Weighted $L^p$ estimates for $bar{partial}$ on a convex domain with piecewise smooth boundary in $mathbb C^2$

J. Korean Math. Soc. 44 (2007), No. 6, pp. 1417–1428

WEIGHTED L ^p ESTIMATES FOR ¯ ∂ ON A CONVEX

DOMAIN WITH PIECEWISE SMOOTH BOUNDARY IN C ²

Hong Rae Cho and Jong-Do Park

Reprinted from the

Journal of the Korean Mathematical Society Vol. 44, No. 6, November 2007

c

°2007 The Korean Mathematical Society

WEIGHTED L ^p ESTIMATES FOR ¯ ∂ ON A CONVEX

DOMAIN WITH PIECEWISE SMOOTH BOUNDARY IN C ²

Hong Rae Cho and Jong-Do Park

Abstract. We obtain weighted L

estimates (1 ≤ p < ∞) for ∂ on convex domains with piecewise smooth boundaries in C

by using explicit formulas of solutions introduced by Berndtsson and Andersson.

1. Introduction and statement of the result

In this paper we investigate weighted L ^p estimates (1 ≤ p < ∞) for solutions of the ¯ ∂-equation on convex domains with piecewise smooth boundaries in C ² . For each j = 1, . . . , N let D j = {z ∈ C ² : ρ j (z) < 0} and assume that ρ j ∈ C ² (U j ), where U j is a neighborhood of bD j . Further assume that dρ j 6= 0 on U j , j = 1, . . . , N . By a convex domain with piecewise C ² -boundary in C ² , we mean a bounded convex domain D = D 1 ∩ · · · ∩ D N in C ² satisfying the transversal condition: ∂ρ i

∧· · ·∧∂ρ i

6= 0 on ∩ ^` _k=1 U i

for 1 ≤ i 1 < · · · < i ` ≤ N . For z ∈ D we define ρ(z) by

1 ρ(z) =

X N j=1

1 ρ j (z) . Then ρ ∈ C ² (D) and

1 N inf

1≤j≤N (−ρ j ) ≤ −ρ ≤ inf

1≤j≤N (−ρ j ).

Hence D = {z ∈ C ² : ρ(z) < 0} and −ρ(z) ∼ δ(z), z ∈ D, where δ(z) is the distance function from z to the boundary of D.

For 1 ≤ p < ∞ and α > 0, we define the weighted L ^p space L ^p,α (D) = {f ; kf k _L

_(D) < ∞},

where k·k _L

_(D) is the weighted L ^p -norm with respect to the weighted measure

|ρ| ^α−1 dV. In the piecewise smooth case D = ∩ ^N _j=1 D j , usually we consider the

Received April 5, 2006.

2000 Mathematics Subject Classification. 32W05, 32A26.

Key words and phrases. weighted L

estimates for ∂, convex domain, piecewise smooth boundary, totally convex.

This work was supported by the Korea Research Foundation Grant funded by Korea Government(MOEHRD, Basic Research Promotion Fund)(KRF-2005-070-C00007).

c

°2007 The Korean Mathematical Society

1417

weighted measure Q _N

j=1 |ρ j | ^α−1 dV . However in this paper we use the weighted measure |ρ| ^α−1 dV . Since |ρ| is equivalent to the distance function, it is perhaps worthwhile to consider our weighted measure. Estimates for the case of weight Q _N

j=1 |ρ j | ^α−1 dV are simple like as them of the smooth case. However, estimates for the case of weight |ρ| ^α−1 dV are difficult (see Lemma 4.1 and Lemma 4.2).

It will be the first time to consider this weighted measure for the weighted L ^p -norms on piecewise smooth domains.

Definition 1.1. We say that a convex domain D = {z ∈ C ⁿ ; ρ(z) < 0} is totally convex at the boundary point ζ ₀ in the complex directions if

D ∩ (H ζ

(∂D) + {ζ 0 }) = {ζ 0 }, where H ζ

(∂D) is the complex tangential space of ∂D at ζ 0 .

In [12], Range introduced the total convexity to study the Carath´eodory metric and holomorphic mappings. A bounded convex domain in C ⁿ with real analytic boundary is totally convex at each boundary point in the complex directions. The following is the main theorem in this paper.

