This paper has been accepted for publication in STUDIA UNIVERSITATIS

BABES-BOLYAI MATHEMATICA on 17.09.2020.

Linear delay-differential operator of a meromorphic function sharing two

arXiv:2109.11257v1 [math.CV] 23 Sep 2021

sets or small function together with values with its c-shift or q-shift

ARPITA ROY AND ABHIJIT BANERJEE

Abstract. The paper is devoted to study the uniqueness problem of linear delay-differential
operator of a meromorphic function sharing two sets or small function together with values
with its c-shift and q-shift operator. Results of this paper drastically improve two recent
results of Meng-Liu [J. Appl. Math. Inform. 37(1-2)(2019), 133-148] and Qi-Li-Yang [Com-
put. Methods Funct. Theo., 18(2018), 567-582]. In addition to this, one of our result
improve and extend that of Qi-Yang [Comput. Methods Funct. Theo., 20(2020), 159-178].

1. Introduction Definitions and Results

Throughout the paper we use standard notations of Nevanlinna theory as stated in [7] and
by any meromorphic function f we always mean that it is defined on C. Let f and g be such
two non constant meromorphic functions. For a ∈ C ∪ {∞}, the following two quantities
N (r, a; f ) m(r, a; f )
δ(a; f ) = 1 − lim sup = lim inf
r−→∞ T (r, f ) r−→∞ T (r, f )

and
N (r, a; f )
Θ(a; f ) = 1 − lim sup
r−→∞ T (r, f )
are respectively known as Nevanlinna deficiency and ramification index of the value a.
In the beginning of the nineteenth century R. Nevanlinna inaugurated the value distribution
theory with his famous Five value and Four value theorems which can be considered as the
backbone of the modern uniqueness theory. Illuminated by these two basic results initially the
research were performed on the value sharing of meromorphic functions. After five decades,
uniqueness theory moved to a new direction led by F. Gross [4], who transformed the traditional
value sharing problem to a more general set up namely to the shared set problems. Now we
recall the definition of set sharing.
Definition 1.1. Let S be a set of S distinct elements of C ∪ {∞}. For a non-constant mero-
morphic function f , let Ef (S) = a∈S {(z, p) ∈ C × N : f (z) = a with multiplicity p}
S

E f (S) = a∈S {(z, 1) ∈ C × N : f (z) = a} . Then we say f , g share the set S CM(IM) if


Ef (S) = Eg (S) E f (S) = E g (S) .
Evidently, if S is a singleton, then it coincides with the traditional definition of CM(IM)
sharing of values, which are known to the readers.
In 2001, due to a revolutionary approach by Lahiri [8, 9], the notion of weighted sharing
of values or sets appeared in the literature and expedite the research work there in. Though
now-a-days the definition is widely circulated, we invoke the definition.

2010 Mathematics Subject Classification: 39A70; 30D35.

Key words and phrases: Meromorphic functions, uniqueness, delay-differential operator, shift, shared set,
weighted sharing.
Type set by AMS-LATEX
1
2 ARPITA ROY AND ABHIJIT BANERJEE

Definition 1.2. [8, 9] Let k be a non-negative integer or infinity. For a ∈ C ∪ {∞} we denote
by Ek (a; f ) the set of all a-points of f , where an a-point of multiplicity m is counted m times
if m ≤ k and k + 1 times if m > k. If Ek (a; f ) = Ek (a; g), we say that f, g share the value a
with weight k and denote it by (a, k). The IM and CM sharing corresponds to (a, 0) and (a, ∞)
respectively.
Definition 1.3. [8] Let S be a set of distinct elements of C ∪ {∞} and k be a non-negative
integer or ∞. We denote by Ef (S, k) the set ∪a∈S Ek (a; f ). Clearly Ef (S) = Ef (S, ∞) and
E f (S) = Ef (S, 0).
If Ef (S, k) = Eg (S, k), then we say that f , g share the set S with weight k and write it as
f , g share (S, k).
By N (r, a; f |< m) we mean the counting function of those a-points of f whose multiplicities
are less than m where each a-point is counted according to its multiplicity and by N (r, a; f |≥
m) we mean the counting function of those a-points of f whose multiplicities are not less than
m where each a-point is counted ignoring multiplicity. We also denote by N2 (r, a; f ) the sum
N (r, a; f ) + N (r, a; f |≥ 2).
Usually, S(r, f ) denotes any quantity satisfying S(r, f ) = o(T (r, f )) for all r outside of a
possible exceptional set of finite linear measure. Also S1 (r, f ) denotes any quantity satisfying
S1 (r, f ) = o(T (r, f )) for all r on a set of logarithmic density 1, where the logarithmic density
of a set F is defined by
1
Z
dt
lim sup .
r→∞ log r [1,r]∩F t

Throughout the paper for a positive integer n, S1 , S1∗ and S2 represents respectively the sets
{1, ω, . . . , ω n−1 }, {α1 , α2 , . . . , αn } and {∞} , where ω = cos 2π 2π
n + i sin n and αi , i = 1, 2, . . . , n
are non zero constants.
Let at−1 (6= 0), at−2 , . . . , a0 and C(6= 0) be complex numbers. We define
(1.1) P (z) = CzQ(z) = Cz(at−1 z t−1 + at−2 z t−2 + . . . + a1 z + a0 ).
For the polynomial P (z) as given in (1.1), let us define the following two functions:
(
t−1 1, if a0 6= 0
χ0 =
0, if a0 = 0
and (
t−1 1, if a0 = 0, a1 6= 0
µ0 =
0, otherwise.
In view of (1.1), corresponding to the set S1∗ , let us consider the polynomial P∗ (z) as follows:
1
(1.2) P∗ (z) = CzQ∗ (z), where C = and
(−1)n+1 α1 α2 . . . αn
n−1
X X
Q∗ (z) = (−1)r α1 α2 . . . αr z n−r−1 ,
r=0
P
α1 α2 . . . αr = sum of the products of the values α1 , α2 , . . . , αn taken r into account. We also
denote by m1 and m2 as the number of simple and multiple zeros of Q∗ (z) respectively.
Next we define linear shift operator, shift-differential operator and differential operator re-
spectively as follows:
L1 (f (z)) = ak f (z + ck ) + ak−1 f (z + ck−1 ) + . . . + a1 f (z + c1 ) + a0 f (z),

