ON THE UNIQUENESS PROBLEM FOR MEROMORPHIC FUNCTIONS WITH HIGHER MULTIPILCITIES OF ZEROS AND POLES

In this paper, we show that the equation P(f1, ..., fs+1) = Q(g1, ..., gs+1), where P, Q are polynomials in a class of homogeneous polynomials of FermatWaring type, has entire solutions f1, · · · , fs+1; g1, · · · , gs+1. Some classes of unique range sets for linearly nondegenerate holomorphic curves are also obtained

ON THE UNIQUENESS PROBLEM FOR MEROMORPHIC FUNCTIONS WITH HIGHER MULTIPILCITIES OF ZEROS AND

POLES

VU HOAI AN, HA HUY KHOAI, AND NGUYEN XUAN LAI

Abstract Assume the polynomial P (z) = (z − a 1 ) (z − a q ), associated to the

set S = {a 1 , , a q } ⊂ C, satisfies Fujimoto’s condition We give some sufficient

conditions for S to be a unique range set for meromorphic(entire) functions With

some additional conditions on multiplicities of zeros and poles, we reduce the

cardinalities of unique range sets for meromorphic functions.

1 Introduction and main results

In this paper, by a meromorphic function we mean a meromorphic function in the complex plane C We assume that the reader is familiar with the notations in the Nevanlinna theory (see [2-4 ],[6-7],[11]) Let f be a non-constant meromorphic function on C For every a ∈ C, define the function νa

f : C → N by

νfa(z) =







0 if f (z) 6= a

d if f (z) = a with multiplicity d, and set νf∞ = ν01

f

, and define the function νaf : C → N by νaf(z) = min {νfa(z), 1}, and set ν∞f = ν0

1

f Let m be a positive integer For every a ∈ C ∪ {∞}, define the function ν(f,a)≤m from C ∪ {∞} into N by

ν(f,a)≤m(z) =







0 if νa

f(z) > m

νa

f(z) ≤ m, and define the function ν≤m(f,a) from C ∪ {∞} into N by ν≤m(f,a)(z) = min {ν≤m(f,a)(z), 1} Similarly, we define ν(f,a)≥m , ν≥m(f,a) Next, define the counting function N≤m(r, f ),

N≤m(r,f −a1 ), N≤m(r, f ), N≤m(r,f −a1 ), N≥m(r, f ), N≥m(r,f −a1 ), N≥m(r, f ),

N≥m(r,f −a1 ) For f ∈ M(C) and S ⊂ C ∪ {∞}, we define

Ef(S) = [

a∈S

{(z, νfa(z)) : z ∈ C}

In case νa

f (i.e., ignoring multiplicity) we denote Ef(S) (this is the preimages of S)

2010 Mathematics Subject Classification Primary 11S80, Secondary 30D35.

1 The work was supported by the National Foundation for Science and Technology Development (NAFOSTED) and the Vietnam Institute for Advanced study in Mathematics (VIASM).

Let m is a positive integer or ∞, we define

Ef≤m(S) = [

a∈S

{(z, ν(f,a)≤m(z)) : z ∈ C}

Note that, if m is ∞, then Ef∞(S) = Ef(S) and if m = 1, then Ef≤1(S) = Ef(S) Let F be a nonempty subset of M(C) Two functions f, g of F are said to share

S, counting multiplicity, (share S CM), if Ef(S) = Eg(S) and to share S, ignoring multiplicity, (share S IM), if Ef(S) = Eg(S) Let a set S ⊂ C ∪ {∞} and f and

g be two non-constant meromorphic (entire) functions If Ef(S) = Eg(S) implies

f = g for any two non-constant meromorphic (entire) functions f, g, then S is called

a unique range set for meromorphic(entire) functions or, in brief, URSM (URSE)

A set S ⊂ C ∪ {∞} is called a unique range set for meromorphic (entire) functions ignoring multiplicity (URSM-IM) (URSE-IM) if Ef(S) = Eg(S) implies f = g for any pair of non-constant meromorphic (entire) functions A set S ⊂ C ∪ {∞}

is called a URSM≤m (URSE≤m) if for any two non-constant meromorphic (entire) functions f, g, Ef≤m(S) = Eg≤m(S), implies f = g

In 1926, R Nevanlinna discovered his famous five-value uniqueness theorem which says that if two non-constant meromorphic functions f and g on the complex plane

