Unicity on entire functions concerning their difference operators and derivatives

• 1.    Introduction and Main Results

The reader is presumed to be familiar with the fundamental notations and conclusions of Nevanlinna’s value distribution theory of meromorphic functions [1, 2]. S (r, f ) means that S(r, f ) = o(T(r, f )) as r ^ ^ outside of a possible exceptional set of finite logarithamic measure, and

A(f) = lim sup r →∞

log + N(r, f log r '

m ( r, -Д-)                  N(r, 1)

f - a                         f

6(P1,f) = lim inf ,   , = 1 - lim sup ,   , r >ra T (r, J)            r vx.       T (r, J)

stand for the exponents of convergence of zero sequence of f and the deficiency of f at the point a , respectively. For a nonconstant meromorphic function h , we denote by T (r, h) the Nevanlinna characteristic of h and by S(r, h) any quantity satisfying S (r, h) = o(T(r, h)) , as r runs to infinity outside of a set E C (0, + to ) of finite linear measure. We say that h is a small

function of f , if T (r, h) = S (r, f ) . In the sequel, we denote by I a set of infinite linear measure not necessarily the same in all its occurrences.

We say that f and g share the value a IM (ignoring multiplicities), if f and g have the same a point. If f and g have the same a point with the same multiplicities, then we say f , g share the value a CM (counting multiplicities).

Definition 1 [3]. Let k be a nonnegative integer or infinity. For a E C U{to} , we denote by E k (a; f ) the set of all a points of f (z) where an a point of mulplicity m is counted m times if m ^ k and k + 1 times if m > k . If E k (a; f ) = E k (a; g) , then we say that f,g share the value a with weight k .

We write f and g share (a, k) to mean that f, g share the value a with weight k.

Rubel and Yang Chung-Chun [4] considered the uniqueness of an entire function and its derivative. They proved the following.

Theorem 1. Let f (z) be a non-constant entire function, let a,b be two finite distinct values. If f (z) and f ‘ (z) share a,b CM, then f (z) = f ‘ (z).

Li Ping and Yang Chung-Chun [5] improved Theorem 1 and proved.

Theorem 2. Let f (z) be a non-constant entire function, and let a, b be two finite distinct complex values. If f(z) and f ^(k) (z) share a CM, and share b IM. Then f(z) = f ^(k) (z).

The value distribution of meromorphic functions of finite order with respect to difference analogue has become a subject of some interests, see [6–16].

Theorem 3 [17] . Suppose that f (z) and g(z~) are nonconstact meromorphic functions. If f, g share 0,1, to CM and N rr, f + N(r, f ) <  (d + o(1))T (r, f ) for r E I and r E to , where d is a positive number satisfying 0 < d <  1 , which I C (0, + to ) is a subset of infinite linear measure, then f = g or f.g = 1.

Theorem 4 [4] . Let f be a nonconstant entire function. If f shares two distinct finite values CM with f ‘ , then f = f ‘ .

More results on uniqueness of f ‘ with its n -th derivative f ⁽ⁿ⁾ were obtained by several authors (see [18–20]). In view of the progress on the difference analogues of classical Nevanlinna theory of meromorphic functions [21, 22], it is quite natural to investigate the uniqueness problems of meromorphic functions and their difference operators (see [23–26]).

Example 1. Let f (z) = e ^Az , where A = 0 is a constant. Then f ^(k) = A k e ^Az and Af = f (z + 1) — f (z) = (e ^A — 1)e ^Az . Clearly, A(f) and f ^(k) share 0 CM and to IM, and that ц = 1 . We can choose A such that e ^A — 1 = A k , and so f = A(f) .

Theorem 5 [27] . Let f (z) be a trancendental entite function of finite order, let n = 0 be a finite complex number, n ^ 1, k ^ 0 be two integers and let a, b be two distinct finite complex values. If f(z) and (A n f (z) ) ^(k) share a CM and share b IM, then f (z) = ( A n f(z) ) (k) .

Lemma 1 [9] . Let Af be a nonconstant meromorphic function of finite order, let n = 0 be a finite complex number. Then

/ Af (z + n) A m V’---Af---) = S (r’ Af)’ for all r outside of a possible exceptional set E with finite logarithmic measure.

