An Inequality for the Beta Function with Application to Pluripotential Theory

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Volume 2009, Article ID 901397,8pages doi:10.1155/2009/901397

Research Article

An Inequality for the Beta Function with

Application to Pluripotential Theory

Per ˚

Ahag

1

_{and Rafał Czy ˙z}

2

1_{Department of Natural Sciences, Engineering and Mathematics, Mid Sweden University,}

871 88 H¨arn¨osand, Sweden

2_{Institute of Mathematics, Jagiellonian University,}_Ł_{ojasiewicza 6, 30-348 Krak´ow, Poland}

Correspondence should be addressed to Per ˚Ahag,[email protected]

Received 4 June 2009; Accepted 22 July 2009

Recommended by Paolo Ricci

We prove in this paper an inequality for the beta function, and we give an application in pluripotential theory.

Copyrightq2009 P. ˚Ahag and R. Czy ˙z. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

A correspondence that started in 1729 between Leonhard Euler and Christian Goldbach was the dawn of the gamma function that is given by

Γx

_∞

0

e−ttx−1dt 1.1

see, e.g.,1,2. One of the gamma function’s relatives is the beta function, which is defined by

Ba, b

1

0

ta−11−tb−1dt . 1.2

The connection between these two Eulerian integrals is

Ba, b ΓaΓb

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Since Euler’s days the research of these special functions and their generalizations have had great impact on, for example, analysis, mathematical physics, and statistics. In this paper we prove the following inequality for the beta function.

Inequality A. For all n∈N and allp≥0_{p /}0,_{p /}1there exists a numberk >0 such that

knpnp/np Bp1, kn> Bp1, n. 1.4

If p0,then we have equality in1.4, and ifp1,then we have the opposite inequality for alln∈N,k >0.

In Section 3 we will give an application of Inequality A within the pluripotential theory.

2. Proof of Inequality A

A crucial tool inLemma 2.2is the following theorem.

Theorem 2.1. Letψx Γx/Γxbe the digamma function. Then forx >0it holds that

ψx> 1 x

1

2x2, ψx>−

1

x2 −

1

x3 −

1

2x4. 2.1

Proof. This follows from3, Theorem 8 see also4,5.

Lemma 2.2. Letα:N×0,∞ → Rbe a function defined by

αn, p 1 n

p

npψn−ψ

np1, 2.2

where ψx Γx/Γxis the digamma function. Thenαn, p/0for alln ∈ Nand allp > 0

(_{p /}1). Furthermore,αn,1 0for alln∈N.

Proof. Sinceψx1 ψx 1/x,we have thatαn,1 0, and

αn, p 1 n

p−1

np ψn−ψ

np. 2.3

From the construction ofαwe also have thatαn,0 0. By using2.3we get that

∂α ∂p

n1

np2 −ψ

_n_p _.

2.4

FromTheorem 2.1it follows that

∂α ∂p <

n1

np2 −

1

np −

1 2np2

1−2p

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Thus,

∂α

∂p <0 forp∈

1 2,∞

. 2.6

Furthermore,

∂2_α

∂p2

−2n1

np3 −ψ

_n_p_, _2.7

and sinceψx>−1/x2₋₁_/x3₋₁_/₂_x4_{Theorem 2.1}_{, we get that}

∂2_α

∂p2 <

−2n1

np3

1

np2

1

np3

1 2np4

−2n2−2n−2p2p21

2np4 ,

2.8

which means that

∂2_α

∂p2 <0 forp∈0,1. 2.9

From2.6,2.9, and the fact thatαn,1 αn,0 0, we conclude thatαn, p/0 for all

n∈Nand allp >0_{p /}1.

Proof of Inequality A.

Case 1p0. The definition

Ba, b

₁

0

ta−11−tb−1dt 2.10

yields thatBa,1 B1, a 1/a. Thus,

kB1, kn 1

n B1, n, 2.11

which is precisely the desired equality.

Case 2p1. We will now prove that for allk >0 it holds that

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Inequality2.12is equivalent to

k2n1/n1 1 kn1

1

kn ≤

1

nn1. 2.13

Hence, to complete this case we need to prove that for allk >0 we have that

kn/n1 1 kn1 ≤

1

n1. 2.14

Leth:0,∞ → Rbe defined by

hk kn1−kn/n1n−kn/n1. 2.15

To obtain2.14it is suﬃcient to prove thath≥0. The definition ofhyields that

h0 1, lim

k→ ∞hk ∞, h

_k _n₁₋_k−1/n1_. _2.16

Thus,

ahhas a minimum point ink1;

bhis decreasing on0,1;

chis increasing on1,∞;

dh1 0.

Thus,hk≥0 fork≥0.

