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Examples on exceptional values of meromorphic functions
Acta Mathematica Scientia 2011,31B(4):1327–1336 http://actams.wipm.ac.cn
EXAMPLES ON EXCEPTIONAL VALUES OF MEROMORPHIC FUNCTIONS∗
)
Peng Yanhong (
Sun Daochun (
)
School of Mathematical Sciences, South China Normal University, Guangzhou 510631, China E-mail: [email protected]; [email protected]
Abstract In this paper, by means of the definition of Borel exceptional value method, another exceptional value of meromorphic function which is a T exceptional value is defined by linking the concept of T direction. And we construct a meromorphic function with zero as Borel exceptional value, but not as T exceptional value; and another meromorphic function with zero as T exceptional value, but not as Borel exceptional value. Key words meromorphic function; Borel exceptional value; T exceptional value 2000 MR Subject Classification
1
30D35
Introduction and Some Results
In the theory of meromorphic functions, the study on singular directions of meromorphic functions makes a great significance to the development of value distribution theory and argument distributed theory of meromorphic functions, particularly it conducts Borel-direction and Julia-direction as forerunner. In 1919, Julia introduced the concept of Julia direction and showed that every transcendental entire function has at least one Julia direction. This result is a refinement of Picard’s theorem. In order to have a similar refinement for Borel’s theorem, a more refined notion of Borel direction was firstly introduced by G.Valiron in 1928, which greatly promoted the development of value distribution theory and argument distribution theory. In 2003, Zheng Jianhua [9] introduced a new singular direction, namely, the T-direction for meromorphic functions in the complex plane, which further promoted the popularization and development of singular directions of meromorphic functions. Furthermore, the exceptional value of meromorphic functions is also one of the main objects studied in the theory of value distribution of meromorphic functions. Several types of exceptional values were introduced in [3]. Their existence and some connections between them were also established. However, exceptional values of meromorphic functions are strictly relative with singular directions. For instance, Picard exceptional value relating with Julia direction and Borel exceptional value relating with Borel direction, and so on. Then we can derive T exceptional value from T direction, ∗ Received
(10161006).
September 27, 2009. This project was supported by the Natural Science Foundation of China
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namely we can call a type of exceptional value relating with T direction T exceptional value. It was known form [3] that the Picard exceptional value of f (z)(where f (z) is a meromorphic function of finite and positive order) must be its Borel exceptional value; and suppose that f (z) is a transcendental meromorphic function, if a is the Picard exceptional value of f (z), then a must be the Nevanlinna exceptional value of f (z). Now, we recall the definition of Borel direction and the definition of T direction as follows. Suppose that f (z) is a meromorphic function of order λ (0 < λ < ∞) in the complex plane. A ray argz = θ is called a Borel direction of order λ for f (z) if for every 0 < ε < π, lim sup r→∞
ln n(r, θ, ε, f = a) =λ ln r
holds for all a in C∞ with at most two possible exceptions, where n(r, θ, ε, f = a) is the number of the solutions of f (z) = a in {z|| arg z − θ| < ε, |z| < r}, counting with multiplicities (see [3]). It’s well known that every λ (0 < λ < ∞) order meromorphic function has at most two Borel exceptional values. A ray arg z = θ is called a T direction for a meromorphic function f (z), if for arbitrary ε (0 < ε < π), N (r, θ, ε, f = a) lim sup >0 T (r, f ) r→∞ holds for any complex value a except at most two possible exceptions, where N (r, θ, ε, f = a) r =a) = 0 n(t,θ,ε,f dt + n(0, f = a) ln r (see [9]). Also the number of T exceptional values of a t meromorphic function is equal to 2 at most. Zhang Qingde discussed in detail the connections between Borel direction and T direction of finite positive order meromorphic functions in [7], he obtained the following result. Let f (z) be a order λ (0 < λ < ∞) meromorphic function in the complex plane, its most Borel-direction for U (r, f ) is also a T-direction of f (z), where U (r, f ) is the type function of f (z). And if the lower order of f (z) u = λ, then its T-direction must be its Borel-direction. But in the normal situation, Borel-direction and T-direction are two different types of singular directions. And in [7] he proved the following results. Theorem A For arbitrary positive number λ, positive integers q1 and q2 , let E1 = {arg z = θj |0 ≤ θ1 < θ2 < · · · < θq1 < 2π} and E2 = {arg z = ϕj |0 ≤ ϕ1 < ϕ2 < · · · < ϕq2 < 2π}, such that E1 E2 = ∅, then there exists a meromorphic function f (z) of order λ in the complex plane, exactly with E1 E2 as its T-direction, and exactly with E2 as its Borel-direction. Theorem B For arbitrary positive number λ, positive integers q1 and q2 , let E1 = {arg z = θj |0 ≤ θ1 < θ2 < · · · < θq1 < 2π} and E2 = {arg z = ϕj |0 ≤ ϕ1 < ϕ2 < · · · < ϕq2 < 2π}, such that E1 E2 = ∅, then there exists a meromorphic function f (z) of order and under-order being both λ in the complex plane, exactly with E1 E2 as its Borel-direction, and exactly with E2 as its T-direction. This suggests the following question, what is the relation between the Borel exceptional value and T exceptional value of meromorphic functions? In most papers, the authors only introduced the related theorems of the connections between a few important exceptional values and some singular directions, and conclusions of related questions. We have found no papers discussing exceptional values separately keeping
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away singular directions, and no papers discussing Borel exceptional values and T exceptional values. In this paper, we initially try to use a special way to construct a meromorphic function with zero as Borel exceptional value, but not as T exceptional value; and another meromorphic function with zero as T exceptional value, but not as Borel exceptional value. Furthmore, this paper introduces exceptional values of meromorphic functions with examples, which is different from some other papers and more significant, and has a better consulted value to us in studying singular directions of meromorphic functions by giving examples in the future. In this paper, let n(r, f = a) be the number of zero points of f (z) − a in |z| < r, counting with multiplicities. Other symbols unexplained are similar to those in [3] without exception. Definition 1 The order of the meromorphic function f (z) is defined by λ = lim sup r→∞
ln T (r, f ) . ln r
where T (r, f ) is the Nevanlinna’s characteristic function of f (z) (see [3]). Let us recall the following two definitions. Definition 2 [3] Suppose that f (z) is a meromorphic function of order λ (0 < λ < ∞) in the complex plane, a complex number a is called a Borel exceptional value of f (z) if lim sup r→∞
ln n(r, f = a) < λ. ln r
Definition 3 Suppose that f (z) is a meromorphic function in the complex plane, a(a ∈ c∞ ) is called a T exceptional value of f (z), if lim sup r→∞
N (r, f = a) = 0. T (r, f )
Our main results are n Example 1 Set bn = 2n , an = bn + un , where un = sm =
m
(1 + un )[
√
2nn ]
1 n 22n
. Suppose that
, sm → s (0 < s < ∞) as m → ∞,
n=1
√ √ where [ 2nn ] is the integral part of 2nn . Let [√2nn ] ∞ an − b n 1+ − 1. f (z) = z − an n=1 √ Set Bn = 2(n+1)(n+1) , An = Bn + βn , where βn = √ 1 (n+1) . 2
(1)
2(n+1)
Also suppose that m (n−1)(n−1) ] tm = , tm → t (0 < t < ∞) as m → ∞ (1 + βn )[2 n=1
and let [ ∞ An − Bn g(z) = 1+ z − An n=1
√
(n−1)
2(n−1)
]
,
(2)
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we set F (z) = f (z)g(z), then F (z) is a meromorphic function in the complex plane, and 0 is a Borel exceptional value of F (z), but not a T exceptional value of F (z). Example 2 Let
√ [√2nn ] 22nnnn ∞ ∞ ln 2 an − b n An − Bn 1+ 1+ f (z) = − 1, g(z) = z − a z − An n n=1 n=1 and set F (z) = f (z)g(z), where an , bn , An , Bn are all the same with those of Example 1. Then 0 is a T exceptional value of F (z), but not a Borel exceptional value of F (z).
