19_739_sorin.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 3, No. 6, 2010, 1150-1164

ISSN 1307-5543 – www.ejpam.com

SPECIAL ISSUE ON COMPLEX ANALYSIS: THEORY AND APPLICATIONS

DEDICATED TO PROFESSOR HARI M. SRIVASTAVA,

ON THE OCCASION OF HIS 70TH
BIRTHDAY

Approximation by Complex Potentials Generated by the Euler’s

Beta Function

Sorin G. Gal

Department of Mathematics and Computer Science, University of Oradea, 410087 Oradea, Roma-

nia

Abstract. In this paper we find the exact orders of approximation of analytic functions by the complex

versions of several potentials generated by the Euler’s Beta function and by some complex singular

integrals.

2000 Mathematics Subject Classifications: 30E10, 41A35, 41A25

Key Words and Phrases: Complex potentials, Beta function, Complex singular integrals, Exact order

of approximation

1. Introduction

Starting from the Flett real potential defined for any f ∈ Lp(R) by [see Flett 1]

Fα( f )(x) =
1

Γ(α)

∫ ∞

0

tα−1e−tQ t( f )(x)d t,

where Q t( f )(x) =
t

π

∫∞
−∞

f (x−u)

u2+t2 du is the classical Poisson-Cauchy real singular integral, in the

recent paper [3] we studied the approximation properties for α ց 0, of its complex version

defined by

FαU ( f )(z) =
1

Γ(α)

∫ ∞

0

tα−1e−tQ t( f )(z)d t,

Email address: galso�uoradea.ro
http://www.ejpam.com 1150 c© 2010 EJPAM All rights reserved.



S. Gal / Eur. J. Pure Appl. Math, 3 (2010), 1150-1164 1151

where Q t( f )(z) =
t

π

∫∞
−∞

f (ze−iu)

u2+t2 du. Also, in the same paper [3], the approximation properties

of following types of complex potentials generated by the Gamma function and some other

singular integrals were studied :

FαU ( f )(z) =
1

Γ(α)

∫ ∞

0

tα−1e−t Ut( f )(z)d t,

with

Ut( f )(z) = Pt( f )(z) =
1

2t

∫ +∞

−∞
f (ze−iu)e−|u|/t du,

Ut( f )(z) = R t( f )(z) =
2t3

π

∫ +∞

−∞

f (ze−iu)

(u2 + t2)2
du,

Ut( f )(z) =W ∗t ( f )(z) =
1p
πt

∫ +∞

−∞
f (ze−iu)e−u2/t du,

representing the complex versions of the Picard, generalized Poisson-Cauchy and Gauss-

Weierstrass singular integrals, respectively.

The goal of the present paper is to find the exact orders of approximation by the complex

potentials generated by the Euler’s Beta function, that is of the form

G
α,β
U ( f )(z) =

1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1Ut( f )(z)d t,

for Q t( f )(z) and for all the Ut( f )(z) defined above.

2. Main Result

For R> 0 let us denote DR = {z ∈ C; |z| < R}. The main result is the following.

Theorem 1. Let us suppose that 0< α≤ β ≤ 1, α+ β ≥ 1 and that f : DR→ C, with R> 1, is

analytic in DR, that is f (z) =
∑∞

k=0 akzk, for all z ∈ DR.

(i) For Ut( f )(z) =
t

π

∫∞
−∞

f (ze−iu)

u2+t2 du we have that G
α,β
U ( f )(z) is analytic in DR and we can

write

G
α,β
U ( f )(z) =

∞
∑

k=0

ak bk(α,β) · zk, z ∈ DR,

where

bk(α,β) =
1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1e−kt d t.

Also, if f is not constant for q = 0, and not a polynomial of degree ≤ q− 1 for q ∈ N, then

for all 1≤ r < r1 < R, q ∈ N∪ {0}, α ∈ (0,β] we have

‖[Gα,β
U ( f )](q)− f (q)‖r ∼ α,



S. Gal / Eur. J. Pure Appl. Math, 3 (2010), 1150-1164 1152

where ‖ f ‖r = sup{| f (z)|; |z| ≤ r} and the constants in the equivalence depend only on f ,

q, r, r1, β .