Theorem 1.2. Let D be a convex domain with piecewise C ² -boundary in C ² . Let f be a ∂-closed (0, 1)-form in L ^p,α _(0,1) (D), 1 ≤ p < ∞, α > 0. For 1 ≤ j ≤ N , we let σ j = {z ∈ C ² ; ρ j (z) = 0, ρ k (z) ≤ 0 for k 6= j}. If D j are totally convex at each point in σ j in the complex direction, then there exists a linear operator S : L ^p,α _(0,1) (D) → L ^p,α (D) satisfying ∂(Sf ) = f and kSf k _p,α ≤ C p,α kf k p,α .

Polking [9] and Range [10] obtained L ^p (1 < p < ∞) and H¨older estimates for ∂ on convex domains with smooth boundaries in C ² , respectively. However, for L ^p estimates of the cases p = 1 and ∞, Range in his survey paper [11]

proposed the problem (Problem 4): Are there L ¹ and L ^∞ estimates for ∂ on convex domains in C ² ? As a special case of Theorem 1.2 our result solves L ¹ estimates for ∂ on convex domains in C ² (see [1], also).

In smooth convex domains we need not the total convexity for estimates for

∂. However, in piecewise smooth convex domains we need the total convexity ¯ for our estimates in the proofs of Lemma 4.1 and Lemma 4.2.

In this paper, for L ^p estimates, we use the weighted Cauchy-Fantappi`e kernel constructed by Berndtsson and Andersson [2], which was also used by Menini [6]

to prove L ^p estimates in domains with piecewise smooth strictly pseudoconvex boundaries. For other cases of the ¯ ∂-problem on domains with piecewise smooth boundaries we can refer ([4], [5], [7], [8], [13], [14]).

2. An example of total convexity For s > 0, let

D s = {(z 1 , z 2 ) ∈ C ² : |z 1 | ² + e ^A e ^−1/|z

^|

< 1},

where A = 1 + 2/s.

Lemma 2.1. Then D s is a smoothly bounded convex domain in C ² .

Before giving the proof we consider the following characterization of the convexity.

Lemma 2.2 ([3]). A real-valued C ² -function λ on an open set U ⊂ C is convex on U if and only if

∆ C λ(z) := ∂ ² λ

∂z∂ ¯ z (z) −

¯ ¯

¯ ¯ ∂ ² λ(z)

∂z ²

¯ ¯

¯ ¯ ≥ 0 for z ∈ U.

(2.1)

Proof of Lemma 2.1. Let σ s (t) = e ^A e ⁻

, 0 ≤ t ≤ ( _s+2 ^s ) ^2/s . Then we have σ _s ⁰ (t) > 0, 0 < t ≤

µ s s + 2

¶ _2/s , (2.2)

2σ ⁰⁰ _s (t)t + σ _s ⁰ (t) > 0, 0 < t ≤ µ s

s + 2

¶ _2/s . (2.3)

Let λ(z) = σ _s (|z| ² ) for z ∈ C with |z| ≤ ( _s+2 ^s ) ^1/s . Then we have

∆ C λ(z) = ∂ ² λ

∂z∂ ¯ z (z) −

¯ ¯

¯ ¯ ∂ ² λ(z)

∂z ²

¯ ¯

¯ ¯

= σ _s ⁰⁰ (|z| ² )|z| ² + σ _s ⁰ (|z| ² ) − |σ ⁰⁰ _s (|z| ² )¯ z ² |

=

( σ _s ⁰ (|z| ² ) if σ _s ⁰⁰ (|z| ² ) ≥ 0

2σ ⁰⁰ _s (|z| ² )|z| ² + σ ⁰ _s (|z| ² ) if σ _s ⁰⁰ (|z| ² ) ≤ 0.

By (2.2) and (2.3), we get ∆ C λ(z) ≥ 0. By Lemma 2.2, D s is convex. ¤ Lemma 2.3. D s is totally convex at each boundary point in the complex di- rections.

Proof. Let ρ s (z) = |z 1 | ² +σ s (|z 2 | ² )−1. Then ρ s is a boundary defining function for D s . Let

φ s (ζ, z) = h∂ρ s (ζ), ζ − zi.

Note that H ζ

(bD) + {ζ 0 } = {z ∈ C ² ; φ s (ζ 0 , z) = 0} and

2 Re φ _s (ζ, z) = ρ _s (ζ) − ρ _s (z) + |ζ ₁ − z ₁ | ² + σ _s (|z ₂ | ² ) − σ _s (|ζ ₂ | ² ) + 2 Re

· ∂σ s

∂ζ ₂ (|ζ 2 | ² )(ζ 2 − z 2 )

¸ . For λ(z) = σ s (|z| ² ), it is enough to prove that

λ(z) − λ(ζ) + 2 Re

· ∂λ

∂ζ (ζ)(ζ − z)

¸

> 0 for ζ 6= z.