2
Some Lemmas Lemma 1 (See [6, Theorem 1.3]) Let fn (z) =
n
[1 + uk (z)]. If for all z in a domain G
k=1
and for every positive integer k we have |uk (z)| ≤ Mk , where Mk do not depend on z, and the ∞ ∞
series Mk is convergent, then the infinite product [1 + uk (z)] is unanimously convergent k=1
k=1
to an analytic function f (z). f (z), g(z), an , bn , An , Bn , un , βn , sm , s, tm , t appeared in the following are all as in Example 1. Lemma 2 2.1) f (z) is defined as (1), then f (z) satisfies: 2.1.1) f (z) is a meromorphic function in the complex plane. 2.1.2) For arbitrary ε > 0, there exists M > 0, such that when |z| > M and |z − an | ≥ 1 (n = 1, 2, · · ·), we have |f (z)| < ε. 2.2) g(z) is defined as (2), then g(z) satisfies: 2.2.1) g(z) is a meromorphic function in the complex plane. 2.2.2) For arbitrary ε > 0, there exists M = 2t ε +1+b > 0, where b = max{An |1 ≤ n ≤ N }, when |z| > M and |z − An | ≥ 1 (n = 1, 2, · · ·), we have |g(z) − 1| < ε. Proof 2.1) Let f1 (z) =
∞ n=1
C=
z [ 1− bn
√ n 2n ]
∞ bn [ n=1
an
, f2 (z) =
∞ n=1
√ n 2n ]
=
∞ n=1
z [ 1− an
un [ 1− an
√ n 2n ]
√ n 2n ]
,
.
Clearly b1 , b2 , · · · , bn , · · · are infinitely many zeros of f1 (z), their moduli satisfy |b1 | < |b2 | < · · · < |bn | < · · · , and we have n
lim |bn | = lim 2n = ∞.
n→∞
n→∞
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Arbitrarily given a positive number R, if bm ≤ R < bm+1 , we consider the following series in |z| < R, ∞ √ z , [ 2nn ] ln 1 − bn n=m+1 where every ln(1 − |z| < R,
z bn )
√ is chosen the main value. Let αn = [ 2nn ] ln 1 −
z bn
(n > m). When
√ √ z z 1 z 2 1 z n = [ 2nn ] − αn = [ 2nn ] ln 1 − − ···− − ··· . bn bn 2 bn n bn n
Since bn = 2n , so |αn | ≤
√ n z z 2 z z 2 z 3 2n + + · · · ≤ √ + √ + √ + · · · . bn bn bn bn bn
When n is sufficiently large, there holds z √ ≤ R = 1. b 2R 2 n Hence |αn | ≤ ∞
Since series
n=1
1 2n
∞ 1 . 2n n=1 ∞
is convergent, by Lemma 1, we know
(1 −
n=m+1
unanimously convergent to an analytic function in |z| < R, so
∞
√ z [ 2nn ] bn )
(1 −
n=1
is absolutely and
√ [ 2nn ]
z bn )
is absolutely
and unanimously convergent to an analytic function in |z| < R, which implies f1 (z) is analytic in |z| < R. Since R is an arbitrary positive number, we know f1 (z) is analytic in the complex plane. Similarly, we can prove that f2 (z) and C are also analytic. Hence f (z) = Cf1 (z)/f2 (z)−1 is meromorphic in the complex plane. When |z − an | ≥ 1,we have n [√2nn ] 0 +T − b a n n 1 + − 1 z − an n=n 0
√
n i [ 2nn ] 0 +T an − b n i√ = 1+ − 1 C[ 2nn ] z − an n=n0
√
i=1
√
n i i 2nn ] 2nn ] n 0 +T [ 0 +T [ an − b n an − bn i√ i√ = C[ 2nn ] + ··· + C[ 2nn ] z − an z − an ≤
n=n0 i=1 √ n 2n ] n 0 +T [ n=n0
i=1
C[i√2nn ] un + · · · +
√ n 2n ] n 0 +T [ n=n0
i=1
n=n0
n0 +T √ n sn +T n = (1 + un )[ 2 ] − 1 = 0 − 1. sn −1 n=n0
0
i=1
C[i√2nn ] un
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Also owing to sn being convergent, there exists M1 > 0, such that when n0 > M1 , no matter s ε how large T is, we always have | snn0 +T − 1| < 2s . Therefore, for fixed N > M1 , −1 0
∞ [√2nn ] − b a ε n n 1+ − 1 < , A = z − an 2s n=N
set b = max{an |1 ≤ n ≤ N }, and let M = 2s ε + 1 + b. Then, when |z| > M, we have √ n N [ 2n ] −1 an − b n 1+ − 1 B = z − an n=1 √ n 2n ] N −1 [
√
nn
2 ] N −1 [ uin uin i√ ≤ C[i√2nn ] + · · · + C n [ 2n ] |z − a |i |z − an |i n n=1 i=1 n=1 i=1 N −1 √ n 1 n ≤ (1 + un )[ 2 ] − 1 (|z| − b) n=1
≤
1 (sN −1 − 1) < ε/2. |z| − b
Thus, when |z| > M and |z − an | ≥ 1 (n = 1, 2, · · ·), we have N −1 N √ nn −1 √ z − bn [ 2 ] [ 2nn ] A + B ≤ A + B < ε. (1 + u ) |f (z)| ≤ n z − an n=1 n=1 The proof of 2.2) is omitted as the method is the same as that of 2.1). So the proof of the lemma is completed. Lemma 3 (See [10, 3.4 Argument Principle]) Let f be meromorphic in G with poles p1 , p2 , · · · , pm and zeros z1 , z2 , · · · , zn counted according to multiplicity. If γ is a closed rectifiable curve in G with γ ≈ 0 and not passing through p1 , p2 , · · · , pm ; z1 , z2 , · · · , zn ; then n m f (z) 1 dz = n(γ; zk ) − n(γ; pj ). 2π γ f (z) j=1 k=1
