This paper has twofolds. The first is to prove that there are at most two meromorphic functions sharing a small function with multiplicities truncated by 2 and other three small functions regardless of multiplicities, where all zeros with multiplicities more than a certain number are not counted. This result is an improvement of the four values theorems of Nevanlinna, Gundersen, Fujimito, Thai Tan and others. The second purpose of this paper is to prove that there are at most three meromorphic functions sharing four small functions ignoring multiplicity, where all zeros with multiplicities more than a certain number are omitted
Trang 1FOUR SMALL FUNCTIONS REGARDLESS OF MULTIPLICITIES
SI DUC QUANG
Abstract This paper has twofolds The first is to prove that there are at most two
meromorphic functions sharing a small function with multiplicities truncated by 2 and
other three small functions regardless of multiplicities, where all zeros with multiplicities
more than a certain number are not counted This result is an improvement of the four
values theorems of Nevanlinna, Gundersen, Fujimito, Thai - Tan and others The second
purpose of this paper is to prove that there are at most three meromorphic functions
sharing four small functions ignoring multiplicity, where all zeros with multiplicities more
than a certain number are omitted.
1 Introduction Let f be a nonzero holomorphic function on C For each z0 ∈ C, expanding f as
f (z) =P∞
i=0bi(z − z0)i around z0, then we define νf0(z0) := min{i : bi 6= 0} Let k be a positive integer or +∞ We set
νf,≤k0 (z) :=
(
νf0(z) if νf0(z) ≤ k,
0 if νf0(z) > k
Similarly, we define ν0
f,>k and ν0
f,=k Let ϕ be a nonconstant meromorphic function on C with a reduced representation
ϕ = (ϕ1 : ϕ2), where ϕ1, ϕ2 are holomorphic functions on C having no common zeros and ϕ = ϕ1
ϕ2 We define ν
0
ϕ := νϕ01, νϕ,≤k0 = νϕ01,≤k, νϕ∞ := νϕ02 and νϕ,≤k∞ = νϕ02,≤k The proximity function of ϕ is defined by:
m(r, ϕ) := 1
2π
2π
Z
0
log+|ϕ(reiθ)|dθ (r > 1),
here log+x = max{0, log x} for x ∈ (0, ∞) Then the Nevanlinna characteristic function
of ϕ is defined by
T (r, ϕ) := m(r, ϕ) + N (r, νϕ∞), where by N (r, ν) we denote the counting function of the divisor ν
Let f, a be two meromorphic functions on C The function a is said to be small (with respect to f ) if || T (r, a) = o(T (r, f )) Here, the notion || P means that the assertion P holds for all r ∈ [1, ∞) outside a finite Borel measure set We denote by Rf the field of all small (with respect to f ) functions on C
2010 Mathematics Subject Classification: Primary 32H30, 32A22; Secondary 30D35.
Key words and phrases: meromorphic function, small function, truncated multiplicity.
1
Trang 2In 1926, Nevanlinna [2] showed that for two nonconstant meromorphic functions f and
g on C, if they share four distinct values with multiplicity, then that g is a m¨obius trans-formation of f In [1] Fujimoto improved the result of Nevanlinna by proving that there are at most two meromorphic functions on C which share four distinct values with multi-plicities truncated by 2 Later, Thai and Tan [7] generalized the result of Fujimoto to the case of small functions In [5] Quang improved the above mentioned results by consider-ing the case where the meromorphic functions share a small functions with multiplicities truncated to level 2 and share other three small functions regardless of multiplicity His result is stated as follows
Theorem A (see [5, Theorem 1.2]) Let f1, f2, f3 be three nonconstant meromorphic functions on C Let a1, , a4 be distinct small (with respect to fi, ∀1 ≤ i ≤ 3) functions
