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On some ultrametric q-difference equations
Bull. Sci. math. 137 (2013) 177–188 www.elsevier.com/locate/bulsci
On some ultrametric q-difference equations Nadjet Boudjerida a,∗ , Abdelbaki Boutabaa b , Samia Medjerab a a Laboratoire de Mathématiques (LMPA), Université de Jijel, Jijel, Algeria b Laboratoire de Mathématiques UMR 6620, Université Blaise Pascal (Clermont-Ferrand), Les Cézeaux,
63177 Aubiere Cedex, France Received 2 April 2010 Available online 31 May 2010
Abstract Let K be a complete ultrametric algebraically closed field and let M(K) be the field of meromorphic functions in all K. Let B(X), A0 (X), . . . , As (X) (s 1) be elements of K(X) such that A0 (X)As (X) = 0. paper is aimed to study functions f ∈ M(K) which are solutions of the functional equation: This s i i=0 Ai (x)(σq f )(x) = B(x), where q ∈ K, 0 < |q| < 1 and (σq f )(x) = f (qx). First we show that, if A0 (X), . . . , As (X), B(X) are constant, then f is a rational function. Next, we examine solutions of the above equation in the general case and give some characterizations of the order of growth of these solutions. © 2010 Elsevier Masson SAS. All rights reserved. MSC: 12J25; 46S10 Keywords: q-Difference equation; Ultrametric Nevanlinna theory
1. Introduction and results In the field C of complex numbers, the above equation was studied by several authors as Bergweiler, Ishizaki, Yanagihara, Hayman, Heittokangas, Laine, Rieppo, Yang, etc. (cf. [9–11,13]). Here we mean to generalize some of their results to meromorphic functions in a p-adic field (cf. [1,12,14]). As in the complex case, our method is based on the (ultrametric) theory of Nevanlinna
* Corresponding author.
E-mail addresses: [email protected] (N. Boudjerida), [email protected] (A. Boutabaa), [email protected] (S. Medjerab). 0007-4497/$ – see front matter © 2010 Elsevier Masson SAS. All rights reserved. doi:10.1016/j.bulsci.2010.05.002
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and follows fairly closely the approach of the article [9]. So we first have to recall some basic notions of this theory (cf. [6–8]). Let K be a complete ultrametric algebraically closed field. Given r > 0, we define the disks d(0, r) and d(0, r − ) by d(0, r) = {x ∈ K | |x| r} and d(0, r − ) = {x ∈ K | |x| < r}. We denote by A(K) the K-algebra of entire functions in K and by M(K) the field of meromorphic functions in K, i.e. the field of fractions of A(K). In the same way, we denote by A(d(0, R − )) the K-algebra of analytic functions inside the disk d(0, R − ), i.e. the set of power series converging inside d(0, R − ) and by M(d(0, R − )) the field of meromorphic functions in d(0, R − ), i.e. the field of fractions of A(d(0, R − )). For x > 0 we put log+ (x) = max(0, log x), where log is the real logarithm function. Let R > 0. For every r ∈ ]0, R[ we define a multiplicative norm | |(r) on A(d(0, R − )) by n n |f |(r) = supn0 |an |r for every function f (x) = n0 an x of A(d(0, R − )). We extend this
g − to M(d(0, R − )) by setting |f |(r) = |g|(r) |h|(r) for every element f = h of M(d(0, R )). Finally, for every f ∈ M(d(0,R − )) \ {0} and every α ∈ d(0, R − ), we denote by ωα (f ) the integer iα of Z such that f (x) = iiα ai (x − α)i and aiα = 0. The following property, which is an immediate consequence of the above definitions, is in fact the starting point of the ultrametric Nevanlinna theory (cf. [6]).
Proposition 1. Let R > 0 and let f ∈ M(d(0, R − )) be such that 0 is neither a zero nor a pole of f . Then, for every r ∈ ]0, R[, we have r ωα (f ) log . log |f |(r) = logf (0) + |α| |α|r
Let f ∈ M(d(0, R − )) be such that 0 is neither a zero nor a pole of f . For every r ∈ ]0, R[, we denote by Z(r, f ) and N (r, f ) the counting functions of zeros and poles of f in the disk d(0, r) defined by r 1 ωα (f ) log , and N (r, f ) = Z r, . Z(r, f ) = |α| f ωα (f )>0, |α|r
The Nevanlinna function T (r, f ) is then defined by T (r, f ) = N(r, f ) + log+ |f |(r). The following properties are easily checked (cf. [6]). Proposition 2. (i) Let f ∈ M(d(0, R − )) be such that 0 is neither a zero nor a pole of f . It is easily seen that, for every r ∈ ]0, R[, we have: 0 Z(r, f ) T (r, f ) and 0 N (r, f ) T (r, f ). (ii) Let f, g ∈ M(d(0, R − )) be such that f , g, f + g and f g have no zero and no pole at 0. Then, for every r ∈ ]0, R[, we have 0 Z(r, f + g) Z(r, f ) + Z(r, g), 0 N (r, f + g) N (r, f ) + N (r, g), 0 T (r, f + g) T (r, f ) + T (r, g),
0 Z(r, fg) Z(r, f ) + Z(r, g); 0 N (r, f g) N (r, f ) + N (r, g); 0 T (r, fg) T (r, f ) + T (r, g).
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(iii) Let f ∈ M(K). Then f belongs to K(X) if and only if T (r, f ) = O(log r), r → +∞. With the above notations, using the fact that log x = log+ x −log+ x1 , for x > 0, we can rewrite Proposition 1: Proposition 3. Let f ∈ M(d(0, R − )) such that f (0) = 0, ∞. Then, for every r ∈ ]0, R[, we have 1 = T (r, f ) + O(1), ∀r ∈ ]0, R[. T r, f Let us now return to our main problem. Let q ∈ K, 0 < |q| < 1 and let B(X), A0 (X), . . . , As (X) (s 1) be elements of K(X) such that A0 (X)As (X) = 0. We want to study the order of growth of meromorphic functions f in all K witch may be solutions of the functional equation (E)
s
Ai (x) σqi f (x) = B(x),
i=0
(σqi f )(x) := f (q i x),
where for i = 0, . . . , s. We will first examine what can happen if we assume that the functions A0 (X), . . . , As (X) above are constant. We will see that in this case every meromorphic solution f of Eq. (E) is a rational function. More precisely, we have Theorem 1. Suppose that, in Eq. (E) above, B(X) is a polynomial and that the coefficients A0 (X), . . . , As (X) are constant. Then every non-constant solution of Eq. (E) in M(K) is a rational function having at most one pole α = 0. This raises a question about the nature of meromorphic solutions of Eq. (E) when A0 (X), . . . , As (X) are not all constant. Does this equation admit non-rational solutions f ∈ M(K) and what can we say about the order of growth of such a solution? Example 1. Let q ∈ K, 0 < |q| < 1 and consider the so-called Tschakaloff function Tq (x) = n(n−1) 2 x n (cf. [4]). It is easily seen that Tq ∈ A(K)\K[X] and is a solution of the equation: n1 q y(x) − xσq y(x) = x. Moreover, by Proposition 2, we have log r = o(log |Tq |(r)), r → +∞, i.e. log r =o(T (r, Tq )), r → +∞. It is even possible to precise more the order of growth of this function. Indeed, let us consider the following sequences of positive numbers (rk )k1 defined, for every 1
k ∈ N∗ , by rk = |q|( 2 −k) so that k = For every fixed k ∈ N∗ , we have: k2
1
log rk −log |q| 2 . − log |q| n(n−2k) n |an |rk = |q| 2 .
