0. For every t ∈ B f (c) ∩ I we can find 0 < h t < ǫ such that Clearly the family of intervals {I t : t ∈ B f (c) ∩ I} is a cover of the set B f (c) ∩ I. By Besicovitch's covering theorem (see Theorem 2.6 in [14]) we can find an integer k, not depending on ǫ, and a subcover which can be represented as a union of k at most countable disjoint subfamilies {I t : t ∈ S i }. More precisely, there are sets Clearly for any i we have where M is the total variation of f on the interval I. The last inequality follows form the fact that the intervals I t , t ∈ S i are disjoint. Now summing the above inequality over Since {I t : t ∈ S i , 1 ≤ i ≤ k} is a cover of B f (c) ∩ I of diameter at most 2ǫ and neither k nor M depend on ǫ, the α-dimensional Hausdorff measure of B f (c) ∩ I is finite. Therefore dim(B f (c) ∩ I) ≤ α and, since the dyadic interval I is arbitrary, the claim follows.
The proof of Theorem 1.5 will be an application of the percolation method due to Hawkes [8], which we now describe. See Chapter 9 in [15] for more on this method. Fix a dyadic interval I, a real number 0 < β < 1 and set p = 2 −β (the construction and the results can be stated for any interval I). Construct the set S β (1) by including in it (as subsets) each of the two dyadic intervals from G 1 (I) with probability p, independently of each other. To construct S β (m + 1), for each dyadic interval J ∈ G m (I) such that J ⊂ S β (m), include each of its dyadic subintervals from G m+1 (I) in S β (m + 1) with probability p, independently of each other. We obtain a decreasing sequence of compact sets (S β (m)) whose intersection we denote by Γ [β]. Comparing this to the Galton-Watson process with binomial offspring distribution B(2, p), we see that P(Γ[β] = ∅) > 0. The following theorem is due to Hawkes (Theorem 6 in [8]). In the above construction of the percolation set we can change the retention probabilities of intervals at each level. If p n = 2 −βn is a retention probability at level n, we denote the union of intervals kept at level m by S (βn) (m), and the limiting percolation set by Γ[(β n )]. We will use the following result which is an easy corollary of Hawkes' theorem. since we can couple two percolation processes so that for m ≥ m 0 , if an interval J ∈ G m (I) is retained in the percolation process with retention probabilities (2 −βn ), it is also retained in the process with retention probability 2 −α . Since P(S α (m 0 ) = B) > 0 we get P(A ∩ Γ[α] = ∅) > 0 and, by Hawkes' theorem, dim(A) ≥ α. Since α < β was arbitrary the claim follows.
Proof of Theorem 1.5. It is enough to prove that the Hausdorff dimension of Z(B − f ) ∩ [1,2] is greater than or equal to 1/2 with positive probability. Let (β n ) be a sequence converging to 1/2 from below, to be chosen later. Consider the percolation process on the interval [1,2] with retention probabilities (2 −βn ), independent of Brownian motion. Fix a positive integer m. For a dyadic interval I ∈ G m ( [1,2]) denote by t I its center and, for a fixed ǫ > 0, consider the event Define Y m,ǫ = I∈Gm([1,2]) 1(F m,ǫ (I)), where 1(F m,ǫ (I)) is the indicator function of the event F m,ǫ (I). Using trivial bounds on the transition density of Brownian motion, the first moment of Y m,ǫ can be estimated simply by for some constant c 1 depending only on max t∈ [1,2] |f (t)|, and where γ m = β 1 + · · · + β m . In the same way, for I < J we can estimate the conditional probability For I, J ∈ G m ([1, 2]) such that I < J let ℓ be the largest integer so that both I and J are contained in a single interval from G ℓ ( [1,2]). In other words there are consecutive intervals I 0 , J 0 ∈ G ℓ+1 ([1, 2]), contained in a single interval from G ℓ ( [1,2]), such that I ⊂ I 0 and J ⊂ J 0 . Then P(I ∪ J ⊂ S (βn) (m)) = 2 −2γm+γ ℓ .
Proof of Corollary 1.7. Assume that |f (t) − f (s)| ≤ C |t − s| for all 0 < s, t < 1. For every positive integer n define a continuous function f n : [1,2] → R as f n (t) = 2 n/2 f (t2 −n ), for all t ∈ [1,2]. Clearly max t∈ [1,2] |f n (t)| ≤ C √ 2 and, by Remark 4.4 and Brownian scaling, there is a constant c > 0 such that Thus by Blumenthal's zero-one law, almost surely, there are infinitely many integers n such that the Hausdorff dimension of the set Z(B−f )∩[2 −n , 2 −n+1 ] is greater or equal than 1/2. The upper bound follows from Theorem 1.6.