A Note On A Conjecture Of R. Brück
Indrajit Lahiri
y, Shubhashish Das
zxReceived 11 March 2020
Abstract
In connection to a conjecture of R. Brück we improve a result of Z. X. Chen and K. H. Shon [4]
concerning value sharing by an entire function with its derivative.
1 Introduction, De…nitions and Results
Let f be an entire function and M(r; f) = maxjzj=rjf(z)j be the maximum modulus function of f. The order (f)and the lower order (f)off are de…ned respectively by
(f) = lim sup
r!1
log logM(r; f) logr and
(f) = lim inf
r!1
log logM(r; f) logr :
The …rst iterated order or hyper order 2(f)and the …rst iterated lower order or hyper lower order 2(f) are de…ned respectively by
2(f) = lim sup
r!1
log log logM(r; f) logr and
2(f) = lim inf
r!1
log log logM(r; f)
logr :
If the Taylor expansion off isf(z) = P1
n=0
anzn, then the power series P1
n=0janjrnconverges for everyr >0 and so for any givenr >0, we havelimr!1janjrn = 0. Hence the maximum term (r; f) = maxn 0janjrn is well de…ned.
Also we de…ne (r; f), the central index off, as the greatest exponentmsuch that (r; f) =jamjrm(see [7, p.50]).
It is well known that
(f) = lim sup
r!1
log (r; f) logr (see [7, p.51]). Similarly it can be veri…ed that
(f) = lim inf
r!1
log (r; f) logr : By Lemma 2 in [3] we see that
2(f) = lim sup
r!1
log log (r; f) logr
Mathematics Sub ject Classi…cations: 30D35.
yDepartment of Mathematics, University of Kalyani, West Bengal 741235, India
zDepartment of Mathematics, University of Kalyani, West Bengal 741235, India
xPresent Address: Department of Mathematics, Bharat Sevak Samaj College, Supaul, Bihar 852131, India.
152
and in a similar fashion we can prove that
2(f) = lim inf
r!1
log log (r; f) logr :
Letu(z)be a nonconstant subharmonic function in the open complex plane. We putB(r; u) = supjzj=ru(z).
The order (u)and the lower order (u)ofuare de…ned by (u) = lim sup
r!1
logB(r; u) logr and
(u) = lim inf
r!1
logB(r; u) logr (see [1]).
LetE [1;1)and Ebe the characteristic function ofE. The upper and the lower logarithmic densities ofE are respectively de…ned by
logdens(E) = lim sup
r!1
Rr 1
E(t) t dt logr and
logdens(E) = lim inf
r!1
Rr 1
E(t) t dt logr : The quantity limr!1
Rr 1
E(t)
t dt is called the logarithmic measure of E. It is easy to note that if logdens(E)>0, thenE has in…nite logarithmic measure.
Let f and g be two entire functions and a be also an entire function, which, in particular, may be a constant. We say thatf andg share the function a CM (counting multiplicities) iff aandg ahave the same set of zeros with counting multiplicities.
L. A Rubel and C. C. Yang [8] were the …rst to consider the uniqueness problem of an entire function sharing two values with its derivative. Afterwards in 1996 R. Brück [2] considered the problem of a single value sharing by an entire function with its derivative and proposed the following conjecture.
Brück’s Conjecture: Letf be a nonconstant entire function with 2(f)<1and 2(f)is not a positive integer. Iff andf(1) share a …nite valueaCM, thenf(1) a=c(f a), where cis a nonzero constant.
Ifa= 0, then the conjecture was resolved by Brück himself [2], but the casea6= 0is not yet fully resolved.
For entire functions of …nite order, G. G. Gundersen and L. Z. Yang [6] resolved the conjecture and proved the following result.
Theorem 1 ([6]) Letf be a nonconstant entire function of …nite order. Iff andf(1) share one …nite value aCM then f(1) a=c(f a)for some nonzero constant c.
Generalizing Theorem1to higher order derivatives, L. Z. Yang [10] proved the following result.