We deduce that |Tq |(rk ) = |q|− 2 and hence 1
log |Tq |(rk ) =
1
k2 (log rk )2 − 2 log |q| 2 log rk + (log |q| 2 )2 . log |q|−1 = 2 2 log |q|−1
So we see that
T (rk , Tq ) = O (log rk )2 ,
k → +∞,
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and
(log rk )2 = O T (rk , Tq ) ,
k → +∞.
As the sequences (rk )k1 is increasing and tends to +∞, we deduce that T (r, Tq ) = O (log r)2 , r → +∞, and
(log r)2 = O T (r, Tq ) ,
r → +∞.
We will see in the following that the example above illustrates a general behavior of meromorphic solutions of Eq. (E). More precisely, we have Theorem 2. If f ∈ M(K) is a solution of Eq. (E), then we have T (r, f ) = O (log r)2 , r → +∞. Theorem 3. If f ∈ M(K) \ K(X) is a solution of Eq. (E), then we have (log r)2 = O T (r, f ) , r → +∞. 2. The proofs We will need the following lemma whose proof is easily checked. Lemma 1. For every f ∈ M(K), every r > 0 and every n ∈ N, we have: (1) (2) (3) (4)
|σqn f |(r) = |f |(|q|n r), m(r, σqn f ) = m(|q|n r, f ), N(r, σqn f ) = N (|q|n r, f ), T (r, σqn f ) = T (|q|r , f ).
Proof of Theorem 1. By hypothesis, we have E
s
Ai σqi f (x) = B(x),
i=0
where A0 , . . . , As are elements of K such that A0 As = 0 and B(X) is an element of K[X]. Let α = 0 be a pole of f . Then Eq. (E ) ensures that, for at least one index j1 among 1, . . . , s, the element α1 = q j1 α of K is a pole of f . Applying this reasoning again to α1 , we deduce that there exists j2 ∈ {1, . . . , s} such that α2 = q j2 α1 . Hence we construct a sequence α1 , α2 , . . . of poles of f tending to 0, a contradiction. Hence f has no poles different from 0. where ∈ N∗ and g ∈ A(K) with g(0) = 0. Suppose that 0 is a pole of f so that f (x) = g(x) x Substituting this in Eq. (E ) shows that g satisfies an equation of the same type: s Ai i x B(x) σ g (x) = . A0 q i q A0 i=0
So, without loss of generality, we may assume that f ∈ A(K) and thus we have
N. Boudjerida et al. / Bull. Sci. math. 137 (2013) 177–188
f (x) =
s
αi f q i x − β(x),
with αi =
i=1
Ai B(X) for i = 1, . . . , s and β(X) = . A0 A0
181
(1)
Hence, for every k ∈ N:
|f | |q|−k max |β| |q|−k , |α1 |f (qx) |q|−k , . . . , |αs |f q s x |q|−k . Using Lemma 1, we obtain
|f | |q|−k max |β| |q|−k , |α1 ||f | |q|1−k , . . . , |αs ||f | |q|s−k . Since |f |(|q|1−k ) · · · |f |(|q|s−k ), deduce that
|f | |q|−k max |β| |q|−k , λ|f | |q|1−k ,
with λ = max |αi |. 1is
In same way, we have the: |f |(|q|1−k ) max{|β|(|q|1−k ), λ|f |(|q|2−k )}. As |β|(|q|1−k ) |β|(|q|−k ), we have
|f | |q|1−k max |β| |q|−k , λ|f | |q|2−k .
(2)
(3)
From (2) and (3), we have |f |(|q|−k ) max{|β|(|q|−k ), λ2 |f |(|q|2−k )}. Repeating this reasoning k times we obtain
|f | |q|−k max |β| |q|−k , λk |f |(1) , and hence
log |f | |q|−k max log |β| |q|−k , k log λ + log |f |(1) .
(4)
Noting rk = |q|−k , the inequality above becomes log λ log |f |(rk ) max log |β|(rk ), log r + log |f |(1) . k log |q|−1
(5)
As (rk ) is an increasing sequence of positive numbers tending to +∞ and as β(X) ∈ K[X], we have log |β|(rk ) = O(log rk ). So it follows, by inequality (5) that log |f |(rk ) = O(log rk ),
k → +∞.
(6)
We deduce finally from (4) that log |f |(r) = O(log r),
r → +∞.
This means that f ∈ K[X] and completes the proof of Theorem 1.
2
Now consider the more general case and suppose that in Eq. (E), the coefficients B(X), A0 (X), . . . , As (X) are rational functions such that A0 (x)As (x) = 0 and A0 (X), . . . , As (X) are not all constant. A first observation is that in the present context, we can take B(X) = 0 withoutloss of generality. Indeed, if B(X) = 0, we have B(x) si=0 Ai (qx)f (q i+1 x) − B(qx) si=0 Ai (x)f (q i x) = 0, which is a nontrivial equation without second member and is
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satisfied by f . We may also assume that the Ai ’s are polynomials. So, from now on, we assume that Eq. (E) is of the form: E
s
Ai (x) σqi f (x) = 0,
i=0
where A0 (X), . . . , As (X) (s 1) are polynomials of K[X] such that A0 (X)As (X) = 0. The following result essentially repeats the statement of Proposition 7.2 in [5] and will be useful for the proof of Theorem 2 (cf. [2,3]). Proposition 4. Let f be a meromorphic solution in K of Eq. (E ). Then, if α is a non-zero pole of f , there exist an integer m ∈ N and a zero θ of A0 different from zero such that α = q −m θ and ωθ (A0 ) + ωθ (f ) 0. Proof. As, in the proof of Theorem 1, we may suppose that f has no pole at 0. If ωα (A0 ) + ωα (f ) 0, we are done because it suffices to take θ = α. Suppose that ωα (A0 ) + ωα (f ) < 0. This means that α is a pole of A0 (x)f (x), then there exists at least an index i1 ∈ {1, . . . , s} such that Ai1 (α)f (q i1 α) = ∞, and particularly α1 = q i1 α is a pole of f . If ωα1 (A0 ) + ωα1 (f ) < 0, we find in the same way an index i2 ∈ {1, . . . , s} such that α2 = q i2 α1 = q i1 +i2 α is a pole of f , etc. As we cannot have a sequence of poles of f with strictly decreasing modules, the above process must stop at a certain rank, and this completes the proof of our assertion. 2 Corollary 1. Suppose that, in Eq. (E ), the polynomial A0 (X) is of the form aX m with a ∈ K \ {0} and m ∈ N . Then every solution f ∈ M(K) of Eq. (E ) (without no pole at the origin) is an entire function. Proof. Indeed, such a solution cannot have non-zero poles, and hence is entire.
