On Meromorphic Functions which are Brody Curves

We discuss meromorphic functions on the complex plane which are Brody curves regarded as holomorphic maps to P_1, i.e., which have bounded spherical derivative.

are rational functions and e z + e λz with λ ∈ C. For these functions we determine completely which ones are Brody.This provides us with some surprising examples: If f, g are Brody neither f + g, nor f g need to be Brody.The Brody condition is not closed: The meromorphic function f t (z) = e z + z zt+1 is Brody if and only if t = 0.
In particular in view of the results of Clunie and Hayman on the order it might be natural to assume that every divisor of sufficiently slow growth can be realized as the zero divisor of an entire Brody function.We show that this is not the case, there are effective divisors of arbitrary slow growth which can not be realized as zero divisors of a Brody function. 1

Basic properties
We start with the observation where Since h(z) ≤ 1  2 for all z ∈ C, we may deduce: If f ′ /f is bounded, then f is Brody.In fact one has the following slightly stronger statement: Lemma 1.Let f be an entire function with Then f is Brody.
Proof.There is a compact subset K ⊂ C and a constant C > 0 such that |f (z)/f ′ (z)| < C for all z ∈ K. Then Concerning rational functions the Brody property follows from the observation below: Lemma 2. Let R be a non-constant rational function on C.
1 A related result has been obtained much earlier by Lehto ([3]).
Proof.Write R as a quotient of polynomials: R = P/Q.Then and the assertion follows from the fact that deg( Corollary 2. Rational functions have the Brody property.
Proof.Each rational function R defines a continuous self-map of P 1 .
The compactness of P 1 implies that for each rational function R there exists a constant C such that the operator norm of the differential map DR : Corollary 3. The trigonometric functions sin, cos, tan, sinh, cosh.. are all Brody.
Proof.These functions can be expressed in the form R(e z ) or R(e iz ) for some rational function R ∈ C(X).

Products with rational functions
Lemma 3. Let f : C → P 1 be Brody and let R be a rational function with R(∞) ∈ {0, ∞}.
Then g = Rf is Brody.
Proof.We have with h(w) = |w| 1+|w| 2 .We will need the following auxiliary fact: This claim is rather immediate: and similarily for the second inequality.Now write the rational function R as quotient of two polynomials: R(z By assumption lim z→∞ R(z) ∈ C * .Therefore there are constants λ > 1 and R 2 > 0 such that As a consequence, we have Hence the assertion.
It is really necessary to assume R(∞) = 0, ∞.For instance, consider the functions f 1 (z) = e z + 1 and f 2 (z) = 1/f 1 (z).Both are Brody functions, but proposition 2 in the next section implies that neither 3. Case study: R(z)e z + Q(z) We will now discuss certain sums.
Proposition 2. Let R, Q be rational functions.Then f (z) = R(z)e z + Q(z) is a Brody function if and only if one of the following conditions is fulfilled: Proof.If R ≡ 0, then f = Q is rational and therefore Brody.Assume now that Q(∞) = ∞ and R ≡ 0. Then there is a rational function Q 0 with Q 0 (∞) ∈ {0, ∞} and a complex number c with But this implies lim sup Hence g is Brody and therefore f is Brody, too.It remains to show that f is not Brody whenever is Brody if and only if g(z) = S(z)e z +z n is Brody.Since R ≡ 0 together with Q 0 (∞) = 0 implies S ≡ 0, the entire function g is transcendental.Therefore there is a sequence of complex numbers z k with lim k→∞ |z k | = ∞ and lim k→∞ g(z k ) = 0. Now We recall that lim z→∞ Thus g is not Brody, which in turn implies that f is not Brody.
Hence the set of all zeros of f is given as the set of all Therefore sup k f # (a k ) = +∞ and consequently f is not Brody.
Case 3) Assume λ > 0. Observe that f (z) = g(λz) for g(w) = e w + e 1 λ w .Hence we may assume without loss of generality that 0 < λ ≤ 1.Since f (z) = 2e z if λ = 1, it suffices to consider the case 0 < λ < 1.We choose Then On the other hand and consequently f is Brody.
As before, we define and obtain Next we observe that the condition ℜ(z) ≤ −C implies As a consequence Finally we observe that Thus an arbitrary entire function f with divisor D can be written as where F is defined as above while g is an entire function.
Claim.There is a constant C > 0 (depending only on λ) such that To prove the claim, we observe that We emphasize that C is independent of n.
We need a second claim.
Claim.For every K > 0 and m ∈ N there is a natural number N ≥ m (depending on K and m) such that for all n ≥ N and p with |p| ≥ √ λ|a n |.
To prove the second claim it suffices to note that Next we choose for each n ∈ N a complex number p n such that Due to the two above claims we know that there is a number M ∈ N such that If we set r n = |a n+1 a n |, we can now deduce: Hence, if f is Brody and therefore T f (r) = O(r), we can deduce that T exp(g) (r) = O(r), which in turn implies that g is a polynomial of degree 1, i.e. affine-linear.Hence condition two implies that there is a subsequence a n j with exp(g(a .
By the first claim and by the second claim for each K > 0 there is a number N ∈ N such that  In the language of Nevanlinna theory: (2) There is no Brody entire function f for which D is the zero divisor.

Growth conditions and characteristic function
Our goal is to show that no bound on the characteristic function T f (r) forces an entire function f to be Brody except when this bound is strong enough to force f to be a polynomial.
In view of the preceding section the crucial point is to verify that for every such bound there exists an entire function f fulfilling this condition on T f (r) and fulfilling simultaneously the condition of theorem 1.
Before stating the result of this section we recall some basic notions.
For an entire function f with f (0) = 0 we may define the characteristic function as Now we can state the result: log t < ∞, then every entire function with T f (r) ≤ ρ(r) for all r ≥ 1 must be a polynomial.
(2) If lim inf t→∞ ρ(t) log t = ∞, then there exists an entire function which is not Brody and such that T f (r) ≤ ρ(r) ∀r ≥ 1.

