Linear differential equations with solutions in weighted Fock spaces

LINEAR DIFFERENTIAL EQUATIONS WITH SOLUTIONS IN WEIGHTED FOCK SPACES

arXiv:1711.06174v1 [math.CV] 16 Nov 2017

GUANGMING HU AND JUHA-MATTI HUUSKO∗ Abstract. This research is concerned with the nonhomogeneous linear complex differential equation f (k) + Ak−1 f (k−1) + · · · + A1 f ′ + A0 f = Ak in the complex plane. In the higher order case, the mutual relations between coefficients and solutions in weighted Fock spaces are discussed, respectively. In particular, sufficient conditions for the solutions of the second order case f ′′ + Af = 0 to be in some weighted Fock space are given by Bergman reproducing kernel and coefficient A.

1. Introduction and main results The growth of solutions of the complex linear differential equation f (k) + Ak−1 f (k−1) + · · · + A1 f ′ + A0 f = Ak

(1)

has drawn wide attention. Abundant results about the growth of fast growing solutions of (1) have been obtained by Nevanlinna theory [3, 6, 7, 8, 10, 17, 25]. However, some other methods are needed in dealing with slow growth analytic solutions [18, 20, 22, 24, 29, 30]. There are some useful and powerful techniques, for instance, Herold’s comparison theorem [19], Gronwall’s lemma [22], Picard’s successive approximations [10, 14] and some methods based on Carleson measures [21, 27, 28, 29]. In [20], J. Heittokangas, R. Korhonen, and J. R¨atty¨a found that if the coefficients Aj (z) (j = 0, · · · , k − 1) of the homogeneous equation (1) belong to certain weighted Bergman or Hardy spaces, then all solutions are finite order of growth. Conversely, if all solutions of the homogeneous equation (1) are finite order of growth, the coefficients Aj (j = 0, · · · , k − 1) belong to weighted Bergman space. In [19], authors gave the growth estimates of solutions of (1). Motivated by the research in [19] and [20], we study similar problems in weighted Fock spaces. These spaces have a long history in mathematics and mathematical physics. They have been given a wide variety of appellations, including many combinations and permutations of the names Fock, Bargmann, Segal and 2010 Mathematics Subject Classification. 34M10; 30H20. Key words and phrases. Linear differential equation, Fock space, weighted Fock space. 1

2

G. M. HU AND J.-M. HUUSKO

Fischer (see [4, 5, 12, 13]). The classical Fock space (see [1, 2, 31]) is a subspace of a special Lebesgue measurable space. Let Lpg be the space of Lebesgue measurable 1 2 functions f on the complex plane C such that the function f (z)e− 2 |z| belongs to Lp (C, dm(z)), for some p ∈ [1, ∞]. The norm of f in the space Lpg is defined by Z 1 2 p kf kp := |f (z)e− 2 |z| |p dm(z), C

for p ∈ [1, ∞), and 1

2

kf k∞ := sup |f (z)|e− 2 |z| , z∈C

for p = ∞, where dm(z) = dxdy denotes the Lebesgue measure in C. The Fock space F p consists of entire functions in Lpg . In particular, the space F 2 is a closed subspace of the Hilbert space L2g with the inner product Z 1 2 f (z)g(z)e−|z| dm(z). hf, gi := π C Recently, a new function space called Fock-Sobolev space attracted the attention of many scholars and was first put forward by Hong Rae Cho and Kehe Zhu in [9]. Fock-Sobolev space F p,m consists of entire functions f on C such that X kf (α) (z)kp < ∞, kf kF p,m := α≤m

where k · kp is the norm in F p , m is a positive integer, and p ∈ [1, ∞]. By Theorem A in [9], we obtain that f ∈ F p,m if and only if z m f (z) is in F p . Namely, there is a positive constant C such that X C −1 kz m f (z)kp ≤ kf (α) (z)kp ≤ Ckz m f (z)kp . α≤m

Moreover, the weighted Fock space [11] is defined as follows. Let φ : [0, ∞) → R+ be a twice continuously differentiable function and extend φ to the complex plane C setting φ(z) = φ(|z|), for z ∈ C. We set the norms of the weighted Fock spaces Z p kf kF p := |f (z)|p e−pφ(z) dm(z), φ