Lemma 4 Let Lp : |z − ap | ≤ p + 1 (p = 1, 2 · · ·). When p > N0 (N0 ≥ N ), then, in the interior of every Lp , the number of zero points of f (z) − d is the same as the number of poles of f (z). Where N is as in 2.1) of Lemma 2. Proof For a point z on the boundary of every circle Lp , choosing arbitrarily a complex ∞ √
number d = 0, let ε = min{ 21 , |d| [ 2nn ]un is convergent. Choosing N , let N0 = 2 }. Since max{N, N }, when p > N0 , such that
n=1
∞ √ 1 1 · [ 2nn ]un < ε(|d| − ε). p + 1 n=1 2
Since p > N, the point z on the boundary of Lp satisfies |z| = ap + p + 1 > M, by 1.2) of Lemma 2, we know |f (z)| < ε < 1. Then √ n f (z) f (z) ∞ [ 2n ]un 1 1 · f (z) − d ≤ f (z) · |f (z) − d| = (z − a )(z − b ) |f (z) − d| n n n=1 √ n ∞ ∞ √ n 1 1 1 1 [ 2n ]un · ≤ · · [ 2n ]un < ε ≤ . ≤ |z − a ||z − b | |d| − ε |d| − ε p + 1 2 n n n=1 n=1
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Y.H. Peng & D.C. Sun: EXAMPLES ON EXCEPTIONAL VALUES
1 2π
Hence
∂Lp
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1 f (z) dz ≤ . f (z) − d 2
By Lemma 3, when p > N0 (N0 ≥ N ), in the interior of each Lp , we know that the number of zero points of f (z) − d is the same as the number of poles of f (z). This concludes the proof of the lemma. Similarly, let Lq : |z − Aq | ≤ q + 1 (q = 1, 2, · · ·), choosing arbitrarily a complex number a = d + 1 (d = 0), by the similar argument of Lemma 4, we can deduce that, when q > N , the number of zero points of g(z) − a is the same as the number of poles of g(z) inside every Lq .
2
Proofs of Examples
Proof of Example 1 By Lemma 2, we know f (z), g(z) are meromorphic, so F (z) = f (z)g(z) is meromorphic in the complex plane. For arbitrary sufficiently large r, there corresponds n such that n
(n+1)
bn = 2n < r ≤ bn+1 = 2(n+1) , n i √ √ n(r, f = −1) = n(r, f ) = [ 2i ] ≤ 2 · [ 2nn ] ≤ 2 · 2nn ,
i=1
√ n(t, f ) dt ≤ n(r, f ) ln r ≤ 2 · 2nn ln r, N (r, f = −1) = N (r, f ) = t 1 ⎧ n √ ⎪ n ⎪ ⎪ [ 2(i−1)( i−1) ], 2n < r ≤ 2(n+1)(n+1) , ⎪ ⎨ i=1 n(r, g) = n+1 √ ⎪ (n+1) ⎪ ⎪ [ 2(i−1)( i−1) ], 2(n+1)(n+1) < r ≤ 2(n+1) , ⎪ ⎩ r
(3)
i=1
namely,
⎧ √ ⎨ 2 · 2(n−1)(n−1) , n(r, g) ≤ √ ⎩ 2 · 2nn ,
√ n 2n < r ≤ 2(n+1)(n+1) , √ (n+1) 2(n+1)(n+1) < r ≤ 2(n+1) .
Thus n(r, g) ≤ 2 · so that
N (r, g) = 1
r
√
√ n(t, g) dt ≤ n(r, g) ln r ≤ 2 · 2nn ln r. t
By a similar argument, we can obtain N (r, g = 0) ≤ 2 · Since T (r, f ) ≤
3
i=1
2nn ,
√ n 2n ln r.
(4)
(5)
N (r, f = ai ) + O{ln(rT (r, f ))}. without harm suppose that ai (i = 1, 2, 3)
is −1, ∞, 5, respectively, by Lemma 4, we have n(r, f = 5) = n(r − (aN +1 − (N + 1)), f = 5) + n((aN +1 − (N + 1)), f = 5) = n(r − (aN +1 − (N + 1)), f = 5) + c n √ [ 2ii ] + c ≤ 2 · 2nn − ( 2N N − 1) + c, = i=N +1
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where c is a finite number depending on N . So that √ N (r, f = 5) ≤ 2 · 2nn − ( 2N N − 1) + c ln r.
(6)
Hence, by (3),(6), we obtain √ T (r, f ) ≤ 6 · 2nn − ( 2N N − 1) + c) ln r + O{ln(rT (r, f ))}. For g(z), we put aj (j = 1, 2, 3) to be 0, ∞, 5, respectively. Similarly, we also have √ N (r, g = 5) ≤ 2 · 2nn − ( 2N N − 1) + c ln r.
(7)
Therefore by, (4), (5), (7), T (r, g) ≤
3
N (r, f = aj ) + O{ln(rT (r, g))}
j=1
√ ≤ 6 · 2nn − ( 2N N − 1) + c ln r + O{ln(rT (r, g))}. Finally, we have T (r, F ) ≤ T (r, f ) + T (r, g) √ ≤ 12 · 2nn − 2 · ( 2N N − 1) + 2 · c ln r + O{ln(rT (r, f ))} + O{ln(rT (r, g))}. Then ln T (r, F ) ln r √ √ ln((12 · 2nn − 2 · ( 2N N − 1) + 2 · c) ln r + O{ln(rT (r, f ))} + O{ln(rT (r, g))}) ≤ lim sup ln r r→∞ √ n (n+1)(n+1) n 1 ln(12 · 2 ln 2 ) = . ≤ lim n→∞ ln 2nn 2 lim sup r→∞
n
On the other hand, choosing a point sequence {rn = 2n + 2n}, we can obtain n(rn , F ) = n(rn , f ) + n(rn , g) =
n n [ 2ii ] + [ 2(i−1)(i−1) ] i=1
i=1
√ √ ≥ [ 2nn ] + [ 2(n−1)(n−1) ] > 2nn − 1, 2rn √ n(t, F ) dt ≥ n(rn , F ) ln 2 ≥ ( 2nn − 1) · ln 2, N (2rn , F ) ≥ t rn √ T (2rn , F ) ≥ N (2rn , F ) ≥ ( 2nn − 1) · ln 2, √ 1 ln T (r, F ) ln T (2rn , F ) ln( 2nn − 1) · ln 2) ≥ lim sup = . lim sup ≥ lim n n n→∞ ln r ln 2rn ln 2 · (2 + 2n) 2 r→∞ rn →∞
(8)
) So that the order of F (z) is λ = lim sup ln Tln(r,F = 12 . r r→∞
Now, we begin to prove that 0 is a Borel exceptional value of F (z), but not a T exceptional value of F (z).