2
Proof of Theorem 2. As indicated above, the whole problem is reduced to the case of equation E
s
Ai (x) σqi f (x) = 0.
i=0
Let f ∈ M(K) be a solution of Eq. (E). We may also suppose that f has no pole at the origin. Let us first estimate N (f, r). If the polynomial A0 (X) is of the form aX m , then by Corollary 1, the function f ∈ M(K) is entire and hence N (f, r) = 0. Suppose then that A0 (X) admits at least one zero different from 0 and let Z = {x ∈ K \ {0} | A0 (x) = 0}, ρ = min{|x| | x ∈ Z}. For every r > 0, large enough, we have r N (r, f ) = − ωα (f ) log . |α| 0<|α|r, f (α)=∞
By Proposition 4, we have ρ min{|x| | f (x) = ∞} and hence r ωα (f ) log . N(r, f ) − ρ 0<|α|r, f (α)=∞
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183
But, by Proposition 4 again, every pole α of f in d(0, r) \ {0} is of the form α = q −n β with some β ∈ Z and some n ∈ N. This implies that
ρ 1 log n log |q| r (where [t] denotes the integer part of the real number t). Hence the α’s occurring in the sum 0<|α|r, f (α)=∞ ωα (f ) are to take among the elements of the set
ρ 1 −n q β β ∈ Z0 and 0 n log log |q| r whose cardinality is upper-bounded by
ρ 1 log × card Z. 1+ log |q| r Using this, we have
r 1 ρ × 1+ log × card Z ρ log |q| r 1 ρ r log log × 1 + × card Z. ρ log |q| r
N(r, f ) log
It follows that N(r, f ) = O (log r)2 ,
r → +∞.
(1)
Let us now estimate log |f |(r). We may, without loss of generality, suppose that f has neither a zero nor a pole at the origin. Hence there exists τ > 0 such that f has no zeros and no poles in d(0, τ ), so that |f |(t) is constant for 0 t τ . From Eq. (E), we have for every r > 0: A1 As A2 |f |(r) max (r)|f | |q|r , (r)|f | |q|2 r , . . . , (r)|f | |q|s r . A0 A0 A0 So there exists λ > 1 such that, for r > 0 large enough, we have
|f |(r) r λ max |f | |q|r , |f | |q|2 r , . . . , |f | |q|s r .
(2)
Now if r > 0 is large enough, so is the integer
log r − log τ +1 k= − log |q| (particularly k s). Hence, by (2) we have
|f |(r) r λ max |f | |q|r , |f | |q|2 r , . . . , |f | |q|s r , . . . , |f | |q|k r .
(3)
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Let us set ⎧ ⎪ μ1 = |f | |q|k r , ⎪ ⎪ ⎪
⎪ ⎪ μ2 = max |f | |q|k−1 r , |f | |q|k r , ⎪ ⎪ ⎨ .. . ⎪ ⎪
⎪ ⎪ ⎪ μk−1 = max |f | |q|2 r , . . . , |f | |q|s r , . . . , |f | |q|k r , ⎪ ⎪ ⎪
⎩ μk = max |f | |q|r , |f | |q|2 r , . . . , |f | |q|s r , . . . , |f | |q|k r .
(4)
Hence, (3) becomes |f |(r) r λ μk .
(5) |q|k r
On the other hand, we have: |q|τ < τ . Hence using the fact that |f |(t) is constant for 0 t τ , we have |f | |q|k r = |f | |q|k+1 r = |f | |q|k+2 r = · · · = μ1 = C = Constant. (6) Then replacing r in (3) successively by |q|r, |q|2 r, . . . , |q|k−1 r we obtain ⎧ λ |f | |q|r |q|r μk−1 , ⎪ ⎪ ⎪ ⎪ λ ⎪ ⎪ |f | |q|2 r |q|2 r μk−2 , ⎪ ⎪ ⎨ .. . ⎪ ⎪ k−2 k−2 λ ⎪ ⎪ ⎪ |f | |q| r |q| r μ2 , ⎪ ⎪ ⎪ ⎩ k−1 k−1 λ |f | |q| r |q| r μ1 . It follows by (4) and (7) that for r > 0 large enough ⎧ μ = C, 1 ⎪ ⎪ λ ⎪ ⎪ ⎪ μ2 |q|k−1 r μ1 , ⎪ ⎪ ⎨ .. . ⎪ ⎪ ⎪ ⎪ μk−1 |q|2 r λ μk−2 , ⎪ ⎪ ⎪ λ ⎩ μk |q|r μk−1 .
(7)
(8)
From (5) and (8) we have λ λ |f |(r) r λ r|q| r|q|2 r|q|k−1 C and hence |f |(r) r kλ |q|
k(k−1) 2 λ
(9)
C.
It easily follows from this that log+ |f |(r) = O (log r)2 ,
r → +∞.
Finally, relations (1) and (10) yield T (r, f ) = O (log r)2 , r → +∞. This completes the proof of Theorem 2.
(10) (11)
2
The following lemmas will be useful for the proof of Theorem 3.
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185
Lemma 2. Let f ∈ A(K) \ K[X] and λ > 1. Then we have |f |(r) = 0. r→+∞ |f |(λr) lim
Proof. Let (an )n0 be the sequence of distinct zeros of f ranged so that, for every n ∈ N, |an | |an+1 | and limn→+∞ |an | = +∞. For every n ∈ N, let kn be the order of an as a zero of f . So f (x) = A n0 (1 − axn )kn . For r > 0 large enough, there exists a unique index as nr such that |anr | < r |anr +1 |. Then we have |f |(r) =
|A| r k0 +···+knr . |a0 . . . anr |
(1)
On the other hand, as λ > 1, we have λr > |anr | and hence |f |(λr)
|A| (λr)k0 +···+knr . |a0 . . . anr |
It follows from (1) that |f |(λr) (λ)k0 +···+knr |f |(r). As k0 1, k1 1, . . . , knr 1, it follows that |f |(λr) λnr +1 |f |(r).
(2)
Hence |f |(r) 1 n +1 , |f |(λr) λ r
and
lim
r→+∞
This completes the proof of Lemma 2.
|f |(r) 1 = lim n +1 = 0. r→+∞ |f |(λr) λ r 2
Lemma 3. Let f ∈ M(K) be a solution of equation
E
s
Ai (x) σqi f (x) = 0.