C

for p ∈ [1, ∞), and kf kFφ∞ := sup |f (z)|e−φ(z) , z∈C

for p = ∞. This definition gives the classical Fock space, when φ(z) = |z|2 /2, and the Fock-Sobolev space, when φ(z) = |z|2 /2 − m log |z|. For the remainder of this paper, we restrict the weight φ to be a so-called rapidly increasing function. Namely, let φ : [0, ∞) → R+ be twice continuously differentiable such that its Laplacian satisfies ∆φ(|z|) > 0. Then there exists a

LDES WITH SOLUTIONS IN WEIGHTED FOCK SPACES

3

positive differentiable function τ : C → R+ , with τ (|z|) = τ (z), and a constant C ∈ (0, ∞) such that τ (z) = C, for 0 ≤ |z| < 1, and 1

1

C −1 (∆φ(|z|))− 2 ≤ τ (z) ≤ C(∆φ(|z|))− 2 ,

|z| ≥ 1.

We have τ (z) → 0 as |z| → ∞ and limr→∞ τ ′ (r) = 0. Furthermore, suppose that either there exists a constant C > 0 such that τ (r)r C increases for large r or lim τ ′ (r) log

r→∞

1 = 0. τ (r)

Denote by I the class of rapidly increasing functions φ satisfying the above mentioned conditions. The class I includes the power functions φ(r) = r α with α > 2 r and exponential type functions such that φ(r) = eβr , β > 0 or φ(r) = ee . Denote the point evaluations by Lζ (f ) = f (ζ), for f ∈ Fφ2 and ζ ∈ C. In [11], it is proved that the point evaluations Lζ are bounded linear functionals in Fφ2 . It follows that, there exists a reproducing kernel Kζ ∈ Fφ2 , with kKζ kFφ2 = kLζ kFφ2 →C , such that Z f (z)Kζ (z)e−2φ(z) dm(z),

f (ζ) = Lζ (f (z)) =

C

for φ in the class I. Moreover, if {en } is an orthonormal basis of Fφ2 , for example, en (z) = z n δn−1 , where δn2 = 2π

R∞ 0

n ∈ N,

r 2n+1 e−2φ(r) dr, then

∞ Kζ (z) = Σ∞ n=0 hKζ , en ien (ζ) = Σn=0 en (ζ)en (z).

By using similar idea with weighted Fock space, Fock-Sobolev weighted space is defined as follows. Let φ : [0, ∞) → R+ be a twice continuously differentiable function and extend φ to the complex plane C setting φ(z) = φ(|z|), for z ∈ C. We set the norms of the weighted Fock-Sobolev spaces Z p1 p −pφ(z) q kf kFφp,q := |f (z)| e φ (z)d(z) , C

for p ∈ [1, ∞) and q ∈ R. The weighted Fock-Sobolev space is the weighted Fock space, when q = 0. Motivated by the result of [23], we can obtain the following two theorems. Theorem 1.1. Suppose that Aj (z) are entire functions, j = 1, · · · , k. If k−1 X |Ai (z)| C Di sup r0 . Then all solutions of (1) belong to Fφp,q . iθ

Next, we give sufficient conditions for some coefficients to be in some weighted Fock space. Theorem 1.4. Suppose that Aj (z) are constant function, j = 0, · · · , k − 1. If there exists a solution of (1) belonging to F p,k . Then the coefficient Ak (z) of (1) belongs to F p . Theorem 1.5. Let φ be in the class I and moreover φ(r) = er . Suppose that Aj (z) belongs to F 1per , j = 1, · · · , k − 1. If there exists a solution of (1) such that 2

lim sup r→∞

and

|f (reiθ )| 0 of depending only on α and a constant r0 = r0 (θ) > 1 such that (m) iθ m α f (re ) ≤ C T (αr, f ) log r log T (αr, f ) , m ∈ N, f (reiθ ) r where r > r0 and θ ∈ [0, 2π)\W .

