Large time unimodality for classical and free Brownian motions with initial distributions

Transcription

1 ALEA, La. Am. J. Probab. Mah. Sa. 5, (08) DOI: 0757/ALEA.v5-5 Large ime unimodaliy for classical and free Brownian moions wih iniial disribuions Takahiro Hasebe and Yuki Ueda Deparmen of Mahemaics, Hokkaido Universiy, Kia 0, Nishi 8, Kia-Ku, Sapporo, Hokkaido, , Japan address: UL: hps:// Deparmen of Mahemaics, Hokkaido Universiy, Kia 0, Nishi 8, Kia-Ku, Sapporo, Hokkaido, , Japan address: Absrac. We prove ha classical and free Brownian moions wih iniial disribuions are unimodal for sufficienly large ime, under some assumpion on he iniial disribuions. The assumpion is almos opimal in some sense. Similar resuls are shown for a symmeric sable process wih index and a posiive sable process wih index /. We also prove ha free Brownian moion wih iniial symmeric unimodal disribuion is unimodal, and discuss srong unimodaliy for free convoluion.. Inroducion A Borel measure µ on is unimodal if here exis a and a funcion f: [0, ) which is non-decreasing on (,a) and non-increasing on (a, ), such ha µ(dx) = µ({a})δ a +f(x)dx. (.) The mos ousanding resul on unimodaliy is Yamazao s heorem (Yamazao, 978) saying ha all classical selfdecomposable disribuions are unimodal. Afer his resul, in Hasebe and Thorbjørnsen (06), Hasebe and Thorbjørnsen proved he free analog of Yamazao s resul: all freely selfdecomposable disribuions are unimodal. The unimodaliy has several oher similariy poins beween he classical and free probabiliy heories Hasebe and Sakuma (07). However, i is no rue ha he unimodaliy shows complee similariy beween he classical and free probabiliy heories. For example, classical compound Poisson processes are likely o be eceived by he ediors Ocober 7h, 07; acceped March 3h, Mahemaics Subjec Classificaion. Primary 46L54; Secondary 60E07; 60G5, 60J65. Key words and phrases. unimodal, srongly unimodal, Brownian moion, Cauchy process, posiive sable process. esearch suppored by JSPS Gran-in-Aid for Young Scieniss (B) 5K7549 (for TH) and JSPS and MAEDI Japan France Inegraed Acion Program (SAKUA) (for boh auhors). 353

2 354 T. Hasebe and Y. Ueda non-unimodal in large ime (Wolfe, 978), while free Lévy processes wih compac suppor become unimodal in large ime (Hasebe and Sakuma, 07). In his paper, we mainly focus on he unimodaliy of classical and free Brownian moions wih iniial disribuions and consider wheher he classical and free versions share similariy poins or no. In free probabiliy heory, he semicircle disribuion is he free analog of he normal disribuion. In random marix heory, i appears as he limi of eigenvalue disribuions of Wigner marices as he size of he random marices goes o infiniy. Free Brownian moion is defined as a process wih free independen incremens, sared a 0 and disribued as he cenered semicircle disribuion S(0,) a ime > 0. Furhermore, we can provide free Brownian moion wih iniial disribuion µ which is a process wih free independen incremens disribued as µ a = 0 and as µ S(0,) a ime > 0. The definiion of, called addiive free convoluion, is provided in Secion.. The addiive free convoluion is he disribuion of he sum of wo random variables which are freely independen (for he definiion of free independence see Nica and Speicher, 006). In Biane (997), Biane gave he densiy funcion formula of free Brownian moion wih iniial disribuion (see Secion.). In our sudies, we firs consider he symmeric Bernoulli disribuion µ := δ + + δ as an iniial disribuion and compue he densiy funcion of µ S(0,). Then we see ha he probabiliy disribuion µ S(0,) is unimodal for 4 (and i is no unimodal for 0 < < 4). This compuaion leads o a naural problem: Problem.. Forwhich classofprobabiliymeasuresµon does he free Brownian moion wih iniial disribuion µ become unimodal for sufficienly large ime? We hen answer o his problem as follows (formulaed as Theorem 3. and Proposiion 3.4 in Secion 3.): Theorem.. () Le µ be a compacly suppored probabiliy measure on and D µ := sup{ x y : x,y supp(µ)}. Then µ S(0,) is unimodal for 4Dµ. () Le f: [0, ) be a Borel measurable funcion. Then here exiss a probabiliy measure µ on such ha µ S(0,) is no unimodal for any > 0 and f(x)dµ(x) <. (.) Noe ha such a measure µ is no compacly suppored by (). The funcion f can grow very fas such as e x, and so, a ail decay of he iniial disribuion does no imply large ime unimodaliy. The corresponding classical problem is naural, ha is, for which class of iniial disribuions on does Brownian moion become unimodal for sufficienly large ime > 0? We prove in his direcion he following resuls (formulaed as Theorem 4.3 and Proposiion 4. in Secion 4, respecively): Theorem.3. () Le µ be a probabiliy measure on such ha α := e εx dµ(x) < (.3) for some ε > 0. Then he disribuion µ N(0,) is unimodal for all 36 log(α) ε.