By the Taylor expansion, there exists 0 < θ < 1 such that λ(z) − λ(ζ) + 2 Re

· ∂λ

∂ζ (ζ)(ζ − z)

¸

= H Z λ(z − ζ),

where H Z λ is the real Hessian of λ at Z and Z = ζ + θ(z − ζ). Then we have H _Z λ(z − ζ) = Re

· ∂ ² λ

∂ζ ² (Z)(z − ζ) ²

¸ + ∂ ² λ

∂ζ∂ζ (Z)|z − ζ| ²

= Re h

σ _s ⁰⁰ (|Z| ² )Z ² (z − ζ) ² i

+ (σ ⁰⁰ _s (|Z| ² )|Z| ² + σ ⁰ _s (|Z| ² ))|z − ζ| ²

≥

( σ _s ⁰ (|Z| ² )|z − ζ| ² if σ _s ⁰⁰ (|Z| ² ) ≥ 0

(σ ⁰ _s (|Z| ² ) + 2|Z| ² σ ⁰⁰ _s (|Z| ² ))|z − ζ| ² if σ _s ⁰⁰ (|Z| ² ) ≤ 0.

By (2.2) and (2.3), we obtain the desired result for the case Z 6= 0.

In case Z = 0, we have ζ = −θ(z − ζ). Thus we have

∂λ

∂ζ (ζ)(ζ − z) = e ^A e ^−1/|ζ|

sζ(ζ − z)

2|ζ| ^s+2 = e ^A e ^−1/|ζ|

s 2θ|ζ| ^s . Since |ζ| ^s ≤ s/(s + 2) and ζ 6= 0, it follows that

λ(z) − λ(ζ) + 2 Re

· ∂λ

∂ζ (ζ)(ζ − z)

¸

≥ −λ(ζ) + 2Re

· ∂λ

∂ζ (ζ)(ζ − z)

¸

= e ^A e ^−1/|ζ|

µ s

θ|ζ| ^s − 1

¶

≥ e ^A e ^−1/|ζ|

µ s

θ + 2 θ − 1

¶

> 0,

if Z = 0 and ζ 6= z. ¤

3. Construction of the solution for the ¯ ∂-equation Before proving Theorem 1.2, we need some notations. Let

s(ζ, z) = X 2 j=1

(ζ j − z j )dz j

and Q ^k (ζ, z) = − ∂ρ k (ζ)

ρ k (ζ) for k = 1, . . . , N . For any r > 1, we define the real- valued function G(t) = (1 + t) ^−r and write G ^(α _k

⁾ = G ^(α

⁾ (hQ ^k , ζ − zi), where G ^(α

⁾ (t) is the α k -th derivative of G(t). For any r > 1 we define

K ^r (ζ, z) = X 2 j=1

X

α

+···+α

=2−j

c α

à _N Y

k=1

G ^(α _k

⁾

! s ∧ (∂s) ^j−1 hs, ζ − zi ^j

^ N k=1

(∂ ζ Q ^k ) ^α

,

which was introduced by Berndtsson and Andersson [2]. Then for a continuous (0, 1)-form f in D with ∂f = 0, this kernel K ^r (ζ, z) gives a solution operator

Sf (z) = Z

ζ∈D

f (ζ) ∧ K ^r (ζ, z), z ∈ D such that

f = ∂(Sf ).

For a smooth form f , this formula holds for any r > 0. Define φ k (ζ, z) = h∂ρ k (ζ), ζ − zi − ρ k (ζ). The following estimate is a well-known consequence of the convexity of D k :

(3.1) 2Re φ k (ζ, z) ≥ −ρ k (ζ) − ρ k (z) for all ζ, z ∈ D k . Note that

(3.2) ∂ ζ Q ^k (ζ, z) = − ∂∂ρ k (ζ)

ρ k (ζ) + ∂ρ k (ζ) ∧ ∂ρ k (ζ) ρ k (ζ) ² and

(3.3) G ^(α _k

⁾ = (−1) ^α

r(r + 1) · · · (r + α k − 1)

µ −ρ k (ζ) φ k (ζ, z)