i=0
Let R > 0 be such that all the zeros of the polynomial A0 (X) belong to d(0, R − ). Then the following statements are equivalent: (i) f admits at least a pole in K \ d(0, R − ), (ii) f admits infinitely many poles, (iii) for every real number r R, f admits at least a pole in d(0, r|q|−s ) \ d(0, r). Proof. Clearly, it is sufficient to show that (i) ⇒ (ii) and (ii) ⇒ (iii). Suppose that α is a pole of f in K \ d(0, R − ) and r R. Let us set β = q −s α. From (E), we have si=0 Ai (β)(σqi f )(β) = 0. But β, qβ, q 2 β, . . . , q s β belong all to K \ d(0, R − ) and q s β is a pole of f . So there exists an index j1 ∈ {0, . . . , s − 1} such that α1 := q j1 β = q j1 −s α is a pole of f . Repeating this reasoning for α1 , we obtain an other pole α2 of f . In this way, we construct | = |q|jn+1 −s |q|−1 > 1. a sequence (αn )n0 of poles of f satisfying, for every n 0, | ααn+1 n Hence (i) ⇒ (ii). Suppose now that f has infinitely many poles. Let r R and let α be a pole of f such that |α| > r. If |α| r|q|−s , we are done. If not, there exists an index j1 , 1 j1 s such that β := q j1 α is a pole of f ; and we have: |α| > |β| = |q|j1 |α| > r|q|j1 −s r. If |β| r|q|−s , we
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are done. If not, we apply this reasoning again to β instead of α. This means that (ii) ⇒ (iii) and completes the proof of Lemma 3. 2 Proof of Theorem 3. We may consider equation E
s
Ai (x) σqi f (x) = 0,
i=0
with polynomials A0 (X), . . . , As (X) such that A0 (X)As (X) = 0. Let R > 0 be such that the disk d(0, R − ) contains all the zeros A0 (X). Let f ∈ M(K) be a solution of Eq. (E). We will distinguish two situations: (I) Suppose first that f admits infinitely many poles. For every integer k 0, we set rk = R|q|−ks so that we have log rk − log R k= . log |q|−s
(1)
By Lemma 3, there exists a sequence (αk )k0 of poles of f such that rk < |αk | rk+1 , for every k ∈ N. Then we have rk rk N(rk , f ) log log = log |q|(i−k)s |αi | ri 0i
It follows that k(k + 1) log |q|−s . 2 By (1) and (2), we deduce that (log rk )2 = O N (rk , f ) , k → +∞. N(rk , f )
(2)
As the sequences (rk )k1 is increasing and tends to +∞, we deduce that (log r)2 = O N (r, f ) , r → +∞. Since N(r, f ) T (r, f ), we finally have (log r)2 = O(T (r, f )), r → +∞. (II) Suppose now that f only has a finite number of poles. Multiplying by a suitable rational function we may assume, without loss of generality, that f is a non-polynomial entire function. We set di = deg Ai (X) for i = 0, . . . , s, so that Ai (X) = ai X di + · · · , where a0 , a1 , . . . , as are elements of K such that a0 as = 0. Let d = max0is di 1 and let be the smallest index among 0, . . . , s, such that d = d. Then, by (E), we have Ai (x) σqi f (x). A (x) σq f (x) = − 0is, i=
Thus, for r > 0 large enough, we have
Ai σ i f (r) A σ f (r) max q
0is, i=
q
N. Boudjerida et al. / Bull. Sci. math. 137 (2013) 177–188
and hence
|a |r d |f | r|q|
max
0is, i=
|ai |r di |f | r|q|i .
187
(3)
If j = is an index among 0, . . . , s for which the above maximum is reached, we have |a |r d |f | r|q| |aj |r dj |f | r|q|j . Hence |f |(r|q| ) |aj | dj −d r . |f |(r|q|j ) |a | As dj d, the quantity > j,
|aj | dj −d |a | r
(4) is bounded when r approaches +∞. So, by Lemma 2, we have
and hence 1.
By (3) and (5), we have |f | r|q|
max
0i−1
|ai | di −d i r |f | r|q| . |a |
(5)
(6)
Replacing r by rk = R|q|−ks in the inequality (6), we obtain |ai | di −d |ai | |f |(rk ) |f |(rk−1 ) max rk |f | rk |q|i max . rk 0i−1 |a | 0i−1 |a | Hence we have |f |(rk ) Ark |f |(rk−1 ), with some positive constant A. k(k+1) It follows that |f |(rk ) Ak ( kj =0 rj )|f |(R) = Ak R k+1 |q|−s 2 |f |(R) and hence log |f |(rk )
k(k + 1) log |q|−s + k log A + (k + 1) log R + log |f |(R). 2
(7)
(8)
We deduce, from (1) and (8) that (log rk )2 = O(log |f |(rk )), k → +∞. Hence (log r)2 = O(log |f |(r)), r → +∞. This completes the proof of Theorem 3. 2 All the results above are not aimed at discussing the existence of non-rational solutions of Eq. (E). However, the proof of Theorem 3 provides some necessary conditions for this to happen. More precisely, we have the following corollary. Corollary 2. Let us consider the equation (E)
s
Ai (x) σqi f (x) = B(x),
i=0
where q ∈ K, 0 < |q| < 1 and B(X), A0 (X), . . . , As (X) (s 1) be elements of K[X] such that A0 (X)As (X) = 0. Suppose that deg A0 (X) = max0is deg Ai (X). Then: (i) Every entire solution of Eq. (E) is a polynomial. (ii) Every meromorphic solution of Eq. (E) which has finitely many poles is a rational function.
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Example 2. Let q ∈ K, 0 < |q| < 1. Then every entire solution of the equation x 2 y(x) + xσq y(x) = x 3 + 1 is a polynomial. Remarks. (1) We can recover Theorem 1 as an immediate consequence of Corollary 2. Indeed, we have seen that f has no poles different from 0. On the other hand, from Eq. (E ), we have B(x)
s i=0
s Ai f q i+1 x − B(qx) Ai f q i x = 0.
s+1
i=0
This can be written as i=0 Ci (x)f (q i x) = 0, with C0 (X) = A0 B(qX), Cs+1 (X) = As B(X) and Ci (X) = Ai−1 B(X) − Ai B(qX) for 1 i s. It is then clear that deg C0 (X) = max{deg Ci (X) | 0 i s + 1}. (2) More generally, more technical calculation may enable us to reduce our study to the case of entire solutions of q-difference equations using only the maximum modulus. Here we have deliberately chosen to use systematically the ultrametric Nevanlinna theory so that the obtained results appear as applications of this latter. References [1] Y. Amice, Les nombres p-adiques, PUF, 1975. [2] J.-P. Bézivin, Sur les équations fonctionnelles aux q-différences, Aequationes Math. 43 (2–3) (1992) 159–176. [3] J.-P. Bézivin, Indépendance linéaire et algébrique de fonctions liées à la fonction q-dzéta, Ann. Fac. Sci. Toulouse 17 (1) (2008) 23–36. [4] J.-P. Bézivin, Fonction de Tschakaloff et Fonction q-exponentielle, preprint. [5] J.-P. Bézivin, A. Boutabaa, Sur les équations fonctionelles p-adiques aux q-différences, Collect. Math. 43 (2) (1992) 125–140. [6] A. Boutabaa, Théorie de Nevanlinna p-adique, Manuscripta Math. 67 (1990) 251–269. [7] A. Boutabaa, Applications de la théorie de Nevanlinna p-adique, Collect. Math. 42 (1) (1991) 75–93. [8] A. Boutabaa, A. Escassut, Applications of the p-adic Nevanlinna theory to functional equations, Ann. Inst. Fourier 50 (3) (2000) 751–766. [9] W. Bergweiler, K. Ishizaki, N. Yanagihara, Meromorphic solutions of some functional equations, Methods Appl. Anal. 5 (1998) 248–258. [10] W. Bergweiler, K. Ishizaki, N. Yanagihara, Growth of meromorphic solutions of some functional equations, I, Aequationes Math. 63 (2002) 140–151. [11] W. Bergweiler, W.K. Hayman, Zeros of solutions of a functional equation, Comput. Methods Funct. Theory 3 (1) (2003) 55–78. [12] A. Escassut, Analytic Elements in p-Adic Analysis, World Scientific Publishing, Singapore, 1995. [13] J. Heittokangas, I. Laine, J. Rieppo, D. Yang, Meromorphic solutions of some linear functional equations, Aequationes Math. 60 (2000) 148–166. [14] A. Robert, A Course in p-Adic Analysis, Grad. Texts in Math., vol. 198, Springer-Verlag, 2000.