Case 3p >0_{, p /}1. Fixn∈N. LetF:0,∞ → Rbe the function defined by

Fk knpnp/npBp1, kn−Bp1, n. 2.17

This construction implies thatF is continuously diﬀerentiable, andF1 0. To prove this case it is enough to show thatF1/0. By rewritingBp1, knwith1.3the functionFcan be written as

Fk knpnp/npΓ

p1Γkn

Γknp1 −B

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and therefore we get that

Fk Γp1 npnp

np k

np/np Γkn

Γknp1

nknpnp/npΓ

_kn_Γ_kn_p₁₋_Γ_kn_Γ_kn_p₁

Γ2_kn_p₁

nknp/np_B_{kn, p}₁1

n p npk

ψkn−ψknp1.

2.19

Thus

F1 nBn, p11

n p

npψn−ψ

np1, 2.20

whereψx Γx/Γxis the digamma function. This proof is then completed by using

Lemma 2.2.

3. The Application

We start this section by recalling some definitions and needed facts. Adomainis an open and connected set, and a bounded domainΩ⊆Cn_is_hyperconvex_{if there exists a plurisubharmonic}

functionϕ :Ω → −∞,0such that the closure of the set{z ∈ Ω :ϕz < c}is compact in

Ω, for everyc∈−∞,0; that is, for everyc < 0 the level set{z ∈Ω:ϕz < c}is relatively compact inΩ. The geometric condition that our underlying domain should be hyperconvex is to ensure that we have a satisfying quantity of plurisubharmonic functions. ByE0Ωwe denote the family of all bounded plurisubharmonic functionsϕdefined onΩsuch that

lim

z→ξϕz 0 for everyξ∈∂Ω,

Ω

ddcϕn<∞, 3.1

where ddc_·n

is the complex Monge-Amp`ere operator. Next letEpΩ,p > 0, denote the

family of plurisubharmonic functions u defined on Ω such that there exists a decreasing sequence{uj},uj∈ E0, that converges pointwise touonΩ, asjtends to∞, and

sup

j≥1

Ω

−uj p

ddcuj n

sup

j≥1

ep

uj

<∞. 3.2

If u ∈ EpΩ, then epu < ∞ 6, 7. It should be noted that it follows from6 that

the complex Monge-Amp`ere operator is well defined onEp. For further information about

pluripotential theory and the complex Monge-Amp`ere operator we refer to8,9.

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Theorem 3.1. Letp >0, andn≥1. Then there exists a constantDn, p≥1, depending only onn

andp, such that for anyu0, u1, . . . , un∈ Epit holds that

Ω−u0 p

ddcu1∧ · · · ∧ddcun≤D

n, pepu0p/pnepu11/pn. . . epun1/pn. 3.3

Moreover,

Dn, p≤

⎧ ⎪ ⎨ ⎪ ⎩

1

p

n/n−p

, if0< p <1,

ppan,p/p−1_, _{if p}_>₁_,

3.4

Dn,1 1and an, p p2p1/pn−1 −p1. If n 1, then one follows [12] and interprets3.3as

Ω−u

p_Δ

v≤D1, p Ω−u

p_Δ

u

p/p1

Ω−v

p_Δ

v

1/p1

. 3.5

IfDn, p 1 for all functions inEp, then the methods in15 would immediately

imply that the vector spaceEp−Ep, with certain norm, is a Banach space. Furthermore, proofs

in15 see also6could be simplified, and some would even be superfluous. Therefore, it is important to know for whichn, pthe constantDn, pis equal or strictly greater than one. With the help of Inequality A we settle this question. InExample 3.2, we show that there are functions such that, for alln∈Nand allp >0_{p /}1, the constantDn, p, in3.3, is strictly greater than 1.

Example 3.2. LetB0,1⊂Cn_{be the unit ball, and for}_{α >}_{0 set}

uαz |z|2α−1. 3.6

Hence,

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wheredλnis the Lebesgue measure onCn. Forβ >0 we then have that

B0,1 −uαp

ddc_u β

n

n!4n_βn1

B0,1

1− |z|2αp|z|2nβ−1dλn

n!4nβn1

∂B0,1dσn 1

0

1−t2αpt2nβ−1t2n−1dt

n!4nβn1σn∂B0,1 1

0

1−t2αpt2nβ−1dt

n!4nβn12 π

n

n−1! 1 2α

1

0

1−spsnβ/α−1ds

n4πnβ

n1

α B

p1,β αn

,

3.8

wheredσnis the Lebesgue measure on∂B0,1. Ifα β, then

B0,1

−uαpddcuαnn4πnαn B

p1, n. 3.9

If we assume thatDn, p 1 inTheorem 3.1, then it holds that

n4πnβ

n1

α B

p1,β αn

≤n4πnαnBp1, np/npn4πnβnBp1, nn/np.

3.10

Hence,

β α

npnp/np

B

p1,β αn

≤Bp1, n ∀α, β >0. 3.11

In particular, ifβ/αk, then we get that

knpnp/npBp1, kn≤Bp1, n. 3.12

This contradicts Inequality A. Thus, there are functions such thatDn, p>1 for alln∈Nand allp >0_{p /}1.

Acknowledgments

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