on C Assume that
(i) min{νf01−ai, 1} = min{νf02−ai, 1} = min{νf03−ai, 1} ∀1 ≤ i ≤ 3,
(ii) min{ν0
f 1 −a4, 2} = min{ν0
f 2 −a4, 2} = min{ν0
f 3 −a4, 2}
Then f1 = f2 or f2 = f3 or f3 = f1
We see that, in the above result, all zeros of the functions (fs − ai) are considered The first purpose of this paper is to improve this result by omitting all such zeros with multiplicity more than a certain number Namely, we will prove the following
Theorem 1.1 Let f1, f2, f3 be three nonconstant meromorphic functions on C Let
a1, , a4 be distinct small (with respect to fi, ∀1 ≤ i ≤ 3) functions on C Let k1, , k4
be four positive integers or +∞ with
14 3
4
X
i=1
1
ki
+ 3
4
X
i=1
1
ki+ 1 <
1
3 +
28 3k0
, where k0 = max1≤i≤4ki Assume that
(i) min{ν0
f 1 −a 1 ,≤k 1, 2} = min{ν0
f 2 −a 1 ,≤k 1, 2} = min{ν0
f 3 −a , ≤k 1, 2}, (ii) min{ν0
f 1 −ai,≤k i, 1} = min{ν0
f 2 −ai,≤k i, 1} = min{ν0
f 3 −ai,≤k i, 1} ∀2 ≤ i ≤ 4
Then f1 = f2 or f2 = f3 or f3 = f1
Let k1 = · · · = k4 = k, we have the following corollary
Corollary 1.2 Let f1, f2, f3 be three nonconstant meromorphic functions on C Let
a1, , a4 be distinct small (with respect to fi, ∀1 ≤ i ≤ 3) functions on C Let k ≥ 24 be
a positive integer or +∞ Assume that
(i) min{ν0
f 1 −a1,≤k, 2} = min{ν0
f 2 −a1,≤k, 2} = min{ν0
f 3 −a,≤k, 2}, (ii) min{νf01−a
i ,≤k, 1} = min{νf02−a
i ,≤k, 1} = min{νf03−a
i ,≤k, 1} ∀2 ≤ i ≤ 4
Then f1 = f2 or f2 = f3 or f3 = f1
However, as far as we know, there has no finiteness theorem for the case meromorphic functions sharing four small functions regardless of multiplicities The second purpose of this paper is to prove a such finiteness theorem For detail, we will show that there are at
Trang 3most three meromorphic functions sharing four small functions regardless of multiplicity.
To state our result, we need the following
Let f be a nonconstant meromorphic function on C Let a1, , a4 be four distinct small (with respect to f ) functions and let k1, , k4 be four positive integers or maybe +∞ We consider the set F (f, {ai, ki}4
i=1, 1) of all meromorphic functions g defined on C satisfying
min{1, νf −a0 i,≤ki(z)} = min{1, νg−a0 i,≤ki(z)} (1 ≤ i ≤ 4) for every z ∈ C
Theorem 1.3 Let f be a nonconstant meromorphic function on C Let a1, , a4 be four meromorphic functions which are small with respect to f Let k1, , k4 be positive integers or maybe +∞ with
25 64
4
X
i=1
1
ki +
17 16
4
X
i=1
1
ki+ 1 <
1
32+
25 64k0. Then ]F (f, {ai, ki}4
i=1, 1) ≤ 3
Letting k1 = · · · = k4 = k, we have the following corollary
Corollary 1.4 Let f be a nonconstant meromorphic function on C Let a1, , a4 be four meromorphic functions which are small with respect to f Let k > 314 be positive integers or maybe +∞ Then ]F (f, {ai, k}4
i=1, 1) ≤ 3
2 Some second main theorems and lemmas For a divisor ν on C, which we may regard as a function on C with values in Z whose support is discrete subset of C, and for a positive integer M (maybe M = ∞) , we define the counting function of ν as follows
n[M ](t) =X
|z|≤t
min{M, ν(z)}
N[M ](r, ν) =
r
Z
1
n(t)
t dt (1 < r < ∞).
For brevity we will omit the character (M ) if M = ∞
For a divisor ν and a positive integer k (maybe k = +∞), we define
ν≤k(z) =
( ν(z) if ν(z) ≤ k
0 otherwise and ν>k(z) =
( ν(z) if ν(z) > k
0 otherwise
Let ϕ be a meromorphic function on C Define:
• ν0
ϕ (resp νϕ∞) the divisor of zeros (resp divisor of poles) of ϕ
• νϕ = νϕ0 − ν∞
ϕ
• ν0
ϕ,≤k = (ν0
ϕ)≤k, ν0
ϕ,>k = (ν0
ϕ)>k
Trang 4Similarly, we define νϕ,≤k∞ , νϕ,≤k∞ , νϕ,≤k, νϕ,≤k and their counting functions.
