ALEA, Lat. Am. J. Probab. Math. Stat. 15, 353–374 (2018) DOI: 10.30757/ALEA.v15-15
Large time unimodality for classical and free Brownian motions with initial distributions
Takahiro Hasebe and Yuki Ueda Department of Mathematics, Hokkaido University, Kita 10, Nishi 8, Kita-Ku, Sapporo, Hokkaido, 060-0810, Japan E-mail address: [email protected] URL: https://www.math.sci.hokudai.ac.jp/~thasebe/ Department of Mathematics, Hokkaido University, Kita 10, Nishi 8, Kita-Ku, Sapporo, Hokkaido, 060-0810, Japan E-mail address: [email protected]
Abstract. We prove that classical and free Brownian motions with initial distri- butions are unimodal for sufficiently large time, under some assumption on the initial distributions. The assumption is almost optimal in some sense. Similar re- sults are shown for a symmetric stable process with index 1 and a positive stable process with index 1/2. We also prove that free Brownian motion with initial sym- metric unimodal distribution is unimodal, and discuss strong unimodality for free convolution.
1. Introduction
A Borel measure µ on R is unimodal if there exist a R and a function f : R [0, ) which is non-decreasing on ( ,a) and non-increasing∈ on (a, ), such that→ ∞ −∞ ∞ µ(dx)= µ( a )δ + f(x) dx. (1.1) { } a The most outstanding result on unimodality is Yamazato’s theorem (Yamazato, 1978) saying that all classical selfdecomposable distributions are unimodal. After this result, in Hasebe and Thorbjørnsen (2016), Hasebe and Thorbjørnsen proved the free analog of Yamazato’s result: all freely selfdecomposable distributions are unimodal. The unimodality has several other similarity points between the classical and free probability theories Hasebe and Sakuma (2017). However, it is not true that the unimodality shows complete similarity between the classical and free prob- ability theories. For example, classical compound Poisson processes are likely to be
Received by the editors October 27th, 2017; accepted March 23th, 2018. 2010 Mathematics Subject Classification. Primary 46L54; Secondary 60E07; 60G52, 60J65. Key words and phrases. unimodal, strongly unimodal, Brownian motion, Cauchy process, positive stable process. Research supported by JSPS Grant-in-Aid for Young Scientists (B) 15K17549 (for TH) and JSPS and MAEDI Japan–France Integrated Action Program (SAKURA) (for both authors). 353 354 T. Hasebe and Y. Ueda non-unimodal in large time (Wolfe, 1978), while free L´evy processes with compact support become unimodal in large time (Hasebe and Sakuma, 2017). In this paper, we mainly focus on the unimodality of classical and free Brownian motions with initial distributions and consider whether the classical and free versions share simi- larity points or not. In free probability theory, the semicircle distribution is the free analog of the normal distribution. In random matrix theory, it appears as the limit of eigenvalue distributions of Wigner matrices as the size of the random matrices goes to infinity. Free Brownian motion is defined as a process with free indepen- dent increments, started at 0 and distributed as the centered semicircle distribution S(0,t) at time t> 0. Furthermore, we can provide free Brownian motion with ini- tial distribution µ which is a process with free independent increments distributed as µ at t = 0 and as µ ⊞ S(0,t) at time t> 0. The definition of ⊞, called additive free convolution, is provided in Section 2.1. The additive free convolution is the distribution of the sum of two random variables which are freely independent (for the definition of free independence see Nica and Speicher, 2006). In Biane (1997), Biane gave the density function formula of free Brownian motion with initial distri- bution (see Section 2.2). In our studies, we first consider the symmetric Bernoulli 1 1 distribution µ := 2 δ+1 + 2 δ 1 as an initial distribution and compute the density function of µ ⊞ S(0,t). Then− we see that the probability distribution µ ⊞ S(0,t) is unimodal for t 4 (and it is not unimodal for 0
2 The function f can grow very fast such as ex , and so, a tail decay of the initial distribution does not imply large time unimodality. The corresponding classical problem is natural, that is, for which class of initial distributions on R does Brownian motion become unimodal for sufficiently large time t > 0? We prove in this direction the following results (formulated as Theo- rem 4.3 and Proposition 4.2 in Section 4, respectively): Theorem 1.3. (1) Let µ be a probability measure on R such that
2 α := eεx dµ(x) < (1.3) R ∞ Z for some ε > 0. Then the distribution µ N(0,t) is unimodal for all t 36 log(2α) ∗ ≥ ε . Unimodality for classical and free Brownian motions 355
(2) There exists a probability measure µ on R satisfying that
A x p e | | dµ(x) < for all A> 0 and 0
2. Preliminaries
2.1. Definition of free convolution. Let µ be a probability measure on R. The Cauchy transform 1 Gµ(z) := dµ(x) (2.1) R z x Z − is analytic on the complex upper half plane C+. We define the truncated cones
Γ± := z C± : Re(z) < α Im(z) , Im(z) >β , α> 0,β R. (2.2) α,β { ∈ | | | | } ∈ In Bercovici and Voiculescu (1993, Proposition 5.4), it was proved that for any + γ < 0 there exist α,β > 0 and δ < 0 such that Gµ is univalent on Γα,β and + 1 Γ− G (Γ ). Therefore the right inverse function G− exists on Γ− . We γ,δ ⊂ µ α,β µ γ,δ define the R-transform of µ by
1 1 R (z) := G− (z) , z Γ− . (2.3) µ µ − z ∈ γ,δ Then for any probability measures µ and ν on R there exists a unique probability measure λ on R such that
Rλ(z)= Rµ(z)+ Rν (z), (2.4) 356 T. Hasebe and Y. Ueda for all z in the intersection of the domains of the three transforms. We denote λ := µ ⊞ ν and call it the (additive) free convolution of µ and ν.