Theorem 2 ([10]) Let f be a nonconstant entire function of …nite order. If f and f(k) share one …nite value aCM, thenf(k) a=c(f a)for some nonzero constant c.
In 2004, J. P. Wang [9] improved Theorem2 in the following manner.
Theorem 3 ([9]) Letf be a nonconstant entire function of …nite order andabe a nonconstant polynomial.
If f andf(k) shareaCM, then f(k) a=c(f a)for some nonzero constantc.
In the same year Z. X. Chen and K. H. Shon [4] extended Theorem 1 to a class of entire functions of unrestricted order and proved the following theorem.
Theorem 4 ([4]) Let f be a nonconstant entire function with 2(f)<12. Iff andf(1) share a …nite value aCM, thenf(1) a=c(f a), wherec is a nonzero constant.
Noting that Brück conjecture remains open for the case 2(f) 12, the purpose of the paper is to improve both Theorem3 and Theorem4and prove the following result. Also our proof is simpler than Z. X. Chen and K. H. Shon [4].
Theorem 5 Let f be a nonconstant entire function with 2(f)<12 and 2(f)<1. Suppose that a=a(z) is a polynomial. Iff andf(k)sharea CM, thenf(k) a=c(f a), wherec is a nonzero constant.
2 Lemmas
In this section we present some necessary lemmas.
Lemma 6 ([7, p.9]) LetP(z) =bnzn+bn 1zn 1+ +b0(bn 6= 0)be a polynomial of degreen. Then for every (>0) there existsR(>0) such that for alljzj=r > R we get
(1 )jbnjrn jP(z)j (1 + )jbnjrn:
Lemma 7 ([7, p.51]) Let f be a transcendental entire function. Then there exists a set E (1;1) with
…nite logarithmic measure such that forjzj=r62[0;1][E andjf(z)j=M(r; f)we get f(k)(z)
f(z) = (1 +o(1)) (r; f) z
k
: (1)
Lemma 8 ([7, p.5]) Let g : (0;+1) ! R and h : (0;+1) ! R be monotone increasing functions such that g(r) h(r)outside of an exceptional set E of …nite logarithmic measure. Then for any >1, there existsR >0 such that g(r) h(r )holds forr > R.
Lemma 9 ([1]) Letu(z)be a nonconstant subharmonic function in the open complex planeCof lower order
;0 <1 . If < <1, then
logdensfr:A(r)>(cos )B(r)g 1 ; whereA(r) = infjzj=ru(z)andB(r) = supjzj=ru(z).
Remark 1 Since for an entire function Q,logjQ(z)j is a subharmonic function in C ([5, p.394]), we can apply Lemma9to the function u(z) = logjQ(z)j.
3 Proof of Theorem 5
Proof. Sincef(k) aandf ashare 0 CM, there exists an entire functionQsuch that f(k) a
f a =eQ: (2)
IfQis a constant, then we are done. So we suppose thatQis nonconstant and consider the following cases.
Case 1. Let (f)<1. Then from (2) we see thatQis a polynomial. Further (f) 1, because if (f)<1, then (2) implies thatQis a constant. Thereforef is transcendental.
Now for any zwithjf(z)j=M(r; f), noting thatf is transcendental, we get by Lemma6 ja(z)
f(z)j M(r; a) M(r; f)
2j jrdega
M(r; f) !0 (3)
asr! 1, where is the leading coe¢ cient ofa=a(z).
From (2) we get
eQ=
f(k) f
a f
1 af : (4)
Now by Lemma 7there exists E (1;1) with …nite logarithmic measure such that for all largejzj=r62 [0;1][E andjf(z)j=M(r; f)we get in view of (3),(4) and (1)
eQ(z)= (1 +o(1)) (r; f) z
k
: (5)
Now from (5) we get for all largejzj=r62[0;1][E withjf(z)j=M(r; f) jQ(z)j = jlogeQ(z)j
= jlog (r; f) z
k
j+o(1)
= jklog (r; f) klogzj+o(1) klog (r; f) +klogr+ 6k
< 2k( (f) + 1) logr+ 6k : (6)
Also by Lemma6 we obtain for all largejzj=r 1
2j jrdegQ jQ(z)j; (7)
where is the leading coe¢ cient ofQ.