Lemma 2.5. Suppose that φ is in the class I and f (z) is an entire function. Then for any R > 0, there is a constant C > 0 such that Z Z p −pφ(z) q |f (z)| e φ (z)dm(z) ≤ C |f (z)|p e−pφ(z) φq (z)dm(z). |z|≥R

C

Proof. Z

p −pφ(z) q

|f (z)| e

φ (z)dm(z) =

R≤|z|≤R+1

≥

min

R≤r≤R+1

e−pφ(r) φq (r)r

Z

Z

R+1 R

Z

R+1 Z

2π

|f (reiθ )|p e−pφ(r) φq (r)rdrdθ 0

2π

|f p (reiθ )|dθdr.

0

R

Since f p (z) is an entire function, |f p (z)| is a subharmonic function. If Z 2π |f p (reiθ )|dθ = 0, 0

R 2π

then f (z) ≡ 0 and the inequality holds. If C1 > 0 such that Z Z p −pφ(z) q |f (z)| e φ (z)dm(z) ≤ C1 |z|≤R

Therefore, Z

0

|f p (reiθ )|dθ > 0, then there exists |f (z)|p e−pφ(z) φq (z)dm(z).

R≤|z|≤R+1

p −pφ(z) q

|f (z)| e

C

where C = C1 + 1.

φ (z)dm(z) ≤ C

Z

|f (z)|p e−pφ(z) φq (z)dm(z),

|z|≥R

Lemma 2.6. Suppose that φ is in the class I. Then φ(r) lim 2 = ∞. r→∞ r Proof. By L’Hospital’s rule, we obtain φ′ (r) rφ′ (r) (rφ′ (r))′ φ(r) = lim = lim lim 2 = lim r→∞ 2r r→∞ 2r 2 r→∞ r→∞ r 4r ′ 1 φ (r) 1 = (φ′′ (r) + ) = ∆φ(r). 4 r 4 1 Since C −1 (∆φ(|z|))− 2 ≤ τ (z) and τ (z) decreases to 0 as |z| → ∞. Then φ(r) lim 2 = ∞. r→∞ r


7

Lemma 2.7. [11] Suppose that φ is in the class I and f, g ∈ Fφ2 . Then Z hf, gi = f (0)g(0) + f ′ (z)g ′ (z)(1 + φ′ (z))−2 e−2φ(z) dm(z). C

Lemma 2.8. [11] Suppose that φ is in the class I and there exists r0 > 0 such that φ′ (r) 6= 0 for r > r0 . Moreover, assume that φ satisfies re−pφ(r) = 0, r→∞ φ′ (r) lim

and

′ ′ r r 1 1 −∞ < lim inf ≤ lim sup < p, ′ r→∞ r φ′ (r) r→∞ r φ (r)

where p ≥ 1. Then for any entire function f (z), Z e−pφ(|z|) p −1 p ′ C kf kF p ≤ |f (0)| + |f (z)| dm(z) ≤ Ckf kpF p , ′ p φ φ (1 + φ (|z|)) C where C only depends on p. 3. Proof of Theorems 1.1 and Theorems 1.2 Proof of Theorem 1.1. Suppose that f is a solution of (1), then f (z) is an entire function by Lemma 2.1. By Lemma 2.2, p p1 X Z (k) 1 1 2 − |z| (α) f (z) e 2 dm(z) kf kp ≤ C |f (0)| + (1 + |z|)k C α≤k−1 p p1 Z (k) 1 − 21 |z|2 = C1 + C dm(z) , f (z) (1 + |z|)k e C P where C1 = C α≤k−1 |f (α) (0)|. Using (1) and Minkowski inequality, ! p1 − 1 |z|2 p Z X k−1 2 e dm(z) kf kp ≤ C1 + C Ai (z)f (i) (z) − Ak (z) k (1 + |z|) C i=0  ! 1p  p Z 1 2 k−1 |z| − X 2 dm(z)  Ai (z)f (i) (z) e ≤ C1 + C  k (1 + |z|) C i=0

p ! p1 Z − 12 |z|2 Ak (z)e + , dm(z) k (1 + |z|) C p1 Z X k−1 − 12 |z|2 p (i) e |Ai (z)| f (z) dm(z) ≤ C1 + C sup k−i i (1 + |z|) (1 + |z|) z∈C C i=0 Z 1 2 p |z| − Ak (z)e 2 dm(z)) 1p . + ( (1 + |z|)k C

8


Using Lemma 2.2 again, kf (z)kp ≤ C1 + C

X k−1 i=0

|Ai | sup{ }Di kf kp + Dk kϕk kp , k−i z∈C (1 + |z|)

where ϕk (z) is the kth primitive function of Ak (z). Therefore !! k−1 X |Ai | kf kp 1 − C sup Di ≤ C1 + Dk kϕk kp k−i (1 + |z|) z∈C i=0

If kf kp = ∞, it contradicts the condition of Theorem 1.1. Therefore, kf (z)kp ≤ and f ∈ F p .