3 Unimodaliy for classical and free Brownian moions 355 () There exiss a probabiliy measure µ on saisfying ha e A x p dµ(x) < for all A > 0 and 0 < p < (.4) such ha µ N(0,) is no unimodal for any > 0. Thus, in he classical case, he ail decay (.3) is sufficien and almos necessary o guaranee he large ime unimodaliy. From hese resuls, we conclude ha he classical and free versions of he problems of unimodaliy for Brownian moions wih iniial disribuions share common feaures as boh become unimodal in large ime under each assumpion. Furher relaed resuls in his paper are as follows. In Secion 3. we prove ha µ S(0,) is always unimodal whenever µ is symmeric around 0 and unimodal (see Theorem 3.6). In Secion 3.3, we define freely srongly unimodal probabiliy measures as a naural free analogue of srongly unimodal probabiliy measures. We conclude ha he semicircle disribuions are no freely srongly unimodal (Lemma 3.9). More generally, we have he following resul (resaed as Theorem 3.0 laer): Theorem.4. Le λ be a probabiliy measure wih finie variance, no being a dela measure. Then λ is no freely srongly unimodal. On he oher hand, here are many srongly unimodal disribuions wih finie variance in classical probabiliy including he normal disribuions and exponenial disribuions. Thus he noion of srong unimodaliy breaks he similariy beween he classical and free probabiliy heories. In Secion 5, we focus on oher processes wih classical/free independen incremens wih iniial disribuions. We consider a symmeric sable process wih index and a posiive sable process wih index /. Then we prove large ime unimodaliy similar o he case of Brownian moion bu under differen ail decay assumpions.. Preliminaries.. Definiion of free convoluion. Le µ be a probabiliy measure on. The Cauchy ransform G µ (z) := dµ(x) (.) z x is analyic on he complex upper half plane C +. We define he runcaed cones Γ ± α,β := {z C± : e(z) < α Im(z), Im(z) > β}, α > 0,β. (.) In Bercovici and Voiculescu (993, Proposiion 5.4), i was proved ha for any γ < 0 here exis α,β > 0 and δ < 0 such ha G µ is univalen on Γ + α,β and Γ γ,δ G µ(γ + α,β ). Therefore he righ inverse funcion G µ exiss on Γ γ,δ. We define he -ransform of µ by µ (z) := G µ (z) z, z Γ γ,δ. (.3) Then for any probabiliy measures µ and ν on here exiss a unique probabiliy measure λ on such ha λ (z) = µ (z)+ ν (z), (.4)

4 356 T. Hasebe and Y. Ueda for all z in he inersecion of he domains of he hree ransforms. We denoe λ := µ ν and call i he (addiive) free convoluion of µ and ν... Free convoluion wih semicircle disribuions. Almos all he maerials and mehods in his secion are following (Biane, 997). The semicircle disribuion S(0,) of mean 0 and variance > 0 is he probabiliy measure wih densiy 4 x, (.5) π suppored on he inerval [, ]. We hen compue is Cauchy ransform and is -ransform (see, for example, Nica and Speicher, 006): G S(0,) (z) = z z 4, z C +, (.6) where he branch of he square roo on C\ + is such ha = i. Moreover S(0,) (z) = z, z C. (.7) Le µ be a probabiliy measure on and le µ = µ S(0,). Then we define he following se: { U := u (x u) dµ(x) > }, (.8) and he following funcion from o [0, ) by seing { v (u) := inf v 0 (x u) +v dµ(x) }. (.9) Then v is coninuous on and one has U = {x v (u) > 0}. Moreover, for every u U, v (u) is he unique soluion v > 0 of he equaion (x u) +v dµ(x) =. (.0) We define Ω,µ := {x+iy C y > v (x)}. (.) Then he map H (z) := z + S(0,) (G µ (z)), defined on C +, is a homeomorphism from Ω,µ o C + and i is conformal from Ω,µ ono C +. Hence here exiss an inverse funcion F : C + Ω,µ. By Biane (997, Lemma, Proposiion ), he domain Ω,µ is equal o he conneced componen of he se H (C + ) which conains iy for large y > 0 and for all z C + one has G µ (z) = G µ (H (F (z))) = G µ ( F (z)+ S(0,) ( Gµ (F (z)) )) = G µ (F (z)). (.) MoreoverG µ hasaconinuousexensiono C +. Then µ is Lebesgueabsoluely coninuous by Sieljes inversion and he densiy funcion p : [0, ) of µ is given by where p (ψ (x)) = π lim Im(G µ (ψ (x)+iy)) = v (x), x, (.3) y +0 π (x u) ψ (x) = H (x+iv (x)) = x+ (x u) dµ(u). (.4) +v (x)

5 Unimodaliy for classical and free Brownian moions 357 I can be shown ha ψ is a homeomorphism of. Furhermore he opological suppor of p is given by ψ (U ), and p exends o a coninuous funcion on and is real analyic in {x : p (x) > 0}. Example.. Le µ := δ + δ + be he symmeric Bernoulli disribuion. ecall ha he opological suppor of he densiy funcion p is equal o ψ (U ). If 0 <, we have U = u + +8 < u < ++ +8, (.5) and if >, we have U = u < u < +8. (.6) If 0 < hen he number of conneced componens of U is wo, and herefore µ is no unimodal. If > hen U is conneced. By he densiy formula (.3), he unimodaliy of µ is equivalen o he unimodaliy of he funcion v (u), he laer being expressed in he form (u v (u) = + )+ +6u. (.7) Calculaing is firs derivaive hen shows ha µ is unimodal if and only if 4; see Figures..6. According o Example., we have a quesion abou wha class of iniial disribuions implies he large ime unimodaliy of free Brownian moion. We will give a resul for his in Secion Figure.. p (x) Figure.. p (x) Figure.3. p (x) Figure.4. p 3 (x) Figure.5. p 4 (x) Figure.6. p 7 (x)