For a discrete subset S ⊂ C, we consider it as a reduced divisor (denoted again by S) whose support is S, and denote by N (r, S) its counting function
Let f be a meromorphic function on C A function h is said to be equal Sf(r) if for every positive number > 0, there exists a Lebesgue subset I ⊂ [1, +∞) of finite measure such that
h(r) ≤ T (r, f ) for every r 6∈ I A function a is said to be small with respect to f if || T (r, a) = o(T (r)) Hence if a is small with respect to f then T (r, a) = Sf(r)
The following second main theorem is due to Yamanoi [8]
Theorem 2.1 (see [8, Corollary 1]) Let f be a nonconstant meromorphic function on
C Let a1, , aq (q ≥ 3) be q distinct meromorphic functions on C Then the following holds
|| (q − 2)T (r, f ) ≤
q
X
i=1
N[1](r, νf −a0 i) + Sf(r),
Lemma 2.2 Let f be a nonconstant meromorphic function on C and let a be a small function with respect to f then for every positive integer k (maybe k = +∞) We have
|| N[1](r, νf −a,>k0 ) ≤ 1
k + 1T (r, f ) + o(T (r, f )) and || N[1](r, νf −ai,≤k) ≥ k + 1
[1](r, νf −a0 ) − 1
kT (r, f ) + o(T (r, f )).
Proof First, we have
|| N[1](r, νf −a,>k0 ) ≤ 1
k + 1N (r, ν
0
f −a,>k) ≤ 1
k + 1T (r, f ) + o(T (r, f )).
Second, we have
|| N[1](r, νf −a i ,≤k) = N[1](r, νf −a i) − N[1](r, νf −a i ,>k)
≥ N[1](r, νf −ai) − 1
k + 1N (r, ν
0
f −a,>k)
= N[1](r, νf −ai) − 1
k + 1(N (r, ν
0
f −a) − N (r, νf −a,≤k0 ))
≥ N[1](r, νf −ai) − 1
k + 1(T (r, f ) − N
[1](r, νf −ai,≤k)) + o(T (r, f )) Thus
|| N[1](r, νf −ai,≤k) ≥ k + 1
[1](r, νf −ai) − 1
kT (r, f ) + o(T (r, f )).
Lemma 2.3 Let f be a meromorphic function on C and let a1, , a4 be four distinct small (with respect to f ) meromorphic functions Let k1, , k4 be four positive integers
or maybe +∞ such that
4
X
i=1
1
ki+ 1 < 2.
Trang 5Then for every g ∈ F (f, {ai, ki}4
i=1, 1), we have || T (r, f ) = O(T (r, g)) and || T (r, g) = O(T (r, f ))
Proof By Lemma 2.1 we have
|| 2T (r, f ) ≤
4
X
i=1
N[1](r, νf −a0 i) + Sf(r)
≤
4
X
i=1
ki
ki+ 1N
[1](r, νf −a0 i,≤ki) +
4
X
i=1
1
ki + 1T (r, f ) + Sf(r)
≤
4
X
i=1
ki
ki+ 1N
[1](r, νg−a0
i ,≤k i) +
4
X
i=1
1
ki+ 1T (r, f ) + Sf(r)
≤
4
X
i=1
ki
ki+ 1T (r, g) +
4
X
i=1
1
ki+ 1T (r, f ) + Sf(r).
Thus
|| 2 −
4
X
i=1
1
ki+ 1T (r, f ) ≤
4
X
i=1
ki
ki+ 1T (r, g) + Sf(r).