¶ _r+α

. By (3.2) and (3.3), we see that

Sf (z) = S 1 f (z) + X N k=1

³

S 2,k f (z) + S 3,k f (z) ´ , where

S ₁ f (z) = Z

ζ∈D

f (ζ) ∧ Y N j=1

|ρ _j (ζ)| ^r K ₁ ^r (ζ, z), (3.4)

S 2,k f (z) = Z

ζ∈D

f (ζ) ∧ Y

j6=k

|ρ j (ζ)| ^r ∂ρ k (ζ) ∧ |ρ k (ζ)| ^r−1 K _2,k ^r (ζ, z), (3.5)

S 3,k f (z) = Z

ζ∈D

f (ζ) ∧ Y N j=1

|ρ j (ζ)| ^r K _3,k ^r (ζ, z) for k = 1, . . . , N.

(3.6)

Here we have the following estimates of the kernels in (3.4), (3.5), and (3.6):

|K ₁ ^r (ζ, z)| . 1

|ζ − z| ³ Q _N

j=1 |φ j (ζ, z)| ^r (3.7)

|K _ν,k ^r (ζ, z)| . 1

|ζ − z| Q

j6=k |φ j (ζ, z)| ^r |φ k (ζ, z)| ^r+1 , (3.8)

for ν = 2, 3, k = 1, . . . , N . In order to estimate (3.4), (3.5), and (3.6), we need two crucial integral estimates which play important parts in the proof of Theorem 1.2. We will prove them in the next section.

4. Integral estimates

Let ˜ D _j = {z ∈ D : −ρ _j (z) = inf _1≤k≤N (−ρ _k (z))}. Then |ρ _j (z)| ∼ δ(z) for z ∈ ˜ D j and D = ∪ ^N _j=1 D ˜ j . We recall that σ j = {z ∈ C ² : ρ j (z) = 0, ρ k (z) ≤ 0 for k 6= j}.

Lemma 4.1. Let 1 ≤ ν, µ ≤ N and let r be large enough and α > 0.

(i) For 0 ≤ ² < α + 1 there exists a constant C α,² such that Z

z∈ ˜ D

|ρ µ (z)| ^α−²

|ζ − z| ³ |φ ν (ζ, z)| ^r dV (z) ≤ C α,² |ρ ν (ζ)| ^{α−²−r+1} , ζ ∈ ˜ D ν , (ii) For 0 ≤ ² < α there exists a constant C α,² such that

Z

z∈ ˜ D

|ρ _µ (z)| ^α−1−²

|ζ − z||φ ν (ζ, z)| ^r dV (z) ≤ C α,² |ρ ν (ζ)| ^{α−²−r+1} , ζ ∈ ˜ D ν .

Lemma 4.2. Let k and r be same as in Lemma 4.1. For any ² > 0, there exists a constant C ² such that

(i) Z

ζ∈ ˜ D

|ρ ν (ζ)| ^r−1−²

|ζ − z| ³ |φ ν (ζ, z)| ^r dV (ζ) ≤ C ² |ρ µ (z)| ^−² , z ∈ ˜ D µ , (ii)

Z

ζ∈ ˜ D

|ρ ν (ζ)| ^r−2−²

|ζ − z||φ _ν (ζ, z)| ^r dV (ζ) ≤ C ² |ρ µ (z)| ^−² , z ∈ ˜ D µ .

Proof of Lemma 4.1. For the proof of (i) of Lemma 4.1, it is enough to prove that for (ζ 0 , z 0 ) ∈ ˜ D ν × ˜ D µ there exist neighborhoods U ζ

of ζ 0 and U z

of z 0

such that

I ₁ (ζ) :=

Z

z∈ ˜ D

∩U

|ρ µ (z)| ^α−²

|ζ − z| ³ |φ ν (ζ, z)| ^r dV (z)

≤C α,² |ρ ν (ζ)| ^{α−²−r+1} for ζ ∈ ˜ D ν ∩ U ζ

.

If φ ν (ζ 0 , z 0 ) 6= 0, then we can assume that |φ ν (ζ, z)| > 1 in small neighbor- hoods U ζ

of ζ 0 and U z

of z 0 . For fixed ζ ∈ U ζ

, we choose a local system t(z) = (t 1 , t 2 , t 3 , t 4 ) in U z

such that t 1 (z) = −ρ µ (z) and t 2 (ζ) = t 3 (ζ) = t 4 (ζ) = 0. If we write t ⁰ = (t 3 , t 4 ) then we have

I 1 (ζ) . Z

|t

|<1

Z ₁

−1

Z ₁

0

|t 1 | ^α−² dt 1 dt 2 dt ⁰ (|t 1 | + |t 2 | + |t ⁰ |) ³ .

