Approximation by Complex Potentials Generated by the Euler's Beta Function

Vol. 3, No. 6, 2010, 1150-1164 ISSN 1307-5543 – www.ejpam.com

SPECIAL

ISSUE ON

COMPLEX

ANALYSIS: THEORY AND

APPLICATIONS

PROFESSOR

HARI

M. SRIVASTAVA,

ON THE OCCASION OF HIS

70

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)

Z ∞

0

tα−1e−tQ_t(f)(x)d t,

whereQt(f)(x) =_πt

R∞

−∞

f(x−u)

u2_+t2 duis the classical Poisson-Cauchy real singular integral, in the

recent paper[3]we studied the approximation properties forα_ց0, of its complex version defined by

F_Uα(f)(z) = 1 Γ(α)

Z ∞

0

tα−1e−tQ_t(f)(z)d t,

Email address:galsouoradea.ro

whereQ_t(f)(z) = _πt R_−∞∞ f_u(2ze₊−_tiu2)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 :

F_Uα(f)(z) = 1 Γ(α)

Z ∞

0

tα−1e−tUt(f)(z)d t,

with

U_t(f)(z) =P_t(f)(z) = 1 2t

Z +∞

−∞

f(ze−iu)e−|u|/tdu,

U_t(f)(z) =R_t(f)(z) =2t 3

π

Z +∞

−∞

f(ze−iu) (u2_+t2₎2du,

U_t(f)(z) =W_t∗(f)(z) = _p1 πt

Z +∞

−∞

f(ze−iu)e−u2/tdu,

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

Gα_U,β(f)(z) = 1

Beta(α,β)

Z 1

0

tα−1(1₋t)β−1Ut(f)(z)d t,

forQ_t(f)(z)and for all theU_t(f)(z)defined above.

2. Main Result

ForR>0 let us denoteD_R={_z∈C_;|_z|_<R}_{. The main result is the following.}

Theorem 1. Let us suppose that0< α_≤β_≤1,α+β_≥1and that f :D_R→C_{, with R>1, is}

analytic inD_{R, that is f(z) =}P∞

k=0akzk, for all z∈DR.

(i) For Ut(f)(z) = _πt

R∞

−∞

f(ze−iu)

u2_+t2 du we have that G

α,β

U (f)(z) is analytic inDR and we can

write

Gα_U,β(f)(z) = ∞

X

k=0

a_kb_k(α,β)·zk,z∈D_R,

where

b_k(α,β) = 1

Beta(α,β)

Z 1

0

tα−1(1−t)β−1e−k td t.

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

for all1≤r <r₁<R, q∈N∪ {₀}_,α∈_{(0,β]we have}

wherekfk_r=sup{|f(z)|;|z| ≤r}and the constants in the equivalence depend only on f , q, r, r₁,β.

(ii) For U_t(f)(z) = ₂1_tR_−∞+∞f(ze−iu)e−|u|/tdu we have that G_Uα,β(f)(z) is analytic inD_{R and}

we can write

Gα_U,β(f)(z) = ∞

X

k=0

ak·bk(α,β)·zk,z∈DR,

where b_k(α,β) = 1 Bet a(α,β)

R1

0

tα−1_(1−t)β−1

1+t2_k2 d t.

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

for all1≤r <r₁<R, q∈N∪ {₀}_,α∈_{(0,β]we have}

k[Gα_U,β(f)](q)₋ f(q)k_r∼α,

where the constants in the equivalence depend only on f , q, r, r1andβ.

(iii) For U_t(f)(z) = 2_πt3R_−∞+∞₍f(ze−iu)

u2_+t2₎2du we have that G

α,β

U (f)(z)is analytic inDRand we can

write

Gα_U,β(f)(z) = ∞

X

k=0

a_k·b_k(α,β)·zk,z∈D_R,

where b_k(α,β) = _{Bet a}1_(α,β)R₀1tα−1(1−t)β−1(1+kt)e−k td t.

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

for all1≤r <r₁<R, q∈N∪ {₀}_,α∈_{(0,β]we have}

k[Gα_U,β(f)](q)₋ f(q)kr∼α,

where the constants in the equivalence depend only on f , q, r, r₁andβ.