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Y.H. Peng & D.C. Sun: EXAMPLES ON EXCEPTIONAL VALUES
Since
⎧ n ⎪ ⎪ ⎪ [ 2(i−1)(i−1) ], ⎪ ⎨
n(r, F = 0) = n(r, g = 0)
i=1
n+1 ⎪ ⎪ ⎪ [ 2(i−1)(i−1) ], ⎪ ⎩
n
2n < r ≤
1335
√ 2(n+1)(n+1) ,
√ (n+1) 2(n+1)(n+1) < r ≤ 2(n+1) ,
i=1
⎧ √ ⎨ 2 · 2(n−1)(n−1) , n(r, F = 0) ≤ √ ⎩ 2 · 2nn ,
so
Thus
√ n 2n < r ≤ 2(n+1)(n+1) , √ (n+1) 2(n+1)(n+1) < r ≤ 2(n+1) .
√ ⎧ ⎪ ln(2 · 2(n−1)(n−1) ) ⎪ ⎪ lim = 0, ln n(r, F = 0) ⎨ n→∞ ln 2nn ≤ lim sup √ ⎪ ln r r→∞ ⎪ ln(2 · 2nn ) ⎪ ⎩ lim √ = 0. n→∞ ln 2(n+1)(n+1)
That is lim sup r→∞
ln n(r, F = 0) 1 =0< . ln r 2
So it is evident that 0 is a Borel exceptional value of F (z). √ (n+1) Also choosing point sequences {rn = 2(n+1)(n+1) + 2(n + 1)} and {rn = 2(n+1) }, then n(rn , F = 0) = n(rn , g = 0) =
n+1
√ √ [ 2(i−1)(i−1) ] > 2nn ≥ 2nn − 1.
i=1
Hence N (rn , F = 0) ≥
rn
rn
√ n(t, F ) r r dt ≥ n(rn , F = 0) ln n ≥ ( 2nn − 1) ln n . t rn rn
So that √ r ( 2nn − 1) · ln rnn N (r, F = 0) N (rn , F = 0) √ ≥ lim sup ≥ lim sup lim sup T (r, F ) T (rn , F ) r→∞ 12 · 2nn · ln rn rn→∞ rn→∞ √ √ (n+1) ( 2nn − 1) · (ln 2(n+1) − ln( 2(n+1)(n+1) + 2(n + 1))) √ ≥ lim n→∞ 12 · 2nn ln 2(n+1)(n+1) 1 > 0. = 12 Therefore we come to that 0 is not a T exceptional value of F (z). Example 1 is proved. Proof of Example 2 By the argument of proof of Example 1 we know the order of F (z) is ln T (r, F ) λ = lim sup = 1. ln r r→∞ n
(n+1)
By a similar argument of proof of Example 1 we obtain, for 2n < r ≤ 2(n+1) , ⎧ √ n √ ⎪ 2n n ⎪ ⎪ 2n < r ≤ 2(n+1)(n+1) , ⎨ 2 · 2 nn , ln 2 n(r, F = 0) = n(r, g = 0) ≤ √ ⎪ (n+1)(n+1) √ ⎪ (n+1) ⎪2 · 2 ⎩ , 2(n+1)(n+1) < r ≤ 2(n+1) . (n+1) 2 (n+1) ln 2
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And n(r, F = 0) = n(r, g = 0) ⎧ √ n √ n ⎪ 2n 2n ⎪ ⎪ ⎪ ⎨ ln2 2nn ≥ ln2 2nn − 1, ≥ √ √ ⎪ (n+1)(n+1) ⎪ 2 2(n+1)(n+1) ⎪ ⎪ ≥ − 1, ⎩ ln2 2(n+1)(n+1) ln2 2(n+1)(n+1)
n
2n < r ≤ √
√ 2(n+1)(n+1) ,
2(n+1)(n+1) < r ≤ 2(n+1)
(n+1)
.
n
Choosing a point sequence {rn = 2n + 2n}, similar to (4), (8) in the proof of Example 1, we have T (2rn , F ) ≥ N (2rn , F ) ≥ N (2rn , f ) + N (2rn , g) √ nn √ n 2 n ≥ ( 2 − 1) · ln 2 + − 1 · ln 2, ln2 2nn √ n 2n N (2rn , F = 0) ≤ 2 · 2 n ln(2rn ). ln 2n So that lim sup r→∞
N (r, F = 0) N (2rn , F = 0) = lim sup T (r, F ) T (2rn , F ) rn →∞ ≤ lim
n→∞
2·
√ ( 2nn
√ n 2n ln2 2nn
n
ln(2 · (2n + 2n)) √ n = 0. n − 1) · ln 2 + ln222nn − 1 · ln 2
Therefore 0 is the T exceptional value of F (z). √ Again we choose {rn = 2 · 2(n+1)(n+1) }, then it yields lim sup r→∞
ln n(r, F = 0) ≥ ln r
ln n(rn , F lim sup →∞ ln rn rn
= 0)
√ ln
= lim
n→∞
(n+1)
2(n+1) n+1) 2(n+1)
ln2
ln(2 ·
−1
√ = 1. 2(n+1)(n+1) )
Thus, we conclude that 0 is not a Borel exceptional value of F (z). This concludes the proof of Example 2. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]
Sun D C. The filling disks on Borel direction. Journal of Mathematics, 1985, 5: 167–173 Hayman W K. Meromorphic Functions. Oxford: Clarendon Press, 1964 Yang L. The Theory of Value Distribution and New Researches. Beijing: Science Press, 1982 Sun D C. Examples on singular direction. Journal of Mathematics, 1994, 14: 176–182 Zhuang Q T. The Singular Directions of Meromorphic Functions. Beijing: Science Press, 1982 Bo S G. Integral Function and Meromorphic Functions. Wuhan: Huazhong Normal University Press, 1989 Zhang Q D. T direction and Borel direction of meromorphic functions of finite and positive order. Acta Mathematica Sinica, 2007, 50(2): 413–420 Yang L, Zhang G H. On zhe construction of meromorphic functions with given zhe singular directions. Science in China, Ser A, 1976, 5(3): 308–319 Zheng J H. Transcendental meromorphic functions with radiant distributed values. Science in China, Series A, 2003, 33(6): 537–550 Conway J B. Functions of One Complex Variable. Beijing: World Publishing Corporation, 1978 Chen T W, Sun D C. Singular directions of quasimeromorphic mappings. Acta Mathematica Scientia, 1999, 19(4): 472–478
EXAMPLES ON EXCEPTIONAL VALUES OF MEROMORPHIC FUNCTIONS∗
)
Peng Yanhong (
Sun Daochun (
)
School of Mathematical Sciences, South China Normal University, Guangzhou 510631, China E-mail: [email protected]; [email protected]
Abstract In this paper, by means of the definition of Borel exceptional value method, another exceptional value of meromorphic function which is a T exceptional value is defined by linking the concept of T direction. And we construct a meromorphic function with zero as Borel exceptional value, but not as T exceptional value; and another meromorphic function with zero as T exceptional value, but not as Borel exceptional value. Key words meromorphic function; Borel exceptional value; T exceptional value 2000 MR Subject Classification
1
30D35
Introduction and Some Results
In the theory of meromorphic functions, the study on singular directions of meromorphic functions makes a great significance to the development of value distribution theory and argument distributed theory of meromorphic functions, particularly it conducts Borel-direction and Julia-direction as forerunner. In 1919, Julia introduced the concept of Julia direction and showed that every transcendental entire function has at least one Julia direction. This result is a refinement of Picard’s theorem. In order to have a similar refinement for Borel’s theorem, a more refined notion of Borel direction was firstly introduced by G.Valiron in 1928, which greatly promoted the development of value distribution theory and argument distribution theory. In 2003, Zheng Jianhua [9] introduced a new singular direction, namely, the T-direction for meromorphic functions in the complex plane, which further promoted the popularization and development of singular directions of meromorphic functions. Furthermore, the exceptional value of meromorphic functions is also one of the main objects studied in the theory of value distribution of meromorphic functions. Several types of exceptional values were introduced in [3]. Their existence and some connections between them were also established. However, exceptional values of meromorphic functions are strictly relative with singular directions. For instance, Picard exceptional value relating with Julia direction and Borel exceptional value relating with Borel direction, and so on. Then we can derive T exceptional value from T direction, ∗ Received