C share ve distinct values IM then f = g A few years later, he showed that when multiplicities are considered, 4 points are sufficient to determine the functions and

in this case either the functions coincide or one is the bilinear transformation of the other In 1982 Gross and Yang [ 8 ] proved the following theorem:

Theorem A [ 8 ] Let S = {z ∈ C: ez+ z = 0} If two entire functions f, g satisfy

Ef(S) = Eg(S), then f = g

It is to be observed that since the range set S given in Theorem A is an innite set, Theorem A cannot be considered as an exact solution to the problem of Gross After the introduction of the novel idea of unique range sets researchers were getting more involved to nd new unique range sets with cardinalities as small as possible In 1994,

Yi [13] exhibited a URSE with 15 elements and in 1995 Li and Yang [12] exhibited

a URSM with 15 elements and a URSE with 7 elements Till date the URSM with 11 elements are the smallest available URSM, obtained by Frank and Reinders [5] The URSM discovered by Frank and Reinders is highlighted by a number of researchers A polynomial P in C, is called a uniqueness polynomial for meromorphic (entire) functions, if for any two non-constant meromorphic (entire) functions f and

g, P (f ) = P (g) implies f = g We say P is a UPM (UPE) in brief A polynomial P

in C, is called a strong uniqueness poly- nomial for meromorphic (entire) functions,

if P (f ) = cP (g) implies f = g for any two non-constant meromorphic (entire) functions f and g , and any nonzero constant c ∈ C We sayP is a SUPM (SUPE)

in brief

H Fujimoto [6] proved the following theorem

Theorem B [6] Suppose that P (z) is a strong uniqueness polynomial of form (1.1) satisfying condition (1.2), and that k ≥ 3, or k = 2 and min{q1, q2} ≥ 2 Then, S is a URSM (resp URSE) whenever q > 2k + 6 (resp q > 2k + 2),while a URSM-IM (resp URSE-IM) whenever q > 2k + 12 (resp.q > 2k + 5)

In 2009 Bai, Han and Chen [2] proved the following truncated sharing version of Theorem B

Theorem C [2] In addition to the hypothesis of Theorem B we suppose that m

is a positive integer or ∞ Let S be the set of zeros of P If

1 m ≥ 3 or ∞ and q > 2k + 6(2k + 2);

2 m = 2 and q > 2k + 7(2k + 2);

3 m = 1 and q > 2k + 10(2k + 4),

then S is a URSM≤m (URSE≤m)

We recall the URSM introduced by Frank and Reinders [5] which is the zero set of

PF R(z) = (n − 1)(n − 2)

n− n(n − 2)zn−1+n(n − 1)

n−2− c(c 6= 0, 1)

Clearly PF R0 (z) has two distinct zeros that is here k = 2 and has a zero at 0 of order

n − 3

Second type of URSM is demonstrated by Yi in [14] which is the zero set of

PY(z) = zn+ azn−r+ b, (a, b 6= 0), where n, r are two positive integers having no common factors, n ≥ 2r + 9, r ≥ 2 and a 6= 0 and b 6= 0 are so chosen so that P has n distinct zeros Clearly PY0(z) has

a zero at 0 of order n − r − 1

In 2014, Banerjee[4 ] introduced the following polynomial

PB(z) =

p

X

i=0

 p i

 (−1)i

n + p + 1 − iz

where

QB(z) =

p

X

i=0

 p i

 (−1)i

n + p + 1 − iz

is a constant Clearly PB0(z) = zn(z − 1)p, and has a zero at 0 of order n From this polynomial, Banerjee[4 ] proved the following theorem

Theorem D [4] Let n, p ≥ 3(p ≥ 2) be two positive integers Now

when a 6= 1, n ≥ p + 3(n ≥ p + 2), or

when a = 1, n ≥ 3(n ≥ 2), then PB given by (1.3), is a SUPM (SUPE)

To reduce the cardinalities of the range sets further in the application part of [4] the following theorem was proved

Theorem E [4] In addition to the hypothesis of Theorem C we suppose that m

is a positive integer or ∞ Let S be the set of zeros of P If

1 m ≥ 3 or ∞ and min{Θ(∞; f ), Θ(∞; g)} > 6+2k−q4 ;

2 m = 2 and min{Θ(∞; f ), Θ(∞; g)} > 14+4k−2q9 ;

3 m = 1 and and min{Θ(∞; f ), Θ(∞; g)} > 10+2k−q6 ,

then S is a URSM≤m

In this paper we consider polynomials: P (z) l of the form (1.1) satisfying the condition (1.2) with P0(z) = q(z − d1)m1(z − d2)m2 (z − dk)mk, where d1 = 0; or