Z ₁

0

|t 1 | ^α−² | log |t 1 ||dt 1 . 1.

Now we consider the case φ ν (ζ 0 , z 0 ) = 0. Since D ν is totally convex at each boundary point of σ ν in the complex direction, it follows that ζ 0 = z 0 ∈ σ ν ∩ σ µ

and

d z ρ ν ∧d z Imφ ν ∧d z ρ µ ∧d z Imφ µ = −∂ z ρ ν ∧ ¯ ∂ z ρ ν ∧∂ z ρ µ ∧ ¯ ∂ z ρ µ 6= 0 at ζ 0 = z 0 . Thus for fixed ζ ∈ U z

we can choose a local coordinate system

t 1 + it 2 = −ρ ν (z) + iIm φ ν (ζ, z) t 3 + it 4 = −ρ µ (z) + iIm φ µ (ζ, z) on z ∈ U z

. Note that

Re φ ν (ζ, z) & −ρ ν (ζ) − ρ ν (z)

≥ −ρ ν (ζ) − ρ µ (z) for ζ ∈ ˜ D ν , z ∈ ˜ D µ .

If we write t ⁰ = (t 1 , t 4 ) and introduce polar coordinates in t ⁰ with ξ = |t ⁰ |, then we have

I 1 (ζ) . Z

|ξ|<1 ξ∈C

Z ₁

0

Z ₁

−1

|t 3 | ^α−² dt 2 dt 3 dξ

(|t 2 | + |t 3 | + |ξ|) ³ (|t 2 | + |t 3 | + |ρ ν (ζ)|) ^r .

Z ₁

0

Z ₁

−1

|t 3 | ^α−² dt 2 dt 3

(|t 2 | + |t 3 |)(|t 2 | + |t 3 | + |ρ ν (ζ)|) ^r .

If we make the change of variables t 2 = |ρ ν (ζ)|t ⁰ ₂ and t 3 = |ρ ν (ζ)|t ⁰ ₃ , and omit the primes, then we have

I 1 (ζ) . |ρ ν (ζ)| ^{α−²−r+1} Z _∞

0

Z _∞

0

t ^α−² ₃ dt 2 dt 3

(t 2 + t 3 )(t 2 + t 3 + 1) ^r .

If α−² ≥ 0, we choose r so that r−α+² > 1 and we introduce polar coordinates in (t 2 , t 3 ) with τ = |(t 2 , t 3 )|. Then we have

I ₁ ⁰ :=

Z _∞

0

Z _∞

0

t ^α−² ₃ dt 2 dt 3

(t 2 + t 3 )(t 2 + t 3 + 1) ^r . Z _∞

0

dτ

(τ + 1) ^r−α+² . 1.

If −1 < α − ² < 0, we have I ₁ ⁰ .

Z _∞

0

Z _∞

0

dt 2 dt 3

t ^²−α+η ₂ t ^1−η ₃ (t 2 + t 3 + 1) ^r .

Z _∞

0

dt 2

t ^²−α+η ₂ (t 2 + 1) ^r/2 Z _∞

0

dt 3

t ^1−η ₃ (t 3 + 1) ^r/2 . 1, where we choose η and r so that 0 < ² − α + η < 1 and r > 2.

(ii) We will prove that for (ζ 0 , z 0 ) ∈ σ ν ∩ σ µ with φ ν (ζ 0 , z 0 ) = 0, there exist neighborhoods U ζ

of ζ 0 and U z

of z 0 such that

I 2 (ζ) :=

Z

z∈ ˜ D

∩U

|ρ µ (z)| ^α−1−²

|ζ − z||φ ν (ζ, z)| ^r dV (z)

≤ C α,² |ρ ν (ζ)| ^{α−²−r+1} , ζ ∈ ˜ D ν ∩ U ζ

. Similarly as the case (i), we have

I 2 (ζ) . Z

|ξ|<1 ξ∈C

Z ₁

−1

Z ₁

0

|t 3 | ^α−1−² dt 2 dt 3 dξ

|ξ|(|t 2 | + |t 3 | + |ρ ν (ζ)|) ^r .

