Fine asymptotics for the consistent maximal displacement of branching Brownian motion

It is well-known that the maximal particle in a branching Brownian motion sits near $\sqrt2 t - \frac{3}{2\sqrt2}\log t$ at time $t$. One may then ask about the paths of particles near the frontier: how close can they stay to this critical curve? Two different approaches to this question have been developed. We improve upon the best-known bounds in each case, revealing new qualitative features including marked differences between the two approaches.


Introduction
A standard branching Brownian motion (BBM) begins with one particle at the origin. This particle moves as a Brownian motion, until an independent exponentially distributed time of parameter 1, at which point it is instantaneously replaced by two new particles. These particles independently repeat the stochastic behaviour of their parent relative to their start position, each moving like a Brownian motion and splitting into two at an independent exponentially distributed time of parameter 1.
Let N (t) be the set of all particles alive at time t, and for a particle v ∈ N (t) let X v (s) represent its position at time s ≤ t (or if v was not yet alive at time s, then the position of the unique ancestor of v that was alive at time s). If we define as t → ∞. This quantity (or rather, again, its analogue for branching random walks) was studied by Fang and Zeitouni [6] and by Faraud, Hu and Shi [7], who showed that there is a critical value a c = 3 1/3 π 2/3 2 −1/2 such that almost surely lim t→∞ Λ(t) t 1/3 = a c .
To summarise, the two approaches to the question give similar results: in each case there appears to be a critical line on the t 1/3 scale above which particles cannot remain.
We shall see, however, that if one peers more closely then the two situations are really quite different. Our first result is that not only is ν(f Ac ) > 0 (which was previously unknown), but in fact particles may stay far above the curve f Ac . Secondly, we are able to give much finer asymptotics for Λ(t), both in distribution and almost surely. These developments are redolent of the results of Bramson [3] and Hu and Shi [10]. The proofs have certain elements in common with those found in [16], but are decidedly more involved, and we must develop several new techniques along the way. We now state our three main theorems, which make precise the discussion above.
Theorem 1. Let A c = 3 4/3 π 2/3 2 −7/6 . Define g : [0, ∞) → R by setting Then P(∀t ≥ 0, ∃v ∈ N (t) : X v (u) > g(u) ∀u ≤ t) > 0. Remarks. Initially we proved a weaker form of Theorem 2, and as a result gained a weaker statement in place of Theorem 3. (Note also that an incorrect version of Theorem 3 previously appeared online.) A few months after this article first appeared, Berestycki, Berestycki and Schweinsberg published an article [1] which revealed the correct asymptotics seen in our new Theorem 2. The key to finding the correct answer lies in Proposition 20 of [1]. Other than that, our methods were developed independently. Even given Theorem 2, new ideas are required to deduce the stated version of Theorem 3.
No upper bound is given in Theorem 1; the only rigorous upper bound we know of is that given by Jaffuel [11]. In fact, as mentioned above, Jaffuel considered branching random walks, but it is not difficult either to adapt his proof, or to apply his result together with standard tightness properties of Brownian motion, to achieve the same upper bound for BBM. A finer upper bound appears to be difficult to prove.
The above results are stated only for standard BBM. There are however now wellknown techniques for transferring the proofs to other cases, where for example each particle might give birth to a random number of new particles when it splits. In order to apply our methods we must only insist that the distribution of this random number has a finite second moment.

Layout of the article
Our main tactic will be to develop detailed estimates for a single Brownian motion, and then to apply standard branching tools to deduce results for the branching system. In Section 2 we develop our main single-particle estimates, on the probability that a Brownian motion stays within a tube about a function f . These will then be used to prove both Theorem 1 in Section 3 and Theorem 2 in Section 4. Finally, in Section 5 we apply Theorem 2 to prove Theorem 3.