This implies that || T (r, f ) = O(T (r, g)) Similarly, we have || T (r, g) = O(T (r, g)) The
3 Proof of Theorem 1.1 Let f1, f2, f3 and {ai}4
i=1be as in Theorem 1.1 By Lemma 2.3, we see that || T (r, fk) = O(T (r, fl)) (1 ≤ k, l ≤ 3) We set
T (r) = T (r, f1) + T (r, f2) + T (r, f3)
For i, j ∈ {1, , 4}, we put Fk
ij = fk− ai
fk− aj
Then
T (r, Fijk) = T (r, fk) + o(T (r))
We define
• S = S
i6=j{z; ai(z) = aj(z)} (then S has counting function small with respect to
fs (s = 1, 2, 3)),
• νi = {z; νf01−ai,≤ki(z) > 0}, Si = ∪3s=1{z; ν0
f s −a i ,>k i(z) > 0},
• T0
i: the set of all z ∈ νi such that ν0
f s −ai,≤k i(z) ≥ ν0
f t −ai,≤k i(z) = ν0
f l −ai,≤k i(z) = 1 for a permutation (s, t, l) of (1, 2, 3),
• Ti = νi\ T0
i
• µi: the set of all z ∈ νi such that ν0
f −a ,≤k (z) = ν0
f −a ,≤k (z) = ν0
f −a ,≤k (z)
Trang 6Now we recall the Cartan’s auxiliary function (see [1, Definition 3.1]) Let F, G, H be three nonzero meromorphic functions, we define Cartan’s auxiliary function by
Φ(F, G, H) := F · G · H ·
1 F
1 G
1 H
(F1)0 (G1)0 (H1)0
= F (
1
H)0
1 H
− (
1
G)0
1 G
! + G (
1
F)0
1 F
− (
1
H)0
1 H
! + H (
1
G)0
1 G
− (
1
F)0
1 F
! (3.1)
It is easy to see that, for every meromorphic function h we have the following property
Φ(hF, hG, hH) = h · Φ(F, G, H)
Lemma 3.2 With the same assumption as in Theorem 1.1, if f1, f2, f3 are distinct then
4
X
i=1
N (r, Ti) ≤ (
4
X
i=1
1
ki − 2
k0)T (r) +
3
X
s=1
Sfs(r)
Proof For k, l ∈ {1, 2, 3}, k 6= l, we have
4
X
i=1
N (r, min{νf0
k −ai,≤k i, νf0
l −ai,≤k i}) ≤ N (r, ν0
f k −fl) ≤ T (r, fk) + T (r, fl)
Therefore
4
X
i=1
X
1≤k<l≤3
N (r, min{νf0
k −ai,≤k i, νf0
l −ai,≤k i}) ≤ 2T (r)
(3.3)
It easy to see that for every z ∈ νi,
• if z ∈ Ti then
X
1≤k<l≤3
min{νf0k−ai,≤ki(z), νf0l−ai,≤ki(z)} ≥ min{1, 2} + min{1, 2} + min{2, 2} = 4
=
3
X
s=1
min{1, νf0s−a
i ,≤k i(z)} + χTi
Here, by χAwe denote the characteristic function of a subset A of C, i.e, χA(z) = 1
if z ∈ A and χA(z) = 0 if z 6∈ A
• if z 6∈ Ti then
X
1≤k<l≤3
min{νf0
k −a i ,≤k i(z), νf0
l −a i ,≤k i(z)} ≥ min{1, 2} + min{1, 2} + min{1, 1} = 3
=
3
X
s=1
min{1, νf0s−ai,≤ki(z)} + χT i Then we have
X
1≤k<l≤3
min{νf0
k −ai,≤k i(z), νf0
l −ai,≤k i(z)} ≥
3
X
s=1
min{1, νf0s−a
i ,≤k i(z)} + χTi,
Trang 7for every z This yields that
X
1≤k<l≤3
N (r, min{νf0k−ai,≤ki, νf0l−ai,≤ki}) ≥
3
X
s=1
N[1](r, νf0s−ai,≤ki) + N (r, Ti)
From (3.3), we have
2T (r) ≥
4
X
i=1
(
3
X
s=1
N[1](r, νf0s−a
i ,≤k i) + N (r, Ti))
≥
4
X
i=1
(
3
X
s=1
(ki+ 1
ki
N[1](r, νf0s−a
i) − 1
ki
)T (r, fs) + N (r, Ti))
≥
4
X
i=1
(
3
X
s=1
k0+ 1
k0 N
[1]
(r, νf0s−ai) −
4
X
i=1
1
kiT (r) +
4
X
i=1
N (r, Ti))