Z ₁

−1

Z ₁

0

|t 3 | ^α−1−² dt 2 dt 3

(|t 2 | + |t 3 | + |ρ ν (ζ)|) ^r . |ρ ν (ζ)| ^{α−²−r+1}

Z _∞

−∞

Z _∞

0

|t 3 | ^α−1−² dt 2 dt 3

(|t 2 | + |t 3 | + 1) ^r . |ρ ν (ζ)| ^{α−²−r+1} ,

since α − ² > 0 and r − α + ² > 1. ¤

Proof of Lemma 4.2. (i) We will prove that for (ζ 0 , z 0 ) ∈ σ ν ∩σ µ with φ ν (ζ 0 , z 0 )

= 0, there exist neighborhoods U ζ

of ζ 0 and U z

of z 0 such that J 1 (z) :=

Z

ζ∈ ˜ D

∩U

|ρ ν (ζ)| ^r−1−²

|ζ − z| ³ |φ ν (ζ, z)| ^r dV (ζ) ≤ C ² |ρ µ (z)| ^−² , z ∈ ˜ D µ ∩ U z

. Since D ν is totally convex at each boundary point of σ ν , it follows that ζ 0 = z 0

and

d ζ ρ ν ∧d ζ Imφ ν ∧d ζ ρ µ ∧d ζ Imφ µ = −∂ ζ ρ ν ∧ ¯ ∂ ζ ρ ν ∧∂ ζ ρ µ ∧ ¯ ∂ ζ ρ µ 6= 0 at ζ 0 = z 0 . Thus for fixed z ∈ U z

we can choose a local coordinate system

u 1 + iu 2 = −ρ ν (ζ) + iIm φ ν (ζ, z) u 3 + iu 4 = −ρ µ (ζ) + iIm φ µ (ζ, z) on ζ ∈ U z

. Then we have

J 1 (z) . Z

|ξ|<1 ξ∈C

Z ₁

−1

Z ₁

0

|u 1 | ^r−1−² du 1 du 2 dξ

(|ξ| + |u 1 | + |u 2 |) ³ (|u 1 | + |u 2 | + |ρ µ (z)|) ^r .

Z ₁

−1

Z ₁

0

du 1 du 2

(|u 1 | + |u 2 |)(|u 1 | + |u 2 | + |ρ µ (z)|) ^1+² . |ρ µ (z)| ^−²

Z

(u

,u

)∈R

du 1 du 2

(|u 1 | + |u 2 |)(|u 1 | + |u 2 | + 1) ^1+² . |ρ µ (z)| ^−²

Z _∞

0

dτ (τ + 1) ^1+² . |ρ µ (z)| ^−² .

(ii) As the case (i), it holds that J 2 (z) :=

Z

ζ∈ ˜ D

∩U

|ρ ν (ζ)| ^r−2−²

|ζ − z||φ ν (ζ, z)| ^r dV (ζ) .

Z

|ξ|<1 ξ∈C

Z ₁

−1

Z ₁

0

|u 1 | ^r−2−² du 1 du 2 dξ

|ξ|(|u 1 | + |u 2 | + |ρ µ (z)|) ^r .

Z ₁

−1

Z ₁

0

|u 1 | ^r−2−² du 1 du 2

(|u 1 | + |u 2 | + |ρ µ (z)|) ^r . |ρ µ (z)| ^−²

Z _∞

−∞

Z _∞

0

|u ₁ | ^r−2−² du ₁ du ₂ (|u 1 | + |u 2 | + 1) ^r . Note that

Z _∞

−∞

Z _∞

0

|u 1 | ^r−2−² du 1 du 2

(|u 1 | + |u 2 | + 1) ^r . Z _∞

0

u ^r−2−² ₁ du 1

(u 1 + 1) ^r−1 .

Z ₁

0

du 1

(u 1 + 1) ^r−1 + Z _∞

1

du 1

(u 1 + 1) ^1+² . 1.

¤

5. L ^p estimates for ¯ ∂

Finally we note that Theorem 1.2 is a direct consequence of the next propo- sition.

Proposition 5.1. For 1 ≤ p < ∞ and α > 0, there exists a constant C p,α > 0 such that

kS 1 f k p,α + X 3 ν=2

X N k=1

kS ν,k f k p,α ≤ C p,α kf k p,α for ν = 2, 3 and k = 1, . . . , N.

