Proofs

(iv) ϕ^∗_u(t)is continuous in (u, t)∈[0,1]×R.

(v) The constants νu,C1,u,C2,u in (158) can be chosen to be independent ofu∈[0,1].

In particular,

sup

ξ∈R sup

u∈[0,1]ρu(ξ)<∞.

注 32 (i) Possibly, the constantb₀ in the theorem can be chosen to be independent of u∈ U, and the convergence in (159) may be uniform inu. This is left open.

(ii) Is it possible that (158) and related assumptions (originally from the second assumption in定理65) would imply P({0}) = 0 and the tightness of {Pn} on (0,∞) (namely, the mass of Pn do not accumulate at {0}), and consequently uniformity of convergence lim

n→∞Gn(s) =G(s)onRe(s)K, which did not hold in general from the assumptions in 定理78?

✸ The following is a direct consequence of定理85,定理87, and定理79.

系 88 Pn,u in定理85 converges weakly asn→ ∞toP_u^∗ uniformly inu∈[0,1].

This and (162) and (161) further imply, for|s|C_∞, λu|gk,u(s/λu)||s|

1 +Mk

λu |s|

C_∞

1 +Mk

λ₋C_∞

C.

Thereforeg_k+1,u(s) =g_1,u(λugk,u(s/λu)) is regular on|s|C_∞ and

|g_1,u(λ_ug_k,u(s/λ_u))−λ_ug_k,u(s/λ_u)| M|λugk,u(s/λu)|²M

|s|+^M_λ^k

u|s|²2

M

1 + ^M_λ^kC∞

−

2

|s|² M(1 +()²|s|²,

(167)

where we also used (162) and C_∞ (λ₋

M_∞ in the last line.

Using (166) and (167) we finally have

|g_k+1,u(s)−s||g_1,u(λugk,u(s/λu))−λugk,u(s/λu)|+λu|gk,u(s/λu)−s/λu|

M_k

λu +M(1 +()²

|s|²

M_k

λ₋ +M(1 +()²

|s|²=M_k+1|s|², |s|C_∞,

which implies (165) forn=k+ 1. By induction, we have (165) for alln. ✷

定理74 の証明. (i) Fixu∈ U and suppress the suffixuhere.

Since定理72 implies thatgn,n= 1,2,3,· · ·, are regular on|s|< C_∞, they have Mclaurin expansion:

gn(s) = ∞ k=1

an,ks^k.

Note that (140) implies an,1 = 1 for all n. Substituting the Mclaurin series in (139) and comparing coefficients ofs^k fork= 2,3,4,· · ·, we have (usingan,1= 1),

a_n+1,2= 1

λa_n,2+a_1,2, (168)

and

an+1,k= 1

λ^k−1an,k+ k

=2

a1,λ

j1+j2+···+j=k

m=1

an,jm

1

λ^k, k= 3,4,5,· · ·. (169) The assumptionλ=λuλ₋ >1 and (168) imply existence of limit

α2= lim

n→∞an,2. We proceed with induction onk and assume the existence of limit

α= lim

n→∞an,, ! < k.

Then the second term in the right hand side of (169) converges as n → ∞, hence (169) implies, with λ >1, existence of limit

α_k = lim

n→∞a_n,k. (170)

By induction, the limit (170) exists for allk= 2,3,4,· · ·.

Applying定理73 to{g_n}, we see that for any subsequence there exists subsubsequence{g_n}, which is uniformly convergent on compact subsets of{s∈C| |s|< C_∞}:

∃g^∗(s) = lim

→∞g_n(s) = lim

→∞

∞ k=1

a_n,ks^k= ∞ k=1

α_ks^k, |s|< C_∞.

(Limit and series is interchangeable because uniform convergence of regular functions implies interchange- ability of derivative and series.) Since the right hand side is independent of subsequences, the limitg^∗is in- dependent of subsequences. Therefore{g_n}converges uniformly on compact subsets of{s∈C| |s|< C_∞} to a regular functiong^∗(s) =

∞ k=1

α_ks^k.

Lettingn→ ∞in (139) and (140), and noting the uniformness of convergence of{gn}, we have (141) and (142).

(ii) As in the above proof, consider the McLaurin expansion h(s) =

∞ k=1

bks^k. Then (142) (forh) impliesbk = 1 and (141) implies

b₂= 1 λu

b₂+a_1,2, (171)

and

bk= 1 λ^ku⁻¹

bk+ k

=2

a_1,λ_u

j₁+j2+···+j=k

m=1

bj_m

1

λ^k_u , k= 3,4,5,· · ·. (172) The assumption λu >1 and (171) imply b₂ = a1,2

1−_λ¹_u , and by induction onk with (172), we find that b_k is uniquely determined for all k= 2,3,4,· · ·. Hence the two equations (141) and (142) determine h

completely, and by definition it must coincide withg^∗_u. ✷

定理78 の証明. Convergence of {Gn} implies that the sequence Gn(−C/2) = _∞

0

e^Cξ/2Pn(dξ), n = 1,2,3,· · ·, is bounded, which further implies, by命題76, that{Pn |n= 1,2,3,· · ·}is tight. Therefore for any subsequence ofPn,n= 1,2,3,· · ·, there exists a weakly converging subsubsequence. 命題77 implies that the limit measure is also supported on non-negative reals.