L2 (f (z)) = bs f (s) (z + cs ) + bs−1 f (s−1) (z + cs−1 ) + . . . + b1 f ′ (z + c1 ),

L3 (f (z)) = dt f (t) (z) + dt−1 f (t−1) (z) + . . . + d1 f ′ (z),

 Linear delay-differential operator of a meromorphic function... 3

where ak , bs and dt are non zero and k, s, t are natural numbers and all c′i s are non zero. For
the sake of convenience we shall call L2 (f (z)) + L3 (f (z)) as delay-differential operator which
is denoted by L̃(f (z)).
As far as the knowledge of the authors are concerned, Qi-Li-Yang [13] were the first authors
who initiated two shared set problems for the derivative of a meromorphic function f (z) with
its shift f (z + c) as follows:
Theorem A. [13] Let f (z) be a non constant meromorphic function of finite order, n ≥ 9 be
an integer and a be a non zero complex constant. If [f ′ (z)]n and f n (z + c) share (a, ∞) and
(∞, ∞), then f ′ (z) = tf (z + c), for a constant t that satisfies tn = 1.
Recently employing the notion of weighted sharing, Meng-Liu [12] further investigated The-
orem A to obtain the following result.
Theorem B. [12] Let f (z) be a non constant meromorphic function of finite order, n ≥ 10
be an integer. If [f ′ (z)]n and f n (z + c) share (1, 2) and (∞, 0), then f ′ (z) = tf (z + c), for a
constant t that satisfies tn = 1.
Considering f (z) = ez and ω = e−c satisfying ω n = 1, it is easy to see that f ′ and f (z + c)
share the sets (S1 , ∞), (∞, ∞) and f ′ (z) = ωf (z + c) for each n. So it is natural to conjecture
that in Theorem A and Theorem B the cardinality of n could further be reduced. To this
end, we have performed our investigations and have been able to reduce the cardinality of n in
Theorem B up to 6. In fact, we have proved our theorem for a more general setting S1∗ rather
tan to consider only the set S1 .
Theorem 1.1. Let f (z) be a non constant meromorphic function of finite order such that
L̃(f (z)) and f (z + c) share (S1∗ , 2) and (S2 , 0). If
15
n > 2(χ0n−1 + µ0n−1 + m1 + 2m2 ) + (χn−1 + m1 + m2 ), then
(2n − 3) 0
n
Y n
Y
(L̃(f (z)) − αi ) ≡ (f (z + c) − αi ).
i=1 i=1

Remark 1.1. From the definitions, we easily can calculate the value of χ0n−1 , µ0n−1 , m1 and
m2 for a particular set S1∗ . Clearly for the set S1 , χ0n−1 = 0; µ0n−1 = 0; m1 = 0 and m2 = 1.
15
Therefore in above theorem for the set S1 if n > 4+ (2n−3) i.e., if n ≥ 6 then L̃(f (z)) = tf (z+c),
for a constant t that satisfies tn = 1. For a particular choices of coefficients of L̃(f (z)) we can
easily make L̃(f (z)) = f ′ .
Corresponding to q-shift Meng-Liu [12] also investigated the same result like Theorem B as
follows :
Theorem C. [12] Let f (z) be a non constant meromorphic function of zero order, n ≥ 10 be
an integer. If [f ′ (z)]n and f n (qz) share (1, 2) and (∞, 0), then f ′ (z) = tf (qz), for a constant t
that satisfies tn = 1.
In connection to Theorem C below we present our result which improve the same.
Theorem 1.2. Let f (z) be a non constant meromorphic function of zero order such that
L̃(f (z)) and f (qz) share (S1∗ , 2) and (S2 , 0). If
15
n > 2(χ0n−1 + µ0n−1 + m1 + 2m2 ) + (χn−1 + m1 + m2 ) then
(2n − 3) 0
n
Y n
Y
(L̃(f (z)) − αi ) ≡ (f (qz) − αi ).
i=1 i=1

In the next theorem we shall show that the lower bound of n can further be reduced at the
expense of allowing both the range sets S1∗ , S2 to be shared CM.
4 ARPITA ROY AND ABHIJIT BANERJEE

Theorem 1.3. Let f (z) be a non constant meromorphic function of finite order such that
L̃(f (z)) and f (z + c) share (S1∗ , ∞) and (S2 , ∞) with T (r, f ) = N r, L̃(f1(z)) + S(r, f ) then
for n > 2(χ0n−1 + m1 + m2 ) + 1,
n
Y n
Y
(L̃(f (z)) − αi ) ≡ (f (z + c) − αi ).
i=1 i=1

Remark 1.2. In connection of Remark 1.1, for the set S1 in Theorem 1.3 the result holds for
n ≥ 4.
Our next theorem is analogous theorem of Theorem 1.3 corresponding to q-shift.
Theorem 1.4. Let f (z) be a non constant meromorphic function of zero order such that
L̃(f (z)) and f (qz) share (S1∗ , ∞) and (S2 , ∞). If n > 2(χ0n−1 + m1 + m2 ) + 1 then
Yn n
Y
(L̃(f (z)) − αi ) ≡ (f (qz) − αi ).
i=1 i=1