(ii) For Ut( f )(z) =
1

2t

∫ +∞
−∞ f (ze−iu)e−|u|/t du we have that G

α,β
U ( f )(z) is analytic in DR and

we can write

G
α,β
U ( f )(z) =

∞
∑

k=0

ak · bk(α,β) · zk, z ∈ DR,

where bk(α,β) = 1

Beta(α,β)

∫ 1

0

tα−1(1−t)β−1

1+t2k2 d t.

Also, if f is not constant for q = 0, and not a polynomial of degree ≤ q− 1 for q ∈ N, then

for all 1≤ r < r1 < R, q ∈ N∪ {0}, α ∈ (0,β] we have

‖[Gα,β
U ( f )](q)− f (q)‖r ∼ α,

where the constants in the equivalence depend only on f , q, r, r1 and β .

(iii) For Ut( f )(z) =
2t3

π

∫ +∞
−∞

f (ze−iu)

(u2+t2)2
du we have that G

α,β
U ( f )(z) is analytic in DR and we can

write

G
α,β
U ( f )(z) =

∞
∑

k=0

ak · bk(α,β) · zk, z ∈ DR,

where bk(α,β) = 1

Beta(α,β)

∫ 1

0
tα−1(1− t)β−1(1+ kt)e−kt d t.

Also, if f is not constant for q = 0, and not a polynomial of degree ≤ q− 1 for q ∈ N, then

for all 1≤ r < r1 < R, q ∈ N∪ {0}, α ∈ (0,β] we have

‖[Gα,β
U ( f )](q)− f (q)‖r ∼ α,

where the constants in the equivalence depend only on f , q, r, r1 and β .

(iv) For Ut( f )(z) =
1p
πt

∫+∞
−∞ f (ze−iu)e−u2/t du we have that G

α,β
U ( f )(z) is analytic in DR and

we can write

G
α,β
U ( f )(z) =

∞
∑

k=0

ak · bk(α,β)zk, z ∈ DR,

where bk(α,β) = 1

Beta(α,β)

∫ 1

0
tα−1(1− t)β−1e−(k

2/4)t d t.

Also, if f is not constant for q = 0, and not a polynomial of degree ≤ q− 1 for q ∈ N, then

for all 1≤ r < r1 < R, q ∈ N∪ {0}, α ∈ (0,β] we have

‖[Gα,β
U ( f )](q)− f (q)‖r ∼ α,

where the constants in the equivalence depend only on f , q, r, r1 and β .



S. Gal / Eur. J. Pure Appl. Math, 3 (2010), 1150-1164 1153

Proof.

(i) By Gal [2, p. 213, Theorem 3.2.5, (i)], Ut( f )(z) is analytic (as function of z) in DR and

we can write

Ut( f )(z) =

∞
∑

k=0

ake−ktzk, for all |z| < R and t ≥ 0.

Since |∑∞k=0 ake−ktzk| ≤∑∞k=0 |ak|·|z|k <∞, this implies that for fixed |z| < R, the series

in t,
∑∞

k=0 ake−ktzk is uniformly convergent on [0,∞), and therefore we immediately

can write

G
α,β
U ( f )(z) =

∞
∑

k=0

ak bk(α,β)zk,

where

bk(α,β) =
1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1e−kt d t.

In other order of ideas, we easily can write

G
α,β
U ( f )(z)− f (z) =

1

Beta(α,β)
·
∫ 1

0

tα−1(1− t)β−1[Ut( f )(z)− f (z)]d t,

which together with the estimate |Ut( f )(z)− f (z)| ≤ Cr( f )t in Gal [2, p. 213, Theorem

3.2.5, (iii)], implies

|Gα,β
U ( f )(z)− f (z)| ≤ 1

Beta(α,β)
·
∫ 1

0

tα−1(1− t)β−1|Ut( f )(z)− f (z)|d t

≤ Cr( f )
1

Beta(α,β)
·
∫ 1

0

tα(1− t)β−1d t = Cr( f ) ·
Beta(α+ 1,β)

Beta(α,β)

= Cr( f ) ·
α

α+ β
≤ Cr( f ) ·α,

for all |z| ≤ r, where Cr( f )> 0 is independent of z (and α, β) but depends on f and r.

