Open Access

An analogue of the Bernstein-Walsh lemma in Jordan regions of the complex plane

Journal of Inequalities and Applications20132013:570

https://doi.org/10.1186/1029-242X-2013-570

Received: 17 May 2013

Accepted: 7 November 2013

Published: 2 December 2013

Abstract

In this paper we continue to study two-dimensional analogues of Bernstein-Walsh estimates for arbitrary Jordan domains.

MSC:Primary 30A10; 30C10; secondary 41A17.

Keywords

algebraic polynomials conformal mapping Bernstein lemma

1 Introduction and main results

Let G C be a finite region, with 0 G , bounded by a Jordan curve L : = G , Δ : = { w : | w | > 1 } , Ω : = ext G ¯ (with respect to C ¯ ). Let w = Φ ( z ) be the univalent conformal mapping of Ω onto the Δ normalized by Φ ( ) = , Φ ( ) > 0 , and Ψ : = Φ 1 .

Let n denote the class of arbitrary algebraic polynomials P n ( z ) of degree at most n N .

Let A p ( G ) , p > 0 , denote the class of functions f which are analytic in G and satisfy the condition
f A p ( G ) : = ( G | f ( z ) | p d σ z ) 1 / p < ,

where σ denotes a two-dimensional Lebesgue measure.

When L is rectifiable, let L p ( L ) , p > 0 , denote the class of functions f which are integrable on L and satisfy the condition
f L p ( L ) : = ( L | f ( z ) | p | d z | ) 1 / p < .
From the well-known Bernstein-Walsh lemma [[1], p.101], we see that
| P n ( z ) | | Φ ( z ) | n P n C ( G ¯ ) , z Ω .
(1.1)
For R > 1 , let us set L R : = { z : | Φ ( z ) | = R } , G R : = int L R , Ω R : = ext L R . Then (1.1) can be written as follows:
P n C ( G ¯ R ) R n P n C ( G ¯ ) .
(1.2)

Hence, setting R = 1 + 1 n , according to (1.2), we see that the C-norm of a polynomial P n ( z ) in G ¯ R and G ¯ is equivalent, i.e., the norm P n C ( G ¯ R ) increases with no more than a constant with respect to P n C ( G ¯ ) .

In the case when L is rectifiable, a similar estimate of (1.2) type in space L p ( L ) was obtained in [2] as follows:
P n L p ( L R ) R n + 1 p P n L p ( L ) , p > 0 .
(1.3)
The Berstein-Walsh type estimation for regions with quasiconformal boundary [[3], p.97] in the space A p ( G ) , p > 0 , is contained in [4]:
P n A p ( G R ) c 2 R n + 1 p P n A p ( G ) , p > 0 ,
(1.4)

where R : = 1 + c 1 ( R 1 ) and c 1 > 0 , c 2 = c 2 ( c 1 , p , G ) > 0 are constants. Therefore, if we choose R = 1 + c 3 n , then (1.4) we can see that the A p -norm of polynomials P n ( z ) in G R and G is equivalent.

In this work, we study a problem similar to (1.4) in A p ( G ) , p > 0 , for regions with arbitrary Jordan boundary.

Now we can state our new result.

Theorem 1.1 Let p > 0 ; G be a Jordan region. Then, for any P n n , R 1 = 1 + 1 n and arbitrary R, R > R 1 , we have
P n A p ( G R ) c 4 R n + 2 p P n A p ( G R 1 ) ,
(1.5)

where c 4 = ( 2 e p 1 ) 1 p [ 1 + O ( 1 n ) ] , n .

The sharpness of (1.5) can be seen from the following remark:

Remark 1.1 For any n = 1 , 2 ,  , there exist a polynomial P n n , region G C and number R > R 1 = 1 + 1 n such that
P n A p ( G R ) ( 2 e p 1 ) 1 p R n + 2 p P n A p ( G R 1 ) .
(1.6)

2 Some auxiliary results

Let G C be a finite region bounded by the Jordan curve L. Let L R : = { z : | Φ ( z ) | = R , R > 1 } , G t : = int L t , Ω t : = ext L t .

We note that, throughout this paper, c 1 , c 2 , (in general, different in different relations) are positive constants.