Recently, corresponding to Theorem A, Qi-Yang [14] obtained the value sharing problem for
entire function as follows:
Theorem D. [14] Let f (z) be a transcendental entire function of finite order and let (a 6= 0) ∈
C. If f ′ (z) and f (z + c) share (0, ∞) and (a, 0), then f ′ (z) ≡ f (z + c).
In view of Theorem 1.1, [14] we know that f (z) actually becomes a transcendental entire
function. Since we are dealing with L̃(f (z)) instead of f ′ , it will be reasonable to consider the
above theorem for meromorphic function under small function sharing category. In this respect
we prove the following theorem.
Theorem 1.5. Let f (z) be a transcendental meromorphic function of finite order and let
a(z)(6≡ 0) ∈ S(f ) be an entire function. If L̃(f (z)) and f (z + c) share (0, ∞), (∞, ∞) and
(a(z), 0) with Θ(0; f ) + Θ(∞; f ) > 0, then L̃(f (z)) ≡ f (z + c).
From Theorem 1.5 we can immediately deduce the following corollary.
Corollary 1.1. Let f (z) be a transcendental entire function of finite order and let a(z)(6≡ 0) ∈
S(f ). If L̃(f (z)) and f (z + c) share (0, ∞) and (a(z), 0), then L̃(f (z)) ≡ f (z + c).
Following example shows that in Theorem 1.5 the CM pole sharing can not be replaced by
IM.
√
2 2iz
√
−8e 2iz +2
√
Example 1.1. Let f (z) = 2e (e√2iz +1)2
and c = 2π. Choose the coefficients of L̃(f (z)) in
 √ √ √ 
2iz 2 2iz
−4e 2iz +1]
such a way that L̃(f (z)) = f ′′ . Then L̃(f (z)) = 24e [e √
(e 2iz +1)4
and f (z + c) share
1
(0, ∞), (1, 0) and (∞, 0) and Θ(0; f ) + Θ(∞; f ) = 2 > 0 but L̃(f (z)) 6≡ f (z + c).
From the next example we can show that in Theorem 1.5 sharing of 0 can not be replaced
by sharing of a non zero value.
s
X t
X
Example 1.2. Let f (z) = (eλz − 1)2 + 1. Choose eλc = 1, bi (2λ)i e2λci + di (2λ)i = 0
i=1 i=1
s t
X X 1
and bi (λ)i eλci + di (λ)i = − . Then f (z + c) = (eλz − 1)2 + 1 and L̃(f (z)) = eλz .
i=1 i=1
2
Clearly f (z + c) and L̃(f (z)) share (2, ∞), (∞, ∞) and (1, 0) with Θ(0; f ) + Θ(∞; f ) > 0. But
L̃(f (z)) 6= f (z + c).
In Theorem 1.5, sharing of the value 0 can be removed at the cost of slightly manipulating
the deficiency condition. In this respect, we state the following theorem for transcendental
meromorphic function.
 Linear delay-differential operator of a meromorphic function... 5

Theorem 1.6. Let f (z) be a transcendental meromorphic function of finite order and let
a(z)(6≡ 0) ∈ S(f ) be an entire function. If L̃(f (z)) and f (z + c) share (a(z), ∞) and (∞, ∞)
with δ(0; f ) > 0, then L̃(f (z)) ≡ f (z + c).
By an example we now show that a(z) CM sharing can not be replaced by IM in Theorem
1.6.
t
−2ez −1
X
Example 1.3. Let f (z) = e2z and c = πi. Choose L̃(f (z)) = L3 (f (z)) with 2 (−1)i+1 di =
i=1
t
z
X
1 2e −1
1 and (−2)i di = 0. Then L̃(f (z)) = ez and f (z + c) = e2z share (1, 0), (∞, ∞) and
i=1
δ(0; f ) = 12 > 0. Clearly L̃(f (z)) 6= f (z + c).
Our next example shows that a(z) 6≡ 0 in Theorem 1.6 can not be dropped as well as (a(z), 0)
sharing in Theorem 1.5 can not be removed.
πiz
Example 1.4. Let f (z) = e c . Choose L̃(f (z)) = f ′ . Then clearly f (z + c) and L̃(f (z))
share (0, ∞), (∞, ∞) and δ(0; f ) > 0. But L̃(f (z)) 6= f (z + c).
Following two examples show that δ(0; f ) > 0 in Theorem 1.6 can not be removed.
Example 1.5. In Example 1.2 though f (z + c) and L̃(f (z)) share (2, ∞), (∞, ∞) but δ(0; f ) =
0. Here L̃(f (z)) 6= f (z + c).
z
Example 1.6. Let f (z) = e 2+z and a(z) = z. Choose L̃(f (z)) = L3 (f (z)) with d1 = 2c and
t
c z
X
dj = 2(ec − c). Then f (z + c) = e e +z+c


2 and L̃(f (z))(= ec ez + c) share (a(z), ∞) and
j=2
(∞, ∞) but δ(0; f ) = 0. Clearly L̃(f (z)) 6= f (z + c).

2. Lemmas
In this section some lemmas will be presented which will be needed in the sequel.
Lemma 2.1. [3] Let f (z) be a meromorphic function of finite order ρ and let c ∈ C \ {0} be
fixed. Then, for each ε > 0, we have
T (r, f (z + c)) = T (r, f (z)) + O(rρ−1+ε ) + O(log r).
Lemma 2.2. [5] Let f (z) be a meromorphic function of finite order and c ∈ C \ {0}. Then
   
f (z + c) f (z)
m r, + m r, = S(r, f ).
f (z) f (z + c)
Lemma 2.3. [6] Let f be a non-constant meromorphic function of finite order and c ∈ C.
Then
   
1 1
N r, ≤ N r, + S(r, f ),
f (z + c) f (z)
N (r, f (z + c)) ≤ N (r, f (z)) + S(r, f ),
   