(10161006).
September 27, 2009. This project was supported by the Natural Science Foundation of China
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namely we can call a type of exceptional value relating with T direction T exceptional value. It was known form [3] that the Picard exceptional value of f (z)(where f (z) is a meromorphic function of finite and positive order) must be its Borel exceptional value; and suppose that f (z) is a transcendental meromorphic function, if a is the Picard exceptional value of f (z), then a must be the Nevanlinna exceptional value of f (z). Now, we recall the definition of Borel direction and the definition of T direction as follows. Suppose that f (z) is a meromorphic function of order λ (0 < λ < ∞) in the complex plane. A ray argz = θ is called a Borel direction of order λ for f (z) if for every 0 < ε < π, lim sup r→∞
ln n(r, θ, ε, f = a) =λ ln r
holds for all a in C∞ with at most two possible exceptions, where n(r, θ, ε, f = a) is the number of the solutions of f (z) = a in {z|| arg z − θ| < ε, |z| < r}, counting with multiplicities (see [3]). It’s well known that every λ (0 < λ < ∞) order meromorphic function has at most two Borel exceptional values. A ray arg z = θ is called a T direction for a meromorphic function f (z), if for arbitrary ε (0 < ε < π), N (r, θ, ε, f = a) lim sup >0 T (r, f ) r→∞ holds for any complex value a except at most two possible exceptions, where N (r, θ, ε, f = a) r =a) = 0 n(t,θ,ε,f dt + n(0, f = a) ln r (see [9]). Also the number of T exceptional values of a t meromorphic function is equal to 2 at most. Zhang Qingde discussed in detail the connections between Borel direction and T direction of finite positive order meromorphic functions in [7], he obtained the following result. Let f (z) be a order λ (0 < λ < ∞) meromorphic function in the complex plane, its most Borel-direction for U (r, f ) is also a T-direction of f (z), where U (r, f ) is the type function of f (z). And if the lower order of f (z) u = λ, then its T-direction must be its Borel-direction. But in the normal situation, Borel-direction and T-direction are two different types of singular directions. And in [7] he proved the following results. Theorem A For arbitrary positive number λ, positive integers q1 and q2 , let E1 = {arg z = θj |0 ≤ θ1 < θ2 < · · · < θq1 < 2π} and E2 = {arg z = ϕj |0 ≤ ϕ1 < ϕ2 < · · · < ϕq2 < 2π}, such that E1 E2 = ∅, then there exists a meromorphic function f (z) of order λ in the complex plane, exactly with E1 E2 as its T-direction, and exactly with E2 as its Borel-direction. Theorem B For arbitrary positive number λ, positive integers q1 and q2 , let E1 = {arg z = θj |0 ≤ θ1 < θ2 < · · · < θq1 < 2π} and E2 = {arg z = ϕj |0 ≤ ϕ1 < ϕ2 < · · · < ϕq2 < 2π}, such that E1 E2 = ∅, then there exists a meromorphic function f (z) of order and under-order being both λ in the complex plane, exactly with E1 E2 as its Borel-direction, and exactly with E2 as its T-direction. This suggests the following question, what is the relation between the Borel exceptional value and T exceptional value of meromorphic functions? In most papers, the authors only introduced the related theorems of the connections between a few important exceptional values and some singular directions, and conclusions of related questions. We have found no papers discussing exceptional values separately keeping
No.4
Y.H. Peng & D.C. Sun: EXAMPLES ON EXCEPTIONAL VALUES
1329
away singular directions, and no papers discussing Borel exceptional values and T exceptional values. In this paper, we initially try to use a special way to construct a meromorphic function with zero as Borel exceptional value, but not as T exceptional value; and another meromorphic function with zero as T exceptional value, but not as Borel exceptional value. Furthmore, this paper introduces exceptional values of meromorphic functions with examples, which is different from some other papers and more significant, and has a better consulted value to us in studying singular directions of meromorphic functions by giving examples in the future. In this paper, let n(r, f = a) be the number of zero points of f (z) − a in |z| < r, counting with multiplicities. Other symbols unexplained are similar to those in [3] without exception. Definition 1 The order of the meromorphic function f (z) is defined by λ = lim sup r→∞
ln T (r, f ) . ln r
where T (r, f ) is the Nevanlinna’s characteristic function of f (z) (see [3]). Let us recall the following two definitions. Definition 2 [3] Suppose that f (z) is a meromorphic function of order λ (0 < λ < ∞) in the complex plane, a complex number a is called a Borel exceptional value of f (z) if lim sup r→∞
ln n(r, f = a) < λ. ln r
Definition 3 Suppose that f (z) is a meromorphic function in the complex plane, a(a ∈ c∞ ) is called a T exceptional value of f (z), if lim sup r→∞
N (r, f = a) = 0. T (r, f )
Our main results are n Example 1 Set bn = 2n , an = bn + un , where un = sm =
m
(1 + un )[
√
2nn ]
1 n 22n
. Suppose that
, sm → s (0 < s < ∞) as m → ∞,
n=1
√ √ where [ 2nn ] is the integral part of 2nn . Let [√2nn ] ∞ an − b n 1+ − 1. f (z) = z − an n=1 √ Set Bn = 2(n+1)(n+1) , An = Bn + βn , where βn = √ 1 (n+1) . 2
(1)
2(n+1)
Also suppose that m (n−1)(n−1) ] tm = , tm → t (0 < t < ∞) as m → ∞ (1 + βn )[2 n=1
and let [ ∞ An − Bn g(z) = 1+ z − An n=1
√
(n−1)
2(n−1)
]
,
(2)
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we set F (z) = f (z)g(z), then F (z) is a meromorphic function in the complex plane, and 0 is a Borel exceptional value of F (z), but not a T exceptional value of F (z). Example 2 Let
√ [√2nn ] 22nnnn ∞ ∞ ln 2 an − b n An − Bn 1+ 1+ f (z) = − 1, g(z) = z − a z − An n n=1 n=1 and set F (z) = f (z)g(z), where an , bn , An , Bn are all the same with those of Example 1. Then 0 is a T exceptional value of F (z), but not a Borel exceptional value of F (z).