Suppose thatPn_kconverges weakly to a Borel probability measureP^∗supported on [0,∞) ask→ ∞. We shall prove thatP^∗is independent of subsequence{n_k}. Put ˜G^∗(s) =

[0,∞)

e^−sξP^∗(dξ). 命題75 then implies that ˜G^∗(s) is defined on complexssatisfyingRe(s)0, regular onRe(s)>0and continuous on Re(s) 0. Note that if Re(s) 0, then |e^−sξ| = e^−Re(s)ξ 1, ξ 0, hence e^−sξ is a bounded continuous function onξ∈[0,∞). Therefore weak convergence ofPn_k to P^∗ and the support properties of the measures imply

klim→∞Gnk(s) = ˜G^∗(s), Re(s)0. (173) Comparing (144) and (173), we see that the regular functionsG^∗ and ˜G^∗ coincide on {s∈C|Re(s)>

0, |s| < C}. Uniqueness of analytic continuation implies that ˜G^∗ is independent of the subsequence {n_k}. HenceG^∗is analytically continued toRe(s)>0, and uniquely continued as a contiguous function toRe(s)0, and satisfies

nlim→∞Gn(s) =G^∗(s), Re(s)0. (174) Characteristic function determines probability measure, hence P^∗ also is independent of the sub- sequence {nk}. This proves that Pn converges weakly as n → ∞ to a Borel probability measure P^∗ supported on non-negative reals.

We turn to a proof that the convergence (174) is uniform on compact sets in{s∈ C|Re(s)0}. LetK, K>0 and consider (174) on J ={s∈C|0Re(s)K, |Im(s)|K}, using an idea in the proof of uniform convergence of characteristic functions [5, Theorem 5.3]. Let( >0 . Since{Pn}is tight, there existsL >0 such that Pn((L,∞))< (,n= 1,2,3,· · ·, and P((L,∞))< (. Sincee^−sξ is uniformly continuous (as a function of complex s and real ξ) on {s ∈ C | |s| 1} ×[0, L], sup

0ξL

|e^−sξ−1| is continuous on|s|1 . Therefore there existsδ >0 such that

sup

0ξL

|e^−sξ−1|< (, |s|< δ.

Hence if|s|< δ,

|Gn(s+s)−Gn(s)| _∞

0

e^−Re(s)ξ|e^−sξ−1|Pn(dξ) _∞

0 |e^−sξ−1|Pn(dξ)(+ 2Pn((L,∞)) 3(, Re(s)0, |s|< δ, n=1,2,3,· · ·,

and a similar estimate forG^∗.

SinceJ is a compact set, there exists a finite number, say!, of pointss1, s2,· · ·, s inJ such that for anys∈J there exists 1j!such that|s−sj|< δ. Choosing an integer n2such that

|Gn(sj)−G^∗(sj)|< (, nn₂, j= 1,2,· · ·, !, we can find 1j!for anys∈J, so that

|Gn(s)−G^∗(s)||Gn(s)−Gn(sj)|+|Gn(sj)−G^∗(sj)|+|G^∗(sj)−G^∗(s)|<7(,

nn2, 0Re(s)K, |Im(s)|K. (175)

which implies uniform convergence ofGn(s) toG^∗(s) onJ. ✷

定理79. の証明. Following the proof in [1, §2.6, §2.5] for the case without parameter u, we prove the theorem along the following outline.

(i) Prove

nlim→∞ sup

u∈[0,1]

f(x)Pu,n(dx)−

f(x)P_u^∗(dx)

= 0, (176)

(a) when f is a Fourier transform f(x) =

e^√−1yxg(y)dy of an integrable function g, using Fubini’s theorem and the assumption on continuity inuof characteristic functions,

(b) whenf is a continuous function of compact support, by uniformly approximatingf by a series of functions in the above considered class,

(c) when f is a bounded continuous functions, by approximating f by a series of continuous functions of compact support, using tightness of{Pu,n}and continuity property of probability measures.

(ii) Approximate A⁰ by closed subsets and ¯A by open neighborhood uniformly in u, to approximate the defining function ofA, and then use (176).

Let us proceed with our proof.

First we prove (176) forf which has a formf(x) =

e^√−1yxg(y)dy with an integrable functiong.

By assumption, characteristic functionφn,u(t) is continuous int and converges asn→ ∞uniformly in uhenceφ^∗_u(t) is continuous int. This implies

sup

u∈[0,1]|φ_u,n(t)−φ_u,∗(t)|= sup

u∈[0,1]∩Q|φ_u,n(t)−φ_u,∗(t)|,

which further implies that this quantity is measurable with respect to t. We can therefore consider integration, and using Fubini’s theorem,

sup

u∈[0,1]

f(x)P_u,n(dx)−

f(x)P_u^∗(dx)

sup

u∈[0,1]|φ_u,n(t)−φ_u,∗(t)||g(t)|dt.