1 1
N r, ≤ N r, + S(r, f )
f (z + c) f (z)
and
N (r, f (z + c)) ≤ N (r, f (z)) + S(r, f ).
Lemma 2.4. [2] Let f (z) be a meromorphic function of zero order and q ∈ C \ {0}. Then
 
f (qz)
m r, = S1 (r, f ).
f (z)
6 ARPITA ROY AND ABHIJIT BANERJEE

Lemma 2.5. [15] Let f (z) be a non constant zero order meromorphic function and q ∈ C\ {0},
then
T (r, f (qz)) = (1 + o(1))T (r, f (z))
and
N (r, f (qz)) = (1 + o(1))N (r, f (z))
on a set of lower logarithmic measure 1.
Using Lemma 2.4 and Lemma 2.5 and by the help of simple transformation one can easily
prove the next lemma.
Lemma 2.6. Let f (z) be a meromorphic function of zero order and q ∈ C \ {0}. Then
 
f (z)
m r, = S1 (r, f ).
f (qz)
Lemma 2.7. [16] Let f (z) be a non constant meromorphic function in the complex plane, and
P (f )
let R(f ) = Q(f ) , where
p
X q
X
k
P (f ) = ak (z)f and Q(f ) = bj (z)f j
k=0 j=0

are two mutually prime polynomials in f . If the coefficients ak (z) for k = 0, 1, . . . , p and bj (z)
for j = 0, 1, . . . , q are small functions of f with ap (z) 6≡ 0 and bq (z) 6≡ 0, then
T (r, P (f )) = max{p, q}T (r, f ) + S(r, f ).
Lemma 2.8. [11] Suppose that h is a non constant meromorphic function satisfying
 
1
N (r, h) + N r, = S(r, h).
h
Let f = a0 hp + a1 hp−1 + . . . + ap , and g = b0 hq + b1 hq−1 + . . . + bq be polynomials in h with
coefficients
 a0 , a1 , . . ., ap ; b0 , b1 , . . ., bq being small functions of h and a0 b0 ap 6≡ 0. If q ≤ p,
g
then m r, f = S(r, h).

Lemma 2.9. [10] If N (r, 0; f (k) | f 6= 0) denotes the counting function of those zeros of f (k)
which are not the zeros of f , where a zero of f (k) is counted according to its multiplicity then
N (r, 0; f (k) | f 6= 0) ≤ kN (r, ∞; f ) + N (r, 0; f |< k) + kN (r, 0; f |≥ k) + S(r, f ).
Lemma 2.10. Let F be a meromorphic function. Then
1
N (r, 1; F |≥ k + 1) ≤ {N (r, 0; F ) + N (r, ∞; F )} + S(r, F ).
k
Since the proof is straight forward, it is omitted.
Lemma 2.11. [1] Let F , G be two meromorphic functions sharing (1, 2) and (∞, k), where
0 ≤ k ≤ ∞. Then one of the following cases holds

(i) T (r, F ) + T (r, G) ≤ 2{N2 (r, 0; F ) + N2 (r, 0; G) + N (r, ∞; F ) + N (r, ∞; G)

+N ∗ (r, ∞; F, G)} + S(r, F ) + S(r, G),
where N ∗ (r, ∞; f, g) is the reduced counting function of those poles of F whose multiplicities
differ from the multiplicities of the corresponding poles of G,
(ii) F ≡ G,
(iii) F G ≡ 1.
 Linear delay-differential operator of a meromorphic function... 7

Lemma 2.12. Let P∗ (f ) and P∗ (g) be defined in (1.2), for two non constant meromorphic
functions f and g. Then
N (r, 0; P∗ (f )) ≤ (χ0n−1 + m1 + m2 )T (r, f );
N2 (r, 0; P∗ (f )) ≤ (χ0n−1 + µ0n−1 + m1 + 2m2 )T (r, f ).
Similar results occur for P∗ (g).
Proof. Rewrite P∗ (f ) and P∗ (g) as
(2.1) P∗ (f ) = Cf (f − β1 ) . . . (f − βm1 )(f − βm1 +1 )nm1 +1 . . . (f − βm1 +m2 )nm1 +m2
and
P∗ (g) = Cg(g − β1 ) . . . (g − βm1 )(g − βm1 +1 )nm1 +1 . . . (g − βm1 +m2 )nm1 +m2 ,
where βi′ s (i = 1, 2, . . . , m1 + m2 ) are distinct complex constants and ni is the multiplicity of
the factor (z − βi ) in P ∗ (z) for i = 1, 2, . . . , m1 + m2 with n1 = n2 = . . . = nm1 = 1 and
nm1 +1 , . . . , nm1 +m2 ≥ 2.
Here we have to consider two cases:
Case 1. Suppose none of βi′ s (i = 1, 2, . . . , m1 + m2 ) be zero. Then
1 +m2
mX
N (r, 0; P∗ (f )) ≤ N (r, 0; f ) + N (r, βi ; f ) ≤ (1 + m1 + m2 )T (r, f );
i=1

m1
X 1 +m2
mX
N2 (r, 0; P∗ (f )) ≤ N (r, 0; f ) + N (r, βi ; f ) + 2 N (r, βi ; f ) ≤ (1 + m1 + 2m2 )T (r, f ).
i=1 i=m1 +1

Case 2: Next let one of βi′ s (i = 1, 2, . . . , m1 + m2 ) be zero.

Subcase 1: Suppose one among βi′ s (i = 1, 2, . . . , m1 ) be zero. Without loss of generality let us
assume that β1 = 0. Then
1 +m2
mX
N (r, 0; P∗ (f )) ≤ N (r, 0; f ) + N (r, βi ; f ) ≤ (m1 + m2 )T (r, f ) ;
i=2

m1
X 1 +m2
mX
N2 (r, 0; P∗ (f )) ≤ 2N (r, 0; f ) + N (r, βi ; f ) + 2 N (r, βi ; f ) ≤ (1 + m1 + 2m2 )T (r, f ).
i=2 i=m1 +1

