Value Distribution for a Class of Small Functions in the Unit Disk

Proposition 2.1. If f and g are in 𝒫 and c is a nonzero complex number, we have

• (i)

  cf is in 𝒫;

• (ii)

  1/f is in 𝒫;

• (iii)

  fⁿ is in 𝒫 for each positive integer n;

• (iv)

  fg may not be in 𝒫;

• (v)

  f + g may not be in 𝒫.

Theorem 2.2. Let f be a meromorphic function in class 𝒫.

• (i)

  If c is a nonzero complex number for which

  $$ (2.4)
  then f + c is in 𝒫.

• (ii)

  If g is a meromorphic function in D which is not identically zero and such that T(r, g) = o(T(r, f)), (r → 1), and m(r, g^′/g) = o(T(r, f)), (r → 1), then fg is in 𝒫.

• (iii)

  There exists a Blaschke product B such that Bf is not in 𝒫.

• (iv)

  There exists a Blaschke product B such that Bf is in 𝒫.

Remark 2.3. In Nevanlinna theory, the Valiron deficiency of a complex value c for a meromorphic function f in D is defined by

It is known (cf. Theorem 2.20 on page 210 in [5]) that if then except for at most a set of a-values of vanishing inner capacity. This fact enables us to show that the function F in 𝒫 defined by has F + c in 𝒫 for all complex numbers c, because for all c ≠ 0.

Remark 2.4. The following example illustrates part (ii) of Theorem 2.2.

Example 2.5. Let f(z) = e^i/(1−z) and g(z) = e^{(1+z)/(1−z)}. Then g is not identically zero, and it is well known that T(r, g) = O(1). Therefore, T(r, g) = o(T(r, f)) as r → 1. Also we have that

since 2/(1 − z) ² is the quotient of two bounded, analytic functions in the unit disk. And so we have that fg = e^{(i+1+z)/(1−z)} ∈ 𝒫.

Proof of Part (i). Let g = f + c. Then g^′ = f^′. We will show that g ∈ 𝒫.

First, by calculation and properties of the Nevanlinna characteristic, note that

Also, by calculation and properties of the proximity function, we get

Now, since f ∈ 𝒫, Also, since Δ(−c, f) = 0, and so

Therefore, g ∈ 𝒫 since m(r, g^′/g) = o(T(r, g)) as r → 1, and T(r, g) is unbounded as r → 1.

Proof of Part (ii). First, T(r, fg) is unbounded, since it can be shown that

since T(r, f) is unbounded.

Now note that (fg)^′/(fg) = g^′/g + f^′/f. Therefore,

Thus fg ∈ 𝒫.

Proof of Part (iii). Let B be the Blaschke product defined in [6, Proposition 6.1, page 273]. This Blaschke product has the feature that for any ϵ > 0 there exists an exceptional set E₁ ⊂ [0,1) satisfying

such that and there exists a set F₁ ∈ [0,1), satisfying and a constant C > 0, such that

Let g(z) = i/(1 − z). Then e^g(z) ∈ 𝒫. Now define q(z) = B(z)e^g(z). We will now show that q(z) ∉ 𝒫.

First, it is easily shown that T(r, e^g(z)) = T(r, q) + O(1) as r → 1.

Note that since $$, we have $$. Therefore, since e^g(z) ∈ 𝒫,

And so Therefore, Using (2.21), we have on a small exceptional set with |z | = r for r ∈ F₁∖E₁, Taking the log   ⁺ of both sides, we get So, calculating the following ratio yields

Therefore, q ∉ 𝒫.

Proof of Part (iv). Let B be a Blaschke product with zeros {z_n} such that |z_n | = 1 − 1/n⁵ for all integers n ≥ 2. Theorem B in Heittokangas [6] shows that B^′ is in H^p for p in (0, 3/4), so B^′/B is of bounded characteristic. Hence Bf is in 𝒫 for f in 𝒫.

Remark 2.6. Since a Blaschke product is a bounded, analytic function, we see from parts (iii) and (iv) above that multiplication by such functions may or may not yield a function in 𝒫.

Theorem 2.7. There exists an analytic function h in 𝒫 such that m(r, h^′′/h) ≠ o(T(r, h)), as r → 1.

Proof. First, we begin by constructing a function which has unbounded characteristic as r → 1, but its derivative is of bounded characteristic. This construction is from [7, page 557].

