Firstly, we present some lemmas which will be needed in the proof of Theorem 1.1.
Lemma 2.1 ()
Let , , and let be a meromorphic function of finite order. Then for any small periodic function with period c, with respect to ,
where the exceptional set associated with is of at most finite logarithmic measure.
Lemma 2.2 ()
Suppose that () and () () are entire functions satisfying
the orders of are less than that of for , ,
Proof of Theorem 1.1 Suppose on the contrary to the assertion that . Note that is a nonconstant entire function of finite order. By Lemma 2.1, for , we have
Since , , and share CM, we have
where and are polynomials.
From (2.1) and (2.2), we get . Then by supposition and (2.1), we see that . By Lemma 2.1, we deduce that
Note that . By using the second main theorem and (2.3), we have
Thus, by (2.3) and (2.4), we have . Similarly, .
Now we divide this proof into the following two steps.
Step 1. Suppose that is not a constant. Now we rewrite the second equation in (2.1) as
where , , .
We deduce that
Let be a finite set of n elements, and denote , where ∅ is an empty set. Then by an argument similar to the above, we deduce that
where A is any element of , , and , for , () are nonzero constants. In particular, , and
Moreover, is a polynomial of and its shifts .
Now set , where are constants satisfying and . Obviously, for , we have
By the above equation and (2.9), we obtain
Here , for , , are polynomials with degree less than m.
Rewrite the first equation in (2.1) as
This together with (2.8) gives
Notice that , , and . If , (2.12) yields
That is impossible.
Hence . This together with (2.11) gives
Now we distinguish three cases as follows:
Case (i). Suppose that . Then, for any , we see that
and for , we have
Since , for , , are polynomials with degree less than m, it is easy to see that, for ,
By Lemma 2.2, we have , which is impossible.
Case (ii). Suppose that . Then, by a similar argument to above, we can also get a contradiction.
Case (iii). Now suppose that . Set , with and . Rewrite (2.13) as
Subcase (i). If , for any , then we have
By this together with (2.14), (2.15), (2.16), and Lemma 2.2, we can get a contradiction.
Subcase (ii). If , for some . Without loss of generality, we assume that . Then we rewrite (2.16) as
By a similar method as the above, we can also get . That is impossible.
Subcase (iii). If , then we rewrite (2.16) as
By a similar argument to the above and Lemma 2.2, we can get
By (2.10) and (2.11), we get
Suppose that . Note that for , we have
where are polynomials with degree less than .
Rewrite (2.19) as
For any , we have
and for , we see that
By this, together with (2.20) and Lemma 2.2, we obtain , which is impossible.
Suppose that , then , with . It is easy to see that
This together with (2.19) gives
which yields . Therefore, for any ,
By the second equation in (2.1) and (2.21), we have
Rewriting (2.22), and combining it with the second equation in (2.1), we obtain
Substituting the first equation in (2.1) into (2.23), we get
If , (2.24) yields
We get a contradiction again.
Hence, . By (2.24), we see that , which implies . That is impossible.
Step 2. Suppose that is a constant. Now we rewrite the second equation in (2.1) as
Since is a periodic function with period c, we have
By Lemma 2.1 and the first equation in (2.1), we deduce that
From (2.27), we have
According to our assumption that and (2.25), it is easy to see that . Now suppose that is a zero of with multiplicity μ. Since , , and share CM, is a zero of and with multiplicity μ. Therefore, is a zero of with multiplicity at least μ. Then by (2.26) and (2.28), we see that
which implies . That is impossible.
Hence, we must have , and Theorem 1.1 is proved. □