Journal of Inequalities in Pure and Applied Mathematics

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Journal of Inequalities in Pure and Applied Mathematics

http://jipam.vu.edu.au/

Volume 7, Issue 5, Article 176, 2006

A NOTE ON THE HÖLDER INEQUALITY

J. PE ˇCARI ´C AND V. ŠIMI ´C FACULTY OFTEXTILETECHNOLOGY

UNIVERSITY OFZAGREB

PRILAZ BARUNAFILIPOVI ´CA30 10000 ZAGREB, CROATIA

[email protected] [email protected]

Received 22 November, 2006; accepted 30 November, 2006 Communicated by W.S. Cheung

ABSTRACT. In the present paper the authors present some new results concerning the Hölder inequality.

Key words and phrases: Hölder inequality, Young’s inequality, Arithmetic-geometric inequality.

2000 Mathematics Subject Classification. 26D15.

In the following,(Ω, F)is a measure space andµis a positive measure onΩ. Letf, g : Ω→ [0,∞) be two measurable functions. Forp, q ≥ 1such that ¹_p + ¹_q = 1, the classical Hölder’s integral inequality is the following one ([2], [3])

(1)

Z

Ω

f(x)g(x)dµ(x)≤ Z

Ω

f^p(x)dµ(x) ¹_pZ

Ω

g^q(x)dµ(x) ¹_q

. Inequality (1) may be written equivalently as

(2) kf gk₁ ≤ kfk_pkgk_q, where

kfk_p = Z

Ω

f^p(x)dµ(x) ¹_p

The classical proof of (1) is based on Young’s inequality

(3) uv ≤ u^p

p +v^q q , whereu, v ≥0and p, q ≥1such that ¹_p + ¹_q = 1.

Moveover, recently the following result about (3) were obtained in ([1]):

ISSN (electronic): 1443-5756

301-06

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2 J. PE ˇCARI ´C ANDV. ŠIMI ´C

Lemma 1. Letu, v≥0and p, q ≥1such that ¹_p +¹_q = 1. Then forp≥2

(4) P(u, v)≤ 1

2u^2−p(v−u^p−1)², where

P(u, v) = u^p p +v^q

q −uv.

Ifp∈(1, 2], then the reverse inequality in (4) is valid. Forp= 2we have the identity in (4).

First, we shall give a new proof of Lemma 1.

Proof. Inequality (4) is equivalent to the following 1

2u^2−pv²+ 1

2− 1 p

u^p− v^q q ≥0, i.e.

(5) u^2−p

q q

2v²+q(p−2)

2p u^2(p−1)−v^qu^p−2

≥0.

Let us denote byQ(u, v)the left-hand side of (5). Observe that q

2+ q(p−2) 2p = 1.

Suppose thatp ≥ 2, that isq ≤ 2. Using the known arithmetic-geometric inequality ([2], [3]) we obtain

q

2v²+q(p−2)

2p u^2(p−1) ≥(v²)^q²(u^2(p−1))

q(p−2)

2p ≡v^qu^p−2.

Thus Q(u, v) ≥ 0 and (5) is valid. For p ∈ (1, 2]applying the reverse arithmetic-geometric

inequality we have the reverse inequality in (5).

We will prove the next theorem.

Theorem 2. Suppose that ¹_p + ¹_q = 1for1< q ≤2≤ p <∞.Then the following inequalities are valid

1 2

g^2−q fkgk

q

qp −g^q−1kfk_p² 1

kfk_p kgk

q p

q

≤ kfk_pkgk_q− kf gk₁ (6)

≤ 1 2

f^2−p gkfk

p q

p −f^p−1kgk_q² 1

kfk

p q

p kgkq

.

Proof. If we set in (4)

(7) u= f(x)

kfk_p, v = g(x) kgk_q, we obtain

1 p

f^p(x) kfk^pp

− f(x)g(x) kfk_p kgk_q + 1

q g^q(x)

kgk^qq

≤ 1 2

f^2−p(x) kfk^2−pp

g(x)

kgk_q − f^p−1(x) kfk^p−1p

2

.

J. Inequal. Pure and Appl. Math., 7(5) Art. 176, 2006 http://jipam.vu.edu.au/

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HÖLDERNOTE 3

Integrating the last inequality, we obtain 1− kf gk₁

kfkp kgkq

≤ 1 2kfk^2−pp

f^2−p g kgkq

− f^p−1 kfk

p

pq

!

2

1,

i.e.,

kfk_p kgk_q− kf gk₁ ≤ 1 2

f^2−p gkfk

p q

p −f^p−1kgk_q² 1

kfk

p q

pkgk_q

, which proves the right-hand side of (6).

For the left-hand side of (6) we use the reverse of the inequality in (4). After the substitutions u→v, v→u, p →qandq→pwe have

P(u, v)≥ 1

2v^2−q(u−v^q−1)².

Foruandv from (7) we can similarly obtain the first inequality in (6).

REFERENCES

[1] O. DOŠLÝANDÁ. ELBERT, Integral charcterization of the principal solution of the half-linear sec- ond order differential equations, Studia Scientiarum Mathematicarum Hungarica, 36 (2000), 455–

469.

[2] G.H. HARDY, J.E. LITTLEWOODANDG. POLYA, Inequalities , 2nd ed., Cambridge Univ. Press, Cambridge, 1959.

[3] D.S. MITRINOVI ´C, J.E. PE ˇCARI ´C ANDA.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht 1993.

J. Inequal. Pure and Appl. Math., 7(5) Art. 176, 2006 http://jipam.vu.edu.au/