Subcase 2: Next suppose one among βi′ s (i = m1 + 1, m1 + 2, . . . , m1 + m2 ) be zero. Without

loss of generality let us assume that βm1 +1 = 0. Then
m1
X 1 +m2
mX
N (r, 0; P∗ (f )) ≤ N (r, 0; f ) + N (r, βi ; f ) + N (r, βi ; f ) ≤ (m1 + m2 )T (r, f );
i=1 i=m1 +2

m1
X 1 +m2
mX
N2 (r, 0; P∗ (f )) ≤ 2N (r, 0; f ) + N (r, βi ; f ) + 2 N (r, βi ; f ) ≤ (m1 + 2m2 )T (r, f ).
i=1 i=m1 +2

Combining all cases we can write

N (r, 0; P∗ (f )) ≤ (χ0n−1 + m1 + m2 )T (r, f );
N2 (r, 0; P∗ (f )) ≤ (χ0n−1 + µ0n−1 + m1 + 2m2 )T (r, f ).
Similarly we can obtain the same conclusions for the function g. 
Lemma 2.13. Let P∗ (f ) and P∗ (g) for two non constant meromorphic functions f and g (as
defined in (1.2)) share (1, 2) and (∞, 0). If
15
n > 2(χ0n−1 + µ0n−1 + m1 + 2m2 ) + (χn−1 + m1 + m2 ),
(2n − 3) 0
then either P∗ (f )(z) ≡ P∗ (g)(z) or P∗ (f )(z).P∗ (g)(z) ≡ 1.
8 ARPITA ROY AND ABHIJIT BANERJEE

Proof. Set
P∗ (f )(P∗ (g) − 1)
Φ= .
P∗ (g)(P∗ (f ) − 1)
Clearly S(r, Φ) can be replaced by S(r, f ) + S(r, g). It is obvious that Φ 6≡ 0. If Φ ≡ 0 then
either P∗ (f ) = 0 or P∗ (g) = 1, which gives f and g are constants, a contradiction.
First suppose that Φ 6≡ 1. So P∗ (f ) 6≡ P∗ (g).
Therefore, using Lemma 2.10 we get
N (r, 0; Φ) + N (r, ∞; Φ)
≤ N (r, 1; P∗ (f ) |≥ 3) + N (r, 0; P∗ (f )) + N (r, 0; P∗ (g))
1

≤ N (r, 0; P∗ (f )) + N (r, ∞; P∗ (f )) + N (r, 0; P∗ (f ))
2
+N (r, 0; P∗ (g)) + S(r, P∗ (f ))
3 1
≤ N (r, 0; P∗ (f )) + N (r, ∞; f ) + N (r, 0; P∗ (g)) + S(r, f ).
2 2
Now,
 
P∗ (g) − P∗ (f ) P∗ (g)′ P∗ (f )′
Φ−1 = and Φ′ = − Φ.
P∗ (g)(P∗ (f ) − 1) P∗ (g)(P∗ (g) − 1) P∗ (f )(P∗ (f ) − 1)
If Φ′ ≡ 0 then
 
P∗ (g)′ P∗ (f )′
− ≡ 0.
P∗ (g)(P∗ (g) − 1) P∗ (f )(P∗ (f ) − 1)
Integrating we have,
P∗ (f ) − 1 P∗ (g) − 1
≡A ,
P∗ (f ) P∗ (g)
where A is non zero constant. i.e.,
1 A
1− ≡A− .
P∗ (f ) P∗ (g)
Since P∗ (f ) and P∗ (g) share (∞, 0) so A = 1. Then P∗ (f ) ≡ P∗ (g) which gives Φ ≡ 1, a
contradiction. Therefore Φ′ 6≡ 0. Clearly all poles of P∗ (f ) and P∗ (g) are multiple poles which
are multiple zeros of Φ − 1 and so zeros of Φ′ with multiplicity at least (n − 1) but not zeros
of Φ. Therefore by Lemma 2.9,
(n − 1)N (r, ∞; f ) = (n − 1)N (r, ∞; P∗ (f )) = (n − 1)N (r, ∞; P∗ (f ) |≥ n)
′
≤ N (r, 0; Φ | Φ 6= 0) ≤ N (r, 0; Φ) + N (r, ∞; Φ) + S(r, Φ).
So,
(2n − 3)N (r, ∞; f ) ≤ 3N (r, 0; P∗ (f )) + 2N (r, 0; P∗ (g)) + S(r, f ).
Applying Lemma 2.12 we obtain
3(χ0n−1 + m1 + m2 ) 2(χ0n−1 + m1 + m2 )
N (r, ∞; f ) ≤ T (r, f ) + T (r, g) + S(r, f ) + S(r, g).
2n − 3 2n − 3
Similarly
3(χ0n−1 + m1 + m2 ) 2(χ0n−1 + m1 + m2 )
N (r, ∞; g) ≤ T (r, g) + T (r, f ) + S(r, f ) + S(r, g).
2n − 3 2n − 3
That is
5(χ0n−1 + m1 + m2 )
(2.2) N (r, ∞; f ) + N (r, ∞; g) ≤ (T (r, f ) + T (r, g))
2n − 3
+S(r, f ) + S(r, g).
 Linear delay-differential operator of a meromorphic function... 9

If possible, we suppose that (i) of Lemma 2.11 holds. Therefore

T (r, P∗ (f )) + T (r, P∗ (g))
≤ 2{N2 (r, 0; P∗ (f )) + N2 (r, 0; P∗ (g)) + N (r, ∞; P∗ (f )) + N (r, ∞; P∗ (g))
+N ∗ (r, ∞; P∗ (f ), P∗ (g))} + S(r, P∗ (f )) + S(r, P∗ (g)).
Then using Lemma 2.7, Lemma 2.12 and (2.2) we have
n (T (r, f ) + T (r, g))
15(χ0n−1 + m1 + m2 )
 