2.2. Free convolution with semicircle distributions. Almost all the materials and methods in this section are following (Biane, 1997). The semicircle distribution S(0,t) of mean 0 and variance t> 0 is the probability measure with density 1 4t x2, (2.5) 2πt − supported on the interval [ 2√t, 2√t].p We then compute its Cauchy transform and its R-transform (see, for example,− Nica and Speicher, 2006): z √z2 4t G (z)= − − , z C+, (2.6) S(0,t) 2t ∈ where the branch of the square root on C R+ is such that √ 1 = i. Moreover \ − R (z)= tz, z C−. (2.7) S(0,t) ∈ Let µ be a probability measure on R and let µt = µ ⊞ S(0,t). Then we define the following set: R 1 1 Ut := u 2 dµ(x) > , (2.8) ∈ R (x u) t Z R − and the following function from to [0, ) by setting ∞ 1 1 vt(u) := inf v 0 2 2 dµ(x) . (2.9) ≥ R (x u) + v ≤ t Z − R R Then vt is continuous on and one has Ut = x vt(u) > 0 . Moreover, for every u U , v (u) is the unique solution v> 0{ of∈ the equation| } ∈ t t 1 1 2 2 dµ(x)= . (2.10) R (x u) + v t Z − We define Ωt,µ := x + iy C y > vt(x) . (2.11) { ∈ | } + Then the map Ht(z) := z + RS(0,t)(Gµ(z)), defined on C , is a homeomorphism + + from Ωt,µ to C R and it is conformal from Ωt,µ onto C . Hence there exists an ∪ + inverse function Ft : C R Ωt,µ. By Biane (1997, Lemma 1, Proposition 1), ∪ → 1 C+ the domain Ωt,µ is equal to the connected component of the set Ht− ( ) which contains iy for large y > 0 and for all z C+ one has ∈ Gµt (z)= Gµt (Ht(Ft(z)))
= Gµt Ft(z)+ RS(0,t) Gµ(Ft(z)) (2.12)
= Gµ(Ft(z)). Moreover G has a continuous extension to C+ R. Then µ is Lebesgue absolutely µt ∪ t continuous by Stieltjes inversion and the density function pt : R [0, ) of µt is given by → ∞
1 vt(x) pt(ψt(x)) = lim Im(Gµ (ψt(x) + iy)) = , x R, (2.13) π y +0 t πt − → ∈ where (x u) ψt(x)= Ht(x + ivt(x)) = x + t 2− 2 dµ(u). (2.14) R (x u) + v (x) Z − t Unimodality for classical and free Brownian motions 357
It can be shown that ψt is a homeomorphism of R. Furthermore the topological support of pt is given by ψt(Ut), and pt extends to a continuous function on R and is real analytic in x R : p (x) > 0 . { ∈ t } 1 1 Example 2.1. Let µ := δ 1 + δ+1 be the symmetric Bernoulli distribution. Recall 2 − 2 that the topological support of the density function pt is equal to ψt(Ut). If 0 < t 1, we have ≤ 2+ t √t2 +8t 2+ t + √t2 +8t U = u R − < u < , (2.15) t ∈ s 2 | | s 2
and if t> 1, we have
2+ t + √t2 +8t 2+ t + √t2 +8t U = u R If 0 (2u2 +2 t)+ √t2 + 16u2 v (u)= − − . (2.17) t s 2 Calculating its first derivative then shows that µt is unimodal if and only if t 4; see Figures 2.1–2.6. ≥ According to Example 2.1, we have a question about what class of initial distri- butions implies the large time unimodality of free Brownian motion. We will give a result for this in Section 3.1. 0.5 0.5 0.5 0.4 0.4 0.4 0.3 0.3 0.3 0.2 0.2 0.2 0.1 0.1 0.1 -6 -4 -2 2 4 6 -6 -4 -2 2 4 6 -6 -4 -2 2 4 6 Figure 2.1. p0.25(x) Figure 2.2. p1(x) Figure 2.3. p2(x) 0.5 0.5 0.5 0.4 0.4 0.4 0.3 0.3 0.3 0.2 0.2 0.2 0.1 0.1 0.1 -6 -4 -2 2 4 6 -6 -4 -2 2 4 6 -6 -4 -2 2 4 6 Figure 2.4. p3(x) Figure 2.5. p4(x) Figure 2.6. p7(x) 358 T. Hasebe and Y. Ueda 2.3. A basic lemma for unimodality. A unimodal distribution is allowed to have a plateau or a discontinuous point in its density such as the uniform distribution or the exponential distribution. If we exclude such cases, then unimodality can be characterized in terms of levels of density. Lemma 2.2. Let µ be a probability measure on R that is Lebesgue absolutely con- tinuous. Suppose that its density p(x) extends to a continuous function on R and is real analytic in x R : p(x) > 0 . Then µ is unimodal if and only if, for any a> 0, the equation{p(∈x)= a has at most} two solutions x. The proof is just to use the intermediate value theorem and the fact that a real analytic function never has plateaux. Note that the idea of the above lemma was first introduced by Haagerup and Thorbjørnsen (2014) to prove that a “free gamma distribution” is unimodal. 