Characteristic functions are bounded (its absolute value is 1 or less) andgis integrable, hence domi- nated convergence theorem can be applied. By assumption that φn(t) converges for eacht uniformly in uasn→ ∞,

nlim→∞ sup

u∈[0,1]

f(x)Pu,n(dx)−

f(x)P_u^∗(dx) lim

n→∞

sup

u∈[0,1]|φu,n(t)−φu,∗(t)||g(t)|dt= 0. which implies (176) for this class of functionf.

Next we prove that a continuous functionfwith compact support, can be approximated (with respect to sup norm) by functions in the class considered above, which implies, that (176) holds also forf. More explicitly, we shall prove that if we put, forn= 1,2,3,· · ·andt∈R,

gn(t) = 1

2πe^−t2/(2n)

e^−√−1tyf(y)dy, hn(x) =

e^√−1txgn(t)dt,

thenhn(x) converges tof(x) as n→ ∞uniformly inx. (Note thatgn is integrable.) Using Fubini’s theorem,

h_n(x) = 1

2π e^{√−1t(x−y)−t2/(2n)}dt f(y)dy= n

2π

e^{−n(x−y)2/2}f(y)dy

= 1

2π

e^−t2/2f(x+ t

√n)dt.

Dividing the range of integration to|t|aand|t|> a, we have

|hn(x)−f(x)| 1

2π

e^−t2/2|f(x+ t

√n)−f(x)|dt

sup

|t|a

|f(x+ t

√n)−f(x)|+

2 sup

x |f(x)| 2π

|t|>a

e^−t2/2dt.

f is a continuous function of compact support, hence sup

x |f(x)|<∞and also is uniformly continuous in x, hence the first term in the right hand side of the above equation converges to 0 asn→ ∞uniformly in x. Lettinga→ ∞, the second term also disappears. Sincea >0 is arbtrary and the left hand side is independent ofa, we therefore have

nlim→∞sup

x |hn(x)−f(x)|= 0. Next we prove that (176) holds for bounded continuous functionf.

LetK >0 . There exists continuous functiongof compact support such that g(x) =f(x), |x|K, g(x) = 0, |x|> K+ 1.

Since f is bounded, c = sup_x|f(x)−g(x)|< ∞. Choose a continuous function hof compact support such that

h(x)





=c, |x|K−1,

∈[0, c], K−1<|x|< K,

= 0, |x|K.

It then holds that|f(x)−g(x)|c−h(x),x∈R. We also have

|f(x)−g(x)|P_u^∗(dx)cP_u^∗([−K, K]^c).

Applying (176) onh, we have

nlim→∞ sup

u∈[0,1]

h(x)P_u,n(dx)−

h(x)P_u^∗(dx) = 0. In particular, for any( >0, we have

∃n0; (∀nn0)(∀u∈[0,1])

(c−h(x))Pu,n(dx)−(

c−

h(x)P_u^∗(dx) =

(c−h(x))P_u^∗(dx)cP_u^∗([−K+ 1, K−1]^c).

This and

|f(x)−g(x)|Pu,n(dx)c−

h(x)Pu,n(dx) imply

nlim→∞ sup

u∈[0,1]

|f(x)−g(x)|Pu,n(dx)c sup

u∈[0,1]P_u^∗([−K+ 1, K−1]^c).

Applying (176) tog, we have lim

n→∞ sup

u∈[0,1]

g(x)P_u,n(dx)−

g(x)P_u^∗(dx) = 0.

Combining all the obtained estimates, we have

nlim→∞ sup

u∈[0,1]

f(x)Pu,n(dx)−

f(x)P_u^∗(dx)

2c sup

u∈[0,1]P_u^∗([−K+ 1, K−1]^c).

SinceKis arbitrary and the left hand side is independent ofK, we can letK→ ∞to find, by tightness assumption, that the right hand sid converges to 0, hence (176) holds.

Finally we prove (145).

PutK = sup

x,u

ρu(x). By assumption, K <∞. Denote the Lebesgue measure byµ. On R, any open set can be written as an increasing limit of a sequence of closed sets, and any closed set can be written as a decreasing limit of a sequence of open sets. Therefore, for any( >0, there exists a closed setF and a open setG, satsfyingF ⊂A⁰⊂A⊂A¯⊂Gand

µ(G)−(/K < µ( ¯A) (=P^∗(A) =)µ(A⁰)< µ(F) +(/K, u∈[0,1].

Therefore we have

P_u^∗(G)−( < P_u^∗( ¯A) (=P^∗(A) =)P^∗(A⁰)< P^∗(F) +(, u∈[0,1].

Define a bounded continuous functionf such that f(x)





= 0, x∈G^c,

∈[0,1], x∈G∩A¯^c, 1, x∈A,¯ is satisfied. Then (176) implies

lim

n

inf

u∈[0,1](P_u^∗(A)−Pu,n(A))lim

n

inf

u∈[0,1](P_u^∗(G)−Pu,n(A)−()

lim

n

inf

u∈[0,1](

f(x)P_u^∗(dx)−

f(x)Pu,n(dx)−()

=−lim

n sup

u∈[0,1](

f(x)Pu,n(dx)−

f(x)P_u^∗(dx))−(

=−(.

Since( >0 is arbitrary,

lim

n

inf

u∈[0,1](P_u^∗(A)−P_u,n(A))0.