On some ultrametric q-difference equations Nadjet Boudjerida a,∗ , Abdelbaki Boutabaa b , Samia Medjerab a a Laboratoire de Mathématiques (LMPA), Université de Jijel, Jijel, Algeria b Laboratoire de Mathématiques UMR 6620, Université Blaise Pascal (Clermont-Ferrand), Les Cézeaux,
63177 Aubiere Cedex, France Received 2 April 2010 Available online 31 May 2010
Abstract Let K be a complete ultrametric algebraically closed field and let M(K) be the field of meromorphic functions in all K. Let B(X), A0 (X), . . . , As (X) (s 1) be elements of K(X) such that A0 (X)As (X) = 0. paper is aimed to study functions f ∈ M(K) which are solutions of the functional equation: This s i i=0 Ai (x)(σq f )(x) = B(x), where q ∈ K, 0 < |q| < 1 and (σq f )(x) = f (qx). First we show that, if A0 (X), . . . , As (X), B(X) are constant, then f is a rational function. Next, we examine solutions of the above equation in the general case and give some characterizations of the order of growth of these solutions. © 2010 Elsevier Masson SAS. All rights reserved. MSC: 12J25; 46S10 Keywords: q-Difference equation; Ultrametric Nevanlinna theory
1. Introduction and results In the field C of complex numbers, the above equation was studied by several authors as Bergweiler, Ishizaki, Yanagihara, Hayman, Heittokangas, Laine, Rieppo, Yang, etc. (cf. [9–11,13]). Here we mean to generalize some of their results to meromorphic functions in a p-adic field (cf. [1,12,14]). As in the complex case, our method is based on the (ultrametric) theory of Nevanlinna
* Corresponding author.
E-mail addresses: [email protected] (N. Boudjerida), [email protected] (A. Boutabaa), [email protected] (S. Medjerab). 0007-4497/$ – see front matter © 2010 Elsevier Masson SAS. All rights reserved. doi:10.1016/j.bulsci.2010.05.002
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and follows fairly closely the approach of the article [9]. So we first have to recall some basic notions of this theory (cf. [6–8]). Let K be a complete ultrametric algebraically closed field. Given r > 0, we define the disks d(0, r) and d(0, r − ) by d(0, r) = {x ∈ K | |x| r} and d(0, r − ) = {x ∈ K | |x| < r}. We denote by A(K) the K-algebra of entire functions in K and by M(K) the field of meromorphic functions in K, i.e. the field of fractions of A(K). In the same way, we denote by A(d(0, R − )) the K-algebra of analytic functions inside the disk d(0, R − ), i.e. the set of power series converging inside d(0, R − ) and by M(d(0, R − )) the field of meromorphic functions in d(0, R − ), i.e. the field of fractions of A(d(0, R − )). For x > 0 we put log+ (x) = max(0, log x), where log is the real logarithm function. Let R > 0. For every r ∈ ]0, R[ we define a multiplicative norm | |(r) on A(d(0, R − )) by n n |f |(r) = supn0 |an |r for every function f (x) = n0 an x of A(d(0, R − )). We extend this
g − to M(d(0, R − )) by setting |f |(r) = |g|(r) |h|(r) for every element f = h of M(d(0, R )). Finally, for every f ∈ M(d(0,R − )) \ {0} and every α ∈ d(0, R − ), we denote by ωα (f ) the integer iα of Z such that f (x) = iiα ai (x − α)i and aiα = 0. The following property, which is an immediate consequence of the above definitions, is in fact the starting point of the ultrametric Nevanlinna theory (cf. [6]).
Proposition 1. Let R > 0 and let f ∈ M(d(0, R − )) be such that 0 is neither a zero nor a pole of f . Then, for every r ∈ ]0, R[, we have r ωα (f ) log . log |f |(r) = logf (0) + |α| |α|r
Let f ∈ M(d(0, R − )) be such that 0 is neither a zero nor a pole of f . For every r ∈ ]0, R[, we denote by Z(r, f ) and N (r, f ) the counting functions of zeros and poles of f in the disk d(0, r) defined by r 1 ωα (f ) log , and N (r, f ) = Z r, . Z(r, f ) = |α| f ωα (f )>0, |α|r
The Nevanlinna function T (r, f ) is then defined by T (r, f ) = N(r, f ) + log+ |f |(r). The following properties are easily checked (cf. [6]). Proposition 2. (i) Let f ∈ M(d(0, R − )) be such that 0 is neither a zero nor a pole of f . It is easily seen that, for every r ∈ ]0, R[, we have: 0 Z(r, f ) T (r, f ) and 0 N (r, f ) T (r, f ). (ii) Let f, g ∈ M(d(0, R − )) be such that f , g, f + g and f g have no zero and no pole at 0. Then, for every r ∈ ]0, R[, we have 0 Z(r, f + g) Z(r, f ) + Z(r, g), 0 N (r, f + g) N (r, f ) + N (r, g), 0 T (r, f + g) T (r, f ) + T (r, g),
0 Z(r, fg) Z(r, f ) + Z(r, g); 0 N (r, f g) N (r, f ) + N (r, g); 0 T (r, fg) T (r, f ) + T (r, g).
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179
(iii) Let f ∈ M(K). Then f belongs to K(X) if and only if T (r, f ) = O(log r), r → +∞. With the above notations, using the fact that log x = log+ x −log+ x1 , for x > 0, we can rewrite Proposition 1: Proposition 3. Let f ∈ M(d(0, R − )) such that f (0) = 0, ∞. Then, for every r ∈ ]0, R[, we have 1 = T (r, f ) + O(1), ∀r ∈ ]0, R[. T r, f Let us now return to our main problem. Let q ∈ K, 0 < |q| < 1 and let B(X), A0 (X), . . . , As (X) (s 1) be elements of K(X) such that A0 (X)As (X) = 0. We want to study the order of growth of meromorphic functions f in all K witch may be solutions of the functional equation (E)
s
Ai (x) σqi f (x) = B(x),
i=0
(σqi f )(x) := f (q i x),
where for i = 0, . . . , s. We will first examine what can happen if we assume that the functions A0 (X), . . . , As (X) above are constant. We will see that in this case every meromorphic solution f of Eq. (E) is a rational function. More precisely, we have Theorem 1. Suppose that, in Eq. (E) above, B(X) is a polynomial and that the coefficients A0 (X), . . . , As (X) are constant. Then every non-constant solution of Eq. (E) in M(K) is a rational function having at most one pole α = 0. This raises a question about the nature of meromorphic solutions of Eq. (E) when A0 (X), . . . , As (X) are not all constant. Does this equation admit non-rational solutions f ∈ M(K) and what can we say about the order of growth of such a solution? Example 1. Let q ∈ K, 0 < |q| < 1 and consider the so-called Tschakaloff function Tq (x) = n(n−1) 2 x n (cf. [4]). It is easily seen that Tq ∈ A(K)\K[X] and is a solution of the equation: n1 q y(x) − xσq y(x) = x. Moreover, by Proposition 2, we have log r = o(log |Tq |(r)), r → +∞, i.e. log r =o(T (r, Tq )), r → +∞. It is even possible to precise more the order of growth of this function. Indeed, let us consider the following sequences of positive numbers (rk )k1 defined, for every 1
k ∈ N∗ , by rk = |q|( 2 −k) so that k = For every fixed k ∈ N∗ , we have: k2
1
log rk −log |q| 2 . − log |q| n(n−2k) n |an |rk = |q| 2 .