P (z) = (n + p + 1)

p

X

i=0

 p i

 (−1)i

n + p + 1 − iz

where

Q(z) = (n + p + 1)

p

X

i=0

 p i

 (−1)i

n + p + 1 − iz

Clearly P0(z) = (n + p + 1)zn(z − b)p, and has a zero at 0 of order n

By unsing these polynomials, we obtained the following results, which improve Theorems B, C, D, E

Theorem 1.1 Let f and g be two non-constant meromorphic(entire) functions and

m be a positive integer, or ∞ Let P (z) be a strong uniqueness polynomial of the form (1.1) satisfying the condition (1.2) with P0(z) = q(z − d1)m1(z − d2)m2 (z − dk)mk, where d1 = 0 Suppose that q ≥ 5, k ≥ 3, or k = 2 and min{m1, m2} ≥ 2, and

Ef≤m(S) = Eg≤m(S), and all of the zeros and poles of f and g are of multiplicities at least s, l, respectively Assume that one of the following conditions is satisfied:

1 q > 2k − 2 + 4s +4l (q > 2k − 2 +4s) when m ≥ 3 or ∞;

2 q > 2k −32 +4s + 2l9 (q > 2k − 32 + 4s) when m = 2;

3 q > 2k +4s + 6l (q > 2k + 4s) when m = 1

Then f = g

Theorem 1.2 Let f and g be two non-constant meromorphic (entire) functions and

P (z) be a polynomial of the form (1.4) satisfying the condition (1.5) Suppose that

P (f ) = cP (g), c 6= 0, and all of the zeros and poles of f and g are of multiplicities at least s, l, respectively, and p > 1 +1

l, np > p + n Assume that one of the following conditions is satisfied:

1 n > 1s + 1l (n > 1s) when a = 1;

2 n > p +1s +1l (n > p + 1s) when a 6= 1

Then f = g

Theorem 1.3 Let f and g be two non-constant meromorphic (entire) functions and P (z) be a polynomial of the form (1.4) satisfying the condition (1.5) and m

be a positive integer, or ∞ Suppose that Ef≤m(S) = Eg≤m(S), and all of the zeros and poles of f and g are of multiplicities at least s, l, respectively, and p > 1 + 1l,

np > p + n Assume that for the following two systems of conditions, one of (I) and one of (II) are satisfied

(I)

1 n + p > 2k − 3 +4s+ 4l (q > 2k − 3 + 4s) when m ≥ 3 or ∞;

2 n + p > 2k − 52 +4s +2l9 (q > 2k − 52 +4s) when m = 2;

3 n + p > 2k − 1 +4s+ 6l (q > 2k − 1 + 4s) when m = 1

(II)

4 n > 1s + 1l (n > 1s) when a = 1;

5 n > p +1s +1l (n > m + 1s) when a 6= 1

Then f = g

Remark 1.4 Take s = l = 1 in Theorem 1.1 (Theorem 1.2), we obtain Theorem

C (Theorem D)

Remark 1.5 Take some values of s, l in Theorem 1.1, we get the unique range sets with 11, 9 and 7 elements

(i) If s = l = 1(s = 1), then q > 10(q > 6);

(ii) If either s = 2 and l = 1 (s = 2) or s = 1 and l = 2, then q > 8 (q > 4); (iii) If s = 2 and l = 2, then q > 6

Remark 1.6 Take in Theorem 1.2 s = 2, l = 2 and p > 32, np > p + n we have:

1 n > 1 (n > 12) when a = 1;

2 n > p + 1 (n > p + 12) when a 6= 1

Remark 1.7 Take in Theorem 1.3 m ≥ 3 or ∞ and s = 2, l = 2 and p > 3

2,

np > p + n we have:

1 n + p > 5 and n > 1 when a = 1;

2 n + p > 5 and n > p + 1 when a 6= 1

2 Preliminaries

H Fujimoto [7] proved the following:

Lemma 2.1 Let P (z) be a polynomial of the form (1.1) satisfying the condition (1.2) Then P (z) is a uniqueness polynomial if and only if

X

1≤l<m≤k

qlqm >

k

X

i=1

ql

In particular, the above inequality is always satisfied whenever k ≥ 4 When k = 3 and max{q1, q2, q3} ≥ 2 or when k = 2, min{q1, q2} ≥ 2 and q1+ q2 ≥ 5, then also the above inequality holds

Lemma 2.2 For any nonconstant meromorphic function f,

N (r, 1

f0) ≤ N (r,

1

f) + N (r, f ) + S(r, f ).