(iv) For U_t(f)(z) = p1

πt

R+∞

−∞ f(ze−iu)e−u

2_/t

du we have that G_Uα,β(f)(z)is analytic inD_Rand

we can write

Gα_U,β(f)(z) = ∞

X

k=0

ak·bk(α,β)zk,z∈DR,

where b_k(α,β) = _{Bet a}1_(α,β)R1 0 t

α−1_(1−t)β−1_e−(k2_/4)t

d t.

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

for all1≤r <r₁<R, q∈N∪ {₀}_,α∈_{(0,β]we have}

k[Gα_U,β(f)](q)₋ f(q)kr∼α,

Proof.

(i) By Gal[2, p. 213, Theorem 3.2.5, (i)],Ut(f)(z)is analytic (as function ofz) inDRand we can write

U_t(f)(z) = ∞

X

k=0

a_ke−k tzk, for all|z|<Randt≥0.

Since_|P∞_k=0ake−k tzk| ≤

P∞

k=0|ak|·|z|k<∞, this implies that for fixed|z|<R, the series in t, P∞_k=0a_ke−k tzk is uniformly convergent on [0,_∞), and therefore we immediately can write

Gα_U,β(f)(z) = ∞

X

k=0

a_kb_k(α,β)zk,

where

b_k(α,β) = 1

Beta(α,β)

Z 1

0

tα−1(1−t)β−1e−k td t.

In other order of ideas, we easily can write

Gα_U,β(f)(z)₋ f(z) = 1

Beta(α,β)·

Z 1

0

tα−1(1₋t)β−1[U_t(f)(z)₋ f(z)]d t,

which together with the estimate|U_t(f)(z)−f(z)| ≤C_r(f)t in Gal[2, p. 213, Theorem 3.2.5, (iii)], implies

|G_Uα,β(f)(z)−f(z)| ≤ 1

Beta(α,β)·

Z 1

0

tα−1(1−t)β−1|U_t(f)(z)−f(z)|d t

≤Cr(f) 1

Beta(α,β)·

Z 1

0

tα(1₋t)β−1d t=Cr(f)·

Beta(α+1,β)

Beta(α,β) =C_r(f)_· α

α+β ≤Cr(f)·α,

for all|z| ≤r, whereCr(f)>0 is independent ofz(andα,β) but depends onf andr. Here we used the well known formula Bet a_{Bet a}(α_(α+1,_,ββ₎)= _α+α_β.

Now, letq ∈ N∪ {₀} _{and 1} ≤ _{r < r1 < R. Denoting byγ the circle of radius r1 and} center 0, since for any|z| ≤r andv∈γwe 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

|[G_Uα,β(f)](q)(z)−f(q)(z)|= q! 2π

Z

γ

Gα_U,β(f)(z)−f(z) (v−z)q+1 d v

≤Cr1(f)α·

q

2π·

2πr₁

(r1−r)q+1

withC∗depending only on f,q,r andr₁.

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], forz=r eiϕ and

p∈N∪ {₀}_{we get} 1 2π

Z π

−π

[f(q)(z)₋[Ut(f)](q)(z)]e−ipϕdϕ

=aq+p(q+p)(q+p−1)...(p+1)rp[1−e−(q+p)t]. Multiplying above with 1

Bet a(α,β)t

α−1_(1−t)β−1 _{an then integrating with respect to t, it} follows

I:= 1

Beta(α,β)·

Z 1

0

¨

1 2π

Z π

−π

[f(q)(z)−[U_t(f)](q)(z)]e−ipϕdϕ

«

tα−1(1−t)β−1d t

=aq+p(q+p)(q+p−1)...(p+1)rp 1

Beta(α,β)

Z 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

Z π

−π

e−ipϕ





1

Beta(α,β)

Z 1

0

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



dϕ

=_|a_q+p|(q+p)(q+p−1)...(p+1)rp

·





1

Beta(α,β)

Z 1

0

tα−1(1₋t)β−1[1₋e−(q+p)t]d t



.