3. Free Brownian motion with initial distribution 3.1. Large time unimodality for free Brownian motion. The semicircular distribu- tion is the free analogue of the normal distribution. Therefore, a process with free independent increments is called (standard) free Brownian motion with initial dis- tribution µ if its distribution at time t is given by µ ⊞ S(0,t). In this section, we firstly prove that free Brownian motion with compactly supported initial distribu- tion is unimodal for sufficiently large time, by using Biane’s density formula (see Section 2.2). Lemma 3.1. Suppose that t> 0. The following statements are equivalent. (1) µ ⊞ S(0,t) is unimodal. (2) For any R> 0 the equation 1 1 2 2 dµ(x)= (3.1) R (x u) + R t Z − has at most two solutions u R. ∈ Proof : We adopt the notations µt = µ ⊞ S(0,t), vt,pt and ψt in Section 2.2. This proof is similar to Hasebe and Thorbjørnsen (2016, Proposition 3.8). Recall that µt is absolutely continuous with respect to Lebesgue measure and the density function R pt is continuous on since its Cauchy transform Gµt has a continuous extension + to C R, and the density function is real analytic in ψt(Ut) and continuous on R by Section∪ 2.2. Suppose (2) first. Since ψt is a homeomorphism of R, by Lemma 2.2 it suffices to show that for any a> 0 the equation v (u) a = p (ψ (u)) = t , u R (3.2) t t πt ∈ has at most two solutions u R. Since vt(u) is a unique solution of the equation (2.10) if it is positive, then we∈ have R vt(u) R 1 1 u a = = u 2 2 dµ(x)= . (3.3) ∈ πt ∈ R (x u) + (πat) t Z − vt(u) R By the assumption (2), the equation a = πt has at most two solutions in . Unimodality for classical and free Brownian motions 359 Conversely, suppose that (2) does not hold. Then for some R> 0 there are three distinct solutions u1,u2,u3 R to (3.1). By (3.3) this shows that vt(ui) = R and R ∈ hence pt(ψt(ui)) = πt for i =1, 2, 3. Again by Lemma 2.2, µt is not unimodal. Now we are ready to prove Theorem 1.2 (1). Theorem 3.2. Let µ be a compactly supported probability measure, and let Dµ be the diameter of the support: Dµ = sup x y : x, y supp(µ) . Then µ ⊞ S(0,t) is unimodal for t 4D2 . {| − | ∈ } ≥ µ Dµ Dµ Proof : Applying a shift we may assume that µ is supported on [ 2 , 2 ]. For fixed R> 0, we define by − 1 R ξR(u) := 2 2 dµ(x), u . (3.4) R (x u) + R ∈ Z − Then we have Dµ 2 2(x u) ξR′ (u)= − dµ(x), Dµ (x u)2 + R2 2 Z− 2 { − } (3.5) Dµ 2 6(x u)2 2R2 ξR′′ (u)= − − dµ(x). Dµ (x u)2 + R2 3 Z− 2 { − } Dµ Dµ Dµ If u< 2 , then x u> 2 ( 2 ) = 0 for all x supp(µ), so that ξR′ (u) > 0. −Dµ − D−µ −Dµ− ∈ If u> , then x u< = 0 for all x supp(µ), so that ξ′ (u) < 0. We 2 − 2 − 2 ∈ R take t 4D2 and consider the form of ξ (u) on u ( Dµ , Dµ ). ≥ µ R ∈ − 2 2 If 0 Such a distribution must have an unbounded support if it exists. Such an example can be constructed with an idea similar to Huang (2015, Proposition 4.13) (see also Hasebe and Sakuma, 2017, Example 5.3). Proposition 3.4. Let f : R [0, ) be a Borel measurable function. Then there exists a probability measure µ→such∞ that µ ⊞ S(0,t) is not unimodal at any t > 0 and f(x) dµ(x) < . (3.7) R ∞ Z Proof : We follow the notations in Section 2.2. Let wn n and an n be sequences in R satisfying { } { } w > 0, ∞ w =1, • n n=1 n an+1 >an, n 1, • P ≥ lim (an+1 an)= . n • →∞ − ∞ ∞ R Consider the probability measure µ := n=1 wnδan on and define the function 1 P ∞ wn Xµ(u) := 2 dµ(x)= 2 . (3.8) R (u x) (u a ) n=1 n Z − X − R ak+1+ak Recall that Ut is the set of u such that Xµ(u) > 1/t. Set bk := 2 for all k N. Then we have ∈ ∈ a a b a k+1 − k , k,n N (3.9) | k − n|≥ 2 ∈ and so 2 2 2 ∞ 2 X (b ) w = 0 µ k ≤ a a n a a → k+1 k n=1 k+1 k − X − as k . This means that for each t > 0 there exists an integer K(t) > 0 such that →X ∞(b ) < 1 for all k K(t). This implies that, for k K(t), the closure of µ k t ≥ ≥ Ut does not contain bk. Note that the set Ut is an open set. Therefore, there exists a sequence ε (t) ∞ of positive numbers such that { k }k=K(t)+1 ∞ U (b , )= (a ε (t),a + ε (t)), (3.10) t ∩ K(t) ∞ k − k k k k=K[(t)+1 where the closures of the intervals are disjoint. Thus the set supp(µt) = ψt(Ut) consists of infinitely many connected components for all t> 0, and hence µt is not unimodal for any t> 0. In the above construction, the weights wn are only required to be positive. There- fore, we may take c w = , (3.11) n n2 max f(a ), 1 { n } where c > 0 is a normalizing constant. Then the integrability condition (3.7) is satisfied. Proposition 3.4 shows that tail decay estimates of the initial distribution do not guarantee the large time unimodality. In Section 4 we shall see that the situation is different for classical Brownian motion. Unimodality for classical and free Brownian motions 361 3.2. Free Brownian motion with symmetric initial distribution. This section proves a unimodality result of different flavor. It is well known that if µ and ν are sym- metric unimodal distributions, then µ ν is also (symmetric) unimodal (see Sato, 2013, Exercise 29.22). The free analogue∗ of this statement is not known. Conjecture 3.5. Let µ and ν be symmetric unimodal distributions. Then µ ⊞ ν is unimodal. We can give a positive answer in the special case when one distribution is a semicircle distribution. Theorem 3.6. Let µ be a symmetric unimodal distribution on R. Then µ ⊞ S(0,t) is unimodal for any t> 0. Remark 3.7. There is a unimodal probability measure µ such that µ⊞S(0, 1) is not unimodal (see Lemma 3.9). Hence we cannot remove the assumption of symmetry of µ. Proof of Theorem 3.6. By Lemma 3.1 it suffices to show that for any R > 0 the equation 1 1 dµ(x)= (3.12) R (u x)2 + R2 t Z − has at most two solutions u R. Up to a constant multiplication, the LHS is the density of µ C , which is unimodal∈ since µ and C are symmetric unimodal. Since ∗ R R the density of µ CR is real analytic, Lemma 2.2 implies that the equation (3.12) has at most two∗ solutions. This shows that µ ⊞ S(0,t) is unimodal. 3.3. Freely strong unimodality. In classical probability, a probability measure is said to be strongly unimodal if µ ν is unimodal for all unimodal distributions ν on R. A distribution is strongly∗ unimodal if and only if the distribution is Lebesgue absolutely continuous, supported on an interval and a version of its density is log-concave (see Sato, 2013, Theorem 52.3). From this characterization, the normal distributions are strongly unimodal. We discuss the free version of strong unimodality. Definition 3.8. A probability measure µ on R is said to be freely strongly unimodal if µ ⊞ ν is unimodal for all unimodal distributions ν on R. Lemma 3.9. The semicircle distributions are not freely strongly unimodal. Proof : The Cauchy distribution is not strongly unimodal since its density is not log-concave. Hence, there exists a unimodal probability measure µ such that µ C ∗ 1 is not unimodal. Since the density of µ C1 is real analytic on R, by Lemma 2.2 there exists t> 0 such that the equation∗ d(µ C1) 1 1 π ∗ (x)= 2 dµ(y)= (3.13) dx R (x y) +1 t Z − has at least three distinct solutions x1, x2, x3 R. Lemma 3.1 then implies that S(0,t)⊞µ is not unimodal. By changing the scaling,∈ we conclude that no semicircle distributions are freely strongly unimodal. Theorem 3.10. Let λ be a probability measure with finite variance, not being a delta measure. Then λ is not freely strongly unimodal. 362 T. Hasebe and Y. Ueda Proof : We may assume that λ has mean 0. Suppose that λ is freely strongly unimodal. A simple induction argument shows that λ⊞n is freely strongly unimodal for all n Z+. We derive a contradiction below. By Lemma 3.9 we can take a unimodal∈ probability measure µ such that µ ⊞ S(0, 1) is not unimodal. Our ⊞ ⊞n hypothesis shows that the measure D√nv(µ) λ is unimodal, where v is the variance of λ and Dc(ρ) is the push-forward of a measure ρ by the map x cx for c R. By the free central limit theorem (Maassen, 1992, Theorem 5.2),7→ ∈ ⊞ ⊞n ⊞ ⊞n the unimodal distributions D1/√nv(D√nv(µ) λ ) = µ D1/√nv(λ ) weakly converge to µ⊞S(0, 1) as n . Since the set of unimodal distributions is weakly closed (see Sato, 2013, Exercise→ ∞ 29.20), we conclude that µ ⊞ S(0, 1) is unimodal, a contradiction. Problem 3.11. Does there exist a freely strongly unimodal probability measure that is not a delta measure? 