2
Some Lemmas Lemma 1 (See [6, Theorem 1.3]) Let fn (z) =
n
[1 + uk (z)]. If for all z in a domain G
k=1
and for every positive integer k we have |uk (z)| ≤ Mk , where Mk do not depend on z, and the ∞ ∞
series Mk is convergent, then the infinite product [1 + uk (z)] is unanimously convergent k=1
k=1
to an analytic function f (z). f (z), g(z), an , bn , An , Bn , un , βn , sm , s, tm , t appeared in the following are all as in Example 1. Lemma 2 2.1) f (z) is defined as (1), then f (z) satisfies: 2.1.1) f (z) is a meromorphic function in the complex plane. 2.1.2) For arbitrary ε > 0, there exists M > 0, such that when |z| > M and |z − an | ≥ 1 (n = 1, 2, · · ·), we have |f (z)| < ε. 2.2) g(z) is defined as (2), then g(z) satisfies: 2.2.1) g(z) is a meromorphic function in the complex plane. 2.2.2) For arbitrary ε > 0, there exists M = 2t ε +1+b > 0, where b = max{An |1 ≤ n ≤ N }, when |z| > M and |z − An | ≥ 1 (n = 1, 2, · · ·), we have |g(z) − 1| < ε. Proof 2.1) Let f1 (z) =
∞ n=1
C=
z [ 1− bn
√ n 2n ]
∞ bn [ n=1
an
, f2 (z) =
∞ n=1
√ n 2n ]
=
∞ n=1
z [ 1− an
un [ 1− an
√ n 2n ]
√ n 2n ]
,
.
Clearly b1 , b2 , · · · , bn , · · · are infinitely many zeros of f1 (z), their moduli satisfy |b1 | < |b2 | < · · · < |bn | < · · · , and we have n
lim |bn | = lim 2n = ∞.
n→∞
n→∞
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Y.H. Peng & D.C. Sun: EXAMPLES ON EXCEPTIONAL VALUES
Arbitrarily given a positive number R, if bm ≤ R < bm+1 , we consider the following series in |z| < R, ∞ √ z , [ 2nn ] ln 1 − bn n=m+1 where every ln(1 − |z| < R,
z bn )
√ is chosen the main value. Let αn = [ 2nn ] ln 1 −
z bn
(n > m). When
√ √ z z 1 z 2 1 z n = [ 2nn ] − αn = [ 2nn ] ln 1 − − ···− − ··· . bn bn 2 bn n bn n
Since bn = 2n , so |αn | ≤
√ n z z 2 z z 2 z 3 2n + + · · · ≤ √ + √ + √ + · · · . bn bn bn bn bn
When n is sufficiently large, there holds z √ ≤ R = 1. b 2R 2 n Hence |αn | ≤ ∞
Since series
n=1
1 2n
∞ 1 . 2n n=1 ∞
is convergent, by Lemma 1, we know
(1 −
n=m+1
unanimously convergent to an analytic function in |z| < R, so
∞
√ z [ 2nn ] bn )
(1 −
n=1
is absolutely and
√ [ 2nn ]
z bn )
is absolutely
and unanimously convergent to an analytic function in |z| < R, which implies f1 (z) is analytic in |z| < R. Since R is an arbitrary positive number, we know f1 (z) is analytic in the complex plane. Similarly, we can prove that f2 (z) and C are also analytic. Hence f (z) = Cf1 (z)/f2 (z)−1 is meromorphic in the complex plane. When |z − an | ≥ 1,we have n [√2nn ] 0 +T − b a n n 1 + − 1 z − an n=n 0
√
n i [ 2nn ] 0 +T an − b n i√ = 1+ − 1 C[ 2nn ] z − an n=n0
√
i=1
√
n i i 2nn ] 2nn ] n 0 +T [ 0 +T [ an − b n an − bn i√ i√ = C[ 2nn ] + ··· + C[ 2nn ] z − an z − an ≤
n=n0 i=1 √ n 2n ] n 0 +T [ n=n0
i=1
C[i√2nn ] un + · · · +
√ n 2n ] n 0 +T [ n=n0
i=1
n=n0
n0 +T √ n sn +T n = (1 + un )[ 2 ] − 1 = 0 − 1. sn −1 n=n0
0
i=1
C[i√2nn ] un
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Also owing to sn being convergent, there exists M1 > 0, such that when n0 > M1 , no matter s ε how large T is, we always have | snn0 +T − 1| < 2s . Therefore, for fixed N > M1 , −1 0
∞ [√2nn ] − b a ε n n 1+ − 1 < , A = z − an 2s n=N
set b = max{an |1 ≤ n ≤ N }, and let M = 2s ε + 1 + b. Then, when |z| > M, we have √ n N [ 2n ] −1 an − b n 1+ − 1 B = z − an n=1 √ n 2n ] N −1 [
√
nn
2 ] N −1 [ uin uin i√ ≤ C[i√2nn ] + · · · + C n [ 2n ] |z − a |i |z − an |i n n=1 i=1 n=1 i=1 N −1 √ n 1 n ≤ (1 + un )[ 2 ] − 1 (|z| − b) n=1
≤
1 (sN −1 − 1) < ε/2. |z| − b