Here we used the well known formula
Beta(α+1,β)

Beta(α,β)
= α

α+β
.

Now, let q ∈ N ∪ {0} and 1 ≤ r < r1 < R. Denoting by γ the circle of radius r1 and

center 0, since for any |z| ≤ r and v ∈ γ we have |v − z| ≥ r1 − r, by using the Cauchy’s

formula, for all |z| ≤ r and 0< α ≤ β ≤ 1, α+ β ≥ 1, we get

|[Gα,β
U ( f )](q)(z)− f (q)(z)| = q!

2π

�

�

�

�

�

∫

γ

G
α,β
U ( f )(z)− f (z)

(v− z)q+1
dv

�

�

�

�

�

≤ Cr1
( f )α · q

2π
· 2πr1

(r1 − r)q+1
= C∗α,



S. Gal / Eur. J. Pure Appl. Math, 3 (2010), 1150-1164 1154

with C∗ depending only on f , q, r and r1.

It remains to prove the lower estimate. For this purpose, reasoning exactly as in the

proof of Theorem 3.2.5, at pages 218-219 in the book Gal [2], for z = reiϕ and

p ∈ N∪ {0} we get

1

2π

∫ π

−π
[ f (q)(z)− [Ut( f )]

(q)(z)]e−ipϕdϕ

= aq+p(q+ p)(q+ p− 1)...(p+ 1)r p[1− e−(q+p)t].

Multiplying above with 1

Beta(α,β)
tα−1(1− t)β−1 an then integrating with respect to t, it

follows

I :=

1

Beta(α,β)
·
∫ 1

0

¨

1

2π

∫ π

−π
[ f (q)(z)− [Ut( f )]

(q)(z)]e−ipϕdϕ

«

tα−1(1− t)β−1d t

= aq+p(q+ p)(q+ p− 1)...(p+ 1)r p
1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1[1− e−(q+p)t]d t.

Applying the Fubini’s result to the double integral I and then passing to modulus, we

easily obtain

�

�

�

�

�

1

2π

∫ π

−π
e−ipϕ





1

Beta(α,β)

∫ 1

0

[ f (q)(z)− [Ut( f )]
(q)(z)]tα−1(1− t)β−1d t



 dϕ

�

�

�

�

�

= |aq+p|(q+ p)(q+ p− 1)...(p+ 1)r p

·




1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1[1− e−(q+p)t]d t



 .

Since
1

Beta(α,β)

∫ 1

0

[ f (q)(z)− [Ut( f )]
(q)(z)]tα−1(1− t)β−1d t

= f (q)(z)− [Gα,β
U ( f )](q)(z),

the previous equality immediately implies

�

�

�

�

�

1

2π

∫ π

−π
e−ipϕ

h

f (q)(z)− (Gα,β
U ( f ))(q)(z)

i

dϕ

�

�

�

�

�

= |aq+p|(q+ p)(q+ p− 1)...(p+ 1)r p

·




1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1[1− e−(q+p)t]d t







S. Gal / Eur. J. Pure Appl. Math, 3 (2010), 1150-1164 1155

and

|aq+p|(q+ p)(q+ p− 1)...(p+ 1)r p

·




1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1[1− e−(q+p)t]d t





≤ ‖ f (q)− (Gα,β
U ( f ))(q)‖r .

First take q = 0. In what follows, denoting

Vα,β = inf
p≥1

 

1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1[1− e−pt]d t

!

,

we clearly get

Vα,β =
1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1[1− e−t]d t.

But denoting g(t) = e−t , by the mean value theorem there exists ξ ∈ (0,1) such that

1− e−t = g(0)− g(t) = te−ξ ≥ t

e
, which immediately implies

Vα,β ≥
1

e · Beta(α,β)

∫ 1

0

tα(1− t)β−1d t =
Beta(α+ 1,β)

e · Beta(α,β)

=
1

e
· α
α+ β

≥ 1

e
· α

2β
≥ α

2e
.

Therefore,

1

2e
· r p · |ap| ≤

‖ f − G
α,β
U ( f )‖r
α

,

for all p ≥ 1 and 0< α≤ β ≤ 1,α+ β ≥ 1.