Lemma 2 [28, Lemma 4.3] . Let Af be a nonconstant meromorphic function. Suppose that the polynomials P j , j = 0, 1,... ,q, q > p, and let P(Af ) = a o (Af) ^p + a ₁ (Af) ^{p 1} + ... + a _p (a o = 0) is a polynomial of degree p with constant coefficient a j , j = 0, 1,... ,p. Then

(          P(Af)(Af)‘        V mV (Af — Pi)(Af — P2)... (Af — Pq))    (r’ f).

Lemma 3. Let Af and Ag be two non constant entire functions, and let P i , P 2 be two polynomials. If

H =

Af ‘ Ag ’ ^{(Af -} P i )(^{Af -} P 2 )   ^(Ag ^- P i ^)(Ag ^- P 2 )

and Af and Ag share Pi CM, and share P2 IM, then either or

2T(r, Af ) ^ N fr,

VP) ⁺ N(r’f^ ) ⁺ S(r^A >

Af = Ag.

<1 Integrating H which leads to

Ag - P 2 ₌ Af - P 2 Ag - P i     Af - P i ^,

where C is a nonzero constant.

If C = 1 , then Af = Ag . If C = 1 , then from above, we have

P i - P 2 ₌ ( C - 1)Af - CP 2 + P i Ag - P i =      Af - P i

and

T (r, Af) = T (r, Ag) + S(r, Af) + S(r, Ag).

Obviously, CcC-P ² = a and Ce ^ - iP ² = b It follows that N

Second Fundamental Theorem,

r _M -   . ' .

= 0 . Then by the

2T(r, Af ) = N (r, Af) + N fr,

Af- P i ) ⁺ N ( r, Af - P 2 ) ⁺ N(r’ Af - CC -^)

+ S(r, Af) ^ N fr,     ¹   ) + N fr,     ¹   ) + S(r, Af),

Af - P 1         Af - P 2

2T (r, Af) ^ N fr, ,/ i + N fr,     ¹      + S (r, Af). ▻

Af - P 1         Af - P 2

Lemma 4. Let Af be a transcendental entire function of finite order, k be positive integer, let P i be a nonzero complex value or constant. If Af and f ^(k) share P i CM, and N (r, fw) = S(r, Af), then one of the following cases must occur:

• •    f ^(k) = He p , where p is a polynomial, and H = 0 is a small function of e ^p .

• •    T(r,e ^p ) = S(r, Af).

< Since Af is a transcendental entire function of finite order, Af and f ^(k) share P i CM, then there is a polynomial p such that

Af - P i = e p (f ^(k) ) - P.

Set g = f ^(k) . It follows by (1) that

g = (ge ^p ) ^(k) - (P^H^W .

Then we rewrite (2) as

_{1 +} (Pe pFl = D_e p , ^g

where

D     (geP ) ^(k)

ge ^p

.

Note that N (r, f ^k) ) = N (r, |) = S(r, f), then by Lemma 1 we have т ( (gep)(k)\

TDTi r

Next we discuss two cases.

Case 1: e-p — D = 0. Rewrite (3) as gep(e-p — D) = (Pi ep)(k).

When D = 0, (5) implies g = Hep.

Here H = 0 is a small function of e ^p .

When D = 0 , it follows from (5) that N (r, e - p - D^ = S(r, f ) . Then use the Second

Fundamental Theorem to e ^p we can obtain

T(r, e^p ) = T(r, e ^p )+O(1) 5 N(r, e^p)+N (r, e -p ^ +N (r, e -p

—

D) +O(1)= S(r, Af).

Case 2: e ^p —D = 0 . It implies that T(r, e ^p ) = T(r, e ^p )+O(1) = S(r, Af ) , a contradiction. From above discussions, we get T(r,e ^p ) = S (r, Af) . >

Theorem 6. Let Af (z) be a transcendental entire functions of finite order, k be integer such that k ^ 0 and let P i and P 2 be two polynomials. If Af (z) and f ^(k) share P i CM and share P ₂ IM, then Af = f ^(k) .

• <1 If Af = f ^(k) , there is nothing to prove. Solve Af = f ^(k) . Since Af is a transcenedental entire function of finite order, Af and f ^(k) share P i CM, then we get

P Q

Af — Pi     ’ where Q is a polynomial.