Thus, when |z| > M and |z − an | ≥ 1 (n = 1, 2, · · ·), we have N −1 N √ nn −1 √ z − bn [ 2 ] [ 2nn ] A + B ≤ A + B < ε. (1 + u ) |f (z)| ≤ n z − an n=1 n=1 The proof of 2.2) is omitted as the method is the same as that of 2.1). So the proof of the lemma is completed. Lemma 3 (See [10, 3.4 Argument Principle]) Let f be meromorphic in G with poles p1 , p2 , · · · , pm and zeros z1 , z2 , · · · , zn counted according to multiplicity. If γ is a closed rectifiable curve in G with γ ≈ 0 and not passing through p1 , p2 , · · · , pm ; z1 , z2 , · · · , zn ; then n m f (z) 1 dz = n(γ; zk ) − n(γ; pj ). 2π γ f (z) j=1 k=1
Lemma 4 Let Lp : |z − ap | ≤ p + 1 (p = 1, 2 · · ·). When p > N0 (N0 ≥ N ), then, in the interior of every Lp , the number of zero points of f (z) − d is the same as the number of poles of f (z). Where N is as in 2.1) of Lemma 2. Proof For a point z on the boundary of every circle Lp , choosing arbitrarily a complex ∞ √
number d = 0, let ε = min{ 21 , |d| [ 2nn ]un is convergent. Choosing N , let N0 = 2 }. Since max{N, N }, when p > N0 , such that
n=1
∞ √ 1 1 · [ 2nn ]un < ε(|d| − ε). p + 1 n=1 2
Since p > N, the point z on the boundary of Lp satisfies |z| = ap + p + 1 > M, by 1.2) of Lemma 2, we know |f (z)| < ε < 1. Then √ n f (z) f (z) ∞ [ 2n ]un 1 1 · f (z) − d ≤ f (z) · |f (z) − d| = (z − a )(z − b ) |f (z) − d| n n n=1 √ n ∞ ∞ √ n 1 1 1 1 [ 2n ]un · ≤ · · [ 2n ]un < ε ≤ . ≤ |z − a ||z − b | |d| − ε |d| − ε p + 1 2 n n n=1 n=1
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Y.H. Peng & D.C. Sun: EXAMPLES ON EXCEPTIONAL VALUES
1 2π
Hence
∂Lp
1333
1 f (z) dz ≤ . f (z) − d 2
By Lemma 3, when p > N0 (N0 ≥ N ), in the interior of each Lp , we know that the number of zero points of f (z) − d is the same as the number of poles of f (z). This concludes the proof of the lemma. Similarly, let Lq : |z − Aq | ≤ q + 1 (q = 1, 2, · · ·), choosing arbitrarily a complex number a = d + 1 (d = 0), by the similar argument of Lemma 4, we can deduce that, when q > N , the number of zero points of g(z) − a is the same as the number of poles of g(z) inside every Lq .
2
Proofs of Examples
Proof of Example 1 By Lemma 2, we know f (z), g(z) are meromorphic, so F (z) = f (z)g(z) is meromorphic in the complex plane. For arbitrary sufficiently large r, there corresponds n such that n
(n+1)
bn = 2n < r ≤ bn+1 = 2(n+1) , n i √ √ n(r, f = −1) = n(r, f ) = [ 2i ] ≤ 2 · [ 2nn ] ≤ 2 · 2nn ,
i=1
√ n(t, f ) dt ≤ n(r, f ) ln r ≤ 2 · 2nn ln r, N (r, f = −1) = N (r, f ) = t 1 ⎧ n √ ⎪ n ⎪ ⎪ [ 2(i−1)( i−1) ], 2n < r ≤ 2(n+1)(n+1) , ⎪ ⎨ i=1 n(r, g) = n+1 √ ⎪ (n+1) ⎪ ⎪ [ 2(i−1)( i−1) ], 2(n+1)(n+1) < r ≤ 2(n+1) , ⎪ ⎩ r
(3)
i=1
namely,
⎧ √ ⎨ 2 · 2(n−1)(n−1) , n(r, g) ≤ √ ⎩ 2 · 2nn ,
√ n 2n < r ≤ 2(n+1)(n+1) , √ (n+1) 2(n+1)(n+1) < r ≤ 2(n+1) .
Thus n(r, g) ≤ 2 · so that
N (r, g) = 1
r
√
√ n(t, g) dt ≤ n(r, g) ln r ≤ 2 · 2nn ln r. t
By a similar argument, we can obtain N (r, g = 0) ≤ 2 · Since T (r, f ) ≤
3
i=1
2nn ,
√ n 2n ln r.
(4)
(5)
N (r, f = ai ) + O{ln(rT (r, f ))}. without harm suppose that ai (i = 1, 2, 3)
is −1, ∞, 5, respectively, by Lemma 4, we have n(r, f = 5) = n(r − (aN +1 − (N + 1)), f = 5) + n((aN +1 − (N + 1)), f = 5) = n(r − (aN +1 − (N + 1)), f = 5) + c n √ [ 2ii ] + c ≤ 2 · 2nn − ( 2N N − 1) + c, = i=N +1
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where c is a finite number depending on N . So that √ N (r, f = 5) ≤ 2 · 2nn − ( 2N N − 1) + c ln r.
(6)
Hence, by (3),(6), we obtain √ T (r, f ) ≤ 6 · 2nn − ( 2N N − 1) + c) ln r + O{ln(rT (r, f ))}. For g(z), we put aj (j = 1, 2, 3) to be 0, ∞, 5, respectively. Similarly, we also have √ N (r, g = 5) ≤ 2 · 2nn − ( 2N N − 1) + c ln r.