C1 + CDk kϕk kp < +∞ Pk−1 |Ai | 1 − C i=0 Di supz∈C { (1+|z|) } k−i

Proof of Theorem 1.2. Suppose that f (z) is a solution of (1), then f (z) is an entire function by Lemma 2.1. Using Lemma 2.2, (1) and Minkowski inequality, X Z k k−1 X 1 1 − 12 |z|2 p (α) (k) e | dm(z)) p kf (z)kF p,k ≤ Ci |f (0)| + ( |f (z) k−i (1 + |z|) C i=0 α=i p1 1 2 k X Z e− 2 |z| p (k−1) | dm(z) = D1 + Ci |(Ak−1f + · · · + A0 (z)f − Ak ) (1 + |z|)k−i C i=0 X p1 1 2 k k−1 Z X e− 2 |z| (j) p Ci |Aj f ≤ D1 + | dm(z) k−i (1 + |z|) C i=0 j=0 p1 Z 1 2 k X Ak e− 2 |z| p + Ci | | dm(z) k−i C (1 + |z|) i=0 Z p1 k k−1 − 21 |z|2 X X |Aj | (j) e p |f | dm(z) ≤ D1 + Ci sup k−i−j j (1 + |z|) (1 + |z|) z∈C C i=0 j=0 1p Z k X Ak e− 12 |z|2 p + Ci dm(z) , k−i C (1 + |z|) i=0 Pk−1 (α) P |f (0)|. Using Lemma 2.2 again, where D1 = ki=0 Ci α=i k k k−1 X X X |Aj | (i) kf kp + Ci Fi kϕk kp Ci sup Ej kf (z)kF p,k ≤ D1 + k−i−j (1 + |z|) z∈C i=0 i=0 j=0 k k−1 k X X X |Aj | (i) ≤ D1 + Ci Ej sup Ci Fi kϕk kp kf kF p,k + k−i−j (1 + |z|) z∈C i=0 j=0 i=0 k k−1 X X |Aj | ≤ D1 + Ci kf kF p,k + Gkϕk kF p,k sup Ej k−i−j (1 + |z|) z∈C i=0 j=0


9

where G = max0≤i≤k {Ci Fi }.So k k−1 X X kf (z)kF p,k 1 − Ci sup Ej i=0

j=0

z∈C

|Aj | ≤ D1 + Gkϕk kF p,k . (1 + |z|)k−i−j

If kf kF p,k = ∞, it contradicts the condition of Theorem 1.2. Therefore, kf (z)kF p,k ≤

1−

and f ∈ F p,k .

D1 + Gkϕk kF p,k < +∞ Pk−1 |Aj | i=0 Ci j=0 Ej supz∈C { (1+|z|)k−i−j }

Pk

4. Proof of Theorem 1.3 Proof of Theorem 1.3. If f (z) is a solution of (1), from Lemma 2.3, Z r 1 iθ iθ |f (re )| ≤ C( max |Ak (xe )| + 1) exp (δ + kc max |Aj (seiθ )| k−j )ds. 0≤x≤r

0≤j≤k−1

0

Since Ak (z) is not a constant function and there exists a nonconstant function among Aj (z), j = 0, · · · , k − 1, Z r 1 iθ iθ |f (re )| ≤ C(2 max |Ak (xe )|) exp (2kc max |Aj (seiθ )| k−j )ds 0≤x≤r

0≤j≤k−1

0

1

for sufficiently large r > r1 . Since |Aj (reiθ )| ≤ φ 2 (r)/r, r > r0 , we have Z r 1 1 1 φ 2 (s) k−j φ 2 (r) iθ ( exp 2kc max ) ds |f (re )| ≤ 2CD 0≤j≤k−1 R r s for r > R = max{r0 , r1 , 1}, where Z R 1 D = exp (2kc max |Aj (seiθ )| k−j )ds. 0