Since Af and f ^(k) share P i CM and share P 2 IM, then by second fundamental theorem and Lemma 1 we have

T (r, Af) 5 N (r, ¹   ) + N fr, ./ i + S(r, Af)

\ ^Af ^— Pi )     \ ^Af ^— P2 /

< ^N (r’ 57—Pg) ⁺ N (r’ ? (k )TP) ⁵ N (r Af - rT i i ) ) ^{+ S(r}' ^{Af 1}

5 T(r, Af — f ^{(k )}) + S(r, Af) 5 m(r, Af — f ^{(k )}) + S(r, Af)

5 m(r, Af) + m

r, 1

( k )

- Af) + S(r,Af)

5 T (rf)+ S(r’f ),

T (r, Af) = N fr,     ¹   ) + N (Af - P ₂ ) + S(r, Af).

V ^Af ^- P i /

According to (7) and (8) we have and

T(r, Af ) = T ( r, Af - f ^(k) ) + S(r, Af ) * N rr, _Af - _{f (k)} ) + S(r, Af )

T ( r,e ^Q ) = m ( r, e ^Q ) = m

(r VP) * ^m ( r V ' P ) ^{+ S(r}' ^Af *•

Then it follows from (7) and (9) that m (r- f=P\ ) = m (r’ >    ■ ) * m (r Af - f(k ) + mr eQ - 1>

* T(r,e^Q) + S(r, Af).

Then by (10) and (11)

T(r,e ^Q ) = m(r,     ¹    ) + S(r, Af ).                        (12)

A f - P i

On the other hand, (1) can be rewritten as

f ^(k) — Af Af - P i

= e ^Q - 1 ,

which implies

N (r, _A f - P ₂ ) * N (r, _e Q - 1 ) = T re Q ) ⁺ S^(r,^Af )•            ⁽¹⁴⁾

Then by (8), (12) and (14)

^m ( r, Af -^) ⁺ N ( r, Af -^) = N ( r, Af -^) ⁺ N ( r, A f P ) ^{+ S(r}- ^Af )

* NCrAf^) ⁺ ''А .)^{+ S(r}'^Af '

* N ( r’ Af-P r ) ⁺ m ( r Af - P \ ) ⁺ S¹', ^Af ),

N (r Afbi) = N (r f-P) + S(r’Af)’ and then

Nfr, 1      = T (r^eQ) + S (r, Af).

Af - P 2

Set

(Af) ‘ (Af - f ^{(k )})

Ф (Af - Pi)(Af - P2)

and                                       f ⁽ k ^{+1) {} ^f - f (k) )

^ф    (f« - P i )(f« - P 2 ) .                         ( )

Easy to know that φ is an entire function by Lemma 1 and Lemma 2 we have

T (r ^ф) = mlr^ф) = m ( r (fPff-Py ) ⁺ S (r,^Af)

< m ( r' Af P ^Af P ) m ( r¹ - ff ) ⁺ S' ^Tf ) = S^(r’ ^Af ¹¹

that is T M) = S(r, Af ).                             (19)

Obiously Let d = P i — k(P i — P 2 ) , k = 0 , by Lemmas 1 and 2, we obtain

( (r A+m(r 1    ( (Af)   JTfT^ (A v ' Af) + V № — Р1)Ф Uf — Pl  Af — Р2Д  Af ) )

• <    _m (r ¹    ( (r ( ^(Af) ^(Af)    + _m (r ^(Af) ^          — Sir A)

• <    ^m^J + m^r ^f — P i ^— Af — P2) Д T Af Д S⁽r’ ^Af )= S⁽r’^A)

and

/    1 V /       (Af )‘(Af — f(k))     \ m V' Af — d) m (r' (Af — Pi)(Af — P2)(Af — d))

<   ( 1 — f ^(k) )^ ( (Af) ‘ ( Af — f ^(k) )       )^cr              л n

< ^m Pr' -AT) ⁺ m V’ (A f — P i )(A f — P 2 )(A f — d ) ) ^{+ S}<^r' ^Af) = ^S(^r' ^Af>•

Set

, =^ (Af) ‘ ( f ^(k) — P i )( f ^(k) — P 2 )    (Af — P i )(Af — P 2 ) ‘

We discuss two cases

Case 1: ф = 0 . Integrating both side of (22) which leads to

Af — P 2 = f ^(k) — P 2

Af — Pi   cf(k) — Pi ’ where c is a non zero constant. Then by Lemma 3 we see that

2T(r’ Af) ^ N fr'     ¹   ) + N fr'     ¹   ) + S(r’ Af)’

Af — P i         Af — P 2

which contradicts with (8).