This implies that if there exists a subsequence (αk)k in (0,β] with limk→∞αk = 0 and

such that limk→∞
‖Gα,β

U ( f )− f ‖r
αk

= 0, then ap = 0 for all p ≥ 1, that is f is constant on Dr .

Therefore, if f is not a constant function, then infα∈(0,β]
‖Gα,β

U ( f )− f ‖r
α

> 0, which implies

that there exists a constant Cr( f )> 0 such that
‖Gα,β

U ( f )− f ‖r
α

≥ Cr( f ), that is

‖Gα,β
U ( f )− f ‖r ≥ Cr( f )α, for all 0< α ≤ β ≤ 1,α+ β ≥ 1.

Now, consider q ≥ 1 and denote

Vq,α,β = inf
p≥0

 

1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1[1− e−(q+p)t]d t

!

.

Evidently that we have Vq,α,β ≥ Vα,β ≥ α · 1

2e
.



S. Gal / Eur. J. Pure Appl. Math, 3 (2010), 1150-1164 1156

Reasoning as in the case of q = 0, we obtain

‖[Gα,β
U ( f )](q)− f (q)‖r

α
≥ |aq+p|

(q+ p)!

p!
· 1

2e
· r p,

for all p ≥ 0 and 0< α≤ β ≤ 1,α+ β ≥ 1.

This implies that if there exists a subsequence (αk)k in (0,β] with limk→∞αk = 0 and

such that limk→∞
‖[Gα,β

U
( f )](q)− f (q)‖r
αk

= 0, then aq+p = 0 for all p ≥ 0, that is f is a

polynomial of degree ≤ q− 1 on Dr .

Therefore, because by hypothesis f is not a polynomial of degree ≤ q − 1, we obtain

infα∈(0,β]
‖[Gα,β

U ( f )](q)− f (q)‖r
α

> 0, which implies that there exists a constant Cr,q( f ) > 0

such that
‖[Gα,β

U ( f )](q)− f (q)‖r
α

≥ Cr,q( f ), for all α ∈ (0,β], that is

‖[Gα,β
U ( f )](q)− f (q)‖r ≥ Cr,q( f )α, for all α ∈ (0,β].

(ii) By Gal [2, p. 206, Theorem 3.2.1, (i)], Ut( f )(z) is analytic (as function of z) in DR and

we can write

Ut( f )(z) =

∞
∑

k=0

ak

1+ t2k2
zk, for all |z| < R and t ≥ 0.

Since |∑∞k=0

ak

1+t2k2 zk| ≤∑∞k=0 |ak|·|z|k <∞, this implies that for fixed |z| < R, the series

in t,
∑∞

k=0

ak

1+t2k2 zk is uniformly convergent on [0,∞), and therefore we immediately

can write

G
α,β
U ( f )(z) =

∞
∑

k=0

akzk
1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1

1+ t2k2
d t.

In other order of ideas, we easily can write

G
α,β
U ( f )(z)− f (z) =

1

Beta(α,β)
·
∫ 1

0

tα−1(1− t)β−1[Ut( f )(z)− f (z)]d t,

which together with the estimate |Ut( f )(z)− f (z)| ≤ Cr( f )t
2 in Gal [2, p. 207, Theorem

3.2.1, (iv)], implies

|Gα,β
U ( f )(z)− f (z)| ≤ 1

Beta(α,β)
·
∫ 1

0

tα−1(1− t)β−1|Ut( f )(z)− f (z)|d t

≤ Cr( f )
1

Beta(α,β)
·
∫ 1

0

tα+1(1− t)β−1d t = Cr( f ) ·
Beta(α+ 2,β)

Beta(α,β)

= Cr( f )
α+ 1

α+β + 1
· α
α+ β

≤ Cr( f )
α(α+ 1)

2
≤ Cr( f )α,



S. Gal / Eur. J. Pure Appl. Math, 3 (2010), 1150-1164 1157

for all |z| ≤ r, where Cr( f )> 0 is independent of z (and α, β) but depends on f and r.

Now, let q ∈ N∪ {0} and 1 ≤ r < r1 < R. By using the Cauchy’s formula and reasoning

as in the proof of the above point (i), we get the upper estimate

‖[Gα,β
U ( f )](q)− f (q)‖r ≤ C∗α,

with C∗ depending only on f , q, r and r1.