Notation
For α ∈ (0, ∞) we use the notation g(t) α h(t) to mean that there exist constants c(α), C(α) ∈ (0, ∞) depending only on α such that c(α)g(t) ≤ h(t) ≤ C(α)g(t) for all t in the specified range. Similarly, when we write g(t) α h(t) we mean that there exists a constant C(α) depending only on α such that g(t) ≤ C(α)h(t) for all t in the specified range. Use of this notation without a specified parameter α (i.e. and rather than α and α ) means that the constants are absolute.
For γ ∈ R and x > 0, we write log γ x to mean (log x) γ .
Throughout the article we shall have two twice continuously differentiable functions f : [0, ∞) → R and L : [0, ∞) → (0, ∞), such that f (0) = −x < 0 and f (0) + L(0) > 0. We suppose that under P, as well as our BBM we have an independent Brownian motion ξ t , t ≥ 0 started from 0. Write (G t , t ≥ 0) for the natural filtration of ξ t . We define We often assume that there exists a constantL > 0 such that for all t, which we call assumption (A).

Single-particle estimates
In this section we are interested in estimating the probability that a Brownian motion stays close to a function f . Set Our main aim for the section is to prove the following result.
If f (t) ≤ 0 then the inequalities are reversed.
Remark. Often we will fix a particular f and L, and be interested in particles near f (t) + L(t) at a chosen time t. For example, if we want to be within a constant distance of f (t) + L(t), then both the upper and lower bounds above collapse, giving In order to prove Proposition 4, we begin with a standard estimate for the probability that a Brownian motion stays within a tube of fixed width for a long time.
where for the last line we used the Dubins-Schwarz theorem. Since under P and using assumption (A), almost surely, so The result now follows from Lemma 5.
Finally we apply Girsanov's theorem, together with standard estimates, to consider tubes about the function f (or, to be precise, f + K) rather than about 0. t 0 (f (s)+K (s)) 2 ds .
where the latter equality follows by standard integration by parts calculations (see for example Lemma 5.1 of [9]). But on A t , assuming (A), the exponent above is almost surely, soP EJP 20 (2015), paper 28.
The result now follows from rearranging (2.2).
For times t < ρ L , the following easy estimate will be sufficient for our needs.
Note that, using similar integration by parts calculations as in the proof of Proposition 4, But underP, (ξ t , t ≥ 0) is Brownian motion started from x. By the reflection principle,

This gives us the bound
applying the same calculations, plus assumption (A), toξ = f (t) + L(t) − ξ t gives the remaining bound.

Proof of Theorem 1
We now apply Proposition 4 and Lemma 7 to prove Theorem 1, which said that if we let A c = 3 4/3 π 2/3 2 −7/6 , then with positive probability there is always a particle above √ 2t where C is any constant such that f (0) < 0 and f (0) + L(0) > 0. We will choose α and β later. We want to count the number of particles staying above f (u) up to time t, but a direct attack via moment estimates will not work because the second moment overcounts the number of particles that go unrealistically high early on. We therefore introduce the upper boundary f (u) + L(u), and count the number of particles that stay in the tube between f (u) and f (u) + L(u) up to time t. Even this, though, does not work on its own.
Our calculation ends up being dominated by particles which have been to the top of the tube early on and then fallen to the bottom by time t, which is not a "sustainable" tactic for survival: particles near the bottom of the tube at time t are very likely to fall below f (u) at some time u slightly after t. We therefore count only particles that stay within the tube up to time t and sit near the top of the tube at time t.
To this end, definẽ We will estimate the first and second moments ofÑ (t). One may easily check that for our choice of L, assumption (A) holds for some constantL. Further, A longer, but still elementary, calculation reveals that Lemma 8. If α = β = 3 1/3 π 2/3 2 −1/6 , then for all t sufficiently large, Proof. From the calculations above, we have Choosing α = β = 3 1/3 π 2/3 2 −1/6 , we see that the coefficients of t 1/3 , t 1/3 / log t, and t 1/3 / log 2 t all disappear, leaving us with Now applying the many-to-one lemma 1 and Proposition 4 (note that ρ L = O(1)), we see that for t sufficiently large, Proof. We apply the many-to-two lemma, which is a tool for calculating second moments (see for example [8] for more details). Suppose that T is an independent exponentially distributed random variable of parameter 2 and, given T , (ξ Then the many-to-two lemma tells us that We separate Θ t into three sections: what happens before time T (when ξ (1) = ξ (2) ); what happens to ξ (1) after time T ; and what happens to ξ (2) after time T . The latter two are independent given ξ (1) T . The plan is to apply Proposition 4 to each section, although we have to worry about s being too close to 0 or t. To this end definē For s > ρ L and j ∈ [0, L(s) − 1], by Proposition 4 and (3.1), 1 This well-known result (see for example [8]) says essentially that the expected number of particles with a certain property at time t is equal to the expected total number of particles -in our case e t -times the probability that one Brownian motion has this property.
Putting these two estimates together, we obtain that for s ∈ (ρ L ,ρ L (t)), For s ≤ ρ L , we can use the trivial bound P(A s ) ≤ 1 together with the above estimate on P(Θ t |A (1) s ∩ · · · ) to get the same bound, so in fact s ∩ · · · ), and apply Lemma 7 together with the fact that when s ≥ρ L (t), we have t 1/3 = s 1/3 + O(1), to get