Proof We have

N (r, 1

f0; f

0

6= 0) = N (r, 1

f0 f

) ≤ T (r,f

0

f ) + S(r, f ) = N (r,

f0

f ) + m(r,

f0

f ) + S(r, f )

≤ N (r, 1

f) + N (r, f ) + S(r, f );

N (r, 1

f0) = N (r,

1

f0; f

0

6= 0) + N≥2(r, 1

f); N (r,

1

f) + N

≥2

(r, 1

f) = N (r,

1

f). So

N (r, 1

f0) ≤ N (r,

1

f) + N (r, f ) + S(r, f ).

Lemma 2.3 Let f, g be two non-constant meromorphic functions and m be a pos-itive integer or ∞ Set

F = 1

f, G =

1

g, L =

F00

F0 −G

00

G0. Suppose that L 6≡ 0, and ν(f,0)≤m = ν(g,0)≤m Then

N (r, L) ≤ N≥2(r, f ) + N≥2(r, g) + N≥m+1(r, 1

f) + N

≥m+1

(r,1

g)+

N (r, 1

f0; f 6= 0) + N (r, 1

g0; g 6= 0)

Proof We consider two cases:

Case 1 m is a positive integer

We have

L = f

”

f0 − 2f

0

f −g

”

g0 + 2

g0

We now consider the poles of L By (2.1), it is clear that all poles of L are of order

1 Write f = f1

f 2 (respectively, g = g1

g 2), where f1, f2 (resp., g1, g2) are entire functions

on C having no common zeros Then

f0 = f

0

1f2− f20f1

f2 2

, f” = (f

”

2f1)f2− 2f20(f10f2− f20f1)

f3 2

;

f”

f0 =

(f”

2f1)f2− 2f20(f10f2− f20f1)

f2(f10f2− f20f1) ,

f0

f =

f10f2− f20f1

f1f2 . (2.2)

g”

g0 =

(g”

2g1)g2− 2g02(g10g2 − g20g1)

g2(g10g2− g20g1) ,

g0

g =

g10g2− g20g1

g1g2

From (2.1), (2.2), (2.3) we see that if a is a pole of L, then f (a) = ∞, or f0(a) = 0,

or f (a) = 1, or g(a) = ∞, or g0(a) = 0, or g(a) = 1 Now let a be a pole of f with

νf∞(a) = 1 Write f = f3

z−a, f3(a) 6= 0, f3(a) 6= ∞ Then F = f1 = z−af

3 , and

F0 = f3− f30.(z − a)

f2 3

, F” = −f”

3.(z − a)f3− 2f30(f3− f30.(z − a))

f3 3

;

F”

F0 =

−f”

3.(z − a)f3− 2(f30)(f3− f30.(z − a))

f3(f3− f30.(z − a)) . (2.4) From (2.4) we get FF”0(a) 6= ∞ Therefore, if a is a common pole of f and g with

νf∞(a) = νg∞(a) = 1, then

L(a) = [F

”

F0(a) −

G”

Now let f (a) = 0 with νf0(a) = l Suppose that l ≤ m From ν(f,0)≤m = ν(g,0)≤m we give

ν1

g(a) = l

Write

F = F1

(z − a)l, F1(a) 6= 0, F1(a) 6= ∞; G = G1

(z − a)l, G1(a) 6= 0, G1(a) 6= ∞ Then

F0 = F

0

1.(z − a) − lF1

(z − a)l+1 ;

F” = [F

”

1.(z − a) + (1 − l)F10](z − a) − (l + 1)(F10.(z − a) − lF1)

G0 = G

0

1.(z − a) − lG1

(z − a)l+1 ;

G” = [G

”

1.(z − a) + (1 − l)G01](z − a) − (l + 1)(G01.(z − a) − lG1)

F”

F0 −G

”

G0 =

1

z − a[

(F”

1.(z − a) + (1 − l)F10)(z − a) − (l + 1)(F10.(z − a) − lF1)

(G”

1.(z − a) + (1 − l)G01)(z − a) − (l + 1)(G01.(z − a) − lG1)

From these, if l = 1, then

F”

F0 − G

”

G0 = 1

z − a

F2.(z − a)2

(F10.(z − a) − F1)(G01.(z − a) − G1) =

F2.(z − a) (F10.(z − a) − F1)(G01.(z − a) − G1);

and if l ≥ 2, then

F”

F0 −G

”

G0 =

1

z − a

F3(z − a) (F10.(z − a) − lF1)(G01.(z − a) − lG1) =

F3 (F10.(z − a) − lF1)(G01.(z − a) − lG1).