0≤j≤k−1

Since max

0≤j≤k−1

Z

r

R

1

φ 2 (s) s

1 ! k−j

1 2

ds ≤ φ (r)

Z

r

1

1

s− k ≤ φ 2 (r)

R

k 1− 1 r k, k−1

we have 1

1 φ 2 (r) 1− k1 2 , |f (re )| ≤ D1 exp D2 φ (r)r r iθ

R < r < ∞,

k where D1 = 2CD and D2 = 2kc k−1 . Since φ is in the class I and Lemma 2.6, ′ there exists R > 0 such that φ(r) > r 2 for r > R′ . Since f (z) is an entire function, by using Lemma 2.5,

10


kf kpF p,q φ

Z

|f (z)|p e−pφ(z) φq (z)dm(z) C Z |f (z)|p e−pφ(z) φq (z)dm(z) ≤M =

|z|≥R1

p 1 1 1 φ 2 (r) 1− ≤M D1 exp D2 φ 2 (r)r k e−pφ(r) φq (r)rdrdθ r 0 R1 Z 2π Z ∞ p +q 1 1 φ 2 (r) p 1− ≤ MD1 exp p(D2 φ 2 (r)r k − φ(r)) drdθ r p−1 0 R1 Z ∞ p +q φ 2 (r) p(D2 φ 21 (r)r1− k1 −φ(r)) p dr, ≤ 2πMD1 e r p−1 R1 Z

2π

Z

∞

1

1

where R1 = max{R, R′ }. Since D2 r 1− k − φ 2 (r) < −1, when r > R1 . Thus Z ∞ p +q φ 2 (r) −pφ 21 (r) p p e kf kF p,q ≤ 2πMD1 dr < ∞. φ r p−1 R

Therefore, f ∈ Fφp,q .

5. Proof of Theorems 1.4 and Theorem 1.5 Proof of Theorem 1.4. By (1), 1p Z 1 2 |(f (k) (z) + Ak−1 (z)f (k−1) (z) + · · · + A0 (z)f (z))e− 2 |z| |p dm(z) kAk kp = C

≤

Z

≤ kf

C (k)

|f

(k) − 12 |z|2 p

e

| dm(z)

p1

+···+

Z

− 21 |z|2 p

|A0 f e C

kp + |Ak−1 |kf (k−1) kp + · · · + |A0 |kf kp

| dm(z)

p1

≤ C(kf (k) kp + kf (k−1) kp + · · · + kf kp ), where C = maxi≤k−1 {|Ai |}. Since there exists a solution f ∈ F p,k , we get Ak ∈ F p. Proof of Theorem 1.5. Using (1), we obtain (k) f′ Ak f + · · · + A1 . |A0 (z)| ≤ + f f f

By Lemma 2.4, for α > 1, there exist sufficient large r1 and a set W of measure zero such that (j) iθ j α f (re ) ≤ C T (αr, f ) log r log T (αr, f ) , j = 1, · · · , k, f (reiθ ) r

for r > r1 and θ ∈ [0, 2π)\W , where T (αr, f ) = log+ M(αr, f ) and M(αr, f ) is the maximum of |f (αreiθ )| at θ ∈ [0, 2π). By the condition of Theorem 1.5, there


11

exists a solution f (z) such that for sufficient large r2 , T (αr, f ) = log+ M(αr, f ) ≤ log ee

αr

= eαr ,

for r > r2 . Thus α α Ak (reiθ ) log r log r iθ αr k iθ αr + C (e |A0 (re )| ≤ αr) + · · · + |A1 (re )|e αr , f (reiθ ) r r

for r > R = max{r1 , r2 } and θ ∈ [0, 2π)\W . Since A0 (z) is an entire function, using Lemma 2.5, Z |z| p kA0 kF p,q = |A0 (z)|p e−pe (z)eq|z| dm(z) er C Z 2π Z ∞ r ≤D |A0 (reiθ )|p e−pe (z)eqr rdrdθ. 0