Thus in this case
Putting this together with (3.3) and plugging back into (3.2), we get the desired bound.
Proof of Theorem 1. It is clearly enough to prove that with probability bounded away from zero there is always a particle above f (t). Note that, by Cauchy-Schwarz and Lemmas 8 and 9, for some constant c, proving the theorem.

Proof of Theorem 2
For x = x t > 0, we now want to estimate The general tactic is to consider instead the set of particles that remain between for some function L t . As in Section 3, the reason for this restriction is that with no upper boundary imposed, events of vanishing probability have a distorting effect on the moments of the number of particles with large position.
However, we have not succeeded in finding an appropriate L t ; it turns out that for most seemingly sensible choices, the probability of staying within the L t -tube is significantly smaller than the probability of staying above for completeness, and since their notation is different from ours, we include a proof translated into our own setup. This will be our Proposition 15. All other estimates are independent of those in [1]. Throughout this section we will use the functions where a = a c = 3 1/3 π 2/3 2 −1/2 . Sometimes we leave out the subscripts, or write f z instead of f t,z , where the parameters are clear from context. Let If s ≥ ρ Lt , then for any z and y, p(t, z; s, y) q(t, z; s, y + 1).
Proof. We note that We now work towards counting the number of particles at the top of the tube, again by estimating moments. Let Proof. By the many-to-one lemma, E[#M t,z (u)] = e u P(ξ ∈ M t,z (u)) = e u p(t, z; u, 1).
The result now follows from Lemma 10.
The last inequality is justified by the fact that for any r ≥ 0, we have ρ Lt−r ≤ ρ Lt , so in particular if r ≤ m − ρ Lt then r ≤ m − ρ Lt−r and hence m − r ≥ ρ Lt−r so by Lemma 10 p(t − r, y; m − r, 1) ≤ q(t − r, y; m − r, 2) for any y. Recalling the definition (4.1) of q, an easy calculation reveals that for r ∈ [ρ Lt , m − ρ Lt ], If r ≤ ρ Lt , then we replace q(t, z; r, y) withq(z; r, y) and get If r ∈ [m − ρ Lt , n − ρ Lt ], noting that ρ Lt = O(ρ Lm ) since m ∈ [t/3, 2t/3], we may instead replace q(t − r, y; m − r, 1) withq(y; m − r, 1) and get If r ≥ (n − ρ Lt ), then we replace both the latter q factors with the appropriateq and get Putting these estimates together, we obtain m 0 e n+m−r P(ξ (1) We also have that when m ≤ n − ρ Lt , and when m ∈ (n − ρ Lt , n], P(ξ ∈ M t,z (m) ∩ M t,z (n)) q(t, z; m, 1)q(1; n − m, 1) By symmetry we get similar results when m > n, and obtain the stated result.
We want to count the number of particles hitting the top of the L-tube whose ancestors have never hit either boundary of the tube. In fact, to get a lower bound we will restrict to those hitting the top of the tube between times t/3 and 2t/3. For a particle v, let σ v be its birth time and τ v its time of death. Let Otherwise said, if we imagine particles being absorbed if they hit either boundary of the tube, then J t,z is the number of particles absorbed at the top of the tube between times t/3 and 2t/3.
It is easy to see that P(J t,z = ∅) P(J t,z ≥ 1).
Applying Lemma 12, But by Lemma 11, as required.
We have established that the probability of a particle hitting the top of the L t -tube at some time s ∈ [t/3, 2t/3], and having never gone below f t,z , behaves as we would like. We now wish to show that such a particle has a positive probability, independent of t and s, of staying above f t,z up to time t. For this we follow [1]. The idea is this: suppose that a particle v has hit the top of the tube. Then v will, with reasonable probability, have a large number of descendants just below the top of the tube a short time later. Each of these descendants then has a reasonable probability of hitting the top of the tube again.
Each descendant of v that hits the top of the tube again we call a tube child of v. We show that this concept of tube children can be used to build a family tree very much like a Galton-Watson process, and calculate that the associated offspring distribution has mean larger than 1. Such a Galton-Watson process survives with strictly positive probability, and the survival of the family tree entails that some particle stays above f t,z up to time t. We now give details of this heuristic.
We use the following result of Neveu [14] to quantify the statement that a particle which hits the top of the tube will, with reasonable probability, have a large number of descendants just below the top of the tube a short time later. Define That is, K(y) counts the number of particles that hit the line ( √ 2s − y, s ≥ 0) whose ancestors have never hit the same line. In other words, if we imagine particles being absorbed when they hit the line, then K(y) counts the total number of particles absorbed.
Later it will also be useful to consider K(y, t) the number of particles in K(y) absorbed before time t.
We now build our family tree and carry out the rest of the above heuristic. Again we stress that the proof of Proposition 15 is based on [1, Proposition 20].