Let a be an integer greater than or equal to 2. Define for m ≥ 1,

where γ is a constant such that 0 < γ < 1. Now define where Shea showed in [7] that f is analytic in the unit disk and satisfies where α and β are constants such that and M(r, f) is the maximum modulus function for f. Choose γ and a such that γ + log   2/log   a < 1, so we can take β < 1. This implies that F is bounded in the unit disk by the following argument: for z = re^iθ, By (2.31), we have that Therefore, since β < 1, we have by a simple integration that

Let $$. Define h(z) = K(z)e^g(z) where g(z) = i/(1 − z). Recall that e^g(z) ∈ 𝒫. We now show that h(z) ∈ 𝒫. First, we see that T(r, K) = O(1) as r → 1, by the following:

and since F is bounded, there exists a constant M such that Thus, T(r, h) ≤ T(r, K) + T(r, e^g(z)) ≤ T(r, e^g(z)) + O(1), as r → 1. On the other hand, Therefore, T(r, h) ~ T(r, e^g(z)) as r → 1.

Now, we bound m(r, h^′/h) from above by using properties of the Nevanlinna characteristic:

And so we have m(r, h^′/h) = o(T(r, h)) as r → 1 and thus h ∈ 𝒫.

Now we show that m(r, h^′′/h) ≠ o(T(r, h)) as r → 1. By a quick calculation, we have

Also we have from the above construction K^′′ = f so there exists an α > 0 such that m(r, K^′′) > αlog   (1/(1 − r)). And so, Combining (2.40) and (2.41), we have

Remark 2.8. If we define A to be the set of functions in F such that (1.4) holds for all positive integers k, Theorem 2.7 shows A is properly contained in F. Further, we note that in the proof of Theorem 2.7 above h = Ke^g can be replaced with h = Kp where p ∈ A. Also the idea of the proof of Theorem 2.7 can be used to show that for k > 1 there exist functions h in F for which

for all integers 1 < j ≤ k, but

The function h in the proof of Theorem 2.7 provides us with further information about 𝒫.

Theorem 2.9. There exists a function h in 𝒫 such that h^′ is not in 𝒫.

Proof. The function h = Ke^g of Theorem 2.7 is in 𝒫. Using the Nevanlinna calculus and properties of K, one can show m(r, h^′′/h^′) ≠ o(T(r, h^′)) as r → 1. We omit the details here (cf. [2]).

Theorem 2.10. If f is an analytic function in class 𝒫, then

• (i)

  T(r, f^′) ≤ T(r, f) + o(T(r, f)) as r → 1;

• (ii)

  α(f^′) ≤ α(f);

• (iii)

  if N(r, 1/f) = o(log   1/(1 − r)) as r → 1, then α(f) = α(f^′).

Proof. For part (i) since f is analytic in 𝒫, we have

from which the result follows. Using the definition of the index of f^′ and of f, part (ii) comes from (i).

To see (iii), we observe

Thus, Dividing both sides of (2.49) by log   1/(1 − r) and taking the limit superior as r approaches one, we get α(f) ≤ α(f^′).

Definition 3.1. We say that an analytic function f in the unit disk is in the weighted Hardy space $$ for 0 < p < ∞ and 0 ≤ q < ∞ if

We say that f is in $$ if

Definition 3.2. We say that an analytic function f in the unit disk, D, is in the weighted Bergman space $$ if the area integral over D satisfies

for 0 < p < ∞ and −1 < q < ∞.

Theorem 3.3 (see [10], page 320.)Let f be a nontrivial solution of (3.1) with analytic coefficients a_j, j = 0, …, n − 1, in the unit disk. Then we have that

• (i)

  if −1 < α < 0 and $$ for all j = 0, …, n − 1, then f ∈ N;

• (ii)

  if a_j ∈ A^1/(n−j) for all j = 0, …, n − 1, or $$ for all j = 0, …, n − 1, then f ∈ N;

• (iii)

  if $$ for all j = 0, …, n − 1, then f ∈ ℱ.

Theorem 3.4 (see [10], page 320.)We have that

• (i)

  if all nontrivial solutions f ∈ N, then the coefficients a_j ∈ ⋂_{0<p<1/(n−j)}A^p for all j = 0, …, n − 1;

• (ii)

  if all nontrivial solutions f ∈ ℱ, then the coefficients $$ for all j = 0, …, n − 1.

Theorem 3.5 (see [9], page 44.)All solutions f of (3.1), where a_j is analytic in D for all j = 0, …, k − 1, satisfy ρ(f) = 0 if and only if a_j ∈ ⋂_{0<p<1/(n−j)}A^p for all j = 0, …, n − 1.