It remains to prove the lower estimate. For this purpose, reasoning exactly as in the

proof of Theorem 3.2.1, at pages 209-210 in the book Gal [2], for z = reiϕ and

p ∈ N∪ {0} we get

1

2π

∫ π

−π
[ f (q)(z)− [Ut( f )]

(q)(z)]e−ipϕdϕ

= aq+p(q+ p)(q+ p− 1)...(p+ 1)r p · t2(q+ p)2

1+ t2(q+ p)2
.

Multiplying above with 1

Beta(α,β)
tα−1(1− t)β−1 an then integrating with respect to t, it

follows

I :=

1

Beta(α,β)
·
∫ 1

0

¨

1

2π

∫ π

−π
[ f (q)(z)− [Ut( f )]

(q)(z)]e−ipϕdϕ

«

tα−1(1− t)β−1d t

= aq+p(q+ p)(q+ p− 1)...(p+ 1)r p

· 1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1

�

t2(q+ p)2

1+ t2(q+ p)2

�

d t.

Applying the Fubini’s result to the double integral I and then passing to modulus, we

easily obtain

�

�

�

�

�

1

2π

∫ π

−π
e−ipϕ





1

Beta(α,β)

∫ 1

0

[ f (q)(z)− [Ut( f )]
(q)(z)]tα−1(1− t)β−1d t



 dϕ

�

�

�

�

�

= |aq+p|(q+ p)(q+ p− 1)...(p+ 1)r p

·




1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1

�

t2(q+ p)2

1+ t2(q+ p)2

�

d t



 .

Since
1

Beta(α,β)

∫ 1

0

[ f (q)(z)− [Ut( f )]
(q)(z)]tα−1(1− t)β−1d t

= f (q)(z)− [Gα,β
U ( f )](q)(z),



S. Gal / Eur. J. Pure Appl. Math, 3 (2010), 1150-1164 1158

the previous equality immediately implies

�

�

�

�

�

1

2π

∫ π

−π
e−ipϕ

h

f (q)(z)− (Gα,β
U ( f ))(q)(z)

i

dϕ

�

�

�

�

�

= |aq+p|(q+ p)(q+ p− 1)...(p+ 1)r p

·




1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1

�

t2(q+ p)2

1+ t2(q+ p)2

�

d t





and

|aq+p|(q+ p)(q+ p− 1)...(p+ 1)r p

·




1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1

�

t2(q+ p)2

1+ t2(q+ p)2

�

d t



≤ ‖ f (q)− (Gα,β
U ( f ))(q)‖r .

First take q = 0. From the previous inequality we immediately obtain

|ap|r p

 

1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1

�

t2p2

1+ t2p2

�

d t

!

≤ ‖ f − G
α,β
U ( f )‖r .

In what follows, denoting

Vα,β = inf
p≥1

 

1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1

�

t2p2

1+ t2p2

�

d t

!

,

we clearly get

Vα,β =
1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1

�

t2

1+ t2

�

d t

=
1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1

�

1− 1

1+ t2

�

d t.

But we have 1− 1

1+t2 ≥ t2

4
, for all t ∈ [0,1]. Indeed, denoting g(t) = 1− 1

1+t2 − t2

4
, we

get g(0) = 0 and g′(t) = 2t

(1+t2)2
− 2t

4
= 2t

�

1

(1+t2)2
− 1

4

�

≥ 0, for all t ∈ [0,1]. It follows

that g(t) is nondecreasing on [0,1] and therefore g(t) ≥ 0 for all t ∈ [0,1].

In conclusion,

Vα,β ≥
1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1
t2

4
d t

=
1

4
· Beta(α+ 2,β)

Beta(α,β)
=

1

4
· α+ 1

α+ β + 1
· α
α+ β



S. Gal / Eur. J. Pure Appl. Math, 3 (2010), 1150-1164 1159

≥ 1

4
· α(α+ 1)

2
≥ α

8
.

Now, by following for q ≥ 0 similar reasonings with those in the above point (i), we get

the desired equivalence in the statement.