Case 2: ф = 0 . By (9), (19) and (22) we can obtain

m(r' ^Af)=^m(r' ^Af ^— f ^(k) ) ⁺_S(r' ^Af)=^m ( ^r' " ( ^ff )) ⁺S ₍r' ^Af, = m(r^J^) + ⁵(r’ Af) ^ Т^’ф—ф) + S(r Af)

< T(r’ ф — ф) + T(r’ ф) + S(r’ Af) ^ T(r’ ф) + N fr’     ¹    ) + S(r’ Af).

Af — P 2

On the other hand

T (r,^ = T

r,

f (k+i) (Af - f ^{(k )})

( f ^(k) — P i )( f ^(k) — P ⁽²⁾ )

)

(      f(k+1) (Af - f(k))    ^ ( m ^r, (f (k) - pi) (f (k) - p(2)) J + S(r, Af)

/    f ^(k+1) \      / Af - f ^(k) \

5 m r, -77т----- + m r, —77т-----

V ^, f ^(k) - P2) V f ^(k) - Pi)

⁵ m(r, Af ¹ P ) ^{+ S} (Г ^Af ) = N (r, Af ¹ P ) ^{+ S} (Г ^Af )• Af - P 1                  Af - P 2

Hence combining (25) and (26) we obtain

T(r, Af ) 5 2N (r,

v P) + S 'r^f '•

Next, case 2 is divided into two subcases.

Subcase 2.1. P i = 0 . Then by (7) and Lemma 1 we get

m(r, e^Q) = m

c f

= S (r, Af ).

Then by (16), (27) and (28) we can have T(r, Af ) = S (r, Af ) a contradiction. Subcase 2.2. P 2 = 0 . Then by (16), (27) and (28) and Lemma 1 we get

T ⁽r,^Af ) <  m^r, _A f - _{P i} ) ^{+ N}(^r'7¹ _k J) ^{+ S(r}, ^Af)

^ ^m(^r'7¹ _k т)⁺ N (^f^r)^{+ S(r},^Af) ^ ( ^T ( ^{r,f (k)} )) ^{+ S(r},^Af)•

From the fact that

T(r,f(k)) ^ T(r, Af)+ S(r, Af), which follows from (29) that

T(r, Af )= T(r,f ^(k) ) + S(r, Af ),

By second Nevanlinna Fundamental theorem, Lemma 1, (8) and (31) we have

2T (r, Af ) ^ 2T ( rf ^(k) ) + S(r, Af )

< N ( ^r, f ck) ¹ -^) ⁺ N ( r 7!b ) ⁺ N ( ^r, f i k T - d ) ^{+ S} (r, ^Af)

⁵ N ( r, Af - P i ) ⁺ N (j’ f ) ⁺ T ( r f^- d ) - m ( r "Й- d l ) ⁺ S (r, ^Af ) ^ 2T (r, Af ) - m rr, f (k) ¹- d ) + S(r, Af).

m ( rfwT-d ) = S^(r,^Af)                       ⁽³²⁾

From the First Fundamental Theorem, Lemma 1, (20) to (21), (31), (32) and Af is a transcendental entire function of finite order, we obtain m (r fd) 5m (r ffd) +m (r fk-d)+S(r Af)

• < T Л⁷! ) - N л.,⁷! ) + S (r,Af )

• ⁵ v • f ^(k) - d) V f ^(k) - d} ⁺ f )

= т^^^) + N^f) — N . ^) +)

• ⁵ N G v   ^- N (r ) ⁺ S(r, ^Af ) = ^T (^r- Af) ^{- T} G fw-d ) ⁺ S^Af>

= T (r, A) - T ( r, f ^(k) ) + S (r, Af) = S (r, Af).

Thus, we get m^Tf-d) = S (r,Af).

It’s easy to see that N (r/f) = S(r, Af ) and ((12)) can be rewritten as

_ Г p - - d (f /k+i)     df (k+1) 1 г f - d _ I

^    [  Pi   f(k) - P1 +  f(k)  ] [f(k) - d ‘                    (3)

Then by (33) and (34) we can get

T(r, ^) = m(r, ^) + N(r, ^) = S(r, Af).