≤ 2(χ0n−1 + µ0n−1 + m1 + 2m2 ) + (T (r, f ) + T (r, g))
2n − 3
+S(r, f ) + S(r, g),
which contradicts our assumption. So by Lemma 2.11 we have
P∗ (f )(z).P∗ (g)(z) ≡ 1.
If Φ ≡ 1, then P∗ (f )(z) ≡ P∗ (g)(z).
Hence the lemma is proved. 
Lemma 2.14. Let f and g be two non constant meromorphic functions of finite order. Let
n ≥ 2, and let 
{a1 (z), a2 (z), . . . , an (z)} ∈ S(f ) be distinct meromorphic periodic functions with
g
period c. If m r, f −a k
= S(r, f ), for k = 1, 2, . . . , n, then
n    
X 1 1
m r, ≤ m r, + S(r, f ),
f − ak g
k=1

where the exceptional set associated with S(r, f ) is of at most finite logarithmic measure.
Proof. Set
n
Y
P (f ) = (f − ak ).
k=1

Rewriting we have,
n
1 X αk
= ,
P (f ) f − ak
k=1

where αk ∈ S(f ) are certain periodic function with period c. Now,

  X n  
g g
m r, ≤ m r, + S(r, f ) = S(r, f ),
P (f ) f − ak
k=1

and so
       
1 g 1 1
m r, = m r, + m r, ≤ m r, + S(r, f ).
P (f ) P (f ) g g
By the first fundamental theorem and using the above inequation we get,
     
1 1 1
m r, ≥ m r, + S(r, f ) = T (r, P (f )) − N r, + S(r, f )
g P (f ) P (f )
n   n  
X 1 X 1
≥ nT (r, f ) − N r, + S(r, f ) = m r, + S(r, f ).
f − ak f − ak
k=1 k=1


Lemma 2.15. If f be a meromorphic
 function
 of finite order
 then L̃(f  (z)) is of finite order
(z)) L̃(f (z)) L̃(f (z))
and m r, L̃(f
f (z+c) = S(r, f ), m r, f (z)−βi = S(r, f ) and m r, f (qz) = S1 (r, f ).
10 ARPITA ROY AND ABHIJIT BANERJEE

Proof. Using logarithmic derivative lemma and Lemma 2.2 we have,

s t
 
X X
!  bj f (j) (z + cj ) + dj f (j) (z) 
L̃(f (z)) 
j=1 j=1

(2.3) m r, = m r,
 
f (z + c) f (z + c)

 
 

s s
f (j) (z + cj ) f (j) (z)
X   X  
≤ m r, + m r,
j=1
f (j) (z) j=1
f (z)
t
f (j) (z)
   
X f (z)
+ m r, + (s + t) m r, + O(1)
j=1
f (z) f (z + c)
= S(r, f ).
Also,
 s t

X X
(j) (j)
!  bj f (z + cj ) + dj f (z) 
L̃(f (z)) 
j=1 j=1

m r, = m r,
 
f (z) − βi f (z) − βi

 
 

s t
f (j) (z + cj ) f (j) (z)
X   X  
≤ m r, + m r,
j=1
f (j) (z) j=1
f (z) − βi
s
f (j) (z)
X  
+ m r, + O(1) = S(r, f ).
j=1
f (z) − βi

Using (2.3) and Lemma 2.1 we have,

s2 + t2 + 3(s + t) + 2
T (r, L̃(f (z))) ≤ T (r, f ) + S(r, f ).
2
As f is of finite order so L̃(f (z)) and f (z + c) is of finite order and S(r, L̃(f (z))) can be replaced
by S(r, f ).
Similarly by using Lemma 2.4, Lemma 2.5 and Lemma 2.6 as and when required we can
prove f (qz) and L̃(f (z)) are zero order when f is of zero order and
!
L̃(f (z))
m r, = S1 (r, f ).
f (qz)


3. Proofs of the theorems

Proof of Theorem 1.1. Since Ef (z+c) (S1∗ , 2) = EL̃(f (z)) (S1∗ , 2) and Ef (z+c) (S2 , 0) = EL̃(f (z))
(S2 , 0), it follows that P∗ (f (z + c)), P∗ (L̃(f (z))) share (1, 2) and (∞, 0). So by Lemma 2.13 we
have either P∗ (f (z + c)) ≡ P∗ (L̃(f (z))) or P∗ (f (z + c)).P∗ (L̃(f (z))) ≡ 1. Suppose that
(3.1) P∗ (f (z + c)).P∗ (L̃(f (z))) ≡ 1.
Noting that P∗ (f (z + c)), P∗ (L̃(f (z))) share (∞, 0), so we can conclude that P∗ (f (z + c)),
P∗ (L̃(f (z))) both are entire functions.
So !
P∗ (L̃(f (z)))
N r, ∞; = N (r, 0; P∗ (f (z + c))).
P∗ (f (z + c))
 Linear delay-differential operator of a meromorphic function... 11

Therefore using Lemma 2.12 and Lemma 2.1, we get

!
P∗ (L̃(f (z)))
N r, ∞; ≤ (χ0n−1 + m1 + m2 )T (r, f (z + c)) ≤ nT (r, f ) + S(r, f ).
P∗ (f (z + c))
Using Lemma 2.2 and Lemma 2.15 we have,
! m +m
!ni !
P∗ (L̃(f (z))) L̃(f (z)) 1Y 2 L̃(f (z)) − βi
m r, = m r,
P∗ (f (z + c)) f (z + c) i=1 f (z + c) − βi
! m1Y+m2
!ni !
L̃(f (z)) L̃(f (z)) − βi
≤ m r, + m r, + O(1)
f (z + c) i=1
f (z + c) − βi
1 +m2
mX
!
L̃(f (z)) − βi
≤ ni m r, + S(r, f )
i=1
f (z + c) − βi
1 +m2
mX
! m +m
1 2
 