(iii) By Gal [2, p. 213, Theorem 3.2.5, (i)], Ut( f )(z) is analytic (as function of z) in DR and

we can write

Ut( f )(z) =

∞
∑

k=0

ak(1+ kt)e−ktzk, for all |z| < R and t ≥ 0.

Since |∑∞k=0 ake−kt(1+kt)zk| ≤ 2
∑∞

k=0 |ak|·|z|k <∞, this implies that for fixed |z| < R,

the series in t,
∑∞

k=0 ak(1+ kt)e−ktzk is uniformly convergent on [0,∞), and therefore

we immediately can write

G
α,β
U ( f )(z) =

∞
∑

k=0

akzk
1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1(1+ kt)e−kt d t,

where denoting bk(α,β) = 1

Beta(α,β)

∫ 1

0
tα−1(1− t)β−1(1+ kt)e−kt d t, we obtain

G
α,α
U ( f )(z) =

∞
∑

k=0

ak · bk(α,β) · zk.

In other order of ideas, we easily can write

G
α,β
U ( f )(z)− f (z) =

1

Beta(α,β)
·
∫ 1

0

tα−1(1− t)β−1[Ut( f )(z)− f (z)]d t,

which together with the estimate |Ut( f )(z)− f (z)| ≤ Cr( f )t
2 in Gal [2, p. 213-214,

Theorem 3.2.5, (iv)], implies

|Gα,β
U ( f )(z)− f (z)| ≤ 1

Beta(α,β)
·
∫ 1

0

tα−1(1− t)β−1|Ut( f )(z)− f (z)|d t

≤ Cr( f )
1

Beta(α,β)
·
∫ 1

0

tα+1(1− t)β−1d t = Cr( f ) ·
Beta(α+ 2,β)

Beta(α,β)
≤ Cr( f )α,

for all |z| ≤ r, where Cr( f ) > 0 is independent of z (and α) but depends on f and r.

We used here the estimate from the above point (ii).

Now, let q ∈ N∪ {0} and 1 ≤ r < r1 < R. By using the Cauchy’s formula and reasoning

as in the proof of the above point (i), we get the upper estimate

‖[Gα,β
U ( f )](q)− f (q)‖r ≤ C∗α,



S. Gal / Eur. J. Pure Appl. Math, 3 (2010), 1150-1164 1160

with C∗ depending only on f , q, r and r1.

It remains to prove the lower estimate. For this purpose, reasoning exactly as in the

proof of Theorem 3.2.5, at pages 219-220 in the book Gal [2], for z = reiϕ and

p ∈ N∪ {0} we get

1

2π

∫ π

−π
[ f (q)(z)− [Ut( f )]

(q)(z)]e−ipϕdϕ

= aq+p(q+ p)(q+ p− 1)...(p+ 1)r p[1− (1+ (q+ p)t)e−(q+p)t].

Multiplying above with 1

Beta(α,β)
tα−1(1− t)β−1 an then integrating with respect to t, it

follows

I :=

1

Beta(α,β)
·
∫ 1

0

¨

1

2π

∫ π

−π
[ f (q)(z)− [Ut( f )]

(q)(z)]e−ipϕdϕ

«

tα−1(1− t)β−1d t

= aq+p(q+ p)(q+ p− 1)...(p+ 1)r p

· 1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1
�

1− (1+ (q+ p)t)e−(q+p)t
�

d t.

Applying the Fubini’s result to the double integral I and then passing to modulus, we

easily obtain

�

�

�

�

�

1

2π

∫ π

−π
e−ipϕ





1

Beta(α,β)

∫ 1

0

[ f (q)(z)− [Ut( f )]
(q)(z)]tα−1(1− t)β−1d t



 dϕ

�

�

�

�

�

= |aq+p|(q+ p)(q+ p− 1)...(p+ 1)r p

·




1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1
�

1− (1+ (q+ p)t)e−(q+p)t
�

d t



 .