Since

1

Beta(α,β)

Z 1

0

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

= f(q)(z)₋[G_Uα,β(f)](q)(z),

the previous equality immediately implies

1 2π

Z π

−π

e−ipϕhf(q)(z)₋(G_Uα,β(f))(q)(z)idϕ

=_|aq+p|(q+p)(q+p−1)...(p+1)rp

·





1

Beta(α,β)

Z 1

0

tα−1(1₋t)β−1[1₋e−(q+p)t]d t



and

|a_q+p|(q+p)(q+p−1)...(p+1)rp

·





1

Beta(α,β)

Z 1

0

tα−1(1₋t)β−1[1₋e−(q+p)t]d t





≤ kf(q)−(G_Uα,β(f))(q)_kr. First takeq=0. In what follows, denoting

Vα,β = inf

p≥1

1

Beta(α,β)

Z 1

0

tα−1(1₋t)β−1[1₋e−p t]d t

!

,

we clearly get

Vα,β =

1

Beta(α,β)

Z 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)

Z 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_{· |a} p| ≤

kf −G_Uα,β(f)_kr

α ,

for allp_≥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 lim_k→∞kG

α,β U (f)−fkr

αk

=0, thena_p=0 for allp≥1, that is f is constant onD_r.

Therefore, if f is not a constant function, then inf_α∈(0,β]kG

α,β U (f)−fkr

α >0, which implies

that there exists a constantC_r(f)>0 such that kG

α,β U (f)−fkr

α ≥Cr(f), that is

kGα_U,β(f)₋fkr≥Cr(f)α, for all 0< α≤β≤1,α+β≥1. Now, considerq≥1 and denote

Vq,α,β= inf

p≥0

1

Beta(α,β)

Z 1

0

tα−1(1₋t)β−1[1₋e−(q+p)t]d t

!

.

Reasoning as in the case ofq=0, we obtain

k[Gα_U,β(f)](q)−f(q)k_r

α ≥ |aq+p|

(q+p)!

p! · 1 2e·r

p_,

for allp≥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 lim_k→∞k[G

α,β U (f)]

(q)_−f(q)_k

r

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

polynomial of degree≤q−1 onD_r.

Therefore, because by hypothesis f is not a polynomial of degree_≤ q₋1, we obtain infα_∈(0,β]k

[Gα_U,β(f)](q)_−f(q)_k

r

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

such that k[G

α,β U (f)]

(q)_−f(q)_k

r

α ≥Cr,q(f), for allα∈(0,β], that is

k[G_Uα,β(f)](q)₋f(q)_kr≥C_r,q(f)α, for allα_∈(0,β].

(ii) By Gal[2, p. 206, Theorem 3.2.1, (i)],U_t(f)(z)is analytic (as function ofz) inD_Rand we can write

U_t(f)(z) = ∞

X

k=0

ak 1+t2_k2z

k_{, for all|z|<Randt≥0.}

Since|P∞_k=0 ak

1+t2_k2zk| ≤

P∞

k=0|ak|·|z|k<∞, this implies that for fixed|z|<R, the series in t, P∞_k=0 ak

1+t2_k2zk is uniformly convergent on[0,∞), and therefore we immediately

can write

Gα_U,β(f)(z) = ∞

X

k=0

a_kzk 1 Beta(α,β)

Z 1

0

tα−1(1−t)β−1 1+t2_k2 d t. In other order of ideas, we easily can write

Gα_U,β(f)(z)− f(z) = 1

Beta(α,β)·

Z 1

0

tα−1(1−t)β−1[U_t(f)(z)− f(z)]d t,

which together with the estimate|U_t(f)(z)₋f(z)_{| ≤}C_r(f)t2in Gal[2, p. 207, Theorem 3.2.1, (iv)], implies

|G_Uα,β(f)(z)₋f(z)_{| ≤} 1

Beta(α,β)·

Z 1

0

tα−1(1₋t)β−1_|Ut(f)(z)−f(z)|d t

≤C_r(f) 1

Beta(α,β) ·

Z 1

0

tα+1(1−t)β−1d t=C_r(f)· Beta(α+2,β)

Beta(α,β)

=Cr(f)

α+1 α+β+1·

α

α+β ≤Cr(f)

α(α+1)

for all|z| ≤r, whereC_r(f)>0 is independent ofz(andα,β) but depends onf andr. Now, letq∈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

k[Gα_U,β(f)](q)₋ f(q)k_r≤C∗α,

withC∗depending only on f,q,r andr1.