6 358 T. Hasebe and Y. Ueda.3. A basic lemma for unimodaliy. A unimodal disribuion is allowed o have a plaeau or a disconinuous poin in is densiy such as he uniform disribuion or he exponenial disribuion. If we exclude such cases, hen unimodaliy can be characerized in erms of levels of densiy. Lemma.. Le µ be a probabiliy measure on ha is Lebesgue absoluely coninuous. Suppose ha is densiy p(x) exends o a coninuous funcion on and is real analyic in {x : p(x) > 0}. Then µ is unimodal if and only if, for any a > 0, he equaion p(x) = a has a mos wo soluions x. The proof is jus o use he inermediae value heorem and he fac ha a real analyic funcion never has plaeaux. Noe ha he idea of he above lemma was firs inroduced by Haagerup and Thorbjørnsen(04) o prove ha a free gamma disribuion is unimodal. 3. Free Brownian moion wih iniial disribuion 3.. Large ime unimodaliy for free Brownian moion. The semicircular disribuion is he free analogue of he normal disribuion. Therefore, a process wih free independen incremens is called (sandard) free Brownian moion wih iniial disribuion µ if is disribuion a ime is given by µ S(0,). In his secion, we firsly prove ha free Brownian moion wih compacly suppored iniial disribuion is unimodal for sufficienly large ime, by using Biane s densiy formula (see Secion.). Lemma 3.. Suppose ha > 0. The following saemens are equivalen. () µ S(0,) is unimodal. () For any > 0 he equaion (x u) + dµ(x) = has a mos wo soluions u. (3.) Proof: We adop he noaions µ = µ S(0,),v,p and ψ in Secion.. This proofis similar o Hasebe and Thorbjørnsen (06, Proposiion 3.8). ecall ha µ is absoluely coninuous wih respec o Lebesgue measure and he densiy funcion p is coninuous on since is Cauchy ransform G µ has a coninuous exension o C +, and he densiy funcion is real analyic in ψ (U ) and coninuous on by Secion.. Suppose () firs. Since ψ is a homeomorphism of, by Lemma. i suffices o show ha for any a > 0 he equaion a = p (ψ (u)) = v (u), u (3.) π has a mos wo soluions u. Since v (u) is a unique soluion of he equaion (.0) if i is posiive, hen we have { u a = v } { (u) = u π (x u) +(πa) dµ(x) = }. (3.3) By he assumpion (), he equaion a = v(u) π has a mos wo soluions in.

7 Unimodaliy for classical and free Brownian moions 359 Conversely, suppose ha () does no hold. Then for some > 0 here are hree disinc soluions u,u,u 3 o (3.). By (3.3) his shows ha v (u i ) = and hence p (ψ (u i )) = π for i =,,3. Again by Lemma., µ is no unimodal. Now we are ready o prove Theorem. (). Theorem 3.. Le µ be a compacly suppored probabiliy measure, and le D µ be he diameer of he suppor: D µ = sup{ x y : x,y supp(µ)}. Then µ S(0,) is unimodal for 4D µ. Proof: Applying a shif we may assume ha µ is suppored on [ Dµ, Dµ ]. For fixed > 0, we define by ξ (u) := (x u) dµ(x), u. (3.4) + Then we have ξ (u) = Dµ Dµ Dµ ξ (u) = Dµ (x u) {(x u) + } dµ(x), 6(x u) {(x u) + } 3 dµ(x). (3.5) If u < Dµ, hen x u > Dµ ( Dµ ) = 0 for all x supp(µ), so ha ξ (u) > 0. If u > Dµ Dµ, hen x u < Dµ = 0 for all x supp(µ), so ha ξ (u) < 0. We ake 4Dµ and consider he form of ξ (u) on u ( Dµ, Dµ ). If 0 < < 3D µ, hen (x u) + < ( Dµ ξ (u) = + Dµ ) +( 3D µ ) = 4D µ, so ha (x u) + dµ(x) > (, u D µ, D ) µ. (3.6) Hence he equaion ξ (u) = has a mos wo soluions u if 4D µ. If 3D µ, hen he funcion ξ (u) saisfies he following hree condiions: ξ (u) > 0 for all u < Dµ and ξ Dµ (u) < 0 for all u >, ξ is coninuous on, ( ) ξ (u) < 0 for all u Dµ, Dµ. By he inermediae value heorem, ξ has a unique zero in [ Dµ, Dµ ] and hence ξ akes a unique local maximum in he inerval [ Dµ, Dµ ] (and herefore i is a unique global maximum on ). Therefore he equaion ξ (u) = has a mos wo soluions if 4Dµ. Hence we have ha he free convoluion µ S(0,) is unimodal if 4Dµ by Lemma 3.. Problem 3.3. Wha is he opimal universal consan C > 0 such ha µ S(0,) is unimodal for all CDµ and all probabiliy measures µ wih compac suppor? We have already shown ha C 4. ecall from Example. ha if µ is he symmeric Bernoulli disribuion on {,}, hen µ S(0,) is unimodal if and only if 4. Since he diameer D µ of he suppor of µ is, we conclude ha C 4. The nex quesion is wheher here exiss a probabiliy measure µ such ha µ S(0,) is no unimodal for any > 0 or a leas for sufficienly large > 0.