R

It follows that

Z 2π Z ∞ (αeαr logα r)pk eqr Ak (reiθ ) p −per qr | e (z)e rdrdθ + D rdr ≤D | k f (reiθ ) eper 0 R 0 R Z 2π Z ∞ αr logα r)p eqr iθ p (αe + · · · + D1 |A1 (re )| rdrdθ. eper 0 R Since p ≥ 1, Z 2π Z ∞ (αeαr logα r)pk eqr rdr < ∞. eper 0 R kA0 kpF p,q er

Since

Ak (reiθ ) f (reiθ )

Z

2π

∞

Z

∈ Fep,q r , Z

0

2π

Z

∞

|

R p F 1 er , 2

Since Aj (z) belong to Z 2π Z 0

Z

Ak (reiθ ) p −per |e (z)eqr rdrdθ < ∞. f (reiθ )

j = 1, · · · , k − 1,

∞

R 2π

0

Therefore, A0 (z) ∈ Fep,q r .

iθ

|Aj (re )| Z

∞

αr p (αe

logα r)jp eqr rdrdθ ≤ eper p r

|Aj (reiθ )|p e− 2 e rdrdθ < ∞.

R

6. Proof of Theorems 1.6 and Theorem 1.7 Proof of Theorem 1.6. Let f be any solution of (2), then Z z ′ f (z) = − f (ζ)A(ζ)dζ + f ′ (0), z ∈ C. 0

If g satisfies the reproducing formula, g ∈ Fφ2 . Since f is an entire function, there Pn P j 2 j exists a finite Taylor expansion f (z) = ∞ j=0 aj z ∈ Fφ . j=0 aj z such that fn = Therefore, Z z

′

f (z) = −

lim fn (ζ)A(ζ)dζ + f ′ (0),

0 n→∞

z ∈ C.

12


By the reproducing formula and Fubini’s theorem, for φ in the class I, we get Z z Z ′ f (z) = − lim fn (η)Kζ (η)e−2φ(η) dm(η) A(ζ)dζ + f ′ (0) n→∞ C Z z Z0 −2φ(η) =− lim fn (η)e Kζ (η)A(ζ)dζ dm(η) + f ′ (0), z ∈ C. C n→∞

0

−2 Since Kζ (0) = Σ∞ n=0 en (ζ)en (0) = δ0 , and using Lemma 2.7, we get Z Z z ′ ′ Kζ′ (η)A(ζ)dζ (1 + φ′ (η))−2 e−2φ(η) dm(η)− f (z) = − lim fn (η) C n→∞ 0 Z z lim fn (0) Kζ (0)A(ζ)dζ + f ′ (0) n→∞ 0 Z z Z ′ = − f (η) Kζ′ (η)A(ζ)dζ (1 + φ′ (η))−2 e−2φ(η) dm(η)− 0 ZC z f (0) δ0−2 A(ζ)dζ + f ′ (0), z ∈ C. 0

It follows that ′

−φ(z)

|f (z)|e

Z Z z ′ −φ(η) | ≤ sup |f (η)|e Kζ′ (η)A(ζ)dζ (1 + φ′ (η))−2 e−φ(η) dm(η)|e−φ(z) η∈C C 0 Z z + |f (0) δ0−2 A(ζ)dζ |e−φ(z) + |f ′ (0)|, z ∈ C. 0

Then, we get ′

kf k

Fφ∞

Z Z z Kζ′ (η)A(ζ)dζ (1 + φ′ (η))−2 e−φ(η) dm(η)|e−φ(z) ≤ kf k sup | z∈C Z zC 0 + sup |f (0) δ0−2 A(ζ)dζ |e−φ(z) + |f ′(0)| ′

Fφ∞

z∈C

Thus ′

kf kFφ∞

0

1 − sup | z∈C

Z

0

z

′ −2 −φ(η) −φ(z) dm(η)|e Kζ′ (η)A(ζ)dζ (1 + φ (η)) e

≤ sup |f (0) z∈C

′

C

Z

Z

0

z

δ0−2 A(ζ)dζ |e−φ(z) + |f ′(0)|

If kf kFφ∞ = ∞, it contradicts the condition of Theorem 1.6. Therefore, Z z −φ(z) 1 −2 ′ ′ sup |f (0) δ0 A(ζ)dζ |e + |f (0)| < ∞ kf kFφ∞ ≤ 1 − XK (A) z∈C 0

and f ′ ∈ Fφ∞ .