Proof. At first we consider the operator S 1 f . We estimate the weighted L ^p norm of S 1 f by using the inequality (3.7) and the formula (3.4). Fix k with 1 ≤ k ≤ N . For the case p = 1, it follows from the inequality (3.1) and (i) of Lemma 4.1 that

kS 1 f k 1,α . Z

ζ∈D

|f (ζ)|

Y N j=1

|ρ j (ζ)| ^r dV (ζ) Z

z∈D

|ρ(z)| ^α−1

|ζ − z| ³ Q _N

j=1 |φ j (ζ, z)| ^r dV (z) .

X N ν,µ=1

Z

ζ∈ ˜ D

|f (ζ)||ρ ν (ζ)| ^r dV (ζ) Z

z∈ ˜ D

|ρ µ (z)| ^α−1

|ζ − z| ³ |φ ν (ζ, z)| ^r dV (z)

. X N ν=1

Z

ζ∈ ˜ D

|f (ζ)||ρ ν (ζ)| ^r |ρ ν (ζ)| ^α−r dV (ζ)

. X N ν=1

Z

ζ∈ ˜ D

|f (ζ)||ρ ν (ζ)| ^α dV (ζ).

If p > 1 and q is the conjugate exponent of p, H¨older inequality applied to S 1 f implies

|S 1 f (z)|

. X N ν=1

Z

ζ∈ ˜ D

|f (ζ)||ρ ν (ζ)| Y

j6=ν

|ρ j (ζ)| ^r |ρ ν (ζ)| ^r−1 |K ₁ ^r (ζ, z)||ρ ν (ζ)| ^² |ρ ν (ζ)| ^−² dV (ζ)

. X N ν=1



 Z

ζ∈ ˜ D

(|f (ζ)||ρ ν (ζ)|) ^p Y

j6=ν

|ρ j (ζ)| ^r |ρ ν (ζ)| ^r−1+²p |K ₁ ^r (ζ, z)|dv(ζ)





1/p

×



 Z

ζ∈ ˜ D

Y

j6=ν

|ρ j (ζ)| ^r |ρ ν (ζ)| ^r−1−²q |K ₁ ^r (ζ, z)|dV (ζ)





1/q

.

By (3.7) and the inequality (ii) of Lemma 4.2, we have Z

ζ∈ ˜ D

Y

j6=ν

|ρ j (ζ)| ^r |ρ ν (ζ)| ^r−1−²q |K ₁ ^r (ζ, z)|dV (ζ)

. Z

ζ∈ ˜ D

|ρ ν (ζ)| ^r−1−²q

|ζ − z| ³ |φ ν (ζ, z)| ^r dV (ζ) . |ρ _µ (z)| ^−²q . If we choose small ² such that ²p < α, then

kS 1 f k p,α

. X N ν,µ=1

Z

ζ∈ ˜ D

(|f (ζ)||ρ ν (ζ)|) ^p Y

j6=ν

|ρ j (ζ)| ^r |ρ ν (ζ)| ^r−1+²p dV (ζ)

× Z

z∈ ˜ D

|K ₁ ^r (ζ, z)||ρ µ (z)| ^α−1−²p dV (z)

. X N ν,µ=1

Z

ζ∈ ˜ D

(|f (ζ)||ρ ν (ζ)|) ^p |ρ ν (ζ)| ^r−1+²p dV (ζ)

× Z

z∈ ˜ D

|ρ µ (z)| ^α−1−²p

|ζ − z| ³ |φ ν (ζ, z)| ^r dV (z) .

X N ν=1

Z

ζ∈ ˜ D

(|f (ζ)||ρ ν (ζ)|) ^p |ρ ν (ζ)| ^α−1 dV (ζ).

Now we consider the operator S 2,k f . We estimate the weighted L ^p norm of S 2,k f by using the inequality (3.8) and the formula (3.5). For the case p = 1 it follows from (3.5) that

kS 2,k f k 1,α . X N ν,µ=1

Z

ζ∈ ˜ D

|f (ζ)|dV (ζ) Z

z∈ ˜ D

1

|ζ − z|

× Y

j6=k

|ρ j (ζ)| ^r

|φ j (ζ, z)| ^r

|ρ k (ζ)| ^r−1

|φ k (ζ, z)| ^r+1 |ρ µ (z)| ^α−1 dV (z).