8 360 T. Hasebe and Y. Ueda Such a disribuion mus have an unbounded suppor if i exiss. Such an example can be consruced wih an idea similar o Huang (05, Proposiion 4.3) (see also Hasebe and Sakuma, 07, Example 5.3). Proposiion 3.4. Le f: [0, ) be a Borel measurable funcion. Then here exiss a probabiliy measure µ such ha µ S(0,) is no unimodal a any > 0 and f(x)dµ(x) <. (3.7) Proof: We follow he noaions in Secion.. Le {w n } n and {a n } n be sequences in saisfying w n > 0, n= w n =, a n+ > a n, n, lim (a n+ a n ) =. n Consider he probabiliy measure µ := n= w nδ an on and define he funcion X µ (u) := (u x) dµ(x) = w n (u a n ). (3.8) ecall ha U is he se of u such ha X µ (u) > /. Se b k := a k++a k for all k N. Then we have and so n= b k a n a k+ a k, k,n N (3.9) ( ) ( ) X µ (b k ) w n = 0 a k+ a k a k+ a k n= as k. This means ha for each > 0 here exiss an ineger K() > 0 such ha X µ (b k ) < for all k K(). This implies ha, for k K(), he closure of U does no conain b k. Noe ha he se U is an open se. Therefore, here exiss a sequence {ε k ()} k=k()+ of posiive numbers such ha U (b K(), ) = k=k()+ (a k ε k (),a k +ε k ()), (3.0) where he closures of he inervals are disjoin. Thus he se supp(µ ) = ψ (U ) consiss of infiniely many conneced componens for all > 0, and hence µ is no unimodal for any > 0. Inheaboveconsrucion,heweighsw n areonlyrequiredobeposiive. Therefore, we may ake c w n = n max{f(a n ),}, (3.) where c > 0 is a normalizing consan. Then he inegrabiliy condiion (3.7) is saisfied. Proposiion 3.4 shows ha ail decay esimaes of he iniial disribuion do no guaranee he large ime unimodaliy. In Secion 4 we shall see ha he siuaion is differen for classical Brownian moion.

9 Unimodaliy for classical and free Brownian moions Free Brownian moion wih symmeric iniial disribuion. This secion proves a unimodaliy resul of differen flavor. I is well known ha if µ and ν are symmeric unimodal disribuions, hen µ ν is also (symmeric) unimodal (see Sao, 03, Exercise 9.). The free analogue of his saemen is no known. Conjecure 3.5. Le µ and ν be symmeric unimodal disribuions. Then µ ν is unimodal. We can give a posiive answer in he special case when one disribuion is a semicircle disribuion. Theorem 3.6. Le µ be a symmeric unimodal disribuion on. Then µ S(0,) is unimodal for any > 0. emark 3.7. There is a unimodal probabiliy measure µ such ha µ S(0,) is no unimodal (see Lemma 3.9). Hence we canno remove he assumpion of symmery of µ. Proof of Theorem 3.6. By Lemma 3. i suffices o show ha for any > 0 he equaion (u x) + dµ(x) = (3.) has a mos wo soluions u. Up o a consan muliplicaion, he LHS is he densiy of µ C, which is unimodal since µ and C are symmeric unimodal. Since he densiy of µ C is real analyic, Lemma. implies ha he equaion (3.) has a mos wo soluions. This shows ha µ S(0,) is unimodal Freely srong unimodaliy. In classical probabiliy, a probabiliy measure is said o be srongly unimodal if µ ν is unimodal for all unimodal disribuions ν on. A disribuion is srongly unimodal if and only if he disribuion is Lebesgue absoluely coninuous, suppored on an inerval and a version of is densiy is log-concave (see Sao, 03, Theorem 5.3). From his characerizaion, he normal disribuions are srongly unimodal. We discuss he free version of srong unimodaliy. Definiion 3.8. Aprobabiliymeasureµon issaid obe freely srongly unimodal if µ ν is unimodal for all unimodal disribuions ν on. Lemma 3.9. The semicircle disribuions are no freely srongly unimodal. Proof: The Cauchy disribuion is no srongly unimodal since is densiy is no log-concave. Hence, here exiss a unimodal probabiliy measure µ such ha µ C is no unimodal. Since he densiy of µ C is real analyic on, by Lemma. here exiss > 0 such ha he equaion π d(µ C ) (x) = dx (x y) + dµ(y) = (3.3) has a leas hree disinc soluions x,x,x 3. Lemma 3. hen implies ha S(0,) µ is no unimodal. By changing he scaling, we conclude ha no semicircle disribuions are freely srongly unimodal. Theorem 3.0. Le λ be a probabiliy measure wih finie variance, no being a dela measure. Then λ is no freely srongly unimodal.