(7)
Therefore by, (4), (5), (7), T (r, g) ≤
3
N (r, f = aj ) + O{ln(rT (r, g))}
j=1
√ ≤ 6 · 2nn − ( 2N N − 1) + c ln r + O{ln(rT (r, g))}. Finally, we have T (r, F ) ≤ T (r, f ) + T (r, g) √ ≤ 12 · 2nn − 2 · ( 2N N − 1) + 2 · c ln r + O{ln(rT (r, f ))} + O{ln(rT (r, g))}. Then ln T (r, F ) ln r √ √ ln((12 · 2nn − 2 · ( 2N N − 1) + 2 · c) ln r + O{ln(rT (r, f ))} + O{ln(rT (r, g))}) ≤ lim sup ln r r→∞ √ n (n+1)(n+1) n 1 ln(12 · 2 ln 2 ) = . ≤ lim n→∞ ln 2nn 2 lim sup r→∞
n
On the other hand, choosing a point sequence {rn = 2n + 2n}, we can obtain n(rn , F ) = n(rn , f ) + n(rn , g) =
n n [ 2ii ] + [ 2(i−1)(i−1) ] i=1
i=1
√ √ ≥ [ 2nn ] + [ 2(n−1)(n−1) ] > 2nn − 1, 2rn √ n(t, F ) dt ≥ n(rn , F ) ln 2 ≥ ( 2nn − 1) · ln 2, N (2rn , F ) ≥ t rn √ T (2rn , F ) ≥ N (2rn , F ) ≥ ( 2nn − 1) · ln 2, √ 1 ln T (r, F ) ln T (2rn , F ) ln( 2nn − 1) · ln 2) ≥ lim sup = . lim sup ≥ lim n n n→∞ ln r ln 2rn ln 2 · (2 + 2n) 2 r→∞ rn →∞
(8)
) So that the order of F (z) is λ = lim sup ln Tln(r,F = 12 . r r→∞
Now, we begin to prove that 0 is a Borel exceptional value of F (z), but not a T exceptional value of F (z).
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Y.H. Peng & D.C. Sun: EXAMPLES ON EXCEPTIONAL VALUES
Since
⎧ n ⎪ ⎪ ⎪ [ 2(i−1)(i−1) ], ⎪ ⎨
n(r, F = 0) = n(r, g = 0)
i=1
n+1 ⎪ ⎪ ⎪ [ 2(i−1)(i−1) ], ⎪ ⎩
n
2n < r ≤
1335
√ 2(n+1)(n+1) ,
√ (n+1) 2(n+1)(n+1) < r ≤ 2(n+1) ,
i=1
⎧ √ ⎨ 2 · 2(n−1)(n−1) , n(r, F = 0) ≤ √ ⎩ 2 · 2nn ,
so
Thus
√ n 2n < r ≤ 2(n+1)(n+1) , √ (n+1) 2(n+1)(n+1) < r ≤ 2(n+1) .
√ ⎧ ⎪ ln(2 · 2(n−1)(n−1) ) ⎪ ⎪ lim = 0, ln n(r, F = 0) ⎨ n→∞ ln 2nn ≤ lim sup √ ⎪ ln r r→∞ ⎪ ln(2 · 2nn ) ⎪ ⎩ lim √ = 0. n→∞ ln 2(n+1)(n+1)
That is lim sup r→∞
ln n(r, F = 0) 1 =0< . ln r 2
So it is evident that 0 is a Borel exceptional value of F (z). √ (n+1) Also choosing point sequences {rn = 2(n+1)(n+1) + 2(n + 1)} and {rn = 2(n+1) }, then n(rn , F = 0) = n(rn , g = 0) =
n+1
√ √ [ 2(i−1)(i−1) ] > 2nn ≥ 2nn − 1.
i=1
Hence N (rn , F = 0) ≥
rn
rn
√ n(t, F ) r r dt ≥ n(rn , F = 0) ln n ≥ ( 2nn − 1) ln n . t rn rn
So that √ r ( 2nn − 1) · ln rnn N (r, F = 0) N (rn , F = 0) √ ≥ lim sup ≥ lim sup lim sup T (r, F ) T (rn , F ) r→∞ 12 · 2nn · ln rn rn→∞ rn→∞ √ √ (n+1) ( 2nn − 1) · (ln 2(n+1) − ln( 2(n+1)(n+1) + 2(n + 1))) √ ≥ lim n→∞ 12 · 2nn ln 2(n+1)(n+1) 1 > 0. = 12 Therefore we come to that 0 is not a T exceptional value of F (z). Example 1 is proved. Proof of Example 2 By the argument of proof of Example 1 we know the order of F (z) is ln T (r, F ) λ = lim sup = 1. ln r r→∞ n
(n+1)
By a similar argument of proof of Example 1 we obtain, for 2n < r ≤ 2(n+1) , ⎧ √ n √ ⎪ 2n n ⎪ ⎪ 2n < r ≤ 2(n+1)(n+1) , ⎨ 2 · 2 nn , ln 2 n(r, F = 0) = n(r, g = 0) ≤ √ ⎪ (n+1)(n+1) √ ⎪ (n+1) ⎪2 · 2 ⎩ , 2(n+1)(n+1) < r ≤ 2(n+1) . (n+1) 2 (n+1) ln 2
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And n(r, F = 0) = n(r, g = 0) ⎧ √ n √ n ⎪ 2n 2n ⎪ ⎪ ⎪ ⎨ ln2 2nn ≥ ln2 2nn − 1, ≥ √ √ ⎪ (n+1)(n+1) ⎪ 2 2(n+1)(n+1) ⎪ ⎪ ≥ − 1, ⎩ ln2 2(n+1)(n+1) ln2 2(n+1)(n+1)
n
2n < r ≤ √
√ 2(n+1)(n+1) ,
2(n+1)(n+1) < r ≤ 2(n+1)
(n+1)
.
n
Choosing a point sequence {rn = 2n + 2n}, similar to (4), (8) in the proof of Example 1, we have T (2rn , F ) ≥ N (2rn , F ) ≥ N (2rn , f ) + N (2rn , g) √ nn √ n 2 n ≥ ( 2 − 1) · ln 2 + − 1 · ln 2, ln2 2nn √ n 2n N (2rn , F = 0) ≤ 2 · 2 n ln(2rn ). ln 2n So that lim sup r→∞
N (r, F = 0) N (2rn , F = 0) = lim sup T (r, F ) T (2rn , F ) rn →∞ ≤ lim
n→∞
2·
√ ( 2nn
√ n 2n ln2 2nn
n
ln(2 · (2n + 2n)) √ n = 0. n − 1) · ln 2 + ln222nn − 1 · ln 2
Therefore 0 is the T exceptional value of F (z). √ Again we choose {rn = 2 · 2(n+1)(n+1) }, then it yields lim sup r→∞
ln n(r, F = 0) ≥ ln r
ln n(rn , F lim sup →∞ ln rn rn
= 0)
√ ln
= lim
n→∞
(n+1)
2(n+1) n+1) 2(n+1)
ln2
ln(2 ·
−1
√ = 1. 2(n+1)(n+1) )
Thus, we conclude that 0 is not a Borel exceptional value of F (z). This concludes the proof of Example 2. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11]
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