Now (6) and (7) together implydegQ= 0, which is a contradiction.
Case 2. Let (f) =1. We note from (2) that (Q) 2(f)<12. We now consider the following subcases.
Subcase 2.1. Let Q be a polynomial. Then from (5) we get for all large jzj = r 62 [0;1][E with jf(z)j=M(r; f)
jQ(z)j klog (r; f) +klogr+ 6k : (8)
From (7) and (8) we obtain for all largejzj=r62[0;1][E withjf(z)j=M(r; f) 1
2j jrdegQ klog (r; f) +klogr+ 6k : So for all largejzj=r62[0;1][E we get
1
2j jrdegQ klog (r; f) +klogr+ 6k : Hence by Lemma8 for given , 1< <32, we get for all large values ofr
1
2j jrdegQ klog (r ; f) +k logr+ 6k and so
rdegQ 1
2j j k logr
rdegQ klog (r ; f) + 6k : This impliesdegQ 2(f)< 2 <1, which is a contradiction.
Subcase 2.2. Let Q be a transcendental entire function. We see by Note 1 that u(z) = logjQ(z)j is a subharmonic function and also (u) = (Q) < 12. Suppose that H = fr : A(r) > (cos )B(r)g, where A(r) = infjzj=rlogjQ(z)j,B(r) = supjzj=rlogjQ(z)jand (Q)< <12.
Then by Lemma 9 H has in…nite logarithmic measure. Also by Lemma 7 for jzj=r2 Hnf[0;1][Eg withjf(z)j=M(r; f)we get (1).
Now by (3), (4) and (1) for all large jzj=r2Hnf[0;1][Eg withjf(z)j=M(r; f)we get (5), whereQ is transcendental entire, and so
jQ(z)j = jlogeQ(z)j
= jlog (r; f) z
k
j+o(1)
= jklog (r; f) klogzj+o(1) klog (r; f) +klogr+ 6k
< 2kr 2(f)+1: (9)
So by (9) and by Lemma9there exists a constantd;0< d 1;such that(M(r; Q))d 2kr 2(f)+1for all large values ofjzj=r2Hnf[0;1][Egandjf(z)j=M(r; f). This is impossible becauseQis transcendental and solimr!1(M(r;Q))d
r 2 (f)+1 =1. This proves the theorem.
Acknowledgment. The authors are thankful to the referee for valuable suggestions. The work of Shubhashish Das was supported by CSIR Fellowship, India.
References
[1] P. D. Barry, On a theorem of Kjellberg, Quart. J. Math. Oxford, 15(1964), 179–191.
[2] R. Brück, On entire functions which share one value CM with their …rst derivative, Result. Math., 30(1996), 21–24.
[3] Z. X. Chen and C. C. Yang, Some further results on the zeros and growths of entire solutions of second order linear di¤erential equations, Kodai Math. J., 22(1999), 273–285.
[4] Z. X. Chen and K. H. Shon, On conjecture of R. Brück concerning the entire function sharing one value CM with its derivative, Taiwanese J. Math., 8(2004), 235–244.
[5] T. W. Gamelin, Complex Analysis, Undergraduate Texts in Mathematics, Springer-Verlag New York, Inc., 2001.
[6] G. G. Gundersen and L. Z. Yang, Entire functions that share one value with one or two of their derivatives, J. Math. Anal. Appl., 223(1998), 88–95.
[7] I. Laine, Nevanlinna Theory and Complex Di¤erential Equations, De Gruyter, Berlin, New York (1993).
[8] L. A. Rubel and C. C. Yang, Values shared by an entire function and its derivative, Lecture Notes in Math., 599(1977), 101–103.
[9] J. P. Wang, Entire functions that share a polynomial with one of their derivatives, Kodai Math. J., 27(2004), 144–151.
[10] L. Z. Yang, Solution of a di¤erential equation and its applications, Kodai Math. J., 22(1999), 458–464.