Since
1

Beta(α,β)

∫ 1

0

[ f (q)(z)− [Ut( f )]
(q)(z)]tα−1(1− t)β−1d t

= f (q)(z)− [Gα,β
U ( f )](q)(z),

the previous equality immediately implies

�

�

�

�

�

1

2π

∫ π

−π
e−ipϕ

h

f (q)(z)− (Gα,β
U ( f ))(q)(z)

i

dϕ

�

�

�

�

�

= |aq+p|(q+ p)(q+ p− 1)...(p+ 1)r p



S. Gal / Eur. J. Pure Appl. Math, 3 (2010), 1150-1164 1161

·




1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1
�

1− (1+ (q+ p)t)e−(q+p)t
�

d t





and

|aq+p|(q+ p)(q+ p− 1)...(p+ 1)r p

·




1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1
�

1− (1+ (q+ p)t)e−(q+p)t
�

d t





≤ ‖ f (q)− (Gα,β
U ( f ))(q)‖r .

First take q = 0. From the previous inequality we immediately obtain

|ap|r p

 

1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1
�

1− (1+ pt)e−pt
�

d t

!

≤ ‖ f − G
α,β
U ( f )‖r .

In what follows, denoting

Vα,β = inf
p≥1

 

1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1
�

1− (1+ pt)e−pt
�

d t

!

,

we immediately get

Vα,β =
1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1
�

1− (1+ t)e−t
�

d t.

But we have 1− (1+ t)e−t ≥ t2

e
, for all t ∈ [0,1]. Indeed, denoting

g(t) = 1− (1+ t)e−t − t2

e
, we have g(0) = 0 and g′(t) = te−t − t

e
= t
�

1

e t − 1

e

�

≥ 0 for

all t ∈ [0,1]. This implies that g(t) is nondecreasing on [0,1] and therefore g(t) ≥ 0

for all t ∈ [0,1].

Therefore,

Vα,β ≥
1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1
t2

2e
d t

=
Beta(α+ 2,β)

2e · B(α,β)
=

1

2e
· α+ 1

α+ β + 1
· α
α+ β

≥ 1

2e
· α(α+ 1)

2
≥ α

4e
.

Now, by following for q ≥ 0 similar reasonings with those in the above point (i), we get

the desired equivalence in the statement.



S. Gal / Eur. J. Pure Appl. Math, 3 (2010), 1150-1164 1162

(iv) By Gal [2, p. 223, Theorem 3.2.8, (i)], Ut( f )(z) is analytic (as function of z) in DR and

we can write

Ut( f )(z) =

∞
∑

k=0

ake−k2 t/4zk, for all |z| < R and t ≥ 0.

Since |∑∞k=0 ake−k2 t/4zk| ≤ ∑∞k=0 |ak| · |z|k < ∞, this implies that for fixed |z| < R,

the series in t,
∑∞

k=0 ake−k2 t/4zk is uniformly convergent on [0,∞), and therefore we

immediately can write

G
α,β
U ( f )(z) =

∞
∑

k=0

akzk
1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1e−(k
2/4)t d t,

where denoting bk(α,β) = 1

Beta(α,β)

∫ 1

0
tα−1(1− t)β−1e−(k

2/4)t d t we can write

G
α,β
U ( f )(z) =

∞
∑

k=0

ak · bk(α,β) · zk.

In other order of ideas, we easily can write

G
α,β
U ( f )(z)− f (z) =

1

Beta(α,β)
·
∫ 1

0

tα−1(1− t)β−1[Ut( f )(z)− f (z)]d t,

which together with the estimate |Ut( f )(z)− f (z)| ≤ Cr( f )t in Gal [2, p. 224, Theorem

3.2.8, (iv)], implies

|Gα,β
U ( f )(z)− f (z)| ≤ 1

Beta(α,β)
·
∫ 1

0

tα−1(1− t)β−1|Ut( f )(z)− f (z)|d t

≤ Cr( f )
1

Beta(α,β)
·
∫ 1

0

tα(1− t)β−1d t = Cr( f ) ·
Beta(α+ 1,β)

Beta(α,β)
≤ Cr( f )α,

for all |z| ≤ r, where Cr( f )> 0 is independent of z (and α) but depends on f and r.

Now, let q ∈ N∪ {0} and 1 ≤ r < r1 < R. By using the Cauchy’s formula and reasoning

as in the proof of the above point (i), we get the upper estimate

‖[Gα,β
U ( f )](q)− f (q)‖r ≤ C∗α,

with C∗ depending only on f , q, r and r1.