10 36 T. Hasebe and Y. Ueda Proof: We may assume ha λ has mean 0. Suppose ha λ is freely srongly unimodal. A simple inducion argumenshows ha λ n is freely sronglyunimodal for all n Z +. We derive a conradicion below. By Lemma 3.9 we can ake a unimodal probabiliy measure µ such ha µ S(0, ) is no unimodal. Our hypohesis shows ha he measure D nv(µ) λ n is unimodal, where v is he variance of λ and D c (ρ) is he push-forward of a measure ρ by he map x cx for c. By he free cenral limi heorem (Maassen, 99, Theorem 5.), he unimodal disribuions D / nv (D nv(µ) λ n ) = µ D / nv (λ n ) weakly converge o µ S(0,) as n. Since he se of unimodal disribuions is weakly closed (see Sao, 03, Exercise 9.0), we conclude ha µ S(0,) is unimodal, a conradicion. Problem 3.. Does here exis a freely srongly unimodal probabiliy measure ha is no a dela measure? 4. Classical Brownian moion wih iniial disribuion This secion discusses large ime unimodaliy for classical sandard Brownian moion wih an iniial disribuion µ, which is a process wih independen incremens disribued as µ N(0,) a ime 0. Before going o he general case, one example will be helpful in undersanding unimodaliy for µ N(0,). Example 4.. Elemenary calculus shows ha (δ +δ ) N(0,) (4.) is unimodal if and only if ; see Figures Figure 4.7. =.0 Figure 4.8. =.0 Figure 4.9. = Figure 4.0. = Figure 4.. = Figure 4.. = 4 This simple example suggess ha Brownian moion becomes unimodal for sufficienly large ime, possibly under some condiion on he iniial disribuion. In some sense, we give an almos opimal condiion on he iniial disribuion for he large ime unimodaliy o hold. We sar by providing an example of iniial disribuion wih which he Brownian moion never becomes unimodal.

11 Unimodaliy for classical and free Brownian moions 363 Proposiion 4.. There exiss a probabiliy measure µ on such ha µ N(0,) is no unimodal a any > 0, and e A x p dµ(x) < (4.) for all A > 0 and 0 < p <. Proof: We consider sequences {w n } n N (0, ) and {a n } n N such ha w n =, (4.3) n= b k := inf a k a n as k. (4.4) n Z +\{k} Moreover we assume ha for all > 0 here exiss k 0 = k 0 () N such ha k k 0 () w k e > bk e b k. (4.5) Define a probabiliy measure µ by seing µ := n= w nδ an and he following funcion: Then we have f (x) := π d(µ N(0,)) (x) = dx n= n= w n e (x an), x. (4.6) f (x) = ( w n x a n )e (x an), x. (4.7) If needed we may replace k 0 () by a larger ineger so ha b k > holds for all k k 0 (). For k k 0 () we have f (a k ) = w ( k e ak a n w n )e (a k an ) n N,n k w k e n N,n k w k e n N,n k w n a k a n w n b k e b ) (w k e bk e b k > 0, k e (a k an ) (4.8) where we used he fac ha x x e x akes he global maximum a x = ± on he hird inequaliy and he assumpion (4.5) on he las line. This implies ha µ N(0,) is no unimodal for any > 0. Nex we akespecific sequences {w n } n and {a n } n saisfying he condiions(4.3) (4.5). Se a k = a k, k N where a. Noe ha here exiss some consan c > 0 such ha b k ca k for all k N. Hence we have ha b k as k. Moreover we se w k := Me b kk, (4.9)

12 364 T. Hasebe and Y. Ueda where M > 0 is a normalized consan, ha is, M = ( k= e b kk ) (noe ha he series converges). Since b k as k, we have b k e b w k k = M b ke b k k b k 0, as k. (4.0) For all > 0 here exiss k 0 = k 0 () N such ha k k 0 implies ha e > b k e b k /w k. Therefore he sequences {w n } n and {a n } n saisfy he condiion (4.5). Finally we show ha µ = n= w nδ an has he propery (4.). For every A > 0 and 0 < p <, using he inequaliy b k ca k shows ha e A x p dµ(x) = w k e A a k p = M e k b k e Aa kp Thus he proof is complee. k= M k= e Aapk (c A k a ( p)k ) <. k= (4.) Noe ha for any sequences {w n } n and {a n } n saisfying he condiions (4.3) (4.5) he disribuion µ = n= w nδ an has he following propery: e εx dµ(x) = for all ε > 0. (4.) Then we have a naural quesion wheher he large ime unimodaliy of µ N(0,) holds or no if he iniial disribuion µ does no saisfy he condiion (4.). We solve his quesion as follows. Theorem 4.3. Le µ be a probabiliy measure on such ha α := e εx dµ(x) <. (4.3) for some ε > 0. Then µ N(0,) is unimodal for 36log(α) ε. emark 4.4. The proof becomes much easier if we assume ha µ has a compac suppor. Proof of Theorem 4.3: For x > 0 we have µ( y > x) = dµ(y) Le Then we have f (x) = = y >x y >x e εy dµ(y) αe εx. (4.4) e εx f (x) := π d(µ N(0,)) (x) = e (x y) dµ(y). (4.5) dx x For x > 0, we have x ( y x )e (y x) dµ(y) y x e (y x) dµ(y) y x e (y x) dµ(y) x x y e (x y) dµ(y). (4.6) e µ((x, )) α e εx. (4.7)