Proposition 15.
There exists δ > 0 such that for all large t, Proof. We construct a sequence of sets Z 0 , Z 1 , . . . whose sizes will behave very much like a Galton-Watson process.
By Lemma 13, we can choose t 0 > 0, ε > 0 such that for all t ≥ t 0 , By Lemma 14, we can choose z 1 > 0, t 1 > 0 such that By increasing t 0 if necessary, we may also assume that a(t 0 + t 1 ) 1/3 − z 1 < at 1/3 0 and L(t−t 0 ) > z 1 . We say that a particle succeeds if it stays above the line ( where v 0 is the initial particle. We now recursively define sets Y j and Z j+1 for each j ≥ 0 (we will give an intuitive description immediately afterwards). Given (v, s) ∈ Z j , let Y v j consist of the set of ordered pairs (v , s ) such that v is a descendant of v that hits the line ( √ 2u − at 1/3 + a(t − s) 1/3 − z 1 , u ≥ s) at some time u ∈ [s, s + t 1 ), and s is the first such time. Let In words, Z j contains particles that hit the top of the tube at appropriate times, and Y j contains descendants of those particles that fall distance z 1 below the top of the tube shortly afterwards. Up until a particle succeeds, we see that each particle in Z j has a number of descendants in Y j with distribution K(z 1 , t 1 ), and each particle in Y j has at least one descendant in Z j+1 with probability at least εz 1 e − √ 2z1 .
To complete the proof, on an auxiliary probability space let where K, B 1 , B 2 , . . . are independent random variables, K is distributed like K(z 1 , t 1 ), and each B j is Bernoulli with parameter εz 1 e − √ 2z1 . Let Z 0 , Z 1 , . . . be a Galton-Watson process with offspring distribution η. Then since the process Z survives forever with positive probability; that is, δ := lim n P(Z n > 0) > 0.
But as described above, either a particle succeeds or the size of Z n stochastically dominates Z n ; otherwise said, Clearly by construction Z n is eventually empty, so the left-hand side above converges to 1, but P(Z n = 0) → 1 − δ , so we must have Now it is a simple task to show that if we stay above √ 2u − at 1/3 up to t − t 0 − t 1 , we have a strictly positive probability (not depending on t) of doing so until time t. Indeed, recall that v 0 is our initial particle, and note that since P(inf s≤1 X v0 (s) ≥ −1, X v0 (1) ≥ √ 2) > 0, there exists δ > 0 such that But any particle above √ 2(t − t 0 − t 1 ) − at 1/3 + 1 at time t − t 0 − t 1 has probability at least Since this does not depend on t, we get the result.
The proof of Theorem 2 is now straightforward.
We begin with the lower bound. The probability that a particle stays above f t,z (u) up to time t is at least the probability that a particle v hits f t,z (s) + L t (s) at some time s ∈ [t/3, 2t/3), having stayed above f t,z (u) for all u ≤ s, and then a descendant of v stays above f t,z (u) for all times u ∈ [s, t]. Applying the Markov property, Lemma 13 tells us that P(J t,z = ∅) ze − √ 2z , and Proposition 15 tells us that the latter probability is bounded below by some fixed δ > 0. We deduce that which is our desired lower bound.
For an upper bound, in order to stay above f t,z (s) for all s ≤ t, clearly a particle must either hit f t,z (s) + L t (s) for some s ≤ t, or stay between f t,z (s) and f t,z (s) + L t (s) for all s ≤ t. That is, By the many-to-one Lemma and the Markov property, this is at most t j=0 Lt(j) 0 e j+1 p(t, z; j, y)P sup s∈[0,1] ξ s ≥ y dy + e t p(t, z; t, 0).
We now apply Lemma 10 to see that which completes the proof.