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], forz=r eiϕ and

p∈N∪ {₀}_{we get} 1 2π

Z π

−π

[f(q)(z)₋[Ut(f)](q)(z)]e−ipϕdϕ

=a_q+p(q+p)(q+p−1)...(p+1)rp· t

2_(q+p)2

1+t2_(q+p)2.

Multiplying above with _{Bet a}1_(α,β)tα−1(1₋t)β−1 _{an then integrating with respect to t, it} follows

I:= 1

Beta(α,β)·

Z 1

0

¨

1 2π

Z π

−π

[f(q)(z)−[U_t(f)](q)(z)]e−ipϕdϕ

«

tα−1(1−t)β−1d t

=a_q+p(q+p)(q+p−1)...(p+1)rp

· 1

Beta(α,β)

Z 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

Z π

−π

e−ipϕ





1

Beta(α,β)

Z 1

0

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



dϕ

=|a_q+p|(q+p)(q+p−1)...(p+1)rp

·





1

Beta(α,β)

Z 1

0

tα−1(1₋t)β−1

–

t2(q+p)2 1+t2(q+p)2

™

d t



.

Since

1

Beta(α,β)

Z 1

0

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

the previous equality immediately implies 1 2π Z π −π

e−ipϕ

h

f(q)(z)−(G_Uα,β(f))(q)(z)

i dϕ

=_|aq+p|(q+p)(q+p−1)...(p+1)rp

·





1

Beta(α,β)

Z 1

0

tα−1(1₋t)β−1

–

t2(q+p)2 1+t2(q+p)2

™

d t





and

|a_q+p|(q+p)(q+p−1)...(p+1)rp

·





1

Beta(α,β)

Z 1

0

tα−1(1−t)β−1

–

t2(q+p)2 1+t2_(q+p)2

™

d t



≤ kf(q)−(G

α,β

U (f)) (q)_k

r. First takeq=0. From the previous inequality we immediately obtain

|ap|rp

1

Beta(α,β)

Z 1

0

tα−1(1₋t)β−1

–

t2p2

1+t2p2

™

d t

!

≤ kf −G_Uα,β(f)_kr. In what follows, denoting

Vα,β= inf

p≥1

1

Beta(α,β)

Z 1

0

tα−1(1₋t)β−1

–

t2p2

1+t2_p2

™

d t

!

,

we clearly get

Vα,β =

1

Beta(α,β)

Z 1

0

tα−1(1₋t)β−1

–

t2

1+t2

™

d t

= 1

Beta(α,β)

Z 1

0

tα−1(1−t)β−1

1− 1 1+t2

d t.

But we have 1−₁₊1_t2 ≥

t2

4, for allt∈[0, 1]. Indeed, denotingg(t) =1− 1 1+t2−

t2 4, we getg(0) =0 andg′(t) = 2t

(1+t2₎2−

2t 4 =2t

1 (1+t2₎2 −

1 4

≥0, for allt∈[0, 1]. It follows that g(t)is nondecreasing on[0, 1]and thereforeg(t)≥0 for all t∈[0, 1].

In conclusion,

V_α,β ≥ 1 Beta(α,β)

Z 1

0

tα−1(1−t)β−1t 2

4d t

= 1 4·

Beta(α+2,β)

Beta(α,β) = 1 4·

α+1 α+β+1·

≥ 1₄·α(α₂+1)≥ α₈.

Now, by following forq≥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 ofz) inDRand we can write

Ut(f)(z) =

∞

X

k=0

ak(1+kt)e−k tzk, for all|z|<Randt≥0.

Since|P∞_k=0a_ke−k t(1+kt)zk| ≤2P∞_k=0|a_k|·|z|k<_∞, this implies that for fixed|z|<R, the series int,P∞_k=0a_k(1+kt)e−k tzkis uniformly convergent on[0,∞), and therefore we immediately can write

G_Uα,β(f)(z) = ∞

X

k=0

a_kzk 1 Beta(α,β)

Z 1

0

tα−1(1−t)β−1(1+kt)e−k td t,

where denotingbk(α,β) = _{Bet a}1_(α,β)

R1

0 t

α−1_(1−t)β−1_(1+kt)e−k t_{d t, we obtain}

Gα_U,α(f)(z) = ∞

X

k=0

a_k·b_k(α,β)·zk.