4. Classical Brownian motion with initial distribution This section discusses large time unimodality for classical standard Brownian motion with an initial distribution µ, which is a process with independent incre- ments distributed as µ N(0,t) at time t 0. Before going to the general case, one example will be helpful∗ in understanding≥ unimodality for µ N(0,t). ∗ Example 4.1. Elementary calculus shows that 1 (δ 1 + δ1) N(0,t) (4.1) 2 − ∗ is unimodal if and only if t 1; see Figures 4.7-4.12. ≥ 1.0 1.0 1.0 0.8 0.8 0.8 0.6 0.6 0.6 0.4 0.4 0.4 0.2 0.2 0.2 -6 -4 -2 2 4 6 -6 -4 -2 2 4 6 -6 -4 -2 2 4 6 Figure 4.7. t =0.25 Figure 4.8. t =0.5 Figure 4.9. t =0.75 1.0 1.0 1.0 0.8 0.8 0.8 0.6 0.6 0.6 0.4 0.4 0.4 0.2 0.2 0.2 -6 -4 -2 2 4 6 -6 -4 -2 2 4 6 -6 -4 -2 2 4 6 Figure 4.10. t =1 Figure 4.11. t =2 Figure 4.12. t =4 This simple example suggests that Brownian motion becomes unimodal for suffi- ciently large time, possibly under some condition on the initial distribution. In some sense, we give an almost optimal condition on the initial distribution for the large time unimodality to hold. We start by providing an example of initial distribution with which the Brownian motion never becomes unimodal. Unimodality for classical and free Brownian motions 363 Proposition 4.2. There exists a probability measure µ on R such that µ N(0,t) is not unimodal at any t> 0, and ∗ A x p e | | dµ(x) < (4.2) R ∞ Z for all A> 0 and 0 Proof : We consider sequences wn n N (0, ) and an n N R such that { } ∈ ⊂ ∞ { } ∈ ⊂ ∞ wn =1, (4.3) n=1 X bk := inf ak an 1 as k . (4.4) n Z+ k | − − | → ∞ → ∞ ∈ \{ } Moreover we assume that for all t> 0 there exists k = k (t) N such that 0 0 ∈ 2 1 bk k k (t) w e− 2t >b e− 2t . (4.5) ≥ 0 ⇒ k k ∞ Define a probability measure µ by setting µ := n=1 wnδan and the following function: P 2 ∞ (x an) d(µ N(0,t)) − f (x) := √2πt ∗ (x)= w e− 2t , x R. (4.6) t · dx n ∈ n=1 X Then we have 2 ∞ (x an) x an − f ′(x)= w − e− 2t , x R. (4.7) t n − t ∈ n=1 X If needed we may replace k0(t) by a larger integer so that bk > √t holds for all k k (t). For k k (t) we have ≥ 0 ≥ 0 2 (a an 1) wk 1 ak an 1 k − − f ′(a 1) = e− 2t w − − e− 2t t k − t − n t n N,n=k ∈X6 2 (a an 1) wk 1 ak an 1 k − − e− 2t w | − − |e− 2t ≥ t − n t n N,n=k ∈X6 (4.8) 2 wk 1 bk bk e− 2t w e− 2t ≥ t − n t n N,n=k ∈X6 2 1 1 bk w e− 2t b e− 2t > 0, ≥ t k − k x2 where we used the fact that x x e− 2t takes the global maximum at x = √t on the third inequality and the7→ assumption | | (4.5) on the last line. This implies± that µ N(0,t) is not unimodal for any t> 0. ∗ Next we take specific sequences wn n and an n satisfying the conditions (4.3)– k { } { } (4.5). Set ak = a , k N where a 2. Note that there exists some constant c> 0 k ∈ ≥ such that bk ca for all k N. Hence we have that bk as k . Moreover we set ≥ ∈ → ∞ → ∞ 2 bk wk := Me− k , (4.9) 364 T. Hasebe and Y. Ueda b2 k k 1 where M > 0 is a normalized constant, that is, M = ( k∞=1 e− )− (note that the series converges). Since bk as k , we have → ∞ → ∞ P b2 k 2 2 bke− 2t 1 bk bk = bke k − 2t 0, as k . (4.10) wk M → → ∞ 1 For all t > 0 there exists k0 = k0(t) N such that k k0 implies that e− 2t > 2 ∈ ≥ bk b e− 2t /w . Therefore the sequences w and a satisfy the condition (4.5). k k { n}n { n}n ∞ Finally we show that µ = n=1 wnδan has the property (4.2). For every A > 0 and 0 2 Peεx dµ(x)= for all ε> 0. (4.12) R ∞ Z Then we have a natural question whether the large time unimodality of µ N(0,t) holds or not if the initial distribution µ does not satisfy the condition (4.12∗ ). We solve this question as follows. Theorem 4.3. Let µ be a probability measure on R such that 2 α := eεx dµ(x) < . (4.13) R ∞ Z for some ε> 0. Then µ N(0,t) is unimodal for t 36 