L̃(f (z)) X 1
≤ ni m r, + ni m r,
i=1
f (z) − βi i=1
f (z) − βi
mX1 +m2  
f (z) − βi
+ ni m r, + S(r, f )
i=1
f (z + c) − βi
1 +m2
mX  
1
≤ ni m r, + S(r, f )
i=1
f (z) − βi
≤ (n1 + n2 + . . . + nm1 +m2 )T (r, f ) + S(r, f )
≤ (n − 1)T (r, f ) + S(r, f ).
By Lemma 2.1, Lemma 2.7 and (3.1),
2nT (r, f ) = 2nT (r, f (z + c)) + S(r, f ) = 2T (r, P∗ (f (z + c))) + S(r, f )
  !
1 P∗ (L̃(f (z)))
≤ T r, + S(r, f ) ≤ T r, + S(r, f )
P∗ (f (z + c))2 P∗ (f (z + c))
≤ (2n − 1)T (r, f ) + S(r, f ),
which is a contradiction.
Therefore P∗ (L̃(f (z))) ≡ P∗ (f (z + c)), which yields
n
Y n
Y
(L̃(f (z)) − αi ) ≡ (f (z + c) − αi ).
i=1 i=1

Proof of Theorem 1.2. By proceeding in a similar way of the proof of Theorem 1.1 we can
prove this theorem using Lemma 2.4, Lemma 2.5 and Lemma 2.6 as and when required instead
of Lemma 2.1 and Lemma 2.2. 

Proof of Theorem 1.3. Since the finite order meromorphic functions f (z + c) and L̃(f (z))
share (S1∗ , ∞), (S2 , ∞), it follows that P∗ (f (z +c)), P∗ (L̃(f (z))) share (1, ∞) and (∞, ∞) which
yields
(3.2) N (r, L̃(f (z))) = N (r, f (z + c))
and
P∗ (L̃(f (z))) − 1
(3.3) = eγ(z) ,
P∗ (f (z + c)) − 1
where γ(z) is a polynomial.
12 ARPITA ROY AND ABHIJIT BANERJEE

Now,
!
P∗ (L̃(f (z))) − 1
T (r, eγ(z) ) = m(r, eγ(z) ) = m r, .
P∗ (f (z + c)) − 1

Using the definition of P∗ (z) we have,

!
γ(z) (L̃(f (z))) − α1 )(L̃(f (z))) − α2 ) . . . (L̃(f (z))) − αn )
T (r, e ) = m r,
(f (z + c) − α1 )(f (z + c) − α2 ) . . . (f (z + c) − αn )
n
!
X L̃(f (z))) − αj
≤ m r, + O(1)
j=1
f (z + c) − αj
n
! n   X n  
X L̃(f (z)) X 1 f (z) − αj
≤ m r, + m r, + m r,
j=1
f (z) − αj j=1
f (z) − αj j=1
f (z + c) − αj
+O(1).
In view of Lemma 2.2, Lemma 2.14, Lemma 2.15 and then by the first fundamental theorem
and (3.2) we have,
n    
X 1 1
T (r, eγ(z)) = m r, + S(r, f ) ≤ m r, + S(r, f )
j=1
f (z) − αj L̃(f (z))
 
1
≤ T (r, L̃(f (z))) − N r, + S(r, f )
L̃(f (z))
!  
L̃(f (z)) 1
≤ m r, + m(r, f (z + c)) + N (r, L̃(f (z))) − N r, + S(r, f )
f (z + c) L̃(f (z))
   
1 1
≤ T (r, f (z + c)) − N r, + S(r, f ) ≤ T (r, f ) − N r, + S(r, f ).
L̃(f (z)) L̃(f (z))
 
According to the given condition T (r, f ) = N r, L̃(f1(z)) + S(r, f ), so T (r, eγ(z)) = S(r, f ).
Now from (3.3) we have,
 
P∗ (L̃(f (z))) = eγ(z) P∗ (f (z + c)) − 1 + e−γ(z) .

Set
P∗ (f (z + c))
W (z) = .
1 − e−γ(z)
If eγ(z) 6≡ 1, then by applying Nevanlinna’s second fundamental theorem to W (z) and using
(3.2) and Lemma 2.12 we obtain,
T (r, P∗ (f (z + c))) ≤ T (r, W ) + S(r, f )
≤ N (r, 0; W ) + N (r, ∞; W ) + N (r, 0; W − 1) + S(r, f )
≤ N (r, 0; P∗ (f (z + c))) + N (r, ∞; P∗ (f (z + c))) + N (r, 0; P∗ (L̃(f (z)))) + S(r, f )
 
≤ (χ0n−1 + m1 + m2 ) T (r, f (z + c)) + T (r, L̃(f (z))) + N (r, ∞; f (z + c)) + S(r, f )
!
n−1 L̃(f (z))
≤ (χ0 + m1 + m2 ) T (r, f (z + c)) + m(r, f (z + c)) + m r,
f (z + c)
+N (r, ∞; f (z + c))) + N (r, ∞; f ) + S(r, f ).
Using Lemma 2.1 and Lemma 2.15 we get,
nT (r, f ) ≤ (2χ0n−1 + 2m1 + 2m2 + 1)T (r, f ) + S(r, f ),
 Linear delay-differential operator of a meromorphic function... 13

which contradicts n > 2(χ0n−1 + m1 + m2 ) + 1. This gives eγ(z) ≡ 1, that yields

n
Y n
Y
(L̃(f (z)) − αi ) ≡ (f (z + c) − αi ).
i=1 i=1

Proof of Theorem 1.4. Here L̃(f (z)) and f (qz) are of zero order. Since f (qz) and L̃(f (z))
share (S1∗ , ∞) and (S2 , ∞), it follows that P∗ (f (qz)) and P∗ (L̃(f (z))) share (1, ∞) and (∞, ∞).
Therefore
P∗ (L̃(f (z))) − 1
= A,
P∗ (f (qz)) − 1
where A is a non zero constant.
This gives
 
1
P∗ (L̃(f (z))) = A P∗ (f (qz)) − 1 + .
A
(f (qz))
Set W1 (z) = P∗1− 1 . If A 6≡ 1, then applying Nevanlinna’s second fundamental theorem to
A
W1 (z) and using Lemmas 2.4 and 2.5 and 2.15 as and when required we can calculate the rest
of the proof similar to Theorem 1.3. 