13 Unimodaliy for classical and free Brownian moions 365 For x, we have x x y x e (x y) 3 dµ(y) 3x Now he funcion g(x) := 4xe 8x x y e (x y) dµ(y) { min 3 e 8,4xe 8x } [ ] µ( ) 3x,x. 3 (4.8) has a local maximum a x = 4, and hence for all x, g(x) e < 3 e 8, (4.9) where e and 3 e Hence we have x x y e (x y) dµ(y) 4x [ ] e 8x µ( ) 3x,x 3 4x [ ] e 8x µ( ) 6, 6 e 8x ( αe ε 36 ), (4.0) where he las inequaliy holds hanks o (4.4) and x. Therefore if x hen f (x) α e εx e 8x ( αe ε 36 ) { = α e εx e 8x α eεx ( αe ε 36 ) }. (4.) If 6 ε hen 8x eεx e εx e ε 8. Hence if 6 ε hen f (x) α e εx { e α e ε 8 ( αe ε 36 ) }. (4.) If 36log(α) ε hen αe ε 36 and we have By aking max{ 6 Similarly, we have e α e ε 8 ( αe ε 36 ) e 9 α 7 < 0. (4.3) ε, 36log(α) ε } = 36log(α) ε, we have f (x) < 0 if x. (4.4) f (x) > 0 if x. (4.5)

14 366 T. Hasebe and Y. Ueda Nex, we will show ha f (x) < 0 for all x wih x <. We hen calculae he following f (x) = = (x y) e (x y) dµ(y) x y > x y (x y) e (x y) dµ(y) (x y) e (x y) dµ(y). (4.6) Since he funcion h(u) := u e u has a local maximum a u = ± 3, we have (x y) e (x y) dµ(y) e 3 µ([x,x+ ] c ). (4.7) x y > For all x wih x <, we have [x,x+ ] c [, ] c, and herefore ] c ) x y > (x y) e (x y) dµ(y) e 3 µ ([, Since he funcion h(u) is decreasing on [0, 3], we have x y (x y) e (x y) dµ(y) x y (x y) e 4 9 ([ µ x = 5 9 e 9 µ ([ x 5 9 e 9 µ ([ 6, e 3 αe ε 4. (4.8) e (x y) dµ(y) 3,x+ ]) 3 3,x+ ]) 3 ]) e 9 ( αe ε 36 ). (4.9) Therefore f (x) e 3 αe ε e 9 ( αe ε 36 ) = { e 3 αe ε e 9 ( αe ε 36 ) }. (4.30) If 36log(α) ε, hen e 3 αe ε e 9 ( αe ε 36 ) e 3 (α) e 9 < 0, (4.3) and herefore f (x) < 0 if x <.

15 Unimodaliy for classical and free Brownian moions 367 To summarize, if 36log(α) ε, hen we have he following properies: x f (x) > 0, x f (x) < 0, (4.3) x < f (x) < 0. ( Hence f (x) has a unique local maximum in, ) when 36log(α) ε, which is a unique global maximum on as well. Therefore µ N(0,) is unimodal for 36log(α) ε. emark 4.5. If µ is unimodal hen µ N(0,) is unimodal for all > 0. This is a consequence of he srong unimodaliy of he normal disribuion N(0, ) (see Secion 3.3), in conras wih he failure of freely srong unimodaliy of he semicircle disribuion (see Lemma 3.9). We close his secion by placing a problem for fuure research. Problem 4.6. Esimae he posiion of he mode of classical Brownian moion wih iniial disribuions saisfying he assumpion (4.3). Our proof shows ha for 36 ε log(α), he mode is locaed in he inerval [ /, /]. How abou free Brownian moion? 5. Large ime unimodaliy for sable processes wih index and index / wih iniial disribuions We invesigae large ime unimodaliy for sable processes wih index (following Cauchy disribuions) and index / (following Lévy disribuions) wih iniial disribuions. 5.. Cauchy process wih iniial disribuion. Le {C } 0 be he symmeric Cauchy disribuion C (dx) := π(x + ) (x)dx, x, C 0 = δ 0, (5.) which forms boh classical and free convoluion semigroups. A Cauchy process wih iniial disribuion µ is a process wih independen incremens ha follows he law µ C a ime 0. I is known ha he Cauchy disribuion saisfies he ideniy µ C = µ C (5.) foranyµand 0, and sohe disribuions µ C can alsobe realizedas he lawsa ime 0 of a process wih free independen incremens wih iniial disribuion µ. The auhors do no know a wrien proof of (5.) in he lieraure, so give a proof below. The Cauchy ransform of he Cauchy disribuion is given by G C (z) = /(z + i) on C + (e.g. by he residue heorem), and hence C (z) = iz. Then µ C (z) = µ (z) iz, and afersome compuaionwe can checkha G µ C (z) = G µ (z+i). The Sieljes inversionformulaimplies ha for > 0he freeconvoluion µ C has he densiy π Im(G µ(x+i)) = π((x y) + dµ(y), (5.3) )