We can repeat the argument by takingF andA⁰ in place ofGand ¯Ato find lim

n

inf

u∈[0,1](Pu,n(A)−P_u^∗(A))0. Therefore, for any( >0, by choosing sufficiently largen, we have

inf

u∈[0,1](P_u^∗(A)−P_u,n(A)) +(0 sup

u∈[0,1]

(P_u^∗(A)−P_u,n(A))−(.

This implies

sup

u∈[0,1]|P_u^∗(A)−Pu,n(A)|(, which implies (145).

✷

定理80 の証明. We basically repeat the idea of proof in [4, Theorem XI.3] forsandu. PutD={s∈C|

|s| < C_∞}, and letK ⊂D be a compact set and ! be a Jordan curve in D surroundingK, such that d, the distance between! andK, is positive. Similarly, let!₁ be a Jordan curve inU surrounding [0,1], such that d₁, the distance between!₁ and [0,1], is positive. Applying Cauchy’s formula twice, we have, for (s, u)∈K×[0,1],

gn,u(s) = -

gn,u(σ) dσ 2π√

−1(σ−s) = -

-

₁

gn,υ(σ) dυ 2π√

−1(υ−u)

dσ 2π√

−1(σ−s).

Hence for (s, u), (s, u)∈K×[0,1],

|g_n,u(s)−g_n,u(s)| -

-

₁

|g_n,υ(σ)||υ−u||s−s|+|σ−s||u−u|

|υ−u||σ−s||υ−u||σ−s| dυ dσ (2π)². By (140) we see that, for (σ, υ)∈!×!₁,

|g_n,υ(σ)|C_∞+M_∞C_∞² LetLandL1be the length of!and!1, respectively, and putC1= sup

u∈U|u|<∞. Then, noting sup

σ∈, s∈K|σ− s|dand sup

υ∈1, u∈[0,1]|υ−u|du, we have

|gn,u(s)−gn,u(s)|LL1(C_∞+M_∞C_∞² )(2C1|s−s|+ 2C_∞|u−u|) (2πdd₁)² .

The coefficients of |s−s| and |u−u| in the right hand side is independent of n, u, u, s, s, which implies that {gn,·| n = 1,2,3,· · ·}is equicontinuous on K×[0,1]. The Ascoli-Arzel`a theorem implies that for any subsequence of{gn,·}, there exists subsubsequence which converges uniformly onK×[0,1].

By 定理74 we know that the limit isg^∗_u(s), which, in particular, is independent of subsequences. Hence

nlim→∞g_n,u(s) =g_u^∗(s), uniformly in (s, u) on K×[0,1].

系81 の証明. For each n, u, and s, put gn,u(s) = −logGn,u(s) . Then (148) implies (139). Note that Gn,u(0) = 1, hence gn,u(0) = 0 . WithG_1,u(0) =−1, we also haveg_1,u(0) = 1 .

By assumptions,G_1,u(s) is regular and non-zero on|s|C₀. Henceg_1,u(s) =−logG_1,u(s) is regular on|s|C0.

Put

M1= sup

|s|=C₀, u∈U|G1,u(s)−1|. By assumption,M1<1 . Then noting that, for|z|<1,

|log(1 +z)| 1

0

z 1 +tz

dt |z| 1− |z|, we have

|g_1,u(s)|=|log(1 + (G_1,u(s)−1))| M₁

1−M₁, |s|=C₀, u∈ U. Therefore

M₀= sup

|s|=C0, u∈U|g_1,u(s)|<∞. (177) As we have seen, g_1,u(s) is regular on |s| C₀, and gn,u(0) = 0 and g_1,u (0) = 1 . Therefore 1

s

g1,u(s) s −1

is regular on|s|C0, Using Cauchy’s theorem, we find, for|s|C0/2, 1

s

g1,u(s) s −1

= -

|σ|=C0

1 σ

g1,u(σ) σ −1

dσ 2π√

−1(σ−s)

sup

|s|=C0, u∈U

1 s

g_1,u(s) s −1

2πC₀

2π(C0−C0/2) 2

C₀²(M₀+C₀), where we used (177) in the last line. This implies (138) with M = 2

C₀²(M₀+C₀) and C =C₀/2. We can apply 定理72 and定理74 to find that there exist positive constantsC_∞ and M_∞ such that for all n= 1,2,3,· · ·andu∈ U,G_n,u(s) =e^−gn,u(s)is regular and uniformly bounded innandu, on|s|C_∞, and that for eachu∈ U there exists a regular functionG^∗_uon|s|< C_∞ such that

nlim→∞G_n,u(s) =G^∗_u(s), |s|< C_∞,

uniformly on compact sets. 定理72 and定理74 also imply, by (140) and (142),

| −logGn,u(s)−s|M_∞|s|², |s|C_∞, u∈ U,

| −logG^∗_u(s)−s|M_∞|s|², |s|< C_∞, u∈ U.

補題 89 |z−1−logz||logz|²e^|logz|,z∈C. 証明. Integrating d²

dt²(e^tw−1−tw) =w²e^tw byttwice, we have e^tw = 1 +tw+w²

t 0

(t−s)e^swds.

Puttingt= 1 andw= logz,

|z−1−logz||logz|² 1

0 |e^slogz|ds|logz|²(1∨e^Re(logz))|logz|²e^|logz|.