Corollary 3.6. Let f be a non-trivial solution of (3.1) with analytic coefficients a₀ and a₁ in the unit disk. Then

• (i)

  if $$ and $$ for −1 < α < 0, then f ∈ N and f ∉ 𝒫;

• (ii)

  if a₁ ∈ A¹ and a₀ ∈ A^1/2 or a₀ and $$, then f ∈ N and f ∉ 𝒫;

• (iii)

  if $$ and $$, then f ∈ ℱ and, therefore, could be in 𝒫;

• (iv)

  if all non-trivial solutions f ∈ ℱ, then $$ and $$.

Example 3.7. We show for a₀ = −2βi/(1 − z) ³ and a₁ = −βi/(1 − z) ², $$ and $$. To see $$, we first note

We integrate and apply a lemma from Tsuji [4, page 226] which states that, in particular, and get that there exists a constant M such that And so which goes to zero as r → 1. Therefore, $$.

A similar calculation shows that $$.

Example 3.8. The function h in Theorem 2.7 is a solution of (3.1) with n = 2 when a₀ = −g^′′ − (g^′) ² and a₁ = −(K^′(2g^′) + K^′′)/(Kg^′ + K^′). Then since

we have that

Now, recall from the proof of Theorem 2.7 that

and since h ∈ 𝒫, by (3.15), we conclude that at least one of a₀ or a₁ has index greater than or equal to α. Therefore, as a consequence of Theorem 2.7, we have a growth estimate for the coefficients of this differential equation.

It can also be shown that $$. However, a₁ is not in $$, and thus the converse of Corollary 3.6(iii) is not true.

Example 3.9 (see [9], Example 10, page 52.)The functions

are linearly independent solutions of where It can be shown that f₂ ∈ 𝒫. However, f₃ is of bounded characteristic and, therefore, f₃ ∉ 𝒫. (It is known that f₁ has order zero but unknown if f₁ ∈ 𝒫.) Also, according to [9], we have that $$, $$, and $$.

Example 3.10 (see [9], Example 11, page 53.)The functions

are linearly independent solutions of where It is known that f₂, f₃, and f₄ are all in 𝒫, and it is known that f₁ has order zero. Also, it can be shown that $$, $$, $$, and $$.

Theorem 3.11. If a function f in 𝒫 satisfies a differential equation of the form (3.1) such that

for all integers j such that 1 ≤ j < n, then at least one of the analytic coefficients a_j has index greater than or equal to α as r → 1 for some α > 0.

Theorem 3.12 (see [9], Theorem 7, page 46.)All solutions f of (3.24) satisfy ρ(f) = 0 if and only if ρ(a_n) = 0 and a_j ∈ ⋂_{0<p<1/(k−j)}A^p for all j = 0, …, n − 1. Therefore, if all solutions f of (3.24) are in 𝒫, then ρ(a_n) = 0 and a_j ∈ ⋂_{0<p<1/(k−j)}A^p for all j = 0, …, n − 1.

Proposition 3.13. Let k be a positive integer. If a₀ is an analytic function in D for which the index of a₀ is α(a₀) > k, then f ∉ 𝒫 for f ∈ ℱ when f^(k) − a₀f = 0.

Proof. Note that since f^(k) − a₀f = 0, we have a₀ = f^(k)/f, and we want to show that

Since α(a₀) > k, there exists a real number s > k such that T(r, a₀) ≥ slog (1/(1 − r)) on some sequence of r′s as r → 1. Also, since f ∈ ℱ, we have T(r, f) ≤ tlog   (1/(1 − r)). Now, since f ∈ ℱ, f^(k) ∈ ℱ and so f⁽ⁱ⁾/f⁽ⁱ⁻¹⁾ ∈ ℱ for 1 ≤ i ≤ k. By Shea and Sons [1], m(r, f⁽ⁱ⁾/f⁽ⁱ⁻¹⁾) ≤ log (1/(1 − r)) + (2 + o(1))log   log   (1/(1 − r)) for 1 ≤ i ≤ k as r → 1. Now, we have the following: So, since a₀ is analytic, we have Therefore, m(r, f^′/f)/T(r, f) ≥ (s − k)/t > 0 as r → 1.