In other order of ideas, we easily can write

Gα_U,β(f)(z)− f(z) = 1

Beta(α,β)·

Z 1

0

tα−1(1−t)β−1[U_t(f)(z)− f(z)]d t,

which together with the estimate |U_t(f)(z)− f(z)| ≤C_r(f)t2 in Gal [2, p. 213-214, Theorem 3.2.5, (iv)], implies

|G_Uα,β(f)(z)−f(z)| ≤ 1

Beta(α,β)·

Z 1

0

tα−1(1−t)β−1|U_t(f)(z)−f(z)|d t

≤C_r(f) 1

Beta(α,β) ·

Z 1

0

tα+1(1₋t)β−1d t=C_r(f)_· Beta(α+2,β)

Beta(α,β) ≤Cr(f)α, for all|z| ≤r, whereC_r(f)>0 is independent ofz (and α) but depends on f and r. We used here the estimate from the above point (ii).

Now, letq∈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

withC∗depending only on f,q,r andr₁.

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], forz=r eiϕ and

p∈N∪ {₀}_{we get} 1 2π

Z π

−π

[f(q)(z)−[U_t(f)](q)(z)]e−ipϕdϕ

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

Multiplying above with 1 Bet a(α,β)t

α−1_(1−t)β−1 _{an then integrating with respect to t, it} follows

I:= 1

Beta(α,β)·

Z 1

0

¨

1 2π

Z π

−π

[f(q)(z)−[U_t(f)](q)(z)]e−ipϕdϕ

«

tα−1(1−t)β−1d t

=aq+p(q+p)(q+p−1)...(p+1)rp

·_Beta1₍

α,β)

Z 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

Z π

−π

e−ipϕ





1

Beta(α,β)

Z 1

0

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



dϕ

=_|aq+p|(q+p)(q+p−1)...(p+1)rp

·





1

Beta(α,β)

Z 1

0

tα−1(1₋t)β−1”1₋(1+ (q+p)t)e−(q+p)t—d t



.

Since

1

Beta(α,β)

Z 1

0

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

= f(q)(z)₋[G_Uα,β(f)](q)(z),

the previous equality immediately implies

1 2π

Z π

−π

e−ipϕ

h

f(q)(z)₋(G_Uα,β(f))(q)(z)idϕ

·





1

Beta(α,β)

Z 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)rp

·





1

Beta(α,β)

Z 1

0

tα−1(1−t)β−1”1−(1+ (q+p)t)e−(q+p)t—d t





≤ kf(q)−(G_Uα,β(f))(q)_kr.

First takeq=0. From the previous inequality we immediately obtain

|a_p|rp 1 Beta(α,β)

Z 1

0

tα−1(1−t)β−1”1−(1+pt)e−p t—d t

!

≤ kf −G_Uα,β(f)k_r. In what follows, denoting

Vα,β = inf

p≥1

1

Beta(α,β)

Z 1

0

tα−1(1₋t)β−1”1₋(1+pt)e−p t—d t

!

,

we immediately get

Vα,β =

1

Beta(α,β)

Z 1

0

tα−1(1−t)β−1”1−(1+t)e−t—d t.

But we have 1−(1+t)e−t ≥ t_e2, for all t∈[0, 1]. Indeed, denoting

g(t) =1₋(1+t)e−t− t_e2, we have g(0) =0 andg′(t) =te−t−_et =t€_e1t −

1 e

Š

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

Therefore,

Vα,β ≥

1

Beta(α,β)

Z 1

0

tα−1(1₋t)β−1t 2

2ed t

= Beta(α+2,β) 2e·B(α,β) =

1 2e·

α+1 α+β+1·

α α+β

≥₂1_e·α(α₂+1) ≥ ₄α_e.

(iv) By Gal[2, p. 223, Theorem 3.2.8, (i)],U_t(f)(z)is analytic (as function ofz) inD_Rand we can write

Ut(f)(z) =

∞

X

k=0

ake−k

2_t/4

zk, for all|z|<Randt≥0.