✷ Then notinig

|Gn,u(s)−1 +s||Gn,u(s)−1−logGn,u(s)|+|logGn,u(s) +s|,

and by using補題89 withz=Gn,u(s) for the first term in the right hand side, we see that there exist C>0 andM>0 such that

|G_n,u(s)−1 +s|M|s|², |s|C, n= 1,2,3,· · ·, u∈ U, and similar uniform estimate forG^∗_u.

LetU be a bounded open complex neighborhood of [0,1], andG1,u(s) be regular inu∈ U for eachs.

LetK⊂ {s∈C| |s|< C_∞} be a compact set.

補題 90 |e^z−e^z||z−z|e^|z|+|z|,z, z∈C. 証明. Put f(t) =z+ (z−z)t. Then

e^z−e^z= 1

0

d

dte^f(t)dt= (z−z) 1

0

e^f(t)dt, hence

|e^z−e^z||z−z| sup

0t1

e^{|(1−t)z+tz|}|z−z|e^|z|+|z|

✷ Since (147) implies that G_1,u(s) = 0 for each u ∈ U and s ∈ C satisfying |s| C₀, g_1,u(s) =

−logG_1,u(s) is regular inu∈ U for eachs. Applying定理80, we see that the convergence ofg_n,u(s) to g_u^∗(s) is uniform in (s, u)∈K×[0,1].

Using補題90 withz=−g_u^∗(s) andz=−gn,u(s), we have

|Gn,u(s)−G^∗_u(s)||g^∗_u(s)−gn,u(s)|e^{|gu∗(s)|+|gn,u(s)|}.

Note that gn,u(s) and g_u^∗(s) are bounded uniformly ins, u,n. Therefore the convergence of Gn,u(s) to

G^∗_u(s) is uniform in (s, u)∈K×[0,1]. ✷

定理82 の証明. The statements on the generating functions in the neighborhood of 0 are direct conse- quences of系81. 系81 and定理78 implies thatP_n,u converges weakly asn→ ∞to a Borel probability measure P_u^∗ supported on non-negative reals. 定理74 and 定理78 imply that the generating function G^∗_u(s) ofP_u^∗ is defined both onRe(s)0and in a neighborhood of 0, and satisfies the defining equations G^∗_u(0) =−1 and (149) in a neighborhood of 0 .

IfRe(s)0then|G^∗_u(s)|1, leading to

Re(−λ_ulogG^∗_u(s/λ_u)) =−λ_ulog|G^∗_u(s/λ_u)|0,

andG1,u(s) is regular onRe(s)>0and continuous onRe(s)0, hence uniqueness of analytic contin- uation implies that (149) holds also onRe(s)0. Similarly, if Re(s)0then |Gn,u(s)|1, leading to

Re(−λ_ulogG_n,u(s/λ_u))0,

hence uniqueness of analytic continuation implies that (148) holds also onRe(s)0.

5Thus we have the corresponding uniform estimates stated in the theorem. Tightness of the family of the corresponding measures is a direct consequence of (151) and命題76.

定理78 also implies that

nlim→∞Gn,u(s) =G^∗_u(s), uniformly on{s∈C| |s|< C_∞ or Re(s)0}.

The claim of the uniformity of convergence ofGn,u(s) toG^∗_u(s) in (s, u) is also a direct consequence of系81.

Assume thatRe(s) 0 and |s| < λ₋C_∞. Using (148) and (149), which we have proved above to hold forRe(s)0,

G_n+1,u(s)−G^∗_u(s) =G_1,u(−λulogGn,u(s/λu))−G_1,u(−λulogG^∗_u(s/λu)). (178) The assumption 1 =−G_1,u(0) =

[0,∞)

ξP_1,u(dξ) implies, with dominated convergence theorem, that G_1,u(s) exists and satisfies

|G_1,u(s)|=|

[0,∞)

ξe^−sξP1,u(dξ)|1, (179)

for anyssatisfyingRe(s)0. (AtRe(s) =0, the direction of differentiation is assumed to be restricted in a natural way.) Let Re(s_i)0, i= 1,2, and for 0 t 1, putF(t) = s₁+t(s₂−s₁). Note that Re(F(t))0.

|G1,u(s2)−G1,u(s1)|

=|G_1,u(F(1))−G_1,u(F(0))|= 1

0

d

dt(G_1,u(F(t)))dt

=|s₂−s₁| 1

0

G_1,u(F(t))dt |s₂−s₁|

1

0 |G_1,u(F(t))|dt |s₂−s₁|,

where we used (179) in the last line.

Applying this formula on (178), we have, sup

u∈[0,1]|G_n+1,u(s)−G^∗_u(s)| sup

u∈[0,1]

λu sup

u∈[0,1]|logGn,u(s/λu))−logG^∗_u(s/λu))|

By assumption,|s/λu|< C_∞, hence we already know (uniform convergence inu) that the right hand side converges to 0 asn→ ∞, hence the left hand side also does, implying uniform convergence inu∈[0,1]

ofGn,u(s) to G^∗_u(s) forssatisfyingRe(s)0and|s|< λ₋C_∞.

We can repeat the argument inductively to conclude the uniform convergence for each s satisfying Re(s)0.