log(2α) . ∗ ≥ ε Remark 4.4. The proof becomes much easier if we assume that µ has a compact support. Proof of Theorem 4.3: For x> 0 we have εy2 e εx2 µ( y > x)= 1 dµ(y) εx2 dµ(y) αe− . (4.14) | | y >x ≤ y >x e ≤ Z| | Z| | Let (x y)2 d(µ N(0,t)) − ft(x) := √2πt ∗ (x)= e− 2t dµ(y). (4.15) · dx R Z Then we have y x (y x)2 −2t ft′(x)= − e− dµ(y) R t Z x (4.16) (y x)2 (x y)2 ∞ y x − x y − = − e− 2t dµ(y) − e− 2t dµ(y). x t − t Z Z−∞ For x> 0, we have 2 (y x) √ 1 1 2 ∞ y x − t α εx − e− 2t dµ(y) e− 2 µ((x, )) e− 2 − . (4.17) t ≤ t ∞ ≤ √t Zx Unimodality for classical and free Brownian motions 365 For x √t , we have ≥ 2 x x √t (x y)2 3 (x y)2 x y − − x y − − e− 2t dµ(y) − e− 2t dµ(y) t ≥ 3x t Z−∞ Z− 1 √t 1 8x2 √t min e− 18 , 4xe− t µ 3x, x . ≥ t 3 − − 3 ! n o h i (4.18) 8x2 √ t t Now the function g(x) := 4xe− has a local maximum at x = 4 , and hence for all x √t , ≥ 2 2 1 1 g(x) 2e− √t< e− 18 √t, (4.19) ≤ 3 2 1 1 where 2e− 0.2706 and e− 18 0.3153. Hence we have ≈ 3 ≈ x 2 (x y) 8x2 x y − 4x √t − e− 2t dµ(y) e− t µ 3x, x t ≥ t − − 3 ! Z−∞ h i 4x 8x2 √t √t (4.20) e− t µ , ≥ t − 6 6 ! h i 2 8x2 εt e− t (1 αe− 36 ), ≥ √t − √t √t where the last inequality holds thanks to (4.14) and x 2 . Therefore if x 2 then ≥ ≥ 2 α 1 εx2 2 8x εt f ′(x) e− 2 − e− t (1 αe− 36 ) t ≤ √t − √t − 1 (4.21) 2 α 1 εx2 2e 2 εx2 8x εt = e− 2 − 1 e − t (1 αe− 36 ) . √t ( − α − ) 2 16 εx2 8x 1 εx2 εt 16 If t then e − t e 2 e 8 . Hence if t then ≥ ε ≥ ≥ ≥ ε 1 α 1 2 2e 2 εt εt 2 εx 8 36 ft′(x) e− − 1 e (1 αe− ) . (4.22) ≤ √t ( − α − ) 36 log(2α) εt 1 If t then 1 αe− 36 and we have ≥ ε − ≥ 2 1 2e 2 εt εt 1 9 7 1 e 8 (1 αe− 36 ) 1 e 2 2 2 α 2 < 0. (4.23) − α − ≤ − · · By taking t max 16 , 36 log(2α) = 36 log(2α) , we have ≥ { ε ε } ε √t f ′(x) < 0 if x . (4.24) t ≥ 2 Similarly, we have √t f ′(x) > 0 if x . (4.25) t ≤− 2 366 T. Hasebe and Y. Ueda R √t Next, we will show that ft′′(x) < 0 for all x with x < 2 . We then calculate the following ∈ | | 2 (x y) t (x y)2 −2t ft′′(x)= − 2 − e− dµ(y) R t Z 2 (x y) t (x y)2 −2t = − 2 − e− dµ(y) (4.26) x y >√t t Z| − | 2 t (x y) (x y)2 −2t − 2− e− dµ(y). − x y √t t Z| − |≤ 2 u2 u t 2t Since the function h(u) := 2− e− has a local maximum at u = √3t, we have t ± 2 2 (x y) 3 (x y) t − 2 c 2t 2 √ √ − 2 − e− dµ(y) e− µ([x t, x + t] ). (4.27) x y >√t t ≤ t − Z| − | c For all x R with x < √t , we have [x √t, x + √t]c √t , √t , and therefore ∈ | | 2 − ⊂ − 2 2 2 2 (x y) t (x y) 2 3 √t √t c 2 3 εt −2t 2 2 4 − 2 − e− dµ(y) e− µ , e− αe− . x y >√t t ≤ t − 2 2 ≤ t Z| − | h i (4.28) Since the function h(u) is decreasing on [0, √3t], we have − 2 2 t (x y) (x y)2 t (x y) (x y)2 −2t −2t − 2− e− dµ(y) − 2− e− dµ(y) x y √t t ≥ x y 2√t t Z| − |≤ Z| − |≤ 3 4 4 t t 9 t 9 2√t 2√t − e− 2t µ x , x + ≥ t2 − 3 3 h i 5 2 2√t 2√t = e− 9 µ x , x + 9t − 3 3 h i 5 2 √t √t e− 9 µ , ≥ 9t − 6 6 5 2 h εt i e− 9 (1 αe− 36 ). ≥ 9t − (4.29) Therefore 2 3 εt 5 2 εt 2 4 9 36 ft′′(x) e− αe− e− (1 αe− ) ≤ t − 9t − (4.30) 1 3 εt 5 2 εt = 2e− 2 αe− 4 e− 9 (1 αe− 36 ) . t − 9 − n o If t 36 log(2α) , then ≥ ε 3 3 εt 5 2 εt e− 2 5 2 2e− 2 αe− 4 e− 9 (1 αe− 36 ) e− 9 < 0, (4.31) − 9 − ≤ (2α)8 − 18 √t and therefore f ′′(x) < 0 if x < . t | | 2 Unimodality for classical and free Brownian motions 367 To summarize, if t 36 log(2α) , then we have the following properties: ≥ ε √t x f ′(x) > 0, ≤− 2 ⇒ t √t x f ′(x) < 0, (4.32) ≥ 2 ⇒ t √t x < f ′′(x) < 0. | | 2 ⇒ t Hence f (x) has a unique local maximum in √t , √t when t 36 log(2α) , which t − 2 2 ≥ ε is a unique global maximum on R as well. Therefore µ N(0,t) is unimodal for ∗ t 36 log(2α) . ≥ ε Remark 4.5. If µ is unimodal then µ N(0,t) is unimodal for all t > 0. This is a consequence of the strong unimodality∗ of the normal distribution N(0,t) (see Sec- tion 3.3), in contrast with the failure of freely strong unimodality of the semicircle distribution (see Lemma 3.9). We close this section by placing a problem for future research. Problem 4.6. Estimate the position of the mode of classical Brownian motion with initial distributions satisfying the assumption (4.13). Our proof shows that for 36 √ √ t ε log(2α), the mode is located in the interval [ t/2, t/2]. How about free Brownian≥ motion? − 5. Large time unimodality for stable processes with index 1 and index 1/2 with initial distributions We investigate large time unimodality for stable processes with index 1 (follow- ing Cauchy distributions) and index 1/2 (following L´evy distributions) with initial distributions. 