We deduce that |Tq |(rk ) = |q|− 2 and hence 1
log |Tq |(rk ) =
1
k2 (log rk )2 − 2 log |q| 2 log rk + (log |q| 2 )2 . log |q|−1 = 2 2 log |q|−1
So we see that
T (rk , Tq ) = O (log rk )2 ,
k → +∞,
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and
(log rk )2 = O T (rk , Tq ) ,
k → +∞.
As the sequences (rk )k1 is increasing and tends to +∞, we deduce that T (r, Tq ) = O (log r)2 , r → +∞, and
(log r)2 = O T (r, Tq ) ,
r → +∞.
We will see in the following that the example above illustrates a general behavior of meromorphic solutions of Eq. (E). More precisely, we have Theorem 2. If f ∈ M(K) is a solution of Eq. (E), then we have T (r, f ) = O (log r)2 , r → +∞. Theorem 3. If f ∈ M(K) \ K(X) is a solution of Eq. (E), then we have (log r)2 = O T (r, f ) , r → +∞. 2. The proofs We will need the following lemma whose proof is easily checked. Lemma 1. For every f ∈ M(K), every r > 0 and every n ∈ N, we have: (1) (2) (3) (4)
|σqn f |(r) = |f |(|q|n r), m(r, σqn f ) = m(|q|n r, f ), N(r, σqn f ) = N (|q|n r, f ), T (r, σqn f ) = T (|q|r , f ).
Proof of Theorem 1. By hypothesis, we have E
s
Ai σqi f (x) = B(x),
i=0
where A0 , . . . , As are elements of K such that A0 As = 0 and B(X) is an element of K[X]. Let α = 0 be a pole of f . Then Eq. (E ) ensures that, for at least one index j1 among 1, . . . , s, the element α1 = q j1 α of K is a pole of f . Applying this reasoning again to α1 , we deduce that there exists j2 ∈ {1, . . . , s} such that α2 = q j2 α1 . Hence we construct a sequence α1 , α2 , . . . of poles of f tending to 0, a contradiction. Hence f has no poles different from 0. where ∈ N∗ and g ∈ A(K) with g(0) = 0. Suppose that 0 is a pole of f so that f (x) = g(x) x Substituting this in Eq. (E ) shows that g satisfies an equation of the same type: s Ai i x B(x) σ g (x) = . A0 q i q A0 i=0
So, without loss of generality, we may assume that f ∈ A(K) and thus we have
N. Boudjerida et al. / Bull. Sci. math. 137 (2013) 177–188
f (x) =
s
αi f q i x − β(x),
with αi =
i=1
Ai B(X) for i = 1, . . . , s and β(X) = . A0 A0
181
(1)
Hence, for every k ∈ N:
|f | |q|−k max |β| |q|−k , |α1 |f (qx) |q|−k , . . . , |αs |f q s x |q|−k . Using Lemma 1, we obtain
|f | |q|−k max |β| |q|−k , |α1 ||f | |q|1−k , . . . , |αs ||f | |q|s−k . Since |f |(|q|1−k ) · · · |f |(|q|s−k ), deduce that
|f | |q|−k max |β| |q|−k , λ|f | |q|1−k ,
with λ = max |αi |. 1is
In same way, we have the: |f |(|q|1−k ) max{|β|(|q|1−k ), λ|f |(|q|2−k )}. As |β|(|q|1−k ) |β|(|q|−k ), we have
|f | |q|1−k max |β| |q|−k , λ|f | |q|2−k .
(2)
(3)
From (2) and (3), we have |f |(|q|−k ) max{|β|(|q|−k ), λ2 |f |(|q|2−k )}. Repeating this reasoning k times we obtain
|f | |q|−k max |β| |q|−k , λk |f |(1) , and hence
log |f | |q|−k max log |β| |q|−k , k log λ + log |f |(1) .
(4)
Noting rk = |q|−k , the inequality above becomes log λ log |f |(rk ) max log |β|(rk ), log r + log |f |(1) . k log |q|−1
(5)
As (rk ) is an increasing sequence of positive numbers tending to +∞ and as β(X) ∈ K[X], we have log |β|(rk ) = O(log rk ). So it follows, by inequality (5) that log |f |(rk ) = O(log rk ),
k → +∞.
(6)
We deduce finally from (4) that log |f |(r) = O(log r),
r → +∞.
This means that f ∈ K[X] and completes the proof of Theorem 1.
2
Now consider the more general case and suppose that in Eq. (E), the coefficients B(X), A0 (X), . . . , As (X) are rational functions such that A0 (x)As (x) = 0 and A0 (X), . . . , As (X) are not all constant. A first observation is that in the present context, we can take B(X) = 0 withoutloss of generality. Indeed, if B(X) = 0, we have B(x) si=0 Ai (qx)f (q i+1 x) − B(qx) si=0 Ai (x)f (q i x) = 0, which is a nontrivial equation without second member and is
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satisfied by f . We may also assume that the Ai ’s are polynomials. So, from now on, we assume that Eq. (E) is of the form: E
s
Ai (x) σqi f (x) = 0,
i=0
where A0 (X), . . . , As (X) (s 1) are polynomials of K[X] such that A0 (X)As (X) = 0. The following result essentially repeats the statement of Proposition 7.2 in [5] and will be useful for the proof of Theorem 2 (cf. [2,3]). Proposition 4. Let f be a meromorphic solution in K of Eq. (E ). Then, if α is a non-zero pole of f , there exist an integer m ∈ N and a zero θ of A0 different from zero such that α = q −m θ and ωθ (A0 ) + ωθ (f ) 0. Proof. As, in the proof of Theorem 1, we may suppose that f has no pole at 0. If ωα (A0 ) + ωα (f ) 0, we are done because it suffices to take θ = α. Suppose that ωα (A0 ) + ωα (f ) < 0. This means that α is a pole of A0 (x)f (x), then there exists at least an index i1 ∈ {1, . . . , s} such that Ai1 (α)f (q i1 α) = ∞, and particularly α1 = q i1 α is a pole of f . If ωα1 (A0 ) + ωα1 (f ) < 0, we find in the same way an index i2 ∈ {1, . . . , s} such that α2 = q i2 α1 = q i1 +i2 α is a pole of f , etc. As we cannot have a sequence of poles of f with strictly decreasing modules, the above process must stop at a certain rank, and this completes the proof of our assertion. 2 Corollary 1. Suppose that, in Eq. (E ), the polynomial A0 (X) is of the form aX m with a ∈ K \ {0} and m ∈ N . Then every solution f ∈ M(K) of Eq. (E ) (without no pole at the origin) is an entire function. Proof. Indeed, such a solution cannot have non-zero poles, and hence is entire.