We will now move on to a proof of continuity ofP_u^∗inu. Assume thatU is a topological space,u∈ U,

ulim→u0G1,u(s) =G1,u₀(s) uniformly on|s|C0, and lim

u→u0λu=λu₀.

定理74 implies thatg^∗_u=−logG^∗_u satisfies (141) and (142). In particular, {g^∗_u|u∈ U}is uniformly bounded, hence is a normal family on |s| < C_∞. Then for any sequence in U converging to u0, there exists a subsequenceu_n,n= 1,2,3,· · ·, such thatg_u^∗

n converges uniformly on compact sets in|s|< C_∞ to a regular functionhdefined on|s|< C_∞. Lettingu=u_n→u0 in (141) and (142),

h(s) =g_1,u0(λ_u0h(s/λ_u0)), |s|< C_∞, (180) and

|h(s)−s|M_∞|s|², |s|< C_∞. (181) 定理74 implies thath=g_u^∗₀, hence, in particular,his independent of the subsequence{u_n}. This proves

nlim→∞g_u^∗

n(s) =g_u^∗₀(s), |s|< C_∞,

uniformly on compact sets. 定理78 then implies that for any sequence{un} inU converging tou0, P_u^∗_n converges weakly toP_u^∗₀ asn→ ∞.

LetA⊂Rbe a Borel set, satisfyingP_u^∗₀(∂A) = 0 . Then we have proved that lim

n→∞P_u^∗_n(A) =P_u^∗₀(A) for any sequence satisfying lim

n→∞un =u0. This implies lim

u→u₀P_u^∗(A) =P_u^∗₀(A). SinceA is an arbitrary Borel set satisfyingP_u^∗₀(∂A) = 0 , We finally see thatP_u^∗ converges weakly toP_u^∗₀ as u→u₀.

Since, for any sequenceu_nconverging to u₀,P_u^∗

n converges weakly toP_u^∗₀, the corresponding charac- teristic function ϕ^∗_u

n(t) converges toϕ^∗_u0(t) for each t, which implies continuity of ϕ^∗_u

n(t) in ufor each

t∈R. ✷

定理84 の証明. The definitions (153) and (154) inductively imply thatx_c,u is a fixed point of Φ_n,ufor all n. The statement on the derivative at the fixed point then inductively follows from (154) and the chain rule.

By definition and assumptions,

0Φ_1,u(x)< x, 0x < xc,u,

which implies that Φn,u(x) is non-negative and decresing inn. Hence the limit exists in [0, xc,u), while the only solution to Φ1,u(x) =x,x∈[0, xc,u) isx= 0 . Hence (155) holds.

We move on to a proof of (156). Since Φ_1,u(x) =

k=2

c_k,ux^k is continuous inx∈[0, r_u), so is R(x) =

k=3

ck,ux^k−3=x⁻³(Φ_1,u(x)−c_2,ux²). (182) Therefore lim

x→0R(x) =R(0) =c_3,u, which, with (155) implies

nlim→∞R(Φn,u(x)) =c3,u, 0x < xc,u. (183) Let 0x < xc,u. Applying (182) to (154),

log(c2,uΦn+1,u(x))−2 log(c2,uΦn,u(x)) = log(1 + Φn,u(x)R(Φn,u(x))/c2,u), n= 1,2,3,· · ·. Iterating,

log(c2,uΦn+m,u(x))−2^mlog(c2,uΦn,u(x)) = m k=1

2^m−klog(1 + Φn+k−1,u(x)R(Φn+k−1,u(x))/c2,u), m= 0,1,2,· · ·, n= 1,2,3,· · ·,

which can easily be checked by induction on m. This proves that 2^−n−mlog(c_2,uΦ_n+m,u(x))2⁻ⁿ(log(c_2,uΦn,u(x)) + sup

kn

Φk,u(x)R(Φk,u(x))/c_2,u), m= 0,1,2,· · ·, n= 1,2,3,· · ·.

(155) implies that the right hand side is negative for sufficiently largen. Therefore

nlim→∞log Φn(x) = lim

m→∞2^−n−mlog(c2,uΦn+m,u(x))<0.

✷ 定理85 の証明. (i) Note that Φn,u(xc,u) =xc,u. Then the right hand side of (152) withzreplaced by Φ_n,u(z) converges for|z|x_c,u. By induction, the McLaurin expansion of Φ_n,u(z) has finite radius of convergence. Put

Φn,u(z) = ∞ k=0

cn,kz^k. (184)

Then

G_n,u(s) = ∞ k=0

c_n,kx^k_c,u⁻¹e^−skλ−n.

Φn,u(xc) = xc impliesGn,u(0) = 1, which implies that a measure Pn,u on {kλ⁻ⁿ |k = 0,1,2,· · ·}

defined by

Pn,u({kλ⁻ⁿ}) =cn,kx^k_c,u⁻¹, k= 0,1,2,· · ·, is a Borel probability measure, and that Gn,u is its generating function.

(ii) As noted in命題83, (146) holds withλ₋2 .

By definitions ofxc,u andλuwith (157), we see thatG1,u(0) = 1 andG_1,u(0) =−1 . The recursion relation (148) also follows directly from (157) and (154).