Theorem 4.1 (see [11], page 48.)Suppose that f₁ and f₂ are meromorphic in the plane and let E_j(a) be the set of points z such that f_j(z) = a(j = 1,2). Then if E₁(a) = E₂(a) for five distinct values of a, f₁(z) ≡ f₂(z) or f₁ and f₂ are both constant.

Theorem 4.2. Let f₁(z) and f₂(z) be meromorphic functions in class ℱ such that α(f₁) ≥ α(f₂) > 0 and let E_j(a) be the set of points z such that f_j(z) = a for j = 1,2. Then if E₁(a) = E₂(a) for q distinct values of a such that q is an integer and q > 4 + 2/α(f₁), then f₁(z) ≡ f₂(z).

Proof. Suppose f₁ and f₂ are not identical and that {a₁, a₂, …, a_q} are q distinct complex numbers such that E₁(a_ν) and E₂(a_ν) are identical for ν = 1,2, …, q and q is an integer greater than or equal to 4 + 2/α(f₁). We write the following notations:

Now using a reformulation of Nevanlinna′s inequality for functions in class ℱ [1], we have the following for all r < 1: Now assume that α(f₁) ≥ α(f₂) > 0. Then from the definition of index we have that for 0 < ϵ < α(f₁), there exists a sequence {r_m} → 1 such that So, combining (4.2) and (4.3) and using (4.4), we get which leads to

Since f₁ and f₂ are not identical, we have

On the other hand, every common root of the equations f_ν(z) = a is a pole of 1/(f₁ − f₂), and so we have which gives a contradiction since q > 4 + 2/α(f₁) implies 2/(q − 2 − (2 + o(1))/(α(f₁) − ϵ)) < 1 as r → 1, unless $$. This, however, cannot occur since T(r, f₁) and T(r, f₂) are unbounded. Therefore, the result follows.

Remark 4.3. Since 𝒫 ⊂ ℱ, Theorem 4.2 gives conditions for when two functions in 𝒫 are identical.

Theorem 5.1 (see [12], Theorem 4.)Let f be a meromorphic function in D which is in class ℱ and for which

Let n be a positive integer, and for k = 0,1, 2, …, n let a_k be a meromorphic function in D for which If ψ is defined in D by and ψ is nonconstant, then ψ assumes every complex number except possibly zero infinitely often provided the index of f is α > 1 + n(n + 1)/2.

Theorem 5.2. Let f be a meromorphic function in D which is in class 𝒫 and for which

Let n be a positive integer, and for k = 0,1, 2, …, n, let a_k be a meromorphic function in D for which Also, define E to be the set {k : m(r, f^(k)/f) = o(T(r, f))  as  r → 1}. If ψ is defined in D by and ψ is nonconstant, then ψ assumes every complex number except possibly zero infinitely often, provided the index of f is α > 1 + n(n + 1)/2 − ∑E, where ∑E is the sum of the values of E.

Proof. Since class ℱ is closed under differentiation, addition, and multiplication, we know that ψ is in class ℱ. Therefore, we can apply the reformulation of the Second Fundamental theorem for class ℱ [1] to ψ. Thus, using 0, ∞, and c, a nonzero complex number, we get

Since poles of ψ come from poles of a_k or f, we have an upper bound for $$: From the hypothesis, we then have Therefore, using (5.9) and the First Fundamental theorem, we get Now, solving for m(r, 1/ψ) in the above calculation, we have the following inequality: Since $$, the N(r, 1/ψ) terms cancel, and so we can say that So, using the First Fundamental theorem, properties of the proximity function and (5.12) give us the following: Noticing the fact that $$ and using the hypothesis that we can say that

We now estimate m(r, ψ/f). By using properties of the proximity function, we get

Recall the set E = {k : m(r, f^(k)/f) = o(T(r, f))  as  r → 1}. Notice that since f ∈ 𝒫, E is not empty. The set E also allows us to split the following sum into two pieces. Indeed, But now we can say that since this is true for each k ∈ E. Therefore, using (5.18) and the hypothesis we can update (5.16) to say that We use (3.26) to say that and, thus, (5.20) becomes We can calculate ∑_k∉Eklog   (1/(1 − r)) by noting that where ∑E is the sum of the elements in E. Note that n(n + 1)/2 − ∑E = β ≥ 0. Thus, (5.20) becomes Therefore, we can now update (5.15) to say

Since the index of f is equal to α > n(n + 1)/2 − ∑E + 1, we have that

where γ > 0. Since T(r, f) is unbounded, we have proved the claim that ψ assumes every complex number except possibly zero infinitely often.