≥ 2(k0 + 1)
k0 −
4
X
i=1
1
ki)T (r) +
4
X
i=1
N (r, Ti) +
3
X
s=1
Sfs(r)
Thus
4
X
i=1
N (r, Ti) ≤ (
4
X
i=1
1
ki − 2
k0)T (r) +
3
X
s=1
Sfs(r)
Lemma 3.4 With the same assumption as in Theorem 1.1, if there exist i, j ∈ {1, , 4},
i 6= j, such that Φ(F1
ij, F2
ij, F3
ij) = 0 then f1 = f2 or f2 = f3 or f3 = f1 Proof Suppose that f1, f2, f3 are distinct Since Φ(F1
ij, F2
ij, F3
ij) = 0, we have
0 =
Fji1 Fji2 Fji3 (F1
ji)0 (F2
ji)0 (F3
ji)0
=
F2
ji− F1
ji F3
ji− F1 ji
(Fji2 − F1
ji)0 (Fji3 − F1
ji)0
= (Fji2 − F1
ji)(Fji3 − F1
ji)0− (F3
ji− F1
ji)(Fji2 − F1
ji)0
It follow that F
3
ji− F1 ji
F2
ji− F1 ji
= λ ∈ C Since f1, f2, f3 are supposed to be distinct, λ 6∈ {0, 1} and we have
(1 − λ)f1− aj
f1− ai + λ
f2− aj
f2− ai =
f3 − aj
f3− ai. This implies
(1 − λ) 1
f1− ai + λ
1
f2− ai =
1
f3− ai. Then for every z ∈ C, one has νf0s−ai(z) = νf0t−ai(z) ≥ νf0
l −a i(z) with a permutation (s, t, l) of (1, 2, 3) We consider the holomorphic mapping φ of C into P1(C) with a reduced representation φ = (f h
1 −ai : f h
2 −ai), where h is a meromorphic function We see that if z is a zero of some h
fs− ai
(1 ≤ s ≤ 3) then z is zero of only one function among
Trang 8{ h
fs− ai}
3
s=1 and one must has ν0
f s −ai(z) = ν0
f t −ai(z) > ν0
f l −ai(z), and hence z ∈ Si or
z ∈ Ti This yields that
3
X
s=1
N[1](r, νh/(f0 s−a
i )) ≤ N (r, Si) + N (r, Ti)
Suppose that φ is not constant By the second main theorem and by Lemma 2.2 we have
T (r, φ) ≤
3
X
s=1
N[1](r, νh/(f0 s−a
i )) ≤ N (r, Si) + N (r, Ti)
≤ 1 3
3
X
s=1
N[1](r, νf0
s −ai,>k i) + N (r, Ti)
≤ 1 3
3
X
s=1
1
ki+ 1N (r, ν
0
f s −ai,>k i) + N (r, Ti)
3(ki+ 1) +
4
X
v=1
1
kv − 2
k0)T (r).
On the other hand, we have
T (r, φ) ≥ N (r, ν0 h
f1−ai−
h f1−ai
) ≥
4
X
v=1 v6=i
N[1](r, νf01−av,≤kv)
= 1 3
4
X
v=1 v6=i
3
X
s=1
N[1](r, νf0s−av,≤kv)
≥ 1 3
4
X
v=1 v6=i
3
X
s=1
(ki+ 1
ki
N[1](r, νf0s−av) + 1
kv
T (r, fs))
≥ 1 3
4
X
v=1 v6=i
3
X
s=1
k0+ 1
k0 N
[1]
(r, νf0s−av) +
4
X
v=1 v6=i
1 3kvT (r)
≥ (k0+ 1 3k0 +
4
X
v=1 v6=i
1 3kv)T (r) +
3
X
s=1
Sfs(r)
Therefore, we have
3(ki + 1) +
4
X
v=1
1
kv − 2
k0)T (r) ≥ (
k0+ 1 3k0 +
4
X
v=1 v6=i
1 3kv)T (r) +
3
X
s=1
Sfs(r)
Trang 9Letting r −→ +∞, we get
1 3(ki+ 1) +
4
X
v=1
1
kv − 2
k0 ≥ k0+ 1
3k0 +
4
X
v=1 v6=i
1 3kv. Then we have
1
3(ki+ 1) +
1 3ki +
2 3
4
X
v=1
1
kv − 7
k0 ≤ 4 3
4
X
v=1
1
kv − 7
k0. This is a contradiction
Hence φ =constant, i.e., f2− ai
f1− ai = a ∈ C If a 6= 1 then
S
j6=iνj = ∅, and hence
0 =X
j6=i
N (r, νj) ≥X
j6=i
(kj+ 1
kj N
[1](r, νf −a0
j) − 1
kjT (r, f ))
≥ (k0+ 1
k0 −X
j6=i
1
ki)T (r, f ) + Sf(r) ≥ (1 +
1
k0 −
4
X
j=1
1
ki)T (r, f ) + Sf(r).