Note that, by 命題83, r_u and x_c,u are continuous in u. Since by assumptions |Φ_1,u(z)−1| is continuous on a compact set{(u, z)∈U ׯ C| |z| ¹₂(x_c,u+r_u)}, it is uniformly continuous. Since Φ_1,u(z) is finite on |z| ¹₂(x_c,u+r_u), it is regular there. So for sufficiently smallC₀, (147) and regularity holds. This proves all the assumptions in系81 (except those assumed at the end of系81 as additional assumptions), hence the assumptions in 定理82.

(iii) 定理82 implies thatG^∗_u(s) =

e^−sξP_u^∗(dξ) is regular ats= 0 and satisfies

| −logG^∗_u(s)−s|M_∞|s|², |s|< C_∞, u∈ U,

for positive constantsM_∞andC_∞. This implies that the meanmu= 1 and the variancevuM_∞ (<∞).

To prove that vu > 0, assume otherwise. Then P_u^∗({1}) = 1, hence G^∗_u(s) = e^s. According to 定理82,G^∗_usatisfies (149), which, combined with (157), becomes

x_c,uG^∗_u(sλ_u) = Φ_1,u(x_c,uG^∗_u(s)).

Therefore

x_c,ue^sλu = Φ_1,u(x_c,ue^s),

in a neighborhood of s= 0, which is impossible because of the assumptions to the coefficients of Φ_1,u(z) in the definition of Φ_1,u.

Using (149) and (157) again, withϕ^∗_u(t) =G^∗_u(−√

−1t), we have ϕ^∗_u(λ_ut) = 1

xc,u

Φ_1,u(ϕ^∗_u(t)x_c,u). (185)

With (152), we have

|ϕ^∗_u(λut)|^∞

k=0

ck,u|ϕ^∗_u(t)|^kx^k_c,u⁻¹

= ∞ k=2

c_k,u|ϕ^∗_u(t)|^kx^k_c,u⁻¹=|ϕ^∗_u(t)|² ∞ k=2

c_k,u|ϕ^∗_u(t)|^k−2x^k_c,u⁻¹ |ϕ^∗_u(t)|² 1

xc,u

∞ k=2

ck,ux^k_c,u=|ϕ^∗_u(t)|² 1 xc,u

∞ k=0

ck,ux^k_c,u=|ϕ^∗_u(t)|² 1 xc,u

Φ(xc,u)

=|ϕ^∗_u(t)|²,

where we used c_1,u=c_2,u= 0, non-negativity ofck,u’s, and |ϕ^∗_u(t)|1 . Therefore, all the assumptions in the 定理65 hold.

✷

定理87 の証明. (i) 定理82 and the existence of density functionρin定理65 imply that

|

R

e^Cξρu(ξ)dξ|1 +C+MC², u∈ U. Sinceρis continuous this implies

sup

ξ∈Re^Cξρ_u(ξ)<∞, u∈ U.

(ii) Using (185) and (152) and the existence of continuous densityρu, we have

λ⁻_u¹ρu(λ⁻_u¹ξ)c_2,uxc,uρu∗ρu(ξ) +c_3,ux²_c,uρu∗ρu∗ρu(ξ), (186)

where ρ^∗_u^k is thek-fold convolution ofρu.

By assumption, the second term in the right hand side of (186) is positive andλu >2 . By same reasoning as λ_u 2, there is a positive term in the right hand side, say at k=k₀, with k₀ > λ.

LetA ⊂[0,∞) be the support ofρ_u. Since mean is 1, there exists x₀ ∈A\ {0}. Then (186) and continuity of ρ implies that 2

λ_ux₀ ∈A, and by induction, 2

λ_u n

x₀ ∈A, n= 1,2,3,· · ·, leading to 0 ∈ A. Then kλ⁻_u¹x₀ ∈A, k= 0,1,2,· · ·, k₀, and by induction kλ⁻_uⁿ ∈A, k = 0,1,2,· · ·, kⁿ₀, n= 1,2,3,· · ·. This impliesA= [0,∞). Soρ(ξ)>0 for dense points in [0,∞). But then (186) and continuity ofρimply thatρ(ξ)>0 for allξ >0 .

(iii) 補題 91 For eachu, there existsn_0,u,Mu>0,Cu>0, and au, such that

|ϕn,u(t)| Cu

x_c,ue^{−Mu(|t|/au)log 2/logλu}, |t| ∈[auλ⁻_u¹, auλⁿ_u^−n0,u], nn_0,u. (187) 証明. Note that (156) implies that if 0x < x_c,u, then there existC_u>0 andM_u>0 such that

Φn,u(x)< Cue^−Mu2n, n0. (188) Since dominated convergence theorem implies that ϕ^∗_u(t) is continuous, 系64 implies that there exist a >0 andt1 satisfyinga/λ|t1|asuch that

sup

a_u/λ_u|t|a_u

|ϕ^∗_u(t)|=|ϕ^∗_u(t₁)|<1.