It remains to prove the lower estimate. For this purpose, reasoning exactly as in the

proof of Theorem 3.2.8, at pages 227-228 in the book Gal [2], for z = reiϕ and

p ∈ N∪ {0} we get

1

2π

∫ π

−π
[ f (q)(z)− [Ut( f )]

(q)(z)]e−ipϕdϕ



S. Gal / Eur. J. Pure Appl. Math, 3 (2010), 1150-1164 1163

= aq+p(q+ p)(q+ p− 1)...(p+ 1)r p[1− e−(q+p)2 t/4].

Multiplying above with 1

Beta(α,β)
tα−1(1− t)β−1 an then integrating with respect to t, it

follows

I :=

1

Beta(α,β)
·
∫ 1

0

¨

1

2π

∫ π

−π
[ f (q)(z)− [Ut( f )]

(q)(z)]e−ipϕdϕ

«

tα−1)1− t)β−1d t

= aq+p(q+ p)(q+ p− 1)...(p+ 1)r p

· 1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1
h

1− e−(q+p)2 t/4
i

d t.

Applying the Fubini’s result to the double integral I and then passing to modulus, we

easily obtain

�

�

�

�

�

1

2π

∫ π

−π
e−ipϕ

�

1

Beta(α,β)

∫ ∞

0

[ f (q)(z)− [Ut( f )]
(q)(z)]tα−1(1− t)β−1d t

�

dϕ

�

�

�

�

�

= |aq+p|(q+ p)(q+ p− 1)...(p+ 1)r p

·




1

Beta(α,β)

∫ 1

0

tα−1e−t
h

1− e−(q+p)2 t/4
i

d t



 .

Since
1

Beta(α,β)

∫ 1

0

[ f (q)(z)− [Ut( f )]
(q)(z)]tα−1(1− t)β−1d t

= f (q)(z)− [Gα,β
U ( f )](q)(z),

the previous equality immediately implies

�

�

�

�

�

1

2π

∫ π

−π
e−ipϕ

h

f (q)(z)− (Gα,β
U ( f ))(q)(z)

i

dϕ

�

�

�

�

�

= |aq+p|(q+ p)(q+ p− 1)...(p+ 1)r p

·




1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1
h

1− e−(q+p)2 t/4
i

d t





and

|aq+p|(q+ p)(q+ p− 1)...(p+ 1)r p

·




1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1
h

1− e−(q+p)2 t/4
i

d t







REFERENCES 1164

≤ ‖ f (q)− (Gα,β
U ( f ))(q)‖r .

First take q = 0. From the previous inequality we immediately obtain

|ap|r p

 

1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1
h

1− e−p2 t/4
i

d t

!

≤ ‖ f − G
α,β
U ( f )‖r .

In what follows, denoting

Vα,β = inf
p≥1

 

1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1
h

1− e−p2 t/4
i

d t

!

,

by simple calculation we get

Vα,β =
1

Beta(α,β)

∫ 1

0

tα−1(1− t)β−1
�

1− e−t/4
�

d t.

But denoting g(t) = e−t/4, by the mean value theorem there exists ξ ∈ (0,1) such that

1− e−t/4 = g(0)− g(t) = t e−ξ/4
4
≥ t

4e1/4 , which immediately implies

Vα,β ≥
1

4e1/4 · Beta(α,β)

∫ 1

0

tα(1− t)β−1d t =
Beta(α+ 1,β)

4e1/4 · Beta(α,β)

=
1

4e1/4
· α
α+ β

≥ 1

4e1/4
· α

2β
≥ α

8e1/4
.

Now, by following for q ≥ 0 similar reasonings with those in the above point (i), we get

the desired equivalence in the statement.

References

[1] T.M. Flett. Temperatures, Bessel Potentials and Lipschitz Space. Proceedings of the London

Mathematical Society, 22(3):385–451, 1971.

[2] S.G. Gal. Approximation by Complex Bernstein and Convolution Type Operators. World

Scientific Publishing Company, New Jersey, London, Singapore, Beijing, Shanghai, Hong

Kong, Taipei, Chennai, 2009.

[3] S.G. Gal. Approximation by Complex Potentials Generated by the Gamma Function. Turk-

ish Journal of Mathematics, accepted for publication.