Since _|P∞_k=0ake−k

2_t/4

zk| ≤ P∞_k=0|ak| · |z|k < ∞, this implies that for fixed |z| < R, the series in t, P∞_k=0a_ke−k2t/4zk is uniformly convergent on[0,∞), and therefore we immediately can write

Gα_U,β(f)(z) = ∞

X

k=0

a_kzk 1 Beta(α,β)

Z 1

0

tα−1(1−t)β−1e−(k2/4)td t,

where denotingb_k(α,β) = 1 Bet a(α,β)

R1

0 t

α−1_(1−t)β−1_e−(k2_/4)t

d twe can write

G_Uα,β(f)(z) = ∞

X

k=0

a_k·b_k(α,β)_·zk.

In other order of ideas, we easily can write

Gα_U,β(f)(z)₋ f(z) = 1

Beta(α,β)·

Z 1

0

tα−1(1₋t)β−1[Ut(f)(z)− f(z)]d t,

which together with the estimate|U_t(f)(z)−f(z)| ≤C_r(f)t in Gal[2, p. 224, Theorem 3.2.8, (iv)], implies

|G_Uα,β(f)(z)−f(z)| ≤ 1

Beta(α,β)·

Z 1

0

tα−1(1−t)β−1|U_t(f)(z)−f(z)|d t

≤C_r(f) 1

Beta(α,β)·

Z 1

0

tα(1−t)β−1d t=C_r(f)· Beta(α+1,β)

Beta(α,β) ≤Cr(f)α, for all_|z_{| ≤}r, whereC_r(f)>0 is independent ofz(andα) but depends on f andr. Now, letq∈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

k[Gα_U,β(f)](q)₋ f(q)kr≤C∗α,

withC∗depending only on f,q,r andr1.

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], forz=r eiϕ and

p∈N∪ {₀}_{we get} 1 2π

Z π

−π

=a_q+p(q+p)(q+p−1)...(p+1)rp[1−e−(q+p)

2_t/4

].

Multiplying above with 1 Bet a(α,β)t

α−1_(1−t)β−1 _{an then integrating with respect to t, it} follows

I:= 1

Beta(α,β)·

Z 1 0 ¨ 1 2π Z π −π

[f(q)(z)−[U_t(f)](q)(z)]e−ipϕdϕ

«

tα−1)1−t)β−1d t

=aq+p(q+p)(q+p−1)...(p+1)rp

·_Beta1₍

α,β)

Z 1

0

tα−1(1₋t)β−1h1−e−(q+p)2t/4

i

d t.

Applying the Fubini’s result to the double integral I and then passing to modulus, we easily obtain 1 2π Z π −π

e−ipϕ

–

1

Beta(α,β)

Z ∞

0

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

™ dϕ

=|a_q+p|(q+p)(q+p−1)...(p+1)rp

·





1

Beta(α,β)

Z 1

0

tα−1e−t

h

1−e−(q+p)2t/4

i d t  . Since 1

Beta(α,β)

Z 1

0

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

= f(q)(z)−[G_Uα,β(f)](q)(z),

the previous equality immediately implies

1 2π Z π −π

e−ipϕ

h

f(q)(z)−(G_Uα,β(f))(q)(z)

i dϕ

=|a_q+p|(q+p)(q+p−1)...(p+1)rp

·





1

Beta(α,β)

Z 1

0

tα−1(1₋t)β−1h1−e−(q+p)2t/4

i

d t





and

|a_q+p|(q+p)(q+p−1)...(p+1)rp

·





1

Beta(α,β)

Z 1

0

tα−1(1₋t)β−1h1−e−(q+p)2t/4

i

d t



≤ kf(q)−(G_Uα,β(f))(q)k_r.

First takeq=0. From the previous inequality we immediately obtain

|a_p|rp 1 Beta(α,β)

Z 1

0

tα−1(1−t)β−1

h

1−e−p2t/4

i

d t

!

≤ kf −Gα_U,β(f)k_r.

In what follows, denoting

Vα,β =inf

p≥1

1

Beta(α,β)

Z 1

0

tα−1(1₋t)β−1h1₋e−p2t/4

i

d t

!

,

by simple calculation we get

Vα,β =

1

Beta(α,β)

Z 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 existsξ∈(0, 1)such that 1−e−t/4= g(0)−g(t) =te−₄ξ/4 ≥ _4et1/4, which immediately implies

Vα,β ≥

1

4e1/4·Beta(α,β)

Z 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 forq≥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.