Proof of Theorem 3
We recall the setup of Theorem 3. For v ∈ N (t), we let λ(v, t) = sup s∈[0,t] { √ 2s−X v (s)} and define Λ(t) = min v∈N (t) λ(v, t). Then we wish to show that almost surely, where a = a c = 3 1/3 π 2/3 2 −1/2 . Showing that lim sup(Λ(t) − at 1/3 )/ log t = 0 is not difficult, simply by applying the estimates from Theorem 2 together with standard branching arguments and Borel-Cantelli exactly as in [16]. However the proof that lim inf( requires a novel approach. We begin with a heuristic explanation before starting on the details. Since the probability of staying above f t (u) := √ 2u − at 1/3 + 1 √ 2 log t up to time t is approximately 1/t, we may naively hope that the probability that a particle stays above f t (u) for all u ≤ t for some t ∈ [n, 2n) is of constant order (does not decay in n). A geometric trials argument would then suggest that the event occurs infinitely often. This is exactly the approach that works when looking at the position of the maximal particle: see [10] or [16]. We might therefore begin by estimating moments of 2n n #H t dt, (5.1) where H t is something like "the set of particles that stay above f t (u) for all u ≤ t".
However, we know as in previous sections that the second moments obtained in such a calculation will be too large; and so, following our strategy from Section 4, we might replace H t with "the set of particles that hit the top of the L t -tube at some time u ∈ [t/3, 2t/3]" for L t (u) = a(t − u) 1/3 . But this approach still yields second moments that are too large. This is a clue that something different is happening, and we will need an alternative strategy.
The key is to realise that if a particle manages to stay above f t (u) up to time t, it will have stayed above f t (u) + δ for some δ > 0. Thus if s ≤ t and f t (0) − f s (0) < δ, the same particle will have stayed above f s (u) up to time s. But there exists η > 0 such that f t (0) − f s (0) < δ for all s ∈ [t − ηt 2/3 , t]. We see, therefore, that the value of 2n n #H t dt is entirely misleading. We should instead work with something like n 1/3 k=1 #H n+kn 2/3 .