Letting r −→ +∞, we get
4
X
j=1
1
ki ≥ 1 + 1
k0. This is a contradiction
Then we have f1 = f2 or f2 = f3 or f3 = f1 The lemma is proved Lemma 3.5 Assume that there exists Φ := Φ(Fij0, Fij1, Fij2) 6≡ 0 Then
N (r, µi) + 2
4
X
s=1 s6∈{i,j}
N (r, νs) ≤ N (r, νΦ0)
and
N (r, νΦ0) ≤ T (r) − 2N (r, νj) + N (r, Si∪ Sj) + o(T (r))
Proof a) We prove the first inequality of the lemma For a fixed point z0 ∈ µi∪S4
s=1 s6∈{i,j}
νs\
S, we consider the following two cases
Case 1 If z0 ∈ µi Then there exists a neighborhood U of z0 and such that all F
k ij
(z − z0)d
(1 ≤ k ≤ 3) are nowhere zero holomorphic functions on U for a positive integer d We rewrite the function Φ on U as follows
Φ = (z − z0)dΦ
F1 ij
(z − z0)d, F
2 ij
(z − z0)d, F
3 ij
(z − z0)d
on U
Then, it yields that
νΦ0(z0) ≥ ν(z−z0
0 ) d(z0) = d ≥ χµi(z0) = χµi(z0) +
4
X
s=1 s6∈{i,j}
χνs(z0)
Trang 10Case 2 If z0 ∈ νt with t 6∈ {i, j} We rewrite the function Φ as follows
Φ =Fij1 · F2
ij· F3
ij ·
1
F 2 ij
− 1
F 1 ij
1
F 3 ij
− 1
F 1 ij
(F12
ij − 1
F 1
ij)0 (F13
ij − 1
F 1
ij)0
=Fij1 · F2
ij· F3
ij ·
(a j −ai)(f 2 −f1) (f 2 −a i )(f 1 −a i )
(a j −ai)(f 3 −f1) (f 3 −a i )(f 1 −a i )
((aj −ai)(f 2 −f1) (f 2 −a i )(f 1 −a i ))0 ((aj −ai)(f 3 −f1)
(f 3 −a i )(f 1 −a i ))0
=(z − z0)2Fij1 · F2
ij · F3
ij ·
(a j −a i )(f 2 −f 1 ) (z−z 0 )(f 2 −ai)(f 1 −ai)
(a j −a i )(f 3 −f 1 ) (z−z 0 )(f 3 −ai)(f 1 −ai)
( (aj −a i )(f 2 −f 1 ) (z−z 0 )(f 2 −ai)(f 1 −ai))0 ( (aj −a i )(f 3 −f 1 )
(z−z 0 )(f 3 −ai)(f 1 −ai))0
We note that all functions (aj −ai)(f k −f1)
(z−z 0 )(f k −a i )(f 1 −a i ) (k = 2, 3) are holomorphic on a neighborhood
of z0 Therefore, it follows that
νΦ0(z0) ≥ 2ν(z−z0 0)(z0) = 2 = χµi(z0) +
4
X
s=1 s6∈{i,j}
χνs(z0)
From the above two cases, we have
νΦ0(z) ≥ χµi +
4
X
s=1 s6∈{i,j}
χνs
for all z outside the analytic set S Integrating both sides of this inequality, we obtain
N (r, νΦ0) ≥ N (r, µi) + 2
4
X
s=1 s6∈{i,j}
N (r, νs)
Then we have the desired inequality
b) We prove the second inequality of the lemma We have
|| N (r, νΦ0) ≤ T (r, Φ) = m(r, Φ) + N (r, νΦ∞)
≤
3
X
k=1
m(r, Fijk) +
3
X
k=1
N (r, νF∞k
ij) + N (r, νΦ∞) −
3
X
k=1
N (r, νF∞k
ij) + o(T (r))
= T (r) + N (r, νΦ∞) −
3
X
k=1
N (r, νF∞k
ij) + o(T (r))
Then, it suffices for us to prove that
N (r, νΦ∞) −
3
X
k=1
N (r, νF∞k
ij) ≤ −2N (r, νj) + N (r, Si∪ Sj) + o(T (r))
... · F 2< /small>ij< /small> · F 3< /small>
ij< /small> ·
1< /small>
F ij< /small>
− 1< /small>
F...
ji< /small> )0− (F 3< /small>
ji< /small> − F 1< /small>
ji< /small> )(Fji2 − F 1< /small>
ji< /small> )0...
ji< /small> − F 1< /small>
ji< /small> F 3< /small>
ji< /small> − F 1 ji< /small>
(Fji2 − F 1< /small>
ji< /small> )0