2
Proof of Theorem 2. As indicated above, the whole problem is reduced to the case of equation E
s
Ai (x) σqi f (x) = 0.
i=0
Let f ∈ M(K) be a solution of Eq. (E). We may also suppose that f has no pole at the origin. Let us first estimate N (f, r). If the polynomial A0 (X) is of the form aX m , then by Corollary 1, the function f ∈ M(K) is entire and hence N (f, r) = 0. Suppose then that A0 (X) admits at least one zero different from 0 and let Z = {x ∈ K \ {0} | A0 (x) = 0}, ρ = min{|x| | x ∈ Z}. For every r > 0, large enough, we have r N (r, f ) = − ωα (f ) log . |α| 0<|α|r, f (α)=∞
By Proposition 4, we have ρ min{|x| | f (x) = ∞} and hence r ωα (f ) log . N(r, f ) − ρ 0<|α|r, f (α)=∞
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183
But, by Proposition 4 again, every pole α of f in d(0, r) \ {0} is of the form α = q −n β with some β ∈ Z and some n ∈ N. This implies that
ρ 1 log n log |q| r (where [t] denotes the integer part of the real number t). Hence the α’s occurring in the sum 0<|α|r, f (α)=∞ ωα (f ) are to take among the elements of the set
ρ 1 −n q β β ∈ Z0 and 0 n log log |q| r whose cardinality is upper-bounded by
ρ 1 log × card Z. 1+ log |q| r Using this, we have
r 1 ρ × 1+ log × card Z ρ log |q| r 1 ρ r log log × 1 + × card Z. ρ log |q| r
N(r, f ) log
It follows that N(r, f ) = O (log r)2 ,
r → +∞.
(1)
Let us now estimate log |f |(r). We may, without loss of generality, suppose that f has neither a zero nor a pole at the origin. Hence there exists τ > 0 such that f has no zeros and no poles in d(0, τ ), so that |f |(t) is constant for 0 t τ . From Eq. (E), we have for every r > 0: A1 As A2 |f |(r) max (r)|f | |q|r , (r)|f | |q|2 r , . . . , (r)|f | |q|s r . A0 A0 A0 So there exists λ > 1 such that, for r > 0 large enough, we have
|f |(r) r λ max |f | |q|r , |f | |q|2 r , . . . , |f | |q|s r .
(2)
Now if r > 0 is large enough, so is the integer
log r − log τ +1 k= − log |q| (particularly k s). Hence, by (2) we have
|f |(r) r λ max |f | |q|r , |f | |q|2 r , . . . , |f | |q|s r , . . . , |f | |q|k r .
(3)
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Let us set ⎧ ⎪ μ1 = |f | |q|k r , ⎪ ⎪ ⎪
⎪ ⎪ μ2 = max |f | |q|k−1 r , |f | |q|k r , ⎪ ⎪ ⎨ .. . ⎪ ⎪
⎪ ⎪ ⎪ μk−1 = max |f | |q|2 r , . . . , |f | |q|s r , . . . , |f | |q|k r , ⎪ ⎪ ⎪
⎩ μk = max |f | |q|r , |f | |q|2 r , . . . , |f | |q|s r , . . . , |f | |q|k r .
(4)
Hence, (3) becomes |f |(r) r λ μk .
(5) |q|k r
On the other hand, we have: |q|τ < τ . Hence using the fact that |f |(t) is constant for 0 t τ , we have |f | |q|k r = |f | |q|k+1 r = |f | |q|k+2 r = · · · = μ1 = C = Constant. (6) Then replacing r in (3) successively by |q|r, |q|2 r, . . . , |q|k−1 r we obtain ⎧ λ |f | |q|r |q|r μk−1 , ⎪ ⎪ ⎪ ⎪ λ ⎪ ⎪ |f | |q|2 r |q|2 r μk−2 , ⎪ ⎪ ⎨ .. . ⎪ ⎪ k−2 k−2 λ ⎪ ⎪ ⎪ |f | |q| r |q| r μ2 , ⎪ ⎪ ⎪ ⎩ k−1 k−1 λ |f | |q| r |q| r μ1 . It follows by (4) and (7) that for r > 0 large enough ⎧ μ = C, 1 ⎪ ⎪ λ ⎪ ⎪ ⎪ μ2 |q|k−1 r μ1 , ⎪ ⎪ ⎨ .. . ⎪ ⎪ ⎪ ⎪ μk−1 |q|2 r λ μk−2 , ⎪ ⎪ ⎪ λ ⎩ μk |q|r μk−1 .
(7)
(8)
From (5) and (8) we have λ λ |f |(r) r λ r|q| r|q|2 r|q|k−1 C and hence |f |(r) r kλ |q|
k(k−1) 2 λ
(9)
C.
It easily follows from this that log+ |f |(r) = O (log r)2 ,
r → +∞.
Finally, relations (1) and (10) yield T (r, f ) = O (log r)2 , r → +∞. This completes the proof of Theorem 2.
(10) (11)
2
The following lemmas will be useful for the proof of Theorem 3.
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185
Lemma 2. Let f ∈ A(K) \ K[X] and λ > 1. Then we have |f |(r) = 0. r→+∞ |f |(λr) lim
Proof. Let (an )n0 be the sequence of distinct zeros of f ranged so that, for every n ∈ N, |an | |an+1 | and limn→+∞ |an | = +∞. For every n ∈ N, let kn be the order of an as a zero of f . So f (x) = A n0 (1 − axn )kn . For r > 0 large enough, there exists a unique index as nr such that |anr | < r |anr +1 |. Then we have |f |(r) =
|A| r k0 +···+knr . |a0 . . . anr |
(1)
On the other hand, as λ > 1, we have λr > |anr | and hence |f |(λr)
|A| (λr)k0 +···+knr . |a0 . . . anr |
It follows from (1) that |f |(λr) (λ)k0 +···+knr |f |(r). As k0 1, k1 1, . . . , knr 1, it follows that |f |(λr) λnr +1 |f |(r).
(2)
Hence |f |(r) 1 n +1 , |f |(λr) λ r
and
lim
r→+∞
This completes the proof of Lemma 2.
|f |(r) 1 = lim n +1 = 0. r→+∞ |f |(λr) λ r 2
Lemma 3. Let f ∈ M(K) be a solution of equation
E
s
Ai (x) σqi f (x) = 0.