(5.2)
But with the choice of f t above, this quantity will decay like n −2/3 , and we realise that we have been working with the wrong function f t all along. Instead we should choose in which case (5.2) will be of constant order. We show that its first two moments behave well and deduce that, infinitely often, there are particles that stay above the line f t (u), To see that there are no particles above g t (u) := √ 2u − at 1/3 + (1+ε) 3 √ 2 log t for ε > 0, we return to a quantity more like (5.1). Its expected value is large, but given that a particle stays above g t (u) for all u ≤ t, it is even larger. We use this observation to complete the proof. (There are other possible approaches. For example, we could work with something like (5.2) again, but then we would have to worry about times in (n+kn 2/3 , n+(k +1)n 2/3 ) for each k. This could be done fairly easily but would rely on some technical estimates.
Using (5.1) will be slicker.) We split the proof of Theorem 3 into four lemmas, each of which represents an upper or a lower bound for a lim sup or a lim inf. We begin with the easier two.

Bounds on the lim sup
Proof. To rephrase the statement of the lemma, we show that for any ε ∈ (0, 1) there are arbitrarily large times t n such that no particles have stayed above √ 2u − at 1/3 n + ε log t n for all u ≤ t n . Choose δ < ε/2, let t 1 = 1 and for n > 1 let t n = exp 1 δ exp(2t n−1 ) . Define We know that F n occurs for all large n, so it suffices to show that For a particle v, let N v (t) be the set of descendants of v at time t, and let E v n be the event that some descendant of v at time t n has stayed above √ 2u − at 1/3 n − ε log t n for all times u ≤ t n . Also let s n = t n − t n−1 . Then if v ∈ N (t n−1 ) and X v (t n−1 ) ≤ √ 2t n−1 , Noting that s n ≥ t n /2, by the upper bound in Theorem 2 the above is at most Thus, since e 2t k−1 = δ log t k , Since we chose δ < ε/2, this tends to zero as k → ∞.  Proof. We show that for large t and any ε > 0, there are always particles that have stayed above √ 2u − at 1/3 − 2ε log t for all times u ≤ t. Let δ = ε/ √ 2. Let A t = { ∃v ∈ N (t) : X v (u) > √ 2u − at 1/3 − ε log t ∀u ≤ t} B t = |N (δ log t)| ≥ t δ/2 , X v (u) > −at 1/3 ∀u ≤ δ log t and |X v (δ log t)| ≤ √ 2δ log t ∀v ∈ N (δ log t) .
As before we write N v (t) for the set of descendants of particle v that are alive at time t. Let l t = t − δ log t. Then for all large t, By the lower bound in Theorem 2 there exists c > 0 such that this is at most Thus by Borel-Cantelli, for any lattice times t n → ∞, P(A tn ∩ B tn infinitely often) = 0. But almost surely for all large t, B t occurs, so we deduce that P(A tn infinitely often) = 0. Then if we choose t n − t n−1 small enough, −at 1/3 − ε log t < −at 1/3 n for all t ∈ (t n−1 , t n ), so the result holds.  Proof. Fix ε > 0. We show that for large t, there are no particles that stay above √ 2u − at 1/3 + ((1 + ε)/(3 √ 2)) log t for all times u ≤ t. Choose M large enough that for any n ≥ M and any t ∈ [n, 2n], we have −at 1/3 + (1 + ε) 3 √ 2 log t ≥ −as 1/3 + (1 + ε) 3 √ 2 log s − 1 ∀s ∈ [t − n 2/3 , t]. Plugging these estimates back into (5.3), we get P(I n > 0) n −ε/3 log n.

A lower bound on the lim inf
By Borel-Cantelli, the probability that there exist infinitely many k with I 2 k > 0 is zero. Since k≥n [2 k , 2 · 2 k ) = [2 n , ∞) and ε > 0 was arbitrary, we deduce the result.

An upper bound on the lim inf
The final lemma in our series requires the most work.
Also define the event A t (u) = {ξ u ∈ A t (u)}, interpreted in the obvious way (imagining that ξ u refers to whichever particle is the spine at time u, rather than to its position). Set D t = [t/3, 2t/3] ∩ N and K n = {n + kn 2/3 : k = 1, . . . , n 1/3 }. Define Our initial aim is to show that P(S n > 0) is bounded away from 0. We do this by estimating moments of S n . We will use notation from Section 4.