By (7), (25) and (35) we get

N(r, AfJ-P1) = S(r, Af )•

Moreover, by (7), (31) and (36), we have

m(r,f1)) = S(r^Af)

which implies

N (r, Af) = m (r, Af - P1) ^ m (r, 71k)) = S(r,Af)•

Then by (7) we obtain T(r, Af ) = S (r, Af ) , a contradiction. So, by (12), (16), (27) and the Second Fundamental Theorem of Nevanlinna, we can get

T (r, Af) < 2m fr, _{A f} - P 1 ) + S(r, Af) < 2m. fr, f^ky) + S(r, Af)

5 ^2T ( r f < ^k) ) - 2N (r, f ^k )) + S (r, А/) 5 N (r, Af-P-) + N (r, Af-p-) + N^-ff k )) - ^2N (r-ffe) + ^S (r^,Af T (r^,Af) - N^f^ k )) + S(r^f).

N ( r-f® ) = S (r^f ).

It follows from the second theorem of Nevanlinna that

T ( r. f ^|k) ) < N (r, f^) + N ('. fj^) + ^S(r ^Af)

• * N G’ _f (k - P l ) + S(r. Af ) * T ( r. f ^№) + S(r. Af ).

which implies that

T ( r. f ^(k) ) = N (r. _{f |k)} - _{p -} ) + S (r. Af).

Similarly

T(r, f ^|k) ) = N (r. _{f |k)} - _p 2 ) + S(r. Af ).

Then, by (27), we get

T(r. Af ) = 2T ( r J ^|k) ) + S(r. Af).

By (25) and (26) we have

T(r.f) = T ( r.f ^|k) ) + S(r. Af).

When case 1 occurs, we apply Lemma 4 and obtain f|k) = Het.

Here H ф 0 is a small function of e t . Rewrite (16) as

, = f ^|k+1) (Af - P - )(Af - P 2 ) - Af ( f ^|k) - P - ) ( f ^|k) - P ₂)

^Ф         (Af - P - )(Af - P 2 ) ( f ^|k) - P - )( f ^|k) - P 2 )

Combining (27) with (44) we get

X = f ^lk+1) Af - P - )(Af - P 2 ) - Af ‘ ( f ^|k) - P l ) ( f ^|k) - P 2 ) = ^2§ i e it .       (46)

i=0

and

Y = (Af - P - )(Af - P 2 ) - Af ‘ ( f ^|k) - P - ) ( f ^|k) - P 2 ) = ^ Y j e jt .         (47)

j =0

where § i and Y j are small functions of e t , 6 5 ф 0 and Y 6 Ф 0 .

If X and Y are two mutually prime polynomials in e t , then we can get T (r. ф) = 6T (r. e t ) +

S (r. Af ) . It follows from (16), (41)-(43) that T (r. Af) , a contradiction.

If X and Y are not two mutually prime polynomials in e ^t , it’s easy to see that the degree of Y is large than X . Then submitting (38) into (12) implies

H = P 2

and t = P- z + P2.

where P l = 0 and P 2 are polynomials.

According to (45), (48), (49) and by simple calculation, we must have

C ф =    -----, f (k) - P2,V 7

where C is a non-zero constant. Put (44) into (16) we have c ((f )(k) - f(k+1) - Pi) _

(f(k) - Pi) (f(k) - P2) = (Af - Pi)(Af - P2).

We claim that f ^(k) = f ^(k+i) .

Otherwise, combining (22), (44) and (51) we can get T(r,e^t ) = S(r, Af ) . It follows from (16) and (27) that T(r, Af ) = S(r, Af) , a contradiction. Hence, it is a easy work to verify that

Pi = 1

and

f(k) = P2ez-P2 = Aez,(53)

where A is a nonzero constant and furthermore

Af = Ae2z - Pi Aez + Pi.

Then rewrite (27) as

Af - f ^(k) f ^(k) - P i

= e t - 1.

Put (49), (52)–(54) into (55) and a direct calculation deduces

A = P2 = eP1 = 1.

It follows from (1), (28), (52) and (56) that

H = -Pi(em - 1)n = 1.

Since Af and f(k) share P2 IM and (41), (42) and (56) we get e2z - Piez + (Pi - 1) = (ez - 1)2,

• i.    e.,

Pi = 2.

It follows from (57) that em = (-2)-i/n + 1.

But we cannot get (2) from (60), a contradiction. When case 2 occurs we know that m(r, e t ) = m(r, e ^Q ) + O(1) = S(r, Af) . Then by (16) and (27) we deduce T (r, Af) = S(r, Af) a contradiction. ⊲