By these equalities and F1(a) 6= 0, F1(a) 6= ∞, G1(a) 6= 0, G1(a) 6= ∞ and L 6≡ 0, we see that if l = 1, then L(a) = 0, and if m ≥ l ≥ 2, then L(a) 6= ∞ From this and (2.1)- (2.5) we can see that if a is a pole of L, then

f (a) = ∞ with νf∞(a) ≥ 2, or f0(a) = 0, or f (a) = 0 with νf(a) ≥ m + 1, or g(a) = ∞ with νg∞(a) ≥ 2, or g0(a) = 0, or g(a) = 0 with νg(a) ≥ m + 1 Case 2 m = ∞ Then N≥m+1(r,1

f) = 0 = N≥m+1(r,1

g) and applying the result

of Case 1 we obtain the proof of Case 2

Thus the proof of Lemma 2.3 is completed

Lemma 2.4 Let f be a non-constant meromorphic function Then

1 N (r,1f) − 12N≤1(r,1f) ≤ 12N (r,f1);

2 N (r,1f) + 12N≥3(r,f1) −12N≤1(r,f1) ≤ 12N (r,1f);

3 N (r,1f) + 12N≥m(r,f1) − 12N≤1(r,f1) ≤ 12N (r,1f) when m ≥ 4 or ∞

Proof We first note that

N (r, 1

f) = N (r,

1

f)+N

≥2

(r, 1

f)+ +N

≥k

(r,1

f)+ ; N (r,

1

f) = N

≤1

(r, 1

f)+N

≥2

(r, 1

f).

1 We have

N (r,1

f) −

1

2N

≤1

(r, 1

f) =

1

2(N (r,

1

f) + N (r,

1

f) − N

≤1

(r, 1

f)) =

1

2(N (r,

1

f)+

N≤1(r, 1

f) + N

≥2

(r, 1

f) − N

≤1

(r,1

f)) =

1

2(N (r,

1

f) + N

≥2

(r, 1

f)) ≤

1

2N (r,

1

f).

2 We have

N (r,1

f) +

1

2N

≥3

(r,1

f) −

1

2N

≤1

(r, 1

f) =

1

2(2N (r,

1

f) + N

≥3

(r, 1

f) − N

≤1

(r, 1

f) = 1

2(N (r,

1

f) + N

≤1

(r,1

f) + N

≥2

(r, 1

f) + N

≥3

(r, 1

f) − N

≤1

(r, 1

f) = 1

2(N (r,

1

f) + N

≥2

(r, 1

f) + N

≥3

(r, 1

f)) ≤

1

2N (r,

1

f).

3 We consider two cases:

Case 1 m is a positive integer

We have

N (r, 1

f) + N

≥m

(r, 1

f) −

1

2N

≤1

(r, 1

f) =

1

2(2N (r,

1

f) + 2N

≥m

(r, 1

f) − N

≤1

(r, 1

f) = 1

2(N (r,

1

f) + N

≤1

(r, 1

f) + N

≥2

(r, 1

f) + 2N

≥m

(r,1

f) − N

≤1

(r,1

f) = 1

2(N (r,

1

f) + N

≥2

(r, 1

f) + 2N

≥m

(r, 1

f)) ≤ 1

2(N (r,

1

f) + N

≥2

(r, 1

f) + N

≥3

(r,1

f) + N

≥m

(r, 1

f)) ≤

1

2N (r,

1

f).

Case 2 m = ∞ Then N≥m(r,1f) = 0 and applying the result of Case 1 we obtain the proof of Case 2

H Fujimoto [6] proved the following:

Lemma 2.5 Under the same situation as in Lemma 2.1, we assume furthermore that q ≥ 5 and there are two meromorphic function f and g such that t

1

P (f ) =

c0

P (f ) + c1 for any two constants c0(6= 0) and c1 If k ≥ 3 or if k = 2, min{q1, q2} ≥ 2 , then

c1 = 0

Lemma 2.6 Under the same situation as in Theorem 1.1 Then there is constant c(6= 0) such that P (f ) = cP (g)