Lemma 2.1 Let p > 0 ; f be an analytic function in | z | > 1 and have a pole of degree at most n, n 1 at z = . Then, for any R 1 and R > R 1 , we have
f A p ( R 1 < | z | < R ) ( R n p + 2 R 1 n p + 2 R 1 n p + 2 1 ) 1 p f A p ( 1 < | z | < R 1 ) .
(2.1)
Proof The function g ( z ) : = f ( z ) z n is analytic in | z | > 1 and continuous in | z | 1 . Applying Hardy’s convexity theorem [[5], p.9: Th.1.5], for any arbitrary R 1 and R ( R > R 1 ), and ρ, s such that R 1 ρ < R , 1 < s R 1 , we can write
| z | = ρ | f ( z ) z n + 1 p | p | d z | | z | = R 1 | f ( z ) z n + 1 p | p | d z | ,
(2.2)
| z | = R 1 | f ( z ) z n + 1 p | p | d z | | z | = s | f ( z ) z n + 1 p | p | d z | ,
(2.3)
respectively. Thus,
| z | = ρ | f ( z ) | p | d z | ρ n p + 1 | z | = R 1 | f ( z ) z n + 1 p | p | d z | ,
(2.4)
s n p + 1 | z | = R 1 | f ( z ) z n + 1 p | p | d z | | z | = s | f ( z ) | p | d z | .
(2.5)
Integrating (2.4) over ρ from R 1 to R, and (2.5) over s from 1 to R 1 , we get
R 1 R | z | = ρ | f ( z ) | p | d z | d ρ 1 n p + 2 ( R n p + 2 R 1 n p + 2 ) | z | = R 1 | f ( z ) z n + 1 p | p | d z | , 1 n p + 2 ( R 1 n p + 2 1 ) | z | = R 1 | f ( z ) z n + 1 p | p | d z | 1 R 1 | z | = s | f ( z ) | p | d z | d s .
After calculation we have
R 1 < | z | < R | f ( z ) | p d σ z R n p + 2 R 1 n p + 2 R 1 n p + 2 1 1 < | z | < R 1 | f ( z ) | p d σ z ,
(2.6)

and we see that (2.1) is true. □

Corollary 2.2 Under the assumptions of Lemma  2.1 for R 1 = 1 + 1 n , we have
f A p ( R 1 < | z | < R ) c 1 R n + 2 p f A p ( 1 < | z | < R 1 ) ,
(2.7)

where c 1 : = c 1 ( p , n ) = ( 1 e p 1 ) 1 p [ 1 + O ( 1 n ) ] , n .

Proof Let us put
S p : = S p ( R , R 1 , n , p ) : = R n p + 2 R 1 n p + 2 R 1 n p + 2 1 = R n p + 2 1 ( R 1 R ) n p + 2 R 1 n p + 2 1 ,
and taking R 1 = 1 + 1 n , we have
S p = R n p + 2 1 ( R 1 R ) n p + 2 ( 1 + 1 n ) n p + 2 1 R n p + 2 ( 1 + 1 n ) n p + 2 1 .
(2.8)
According to the right-hand side of the well-known estimation (see, for example, [[6], p.52 (Problem 170)])
e 2 n + 2 < e ( 1 + 1 n ) n < e 2 n + 1 , n = 1 , 2 , ,
(2.9)
we have
( 1 + 1 n ) n p + 2 ( 1 + 1 n ) n p ( e e 2 n + 1 ) p = e p ( 1 1 2 n + 1 ) p ( ε n e ) p ,
where
2 3 ε n : = 1 1 2 n + 1 1 , n .
(2.10)
Therefore
S p 1 ( ε n e ) p 1 R n p + 2 = R n p + 2 1 e p 1 [ 1 + O ( 1 n ) ] , n .
(2.11)

From (2.8) and (2.11) we complete the proof. □

Remark 2.1 For the polynomial Q n ( z ) = z n , R 1 = 1 + 1 n and any R > R 1 ,
Q n A p ( R 1 < | z | < R ) c 2 R n + 2 p Q n A p ( 1 < | z | < R 1 ) ,
(2.12)

where c 2 : = c 2 ( p , n ) : = ( 1 e p 1 ) 1 p [ 1 O ( 1 n ) ] , n .

Proof Really, from (2.6) we get
S p = R n p + 2 1 ( R 1 R ) n p + 2 ( 1 + 1 n ) n p + 2 1 = R n p + 2 1 δ n ( 1 + 1 n ) n p + 2 1 ,
(2.13)
where
δ n : = ( R 1 R ) n p + 2 0 , n .
(2.14)
According to the left-hand side of (2.9), we obtain
( 1 + 1 n ) n p + 2 = ( 1 + 1 n ) n p ( 1 + 1 n ) 2 ( e e 2 n + 2 ) p η n = e p ( 1 1 2 n + 2 ) p η n e p η n ,
where
η n : = ( 1 + 1 n ) 2 1 , n .
Therefore,
S p R n p + 2 1 δ n η n e p 1 = R n p + 2 [ 1 η n e p 1 δ n η n e p 1 ] = R n p + 2 { 1 e p 1 [ 1 O ( 1 n ) ] O ( δ n ) } = R n p + 2 1 e p 1 [ 1 O ( 1 n ) ] , n .