16 368 T. Hasebe and Y. Ueda which is exacly he densiy of µ C. As in he cases of free and classical BMs, aking µ o be he symmeric Bernoulli (δ +δ ) is helpful. By calculus we can show ha µ C is unimodal if and only if 3; see Figures Thus i is again naural o expec ha a Cauchy process becomes unimodal for sufficienly large ime, under some condiion on he iniial disribuion. We sar from he following counerexample Figure 5.3. = 0.7 Figure 5.4. = 0.7 Figure 5.5. = Figure 5.6. = Figure 5.7. = 3 Figure 5.8. = 3 Proposiion 5.. There exiss a probabiliy measure µ on such ha µ C is no unimodal for any > 0, and x p dµ(x) <, 0 < p < 3. (5.4) Proof: Le {w n } n be a sequence of posiive numbers such ha n= w n = and {a n } n be a sequence of real numbers. Consider he probabiliy measure µ = w n δ an. (5.5) Suppose ha he sequence b k = n= inf a k a n, k Z + (5.6) n Z +\{k} saisfies he condiion lim w kb 3 k =. (5.7) k When {a n } is increasing, his condiion means ha he disance beween a n and a n+ grows sufficienly fas. Le f (x) := π d(µ C ) w n (x) = dx (x a n ) +. (5.8) Then we obain f (x) = n= n= w n (x a n ) [(x a n ) + ], (5.9)

17 Unimodaliy for classical and free Brownian moions 369 and so for sufficienly large k such ha b k > 0 and for each > 0 f (a k ) = w k (+ ) + n,n k w k (+ ) n,n k w k (+ ) w k (+ ) b 3 k w n b 3 k ( = w k n,n k w n (a k a n ) [(a k a n ) + ] w n a k a n 3 (+ ) w k b 3 k ). (5.0) The condiion (5.7) shows ha f (a k ) is posiive for sufficienly large k Z +. This shows ha µ C is no unimodal for any > 0. If we ake he paricular sequences a n = a n and w n = cn r a 3n, where a,r > 0 and c > 0 is a normalizing consan, hen he sequence b n saisfies b n Ca n for some consan C > 0 independen of n. Then he condiions (5.7) and (5.4) hold rue. In he above consrucion, for any posiive weighs {w n } n and any sequence {a n } n ha saisfies (5.7), he hird momen of µ is always infinie, x 3 dµ(x) =, (5.) due o he inequaliy a k b k a and he condiion (5.7). The nex quesion is hen wheher here exiss a probabiliy measure µ wih a finie hird momen such ha µ C is no unimodal for any > 0, or a leas for sufficienly large > 0. The complee answer is given below. Theorem 5.. Le µ be a probabiliy measure on which has a finie absolue hird momen β := x 3 dµ(x) <. Then µ C is unimodal for 0β 3. Proof: The Markov inequaliy implies he ail esimae I suffices o prove ha he funcion y>x f (x) := π µ([ x,x] c ) β x3, x > 0. (5.) d(µ C ) (x) = dx (x y) dµ(y) (5.3) + has a unique local maximum for large > 0. Suppose firs ha x > 0. The derivaive f splis ino he posiive and negaive pars f (x) (y x) = [(x y) + ] dµ(y) (x y) [(x y) + dµ(y). (5.4) ] y x

18 370 T. Hasebe and Y. Ueda By calculus he funcion u u (u + ) akes a global maximum a he unique poin u = / 3. Using he ail esimae (5.) yields he esimae of he posiive par y>x y x (y x) [(x y) + ] dµ(y) y>x 3 ( 3 + ) dµ(y) β 3 x 3. (5.5) On he oher hand he negaive par can be esimaed as (x y) [(x y) + ] dµ(y) (x y) [(x y) + dµ(y). ] (5.6) x<y<x/ Elemenary calculus shows ha for x < y < x/, { (x y) [(x y) + ] min and if we furher resric o he case x /4, hen { } 4x x min (4x + ), (x /4+ ) { min 4x x (4x + ), (x /4+ ) 4x x (4x +6x ), (x /4+6x ) } 0 3 x 3. }, (5.7) Thus we obain ( (x y) y x [(x y) + dµ(y) µ(( x/,x/)) ] x3 x 3 β ) (x/) 3 ( ) 0 3 x 3 83 β 3, x /4. (5.8) (5.9) Comparing (5.5) and (5.9), aking 0β /3 guaranees ha he posiive par is smaller han he negaive par, and hence Similarly, if 0β /3 hen f (x) < 0, x 4. (5.0) f (x) > 0, x 4. (5.) In order o show ha f has a unique zero, i suffices o show ha f (x) < 0 for x /4. Now we have y x >/ 3 f (x) = y x >/ 3 y x / 3 [3(y x) ] [(x y) + ] 3 dµ(y) [ 3(y x) ] [(x y) + ] 3 dµ(y). (5.) The funcion u (3u )/(u + ) 3 aains a global maximum a u = ± and a global minimum a u = 0. Therefore, he posiive par can be esimaed as follows: [3(y x) ] [(x y) + ] 3 dµ(y) (3 ) ( + ) 3 dµ(y) y x >/ 3 = 4µ ([ x 3,x+ 3 ] c ). (5.3)