5.1. Cauchy process with initial distribution. Let Ct t 0 be the symmetric Cauchy distribution { } ≥ t C (dx) := 1R(x) dx, x R, C = δ , (5.1) t π(x2 + t2) · ∈ 0 0 which forms both classical and free convolution semigroups. A Cauchy process with initial distribution µ is a process with independent increments that follows the law µ C at time t 0. It is known that the Cauchy distribution satisfies the identity ∗ t ≥ µ ⊞ C = µ C (5.2) t ∗ t for any µ and t 0, and so the distributions µ Ct can also be realized as the laws at time t 0 of a process≥ with free independent increments∗ with initial distribution µ. The authors≥ do not know a written proof of (5.2) in the literature, so give a proof below. The Cauchy transform of the Cauchy distribution is given by GCt (z) = 1/(z + it) on C+ (e.g. by the residue theorem), and hence R (z) = itz. Then Ct − R ⊞ (z)= R (z) itz, and after some computation we can check that G ⊞ (z)= µ Ct µ − µ Ct Gµ(z+it). The Stieltjes inversion formula implies that for t> 0 the free convolution µ ⊞ Ct has the density 1 t Im(Gµ(x + it)) = 2 2 dµ(y), (5.3) − π R π((x y) + t ) Z − 368 T. Hasebe and Y. Ueda which is exactly the density of µ Ct. As in the cases of free and classical∗ BMs, taking µ to be the symmetric Bernoulli 1 (δ 1 + δ1) is helpful. By calculus we can show that µ Ct is unimodal if and only 2 − ∗ if t √3; see Figures 5.13-5.18. Thus it is again natural to expect that a Cauchy process≥ becomes unimodal for sufficiently large time, under some condition on the initial distribution. We start from the following counterexample. 0.7 0.7 0.7 0.6 0.6 0.6 0.5 0.5 0.5 0.4 0.4 0.4 0.3 0.3 0.3 0.2 0.2 0.2 0.1 0.1 0.1 -6 -4 -2 2 4 6 -6 -4 -2 2 4 6 -6 -4 -2 2 4 6 Figure 5.13. t =0.25 Figure 5.14. t =0.5 Figure 5.15. t =1 0.7 0.7 0.7 0.6 0.6 0.6 0.5 0.5 0.5 0.4 0.4 0.4 0.3 0.3 0.3 0.2 0.2 0.2 0.1 0.1 0.1 -6 -4 -2 2 4 6 -6 -4 -2 2 4 6 -6 -4 -2 2 4 6 Figure 5.16. t = √2 Figure 5.17. t = √3 Figure 5.18. t =3 Proposition 5.1. There exists a probability measure µ on R such that µ Ct is not unimodal for any t> 0, and ∗ x p dµ(x) < , 0 bk = inf ak an 1 , k Z+ (5.6) n Z+ k | − − | ∈ ∈ \{ } satisfies the condition 3 lim wkbk = . (5.7) k ∞ →∞ When a is increasing, this condition means that the distance between a and { n} n an+1 grows sufficiently fast. Let π d(µ C ) ∞ w f (x) := ∗ t (x)= n . (5.8) t t dx (x a )2 + t2 n=1 n X − Then we obtain ∞ 2wn(x an) f ′(x)= − − , (5.9) t [(x a )2 + t2]2 n=1 n X − Unimodality for classical and free Brownian motions 369 and so for sufficiently large k such that bk > 0 and for each t> 0 2wk 2wn(ak an 1) ft′(ak 1) = 2 2 + − − 2 − 2 2 − (1 + t ) [(ak an 1) + t ] n 1,n=k ≥X6 − − 2wk 2wn 2 2 3 ≥ (1 + t ) − ak an 1 n 1,n=k ≥X6 | − − | (5.10) 2wk 2wn ≥ (1 + t2)2 − b3 n 1,n=k k ≥X6 2w 2 1 1 k =2w . ≥ (1 + t2)2 − b3 k (1 + t2)2 − w b3 k k k The condition (5.7) shows that f ′(a 1) is positive for sufficiently large k Z . t k − ∈ + This shows that µ Ct is not unimodal for any t> 0. ∗ n r 3n If we take the particular sequences an = a and wn = cn a− , where a 2,r> ≥ n 0 and c> 0 is a normalizing constant, then the sequence bn satisfies bn Ca for some constant C > 0 independent of n. Then the conditions (5.7) and≥ (5.4) hold true. In the above construction, for any positive weights wn n and any sequence a that satisfies (5.7), the third moment of µ is always{ infinite,} { n}n x 3 dµ(x)= , (5.11) R | | ∞ Z due to the inequality ak bk a1 1 and the condition (5.7). The next question is then whether there| exists|≥ a−| probability|− measure µ with a finite third moment