 □

Corollary 2.3 For f P n , we have
P n A p ( | z | < R ) c 3 R n + 2 p P n A p ( | z | < R 1 ) ,

where c 3 : = c 3 ( p , n ) : = ( 2 e p 1 ) 1 p [ 1 + O ( 1 n ) ] , n .

Proof Really, (2.1) implies, for any f P n ,
P n A p ( R 1 < | z | < R ) p S p P n A p ( 1 < | z | < R 1 ) p .
Adding P n A p ( | z | < R 1 ) p to the both sides, we obtain
P n A p ( | z | < R ) p S p P n A p ( 1 < | z | < R 1 ) p + P n A p ( | z | < R 1 ) p 2 max { S p , 1 } P n A p ( | z | < R 1 ) p .
Passing to the limit as R 1 = 1 + 1 n 1 , from (2.11) we obtain
P n A p ( | z | < R ) p 2 e p 1 [ 1 + O ( 1 n ) ] R n p + 2 P n A p ( | z | < R 1 ) p .

 □

3 Proof of the theorem

Proof First of all, let us convince ourselves that for the proof of (1.5) it is sufficient to show the fulfilment of estimation
P n A p ( G R G R 1 ) c R n + 2 p P n A p ( G R 1 G )
(3.1)
for some constant c = c ( p , R 1 ) > 0 independent of R and n. Really, let (3.1) be true. Then
P n A p ( G R G R 1 ) p c p R n p + 2 P n A p ( G R 1 G ) p .
(3.2)
Now, we will add to both sides P n A p ( G R 1 ) p :
P n A p ( G R ) p c p R n p + 2 P n A p ( G R 1 G ) p + P n A p ( G R 1 ) p c p R n p + 2 P n A p ( G R 1 G ) p + c p R n p + 2 P n A p ( G R 1 ) p = 2 c p R n p + 2 P n A p ( G R 1 ) p .
(3.3)
Therefore,
P n A p ( G R ) 2 1 p c R n + 2 p P n A p ( G R 1 ) .

Now, let us make a proof of (3.1).

For the p > 0 , let us set
f n ( w ) : = P n ( Ψ ( w ) ) [ Ψ ( w ) ] 2 p , w = Φ ( z ) .
The function f n is analytic in Δ and has a pole of degree at most n at w = . Then, according to Lemma 2.1, we have
f n A p ( R 1 < | w | < R ) p S ( R , R 1 , n , p ) f n A p ( 1 < | w | < R 1 ) p ,
where
S p : = R n p + 2 R 1 n p + 2 R 1 n p + 2 1 = R n p + 2 1 ( R 1 R ) n p + 2 R 1 n p + 2 1 .
Then
G R G R 1 | P n ( z ) | p d σ z = R 1 < | w | < R | f n ( w ) | p d σ w S p 1 < | w | < R 1 | f n ( w ) | p d σ w R n p + 2 1 R 1 n p + 2 1 G R 1 G | P n ( z ) | p d σ z .
Therefore,
G R | P n ( z ) | p d σ z 2 R n p + 2 1 R 1 n p + 2 1 G R 1 | P n ( z ) | p d σ z .
(3.4)
Taking R 1 = 1 + 1 n , from (2.9) and (2.11) we get
1 R 1 n p + 2 1 = 1 e p 1 [ 1 + O ( 1 n ) ] , n .
(3.5)

Now, from (3.4) and (3.5) we complete the proof. □

3.1 Proof of the remark

Proof Let P n = z n , G = B : = { z : | z | < 1 } and R 8 e p e p 1 . Then
P n p A p ( G R ) = | z | < R | z n | p d σ z = R n p + 2 R 1 ( n p + 2 ) P n p A p ( G R 1 ) = R R 1 2 R 1 n p R n p + 2 P n p A p ( G R 1 ) .
(3.6)
For R 1 = 1 + 1 n , from (2.9) we obtain
( 1 + 1 n ) n p ( e e 2 n + 2 ) p e p , ( 1 + 1 n ) 2 4 .
Then
R R 1 n p + 2 R 4 e p
and
P n p A p ( G R ) R 4 e p R n p + 2 P n p A p ( G R 1 ) .
In particular, for R = 8 e p e p 1 we have
P n p A p ( G R ) 2 e p 1 R n p + 2 P n p A p ( G R 1 ) .

 □

Declarations

Authors’ Affiliations

(1)
Faculty of Arts and Science, Department of Mathematics, Mersin University

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Copyright

© Abdullayev and Özkartepe; licensee Springer. 2013

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