Proof Set

P (f ), G =

1

P (g), L =

F00

F0−G

00

G0, S(r) = S(r, f )+S(r, g), T (r) = T (r, f )+T (r, g). Then T (r, P (f )) = qT (r, f ) + O(1) and T (r, P (g)) = qT (r, g) + O(1), and hence S(r, P (f )) = S(r, f ) and S(r, P (g)) = S(r, g), since P (f ) and f , and P (g) and g have the same growth estimates, respectively Suppose that L 6≡ 0 By the lemma

on logarithmic derivative, m(r, L) = S(r) From P (z) = (z − a1) (z − aq), P0(z) = q(z − d1)m 1(z − d2)m 2 (z − dk)m k, where d1 = 0 we have

P0(f ) = qfm1(f − d2)m2 (f − dk)mkf0, P0(g) = qgm1(g − d2)m2 (g − dk)mkf0 Then, applying the second main theorem to the functions f, g and the values a1, a2, , aq,

d1 = 0, d2, , dk, ∞, respectively, and note that

q

X

i=1

N (r, 1

f − ai) = N (r,

1

P (f )),

q

X

i=1

N (r, 1

g − ai) = N (r,

1

P (g))

we obtain

(q + k − 1)T (r) ≤ N (r, f ) + N (r, g) + N (r, 1

P (f )) + N (r,

1

P (g)) + N (r,

1

f) + N (r,

1

g)+

X

i=2

N (r, 1

f − di) +

k

X

i=2

N (r, 1

g − di) − N (r,

1

f0; C1) − N (r, 1

g0; C2), (2.6) where C1is the condition (f −a1) (f −aq)f (f −d2) (f −dk) 6= 0 and C2is similarly defined Since the zeros (poles) of f and g are of multiplicities at least s (respect., l),

N (r, 1

f)+N (r,

1

g) ≤

1

s(N (r,

1

f)+N (r,

1

g)) ≤

1

s(T (r, f )+T (r, g)+S(r) =

1

sT (r)+S(r)

N (r, f )+N (r, g) ≤ 1

l(N (r, f )+N (r, g)) ≤

1

l(T (r, f )+T (r, g)+S(r) =

1

lT (r)+S(r).

(2.7)

On the other hand

k

X

i=2

N (r, 1

f − di) +

k

X

i=2

N (r, 1

g − di) ≤ (k − 1)T (r) + S(r). (2.8)

By the hypothesis we have N≤1(r,P (f )1 ) = N≤1(r,P (g)1 ) From Lemma 3.2 we see that if a is a common zero of P (f ) and Q(g) with ν0

L(a) = 0 Therefore,

N≤1(r, 1

P (f )) = N

≤1

(r, 1

P (g)) ≤ N (r,

1

L) ≤ T (r, L) + S(r) ≤ N (r, L) + S(r). Hence, by Lemma 2.3 we obtain

N (r, L) ≤ N (r, f )+N (r, g)+N≥m+1(r, 1

P (f ))+N

≥m+1

(r, 1

P (g))+N (r,

1

f)+N (r,

1

g)+

N (r, 1

f − di) +

k

X

i=2

N (r, 1

g − di) + N (r,

1

f0; C1) + N (r,

1

g0; C2)

≤ 1

lT (r) +

1

sT (r) + N

≥m+1

(r, 1

P (f )) + N

≥m+1

(r, 1

P (g)) + (k − 1)T (r)+

N (r, 1

f0; C1) + N (r,

1

g0; C2) + S(r), (2.9) where m is a positive integer or ∞ From (2.6)- (2.9) we have

(q + k − 1)T (r) ≤ (1

l +

1

s)T (r) + N (r,

1

P (f )) + N (r,

1

P (g)) + (k − 1)T (r)

−N (r, 1

f0; C1) − N (r,

1

g0; C2) + S(r) ≤ (

1

l +

1

s + k − 1)T (r) + N (r,

1

P (f ))+

N (r, 1

P (g)) + N

≤1

(r, 1

P (f )) − N

≤1

(r, 1

P (f )) − N (r,

1

f0; C1) − N (r,

1

g0; C2) + S(r)

≤ (1

l +

1

s + k − 1)T (r) + N (r,

1

P (f )) + N (r,

1

P (g)) + N (r, L) − N

≤1

(r, 1

P (f ))

−N (r, 1

f0; C1) − N (r,

1

g0; C2) + S(r) ≤ (

1

l +

1

s+ k − 1)T (r) + N (r,

1

P (f )) + N (r,

1

P (g))