Proof of Theorem 1.7. By the proof of Theorem 1.6, we get Z z Z ′ ′ f (z) = − f (η) Kζ′ (η)A(ζ)dζ (1 + φ′ (η))−2e−2φ(η) dm(η)− 0 ZC z f (0) δ0−2 A(ζ)dζ + f ′ (0), z ∈ C. 0


13

It follows that − 12 |z|2

′

|f (z)|e

Z

′

≤ kf k∞ | Z + |f (0)

C z

0

Then, we get

Z

z

0

1 1 2 2 Kζ′ (η)A(ζ)dζ (1 + φ′ (η))−2 e−2φ(η)+ 2 |η| dxdy|e− 2 |z|

1 2 δ0−2 A(ζ)dζ |e− 2 |z| + |f ′ (0)|,

z ∈ C.

Z Z z ′ −2 −2φ(η)+ 12 |η|2 − 12 |z|2 ′ kf k∞ ≤ kf k∞ sup | Kζ (η)A(ζ)dζ (1 + φ (η)) e dm(η)|e z∈C C 0 Z z 1 2 δ0−2 A(ζ)dζ |e− 2 |z| + |f ′ (0)| + sup |f (0) ′

′

z∈C

0

Thus

Z kf k∞ 1 − sup | ′

z∈C

C

Z

0

Kζ′ (η)A(ζ)dζ

≤ sup |f (0) z∈C

′

z

Z

z 0

1 1 2 2 (1 + φ′(η))−2 e−2φ(η)+ 2 |η| dm(η)|e− 2 |z|

1 2 δ0−2 A(ζ)dζ |e− 2 |z| + |f ′ (0)|

If kf k∞ = ∞, it contradicts the condition of Theorem 1.7. Therefore, Z z − 1 |z|2 1 −2 ′ ′ 2 sup |f (0) δ0 A(ζ)dζ |e + |f (0)| < ∞ kf k∞ ≤ 1 − YK (A) z∈C 0

and f ′ ∈ F ∞ .

7. Proof of Theorems 1.8 Proof of Theorem 1.8. By the proof of Theorem 1.6, we get Z z Z ′ −2φ(η) Kζ (η)A(ζ)dζ dm(η) + f ′ (0) f (z) = − lim fn (η)e n→∞ 0 ZC Z z = − f (η)e−2φ(η) Kζ (η)A(ζ)dζ dm(η) + f ′ (0), z ∈ C. C

0

Since the condition of Theorem 1.8 satisfies Lemma 2.8, we get Z e−2φ(|z|) 2 ′ 2 kf kF 2 ≤ C |f (0)| + |f (z)| dm(z) φ (1 + φ′ (|z|))2 C Z Z Z z e−2φ(|z|) 2 −2φ(η) Kζ (η)A(ζ)dζ dm(η)| ≤ C |f (0)| + | f (η)e dm(z)+ (1 + φ′ (|z|))2 C C 0 Z e−2φ(|z|) ′ dm(z) , z ∈ C. |f (0)| ′ 2 C (1 + φ (|z|)) By Lemma 2.8, we get Z e−2φ(|z|) dm(z) ≤ Ckzk2F 2 . ′ 2 φ C (1 + φ (|z|))

14


Using Lemma 2.6, it is easy to get that there exists a positive number M such that kzk2F 2 < M. Therefore, φ Z Z z e−2φ(|z|) 2 2 kf kF 2 ≤ P + kf kF 2 sup dm(z) Kζ (η)A(ζ)dζ φ φ (1 + φ′ (|z|))2 η∈C C 0 where P = C(|f (0)|2 + |f ′ (0)|CM). Then Z Z z 2 Kζ (η)A(ζ)dζ kf kF 2 1 − sup φ

η∈C

C

0

e−2φ(|z|) dm(z) (1 + φ′ (|z|))2

≤ P.

If kf k2F 2 = ∞, it contradicts the condition of Theorem 1.8. Therefore, φ

kf k2F 2 < φ

P