such that µ Ct is not unimodal for any t> 0, or at least for sufficiently large t> 0. The complete∗ answer is given below. Theorem 5.2. Let µ be a probability measure on R which has a finite absolute third moment β := x 3 dµ(x) < . R | | ∞ Z 1 Then µ C is unimodal for t 20β 3 . ∗ t ≥ Proof : The Markov inequality implies the tail estimate β µ([ x, x]c) , x> 0. (5.12) − ≤ x3 It suffices to prove that the function π d(µ Ct) 1 ft(x) := ∗ (x)= 2 2 dµ(y) (5.13) t dx R (x y) + t Z − has a unique local maximum for large t > 0. Suppose first that x > 0. The derivative ft′ splits into the positive and negative parts 2(y x) 2(x y) ft′(x)= −2 2 2 dµ(y) −2 2 2 dµ(y). (5.14) y>x [(x y) + t ] − y x [(x y) + t ] Z − Z ≤ − 370 T. Hasebe and Y. Ueda By calculus the function u 2u takes a global maximum at the unique point 7→ (u2+t2)2 u = t/√3. Using the tail estimate (5.12) yields the estimate of the positive part t 2(y x) 2 √ β − dµ(y) 3 dµ(y) . (5.15) [(x y)2 + t2]2 ≤ ( t2 + t2)2 ≤ t3x3 Zy>x − Zy>x 3 On the other hand the negative part can be estimated as 2(x y) 2(x y) −2 2 2 dµ(y) −2 2 2 dµ(y). (5.16) y x [(x y) + t ] ≥ x In order to show that ft′ has a unique zero, it suffices to show that ft′′(x) < 0 for x t/4. Now we have | |≤ 2[3(y x)2 t2] ft′′(x)= − 2 −2 3 dµ(y) y x >t/√3 [(x y) + t ] Z| − | − (5.22) 2[t2 3(y x)2] − 2 − 2 3 dµ(y). − y x t/√3 [(x y) + t ] Z| − |≤ − The function u 2(3u2 t2)/(u2 +t2)3 attains a global maximum at u = t and a global minimum7→ at u = 0.− Therefore, the positive part can be estimated as± follows: 2[3(y x)2 t2] 2(3t2 t2) − 2 −2 3 dµ(y) 2 −2 3 dµ(y) y x >t/√3 [(x y) + t ] ≤ y x >t/√3 (t + t ) Z| − | − Z| − | (5.23) 1 t t c = µ x , x + . 2t4 − √3 √3 Unimodality for classical and free Brownian motions 371 For all x such that x t/4, we have the inclusion | |≤ t t c t t c x , x + , , (5.24) − √3 √3 ⊆ −5 5 and hence we obtain 2[3(y x)2 t2] 1 t t c 53β − 2 −2 3 dµ(y) 4 µ , 7 . (5.25) y x >t/√3 [(x y) + t ] ≤ 2t −5 5 ≤ 2t Z| − | − On the other hand, the negative part has the estimate 2[t2 3(y x)2] 2[t2 3(y x)2] − 2 − 2 3 dµ(y) − 2 − 2 3 dµ(y). (5.26) y x t/√3 [(x y) + t ] ≥ y x t/2 [(x y) + t ] Z| − |≤ − Z| − |≤ − By calculus, the function u (t2 3u2)/(u2 + t2)3 is decreasing on [0,t], and so 7→ − 2[t2 3(y x)2] − 2 − 2 3 dµ(y) y x t/2 [(x y) + t ] Z| − |≤ − 2 2(t2 3 t ) 32 t t − 4 dµ(y)= µ x , x + (5.27) t2 2 3 4 ≥ y x t/2 ( + t ) 125t − 2 2 Z| − |≤ 4 32 t t 1 43β µ , 1 ≥ 125t4 −4 4 ≥ 4t4 − t3 for all x t/4. The positive part (5.25) is smaller than the negative part (5.27) | | ≤ 1/3 if we take t in such a way that t 10β . Thus f ′′(x) < 0 for all x t/4 and ≥ t | | ≤ t 10β1/3. ≥ 5.2. Positive stable process with index 1/2 with initial distribution. A positive stable process with index 1/2 has the distribution t2 t e− 2x Lt(dx) := 1(0, )(x) dx, x R, (5.28) √2π · x3/2 · ∞ ∈ at time t 0 which is called the L´evy distribution. We restrict to the case where the initial≥ distribution is compactly supported. Theorem 5.3. If µ is a compactly supported on R with diameter Dµ then µ Lt 1/4 1/2 ∗ is unimodal for all t (90/4) Dµ . ≥ Proof : By performing a translation we may assume that µ is supported on [0,γ], where γ = Dµ. We set the following function: t2 2(x y) e− − gt,y(x) := 1(0, )(x y), x,y R. (5.29) (x y)3/2 ∞ − ∈ − Note that this function is C∞ with respect to x. Consider the following function √2π d(µ L ) γ f (x) := ∗ t (x)= g (x) dµ(y), x R, (5.30) t t · dx t,y ∈ Z0 which is supported on (0, ) and has the derivative ∞ γ 2 d t 3(x y) t2 2(x y) ft(x)= − − e− − 1(0, )(x y) dµ(y). (5.31) dx 2(x y)7/2 ∞ − Z0 − If 0 2 2 t 3(t /3 + γ)+3γ = 0 for all 0 y γ, and therefore ft′(x) < 0. For 2 − 2 ≤ ≤ t /3 Proposition 5.4. There exists a probability measure µ on R such that µ Lt is not unimodal for any t> 0, and ∗ 5 x p dµ(x) < , 0 ∞ µ = wnδan . (5.37) n=1 X Let t2 2(x an) √2π d(µ L ) e− − f (x) := ∗ t (x)= w . (5.38) t t dx n (x a )3/2 n 1,a 2 54√6 3 2 f ′(a + t /6) = w e− w h(a a + t /6). t k k · t5 − n k − n (5.40) n:n=k X6 2 an