19 Unimodaliy for classical and free Brownian moions 37 For all x such ha x /4, we have he inclusion [ x,x+ ] c [ ], c, (5.4) and hence we obain y x >/ 3 y x / 3 [3(y x) ] [(x y) + ] 3 dµ(y) 4µ ([ 5, ] c ) 53 β 5 7. (5.5) On he oher hand, he negaive par has he esimae [ 3(y x) ] [(x y) + ] 3 dµ(y) [ 3(y x) ] [(x y) + dµ(y). (5.6) ] 3 y x / By calculus, he funcion u ( 3u )/(u + ) 3 is decreasing on [0,], and so [ 3(y x) ] y x / [(x y) + ] 3 dµ(y) ( 3 4 ) ([ 3 dµ(y) = y x / ( 4 + ) 3 5 4µ x,x+ ]) (5.7) 3 ([ 5 4µ 4, ]) ( ) β 3 for all x /4. The posiive par (5.5) is smaller han he negaive par (5.7) if we ake in such a way ha 0β /3. Thus f (x) < 0 for all x /4 and 0β / Posiive sable process wih index / wih iniial disribuion. A posiive sable process wih index / has he disribuion L (dx) := e x π x (0, )(x)dx, x, (5.8) 3/ a ime 0 which is called he Lévy disribuion. We resric o he case where he iniial disribuion is compacly suppored. Theorem 5.3. If µ is a compacly suppored on wih diameer D µ hen µ L is unimodal for all (90/4) /4 D / µ. Proof: By performing a ranslaion we may assume ha µ is suppored on [0,γ], where γ = D µ. We se he following funcion: g,y (x) := e (x y) (x y) 3/ (0, )(x y), x,y. (5.9) Noe ha his funcion is C wih respec o x. Consider he following funcion π f (x) := d(µ L γ ) (x) = g,y (x)dµ(y), x, (5.30) dx which is suppored on (0, ) and has he derivaive d dx f (x) = γ 0 0 3(x y) e (x y) (0, ) (x y)dµ(y). (5.3) (x y) 7/ If 0 < x < /3 hen 3(x y) > 3 /3 = 0 for all 0 y γ, and herefore f (x) > 0. Moreover, if /3+γ < x hen we have ha 3(x y) <

20 37 T. Hasebe and Y. Ueda 3( /3 + γ) + 3γ = 0 for all 0 y γ, and herefore f (x) < 0. For /3 < x < /3+γ, he second derivaive of f is given by f (x) = γ 0 5(x y) 0 (x y)+ 4 4(x y) / Noe ha /3 γ < x y < /3+γ. Since for X if 3 0 e (x y) (0, ) (x y)dµ(y). (5.3) 5X 0 X + 4 < 0 iff < X < , (5.33) γ hen f (x) < 0. To summarize, we have obained ha if 3 0 γ hen f (x) > 0, x < /3, f (x) < 0, x > /3+γ, f (x) < 0, /3 < x < /3+γ. Hence f has a unique local maximum in [ /3, /3+γ], which is a unique global maximum on as well. Therefore µ L is unimodal for all 3 0 γ. We give a counerexample for large ime unimodaliy for posiive sable processes of index / when he iniial disribuion is no compacly suppored. Proposiion 5.4. There exiss a probabiliy measure µ on such ha µ L is no unimodal for any > 0, and x p dµ(x) <, 0 < p < 5. (5.34) Proof: Le {w n } n be a sequence of posiive numbers such ha n= w n = and {a n } n be a sequence of real numbers such ha he sequence b k = inf n N\{k} a k a n, k N (5.35) saisfies lim w kb 5/ k k =. (5.36) Consider he probabiliy measure µ = w n δ an. (5.37) Le Then we obain f (x) := where h(x) = 3x e x 7/ π n= d(µ L ) (x) = dx n,a n<x f (x) = f (a k + /6) = w k e 3 n,a n<x x, and for each k N and each > 0 n:n k a n<a k + /6 (x an) e w n (x a n ) 3/. (5.38) w n h(x a n ), (5.39) w n h(a k a n + /6). (5.40)

21 Unimodaliy for classical and free Brownian moions 373 Since b k as k, here exiss a posiive ineger k() such ha b k + /6 > for all k k(). For k k() and n k such ha a n < a k + /6, we can show ha a k a n b k ; oherwise, by he definiion of b k i mus hold ha a k a n b k < , which conradics a n < a k + /6. Since he map h is posiive and sricly decreasing on ( , ), for k k() we have f (a k + /6) w k e 3 n:n k a n<a k + /6 w n h(b k + /6) (5.4) w k e 3 h(b k + /6) (5.4) = w k e 3 3b k / (b k + /6) 7/e (b k + /6). (5.43) The condiion (5.36) shows ha f (a k+ /6) is posiive for sufficienly large k N. This shows ha µ L is no unimodal for any > 0. If we ake he paricular sequences a k = a k and w k = cka 5 k where a > and c > 0 is a normalizing consan, hen he sequence b k saisfies b k Ca k for some consan C > 0. Then he condiion (5.36) holds rue and x p dµ(x) = w k a k p = c k k ka (p 5/)k. (5.44) Hence he above inegral is finie if and only if 0 < p < 5/. In he above consrucion, for any posiive weighs {w n } n and any sequence {a n } n ha saisfies (5.36), he 5/-h momen of µ is always infinie, ha is, x 5/ dµ(x) =. (5.45) We conjecure ha if he inegral in (5.45) is finie hen µ L is unimodal in large ime. More generally, considering resuls on Cauchy processes in Secion 5., i is naural o expec he following. Conjecure 5.5. Suppose ha S (resp. T ) is he law a ime 0 of a classical (resp. free) sricly sable process of index α (0,). If µ is a probabiliy measure such ha x +α dµ(x) <, (5.46) hen µ S (resp. µ T ) is unimodal for sufficienly large > 0. Acknowledgemens The auhors express sincere hanks o he referee who gave useful commens o improve he manuscrip. eferences H. Bercovici and D. Voiculescu. Free convoluion of measures wih unbounded suppor. Indiana Univ. Mah. J. 4 (3), (993). M546.

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