We will show that A k := Y

j6=k

|ρ j (ζ)| ^r

|φ j (ζ, z)| ^r

|ρ k (ζ)| ^r−1

|φ k (ζ, z)| ^r+1 . |ρ ν (ζ)| ^r−2

|φ ν (ζ, z)| ^r . (5.1)

If k = ν, then

A ν . |ρ ν (ζ)| ^r−1

|φ ν (ζ, z)| ^r+1 . |ρ ν (ζ)| ^r−2

|φ ν (ζ, z)| ^r for any r > 2. If k 6= ν, then

A k . |ρ ν (ζ)| ^r

|φ ν (ζ, z)| ^r

|ρ k (ζ)| ^r−1

|φ k (ζ, z)| ^r+1 . |ρ ν (ζ)| ^r

|φ ν (ζ, z)| ^r 1

|ρ k (ζ)| ² . |ρ ν (ζ)| ^r−2

|φ ν (ζ, z)| ^r . Thus we have

kS 2,k f k 1,α . X N ν,µ=1

Z

ζ∈ ˜ D

|f (ζ)||ρ ν (ζ)| ^r−2 dV (ζ) Z

z∈ ˜ D

|ρ µ (z)| ^α−1

|ζ − z||φ ν (ζ, z)| ^r dV (z)

. X N ν=1

Z

ζ∈ ˜ D

|f (ζ)||ρ ν (ζ)| ^r−2 |ρ ν (ζ)| ^α−r+1 dV (ζ)

= X N ν=1

Z

ζ∈ ˜ D

|f (ζ)||ρ ν (ζ)| ^α−1 dV (ζ).

If p > 1 and q is the conjugate exponent of p, H¨older inequality applied to S 2,k f implies

|S 2,k f (z)|

. X N ν=1

Z

ζ∈ ˜ D

|f (ζ)| Y

j6=k

|ρ j (ζ)| ^r |ρ k (ζ)| ^r−1 |K _2,k ^r (ζ, z)||ρ ν (ζ)| ^² |ρ ν (ζ)| ^−² dV (ζ)

. X N ν=1



 Z

ζ∈ ˜ D

|f (ζ)| ^p Y

j6=k

|ρ j (ζ)| ^r |ρ k (ζ)| ^r−1 |K _2,k ^r (ζ, z)||ρ ν (ζ)| ^²p dV (ζ)





1/p

×



 Z

ζ∈ ˜ D

Y

j6=k

|ρ j (ζ)| ^r |ρ k (ζ)| ^r−1 |K _2,k ^r (ζ, z)||ρ ν (ζ)| ^−²q dV (ζ)





1/q

.

By using (3.8) and (5.1), we have Z

ζ∈ ˜ D

Y

j6=k

|ρ j (ζ)| ^r |ρ k (ζ)| ^r−1 |K _2,k ^r (ζ, z)||ρ ν (ζ)| ^−²q dV (ζ)

. Z

ζ∈ ˜ D

1

|ζ − z|

Y

j6=k

|ρ j (ζ)| ^r

|φ j (ζ, z)| ^r

|ρ k (ζ)| ^r−1

|φ k (ζ, z)| ^r+1 |ρ ν (ζ)| ^−²q dV (ζ) .

Z

ζ∈ ˜ D

|ρ ν (ζ)| ^r−2−²q

|ζ − z||φ ν (ζ, z)| ^r dV (ζ) . |ρ µ (z)| ^−²q . If we choose small ² such that ²p < α, then

kS 2,k f k ^p _p,α .

X N ν,µ=1

Z

ζ∈ ˜ D

|f (ζ)| ^p |ρ ν (ζ)| ^r−2+²p dV (ζ) Z

z∈ ˜ D

|ρ µ (z)| ^α−1−²p

|ζ − z||φ ν (ζ, z)| ^r dV (z)

. X N ν=1

Z

ζ∈ ˜ D

|f (ζ)| ^p |ρ ν (ζ)| ^r−2+²p |ρ ν (ζ)| ^{α−²p−r+1} dV (ζ)

= X N ν=1

Z

ζ∈ ˜ D

|f (ζ)| ^p |ρ ν (ζ)| ^α−1 dV (ζ).

The estimate of S 3,k f easily can be reduced to that of S 2,k f . ¤

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Hong Rae Cho

Department of Mathematics Pusan National University Pusan 609-735, Korea

E-mail address: [email protected] Jong-Do Park

Department of Mathematical Sciences Seoul National University

Seoul 151-747, Korea

E-mail address: [email protected] current address:

Department of Mathematics

Pohang University of Science and Technology Pohang 790-784, Korea

E-mail address: [email protected]