Usingϕ^∗_u(t) =G^∗_u(−√

−1t), and noting that lim

n→∞Gn,u(s) =G^∗_u(s) uniformly inson a neghborhood of 0, and then using (157), we see that there exist ( >0 andn_0,usuch that

|Φn,u(xc,ue^{√−1λ−nu t})|< xc,u−(, nn0,u, au/λ|t|au. (189) The two estimates (188) and (189) imply, for m0 andNn_0,uanda_u/λ|t|a_u,

|ΦN+m,u(xc,ue^{√−1λ−Nu t})|=|Φm,u(ΦN,u(xc,ue^{√−1λ−Nu t}))|Φm,u(|ΦN,u(xc,ue^{√−1λ−Nu t})|) Φm,u(xc,u−()Cue^−Mu2m.

(190) Let|t| ∈[auλ⁻_u¹, auλⁿu^−n0,u], and choosemsuch thatau/λ|t|λ^−m< au. Then 0mn−n0,u. Put, in (190),N =n−mn0,u andt=tλ^−m. Then noting that

ϕn,u(t) = 1

x_c,uΦn,u(xc,ue^{√−1λ−nt}),

we have (187) from (190). ✷

We proceed with a proof of (159). Put

˜

ϕn,u(t) =

R

e^√−1ξtPn,u∗gn(ξ)dξ=

e^√−1ξtgn(ξ−η)Pn,u(dη).

Then Fubini’s theorem implies

˜

ϕ_n,u(t) =e^−12t2h2nϕ_n,u(t).

For a set A define χ_A byχ_A(t) = 1, if t ∈A, and 0 otherwise. Then 補題91 implies a following bound uniform inn:

|χ

[0,auλⁿ⁻ⁿ_u ^0,u](|t|) ˜ϕn,u(t)||χ

[0,auλⁿ⁻ⁿ_u ^0,u](|t|)ϕn,u(t)|1∧ Cu

x_c,ue^{−Mu(|t|/au)log 2/logλu}, t∈R. Therefore, dominated convergence theorem implies

nlim→∞

|χ_[0,a

uλ^n−n0,u](|t|) ˜ϕn,u(t)−ϕ^∗_u(t)|dt= 0.

On the other hand,

nlim→∞

|χ_[0,a

uλ^n−n0,u](|t|) ˜ϕ_n,u(t)−ϕ˜_n,u(t)|dt lim

n→∞2 _∞

a_uλ^n−n0,u

e^−h2nt22dt

lim

n→∞2a⁻_u²λⁿ_u^0,u−2b⁻²λⁿ_un⁻¹exp(−a²_uλ⁻_u^2n0,ub²n/2) = 0, ifb >

√2 logλu

a_u λⁿ_u^0,u. Therefore we have

nlim→∞

|ϕ˜n,u(t)−ϕ^∗_u(t)|dt= 0,

which implies that the Fourier transform ofPn,u∗gn(ξ) converges inL¹ to that ofρu(ξ). Therefore (159) holds.

(iv) Since

|ϕ^∗_u(t)−ϕ^∗_u(t)||ϕ^∗_u(t)−ϕ^∗_u(t)|+|ϕ^∗_u(t)−ϕ^∗_u(t)|, t, t ∈R, u, u∈ U, it is sufficient to prove thatϕ^∗_u(t) is

(a) continuous intuniformly inu∈ U, and (b) continuous inufor eacht∈R.

(a) As we have noted before,

[0,∞)

ξP_u^∗(dξ) =mu= 1, hence

|ϕ^∗_u(t)−ϕ^∗_u(t)|

[0,∞)|2 sinξ(t−t)

2 |P_u^∗(dξ)|t−t|

[0,∞)

ξP_u^∗(dξ) =|t−t|, which implies thatϕ^∗_u(t) is continuous intuniformly inu∈ U.

(b) As we noted above,ru andxc,u are continuous inu, by命題83, and Φ1,u(z) is continuous on {(u, z)∈U ׯ C | |z| < ru}, hence uniformly continuous on a compact set {(u, z)∈ U ׯ C|

|z| ¹₂(x_c,u+r_u)}. Also by definition (157),

G_1,u(s) =x⁻_c,u¹Φ_1,u(e^−s/λuxc,u).

Hence G_1,u(s) is uniformly continuous on (u, s) ∈[0,1]× {|s| C₀} for a positive constant C0. In particular, it is continuous inu∈[0,1], uniformly inson|s|C0.

Note also that by 命題83,λu is continuous inu.

We have proved that the additional assumptions in the last part of 定理82 hold. Therefore, by the last statement in定理82,ϕ^∗_u(t) is continuous inufor eacht∈R.

(v) As we noted before, 定理82 implies that for family of measures P_u^∗, u∈ U, the mean m_u= 1 and the variance sup

u∈Uv_uM_∞(<∞). With命題76 we further have uniform bound of moments to all order. We also know by命題86 that inf_u∈[0,1]v_u>0 . These facts are sufficient to proveu-uniform decay estimate 定理63. As we have proved already,ϕ^∗_u(t) is continuous in (u, t)∈[0,1]×R. Then 定理66 implies the claim.

✷

C 指数型の Tauber 型定理．

この節は確率論講義録[17]とregular.tex [19] (§Bに引用)に入れるべき内容⁶⁰だが，講義録執筆時点で入 れ損なっている⁶¹ので，ここに [40]の主結論のみ挟んでおく．

6020010105–06;12–17．

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