Proof of Theorem 1.5. Here f (z + c) and L̃(f (z)) are of finite order. Since f (z + c) and
L̃(f (z)) share (0, ∞) and (∞, ∞), so
L̃(f (z))
(3.4) = eδ(z),
f (z + c)
where δ(z) is a polynomial.
Clearly by Lemma 2.15 we get,
T (r, eδ(z) ) = S(r, f ).
When eδ(z) ≡ 1 then L̃(f (z)) ≡ f (z + c).
When eδ(z) 6≡ 1, using the fact that f (z + c) and L̃(f (z)) share (a(z), 0) we have,
       
1 1 1 1
N r, = N r, ≤ N r, δ(z) + N r,
L̃(f (z)) − a(z) f (z + c) − a(z) e −1 a(z)
≤ T (r, eδ(z) ) + S(r, f ) = S(r, f ).
Rewriting (3.4) we get,
L̃(f (z)) − a(z) = eδ(z) (f (z + c) − a(z)e−δ(z) ).
Clearly a(z)e−δ(z) ≡
6 a(z). So,
   
1 1
N r, = N r, = S(r, f ).
f (z + c) − a(z)e−δ(z) L̃(f (z)) − a(z)
Using Lemma 2.1, 2.3 and the second fundamental theorem we obtain,
2T (r, f ) = 2T (r, f (z + c)) + S(r, f )
   
1 1
≤ N (r, f (z + c)) + N r, + N r,
f (z + c) f (z + c) − a(z)
   
1 1
+N r, + S(r, f ) ≤ N (r, f ) + N r, + S(r, f ),
f (z + c) − a(z)e−δ(z) f
which is a contradiction to Θ(0; f ) + Θ(∞; f ) > 0. Hence L̃(f (z)) ≡ f (z + c). 
14 ARPITA ROY AND ABHIJIT BANERJEE

Proof of Theorem 1.6. Here f (z + c) and L̃(f (z)) are of finite order. Since f (z + c) and
L̃(f (z)) share (a(z), ∞) and (∞, ∞), so
L̃(f (z)) − a(z)
(3.5) = eζ(z),
f (z + c) − a(z)
where ζ(z) is a polynomial. Using logarithmic derivative lemma, Lemma 2.1 and Lemma 2.2
we get,
!
ζ(z) ζ(z) L̃(f (z)) − a(z)
T (r, e ) = m(r, e ) = m r,
f (z + c) − a(z)
! !
L̃(f (z)) − L̃(a(z − c)) L̃(a(z − c)) − a(z)
≤ m r, + m r,
f (z + c) − a(z) f (z + c) − a(z)
!  
L̃(f (z)) − L̃(a(z − c)) f (z) − a(z − c)
≤ m r, + m r,
f (z) − a(z − c) f (z + c) − a(z)
 
1
+m r, + S(r, f )
f (z + c) − a(z)
 s t

X X
(j) (j) (j) (j)
 bj (f (z + cj ) − a (z − c + cj )) + dj (f (z) − a (z − c)) 
 
 j=1 j=1
≤ m r,

f (z) − a(z − c)

 
 

+T (r, f (z + c)) + S(r, f )

s t
f (j) (z + cj ) − a(j) (z − c + cj ) f (j) (z) − a(j) (z − c)
X   X  
≤ m r, + m r,
j=1
f (j) (z) − a(j) (z − c) j=1
f (z) − a(z − c)
s
f (j) (z) − a(j) (z − c)
X  
+ m r, + T (r, f ) + S(r, f )
j=1
f (z) − a(z − c)
≤ T (r, f ) + S(r, f ).
So S(r, eζ(z) ) can be replaced by S(r, f ). When eζ(z) ≡ 1 then L̃(f (z)) ≡ f (z + c).
Suppose eδ(z) 6≡ 1. Now rewriting (3.5) we can obtain,
1 L̃(f (z)) eζ(z)
=− + .
f (z + c) f (z + c)(eζ(z) − 1) a(z)(eζ(z) − 1)
Therefore in view of Lemma 2.15 we have,
   
1 1
m r, ≤ 2 m r, ζ(z) + S(r, f ).
f (z + c) e −1
 
1
If ζ(z) is constant then automatically m r, f (z+c) = S(r, f ). If ζ(z) is non constant then by
Lemma 2.8 we get,
 
1
m r, = S(r, eζ(z) ) = S(r, f ).
f (z + c)
By Lemma 2.1 and Lemma 2.3 we have,
 
1
T (r, f ) = T (r, f (z + c)) + S(r, f ) = T r, + S(r, f )
f (z + c)
   
1 1
≤ N r, + S(r, f ) ≤ N r, + S(r, f ) ≤ T (r, f ) + S(r, f ).
f (z + c) f
 Linear delay-differential operator of a meromorphic function... 15

Therefore,
 
1
N r, = T (r, f ) + S(r, f ),
f
which contradicts the fact that δ(0, f ) > 0. Hence L̃(f (z)) ≡ f (z + c). 

4. Observation
2 t
Take L̃(f (z)) = L3 with all coefficients are 1. Then we see that choosing c = log(α+αα+...+α ) ,
where 1 + α + . . . + αt−1 6= 0, we somehow get a solution f (z) = eαz (α 6= 0) of
(4.1) L̃(f (z)) = f (z + c).
π
However choosing c = we can present the solution of f ′ = f (z + c) as the linear combination
2,
of two independent solutions. e.g., f (z) = d1 eiz + d2 e−iz . So it is a matter of concern that how
the solutions of (4.1) looks like. Unfortunately we can not elucidated in this matter.

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Department of Mathematics, University of Kalyani, West Bengal 741235, India.

Email address: arpita140793@gmail.com

Department of Mathematics, University of Kalyani, West Bengal 741235, India.

Email address: abanerjee kal@yahoo.co.in, abanerjeekal@gmail.com