Proof. (i). If there is a constant
for every entire function f with T f (r) ≤ ρ(r) and for all a ∈ C. It follows that each fiber f −1 {a} has cardinality ≤ C and that f is a polynomial of degree bounded by C.
(2).As a first preparation, we observe that We will construct f as follows: We will choose a sequence The conditions on the c k ensure that P k converges locally uniformly to an entire function f whose zeroes are precisely the points c k .Let k ∈ N and let z be a complex number with which in turn implies |P j (z)| ≥ 1 for all j and all z with Since T f (r) is increasing, we have Summarizing, we have shown that T f (r) ≤ k log(4r) for all r with r ≤ 2|c k |.
Thus it suffices to choose the c k such that k log(4r This is possible: We assumed lim inf t→∞ ρ(t) Thus we have established: Finally, we note that in our construction we choose the c k such that Proof.We pose It is easily verified that this is convergent.An explicit calculation shows: In particular, |f ′ (a)| ≤ 1 2 for all a ∈ D. Now let z ∈ |D|.We will deduce an estimate for for m ∈ N.
We are now ready to show that f # is bounded.We start with the case As a preparation we discuss Using |b − z| ≤ 1 we obtain the following bound: .
We know: Combined these two equations yield: for all z ∈ C \ D with d(z, D) ≤ 1.

Discussion
Using the special cases of entire functions studied above we see that the class of entire functions which are Brody is not closed neither under addition nor under multiplication: the entire functions z, e z + 1 and ze z are all Brody, but ze z + z is not, although ze z + z = z(e z + 1) = (ze z ) + z.
We see also that the Brody condition is neither closed nor open nor complex: For (s, t) ∈ C 2 let us consider Moreover, proposition 3 provides an example of a family of entire functions depending holomorphically on a complex parameter λ such that the function is Brody if and only if λ is real.
All this properties are in stark contrast to the situation for Brody curves with values in abelian varieties.If A is an abelian variety, its universal covering is isomorphic to some C g .Since every holomorphic map from C to A lifts to a holomorphic map with values in the universal covering of A, the classical theorem of Liouville implies that an entire curve with values in A is Brody if and only if it can be lifted to an affine-linear map from C to C g .As a consequence one obtains: • If f, g : C → A are Brody curves, so is z → f (z) + g(z).
• If f t : C → A is a family of entire curves depending holomorphically on a parameter t ∈ P where P is a complex manifold, then the set of all t ∈ P for which f t is Brody forms a closed complex analytic subset of P .

Corollary 1 .
The exponential function z → e z is Brody.

4 .
Case study: e z + e λz Proposition 3. The entire function f (z) = e z + e λz is Brody if and only

5 .
Divisors of slow growth Theorem 1.Let D be the divisor defined by D = {a k : k ∈ N} (all multiplicities being one).Assume that there is a number λ > 1 such that (1) |a k+1 | > λ|a k | > 0 for all k, and (2) The origin 0 is contained in the interior of the convex hull of the set of accumulation points of the sequence a k |a k | .Then there does not exist any Brody function with D as zero divisor.Proof.The first condition implies the absolute convergence of k 1 |a k | which implies the convergence of k log 1 − 1 |a k | which in turn implies the convergence of all n ≥ N. Combined these assertions show that for each K > 0 there is a number N such that|F ′ (a n )| ≥ C 2|a 0 | K for all n ≥ N.Thus lim sup |f # (a n )| = +∞ and f is not Brody.

Corollary 4 . 1 ) 2 )
Let ζ : R + → R + be an unbounded monotone increasing function.Then there exists an effective reduced divisor D such that (For every r ∈ R + the inequality deg(D r ) ≤ ζ(r) holds where D r denotes the restriction of D to the open disc with radius r. (There is no Brody entire function f for which D is the zero divisor.

Corollary 5 .
Let ζ : R + → R + be an unbounded monotone increasing function.Then there exists an effective reduced divisor D such that (1) For every r ∈ R + the inequality N(r, D) ≤ ζ(r) holds.
Furthermore we may choose the c k such that the set c k |c k | : k ∈ N is dense in S 1 = {z ∈ C : |z| = 1}.Then theorem 1 implies such an entire function f ist not Brody.7. The divisor D = {k 2 : k ∈ N} Proposition 4. Let a k = k 2 .Then there is a Brody function with divisor {a k : k ∈ N}.

First
we note: If a subset of C contains at least m numbers in |D| = {k 2 : k ∈ N}, then its diameter must be at least m 2 − 1.Let b k be a renumbering of the points in |D| such that |b k+1 − z| ≥ |b k − z| for all k.If r = |b k − z| for some k ∈ N, then b 1 , . . ., b k ∈ D r (z) = {w ∈ C : |z − w| ≤ r} and therefore r ≥ k 2 − 1.Thus 1 |b k − z| ≤ 1 k 2 − 1 for all k > 1.We deduce: For every z ∈ |D| there is a point b ∈ |D| with d(z, D) = |z − b| and (1) k

f
s,t (z) = se z + z tz − 1 By proposition 2 the function f s,t is Brody iff (s, t) ∈ {(x, y) ∈ C 2 : x = 0 or y = 0}which is neither a closed nor an open set.
• An entire curve f : C → A is Brody if and only if