i=0
Let R > 0 be such that all the zeros of the polynomial A0 (X) belong to d(0, R − ). Then the following statements are equivalent: (i) f admits at least a pole in K \ d(0, R − ), (ii) f admits infinitely many poles, (iii) for every real number r R, f admits at least a pole in d(0, r|q|−s ) \ d(0, r). Proof. Clearly, it is sufficient to show that (i) ⇒ (ii) and (ii) ⇒ (iii). Suppose that α is a pole of f in K \ d(0, R − ) and r R. Let us set β = q −s α. From (E), we have si=0 Ai (β)(σqi f )(β) = 0. But β, qβ, q 2 β, . . . , q s β belong all to K \ d(0, R − ) and q s β is a pole of f . So there exists an index j1 ∈ {0, . . . , s − 1} such that α1 := q j1 β = q j1 −s α is a pole of f . Repeating this reasoning for α1 , we obtain an other pole α2 of f . In this way, we construct | = |q|jn+1 −s |q|−1 > 1. a sequence (αn )n0 of poles of f satisfying, for every n 0, | ααn+1 n Hence (i) ⇒ (ii). Suppose now that f has infinitely many poles. Let r R and let α be a pole of f such that |α| > r. If |α| r|q|−s , we are done. If not, there exists an index j1 , 1 j1 s such that β := q j1 α is a pole of f ; and we have: |α| > |β| = |q|j1 |α| > r|q|j1 −s r. If |β| r|q|−s , we
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are done. If not, we apply this reasoning again to β instead of α. This means that (ii) ⇒ (iii) and completes the proof of Lemma 3. 2 Proof of Theorem 3. We may consider equation E
s
Ai (x) σqi f (x) = 0,
i=0
with polynomials A0 (X), . . . , As (X) such that A0 (X)As (X) = 0. Let R > 0 be such that the disk d(0, R − ) contains all the zeros A0 (X). Let f ∈ M(K) be a solution of Eq. (E). We will distinguish two situations: (I) Suppose first that f admits infinitely many poles. For every integer k 0, we set rk = R|q|−ks so that we have log rk − log R k= . log |q|−s
(1)
By Lemma 3, there exists a sequence (αk )k0 of poles of f such that rk < |αk | rk+1 , for every k ∈ N. Then we have rk rk N(rk , f ) log log = log |q|(i−k)s |αi | ri 0i
It follows that k(k + 1) log |q|−s . 2 By (1) and (2), we deduce that (log rk )2 = O N (rk , f ) , k → +∞. N(rk , f )
(2)
As the sequences (rk )k1 is increasing and tends to +∞, we deduce that (log r)2 = O N (r, f ) , r → +∞. Since N(r, f ) T (r, f ), we finally have (log r)2 = O(T (r, f )), r → +∞. (II) Suppose now that f only has a finite number of poles. Multiplying by a suitable rational function we may assume, without loss of generality, that f is a non-polynomial entire function. We set di = deg Ai (X) for i = 0, . . . , s, so that Ai (X) = ai X di + · · · , where a0 , a1 , . . . , as are elements of K such that a0 as = 0. Let d = max0is di 1 and let be the smallest index among 0, . . . , s, such that d = d. Then, by (E), we have Ai (x) σqi f (x). A (x) σq f (x) = − 0is, i=
Thus, for r > 0 large enough, we have
Ai σ i f (r) A σ f (r) max q
0is, i=
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and hence
|a |r d |f | r|q|
max
0is, i=
|ai |r di |f | r|q|i .
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(3)
If j = is an index among 0, . . . , s for which the above maximum is reached, we have |a |r d |f | r|q| |aj |r dj |f | r|q|j . Hence |f |(r|q| ) |aj | dj −d r . |f |(r|q|j ) |a | As dj d, the quantity > j,
|aj | dj −d |a | r
(4) is bounded when r approaches +∞. So, by Lemma 2, we have
and hence 1.
By (3) and (5), we have |f | r|q|
max
0i−1
|ai | di −d i r |f | r|q| . |a |
(5)
(6)
Replacing r by rk = R|q|−ks in the inequality (6), we obtain |ai | di −d |ai | |f |(rk ) |f |(rk−1 ) max rk |f | rk |q|i max . rk 0i−1 |a | 0i−1 |a | Hence we have |f |(rk ) Ark |f |(rk−1 ), with some positive constant A. k(k+1) It follows that |f |(rk ) Ak ( kj =0 rj )|f |(R) = Ak R k+1 |q|−s 2 |f |(R) and hence log |f |(rk )
k(k + 1) log |q|−s + k log A + (k + 1) log R + log |f |(R). 2
(7)
(8)
We deduce, from (1) and (8) that (log rk )2 = O(log |f |(rk )), k → +∞. Hence (log r)2 = O(log |f |(r)), r → +∞. This completes the proof of Theorem 3. 2 All the results above are not aimed at discussing the existence of non-rational solutions of Eq. (E). However, the proof of Theorem 3 provides some necessary conditions for this to happen. More precisely, we have the following corollary. Corollary 2. Let us consider the equation (E)
s
Ai (x) σqi f (x) = B(x),
i=0
where q ∈ K, 0 < |q| < 1 and B(X), A0 (X), . . . , As (X) (s 1) be elements of K[X] such that A0 (X)As (X) = 0. Suppose that deg A0 (X) = max0is deg Ai (X). Then: (i) Every entire solution of Eq. (E) is a polynomial. (ii) Every meromorphic solution of Eq. (E) which has finitely many poles is a rational function.
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Example 2. Let q ∈ K, 0 < |q| < 1. Then every entire solution of the equation x 2 y(x) + xσq y(x) = x 3 + 1 is a polynomial. Remarks. (1) We can recover Theorem 1 as an immediate consequence of Corollary 2. Indeed, we have seen that f has no poles different from 0. On the other hand, from Eq. (E ), we have B(x)
s i=0
s Ai f q i+1 x − B(qx) Ai f q i x = 0.
s+1
i=0
This can be written as i=0 Ci (x)f (q i x) = 0, with C0 (X) = A0 B(qX), Cs+1 (X) = As B(X) and Ci (X) = Ai−1 B(X) − Ai B(qX) for 1 i s. It is then clear that deg C0 (X) = max{deg Ci (X) | 0 i s + 1}. (2) More generally, more technical calculation may enable us to reduce our study to the case of entire solutions of q-difference equations using only the maximum modulus. Here we have deliberately chosen to use systematically the ultrametric Nevanlinna theory so that the obtained results appear as applications of this latter. References [1] Y. Amice, Les nombres p-adiques, PUF, 1975. [2] J.-P. Bézivin, Sur les équations fonctionnelles aux q-différences, Aequationes Math. 43 (2–3) (1992) 159–176. [3] J.-P. Bézivin, Indépendance linéaire et algébrique de fonctions liées à la fonction q-dzéta, Ann. Fac. Sci. Toulouse 17 (1) (2008) 23–36. [4] J.-P. Bézivin, Fonction de Tschakaloff et Fonction q-exponentielle, preprint. [5] J.-P. Bézivin, A. Boutabaa, Sur les équations fonctionelles p-adiques aux q-différences, Collect. Math. 43 (2) (1992) 125–140. [6] A. Boutabaa, Théorie de Nevanlinna p-adique, Manuscripta Math. 67 (1990) 251–269. [7] A. Boutabaa, Applications de la théorie de Nevanlinna p-adique, Collect. Math. 42 (1) (1991) 75–93. [8] A. Boutabaa, A. Escassut, Applications of the p-adic Nevanlinna theory to functional equations, Ann. Inst. Fourier 50 (3) (2000) 751–766. [9] W. Bergweiler, K. Ishizaki, N. Yanagihara, Meromorphic solutions of some functional equations, Methods Appl. Anal. 5 (1998) 248–258. [10] W. Bergweiler, K. Ishizaki, N. Yanagihara, Growth of meromorphic solutions of some functional equations, I, Aequationes Math. 63 (2002) 140–151. [11] W. Bergweiler, W.K. Hayman, Zeros of solutions of a functional equation, Comput. Methods Funct. Theory 3 (1) (2003) 55–78. [12] A. Escassut, Analytic Elements in p-Adic Analysis, World Scientific Publishing, Singapore, 1995. [13] J. Heittokangas, I. Laine, J. Rieppo, D. Yang, Meromorphic solutions of some linear functional equations, Aequationes Math. 60 (2000) 148–166. [14] A. Robert, A Course in p-Adic Analysis, Grad. Texts in Math., vol. 198, Springer-Verlag, 2000.