Journal of Inequalities in Pure and Applied Mathematics

Journal of Inequalities in Pure and Applied Mathematics

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

Volume 7, Issue 2, Article 62, 2006

A PROOF OF HÖLDER’S INEQUALITY USING THE CAUCHY-SCHWARZ INEQUALITY

YUAN-CHUAN LI AND SEN-YEN SHAW DEPARTMENT OFAPPLIEDMATHEMATICS

NATIONALCHUNG-HSINGUNIVERSITY

TAICHUNG, 402 TAIWAN

[email protected] GRADUATESCHOOL OFENGINEERING

LUNGHWAUNIVERSITY OFSCIENCE ANDTECHNOLOGY

TAOYUAN, 333 TAIWAN

[email protected]

Received 14 October, 2005; accepted 15 November, 2005 Communicated by H.M. Srivastava

ABSTRACT. In this note, Hölder’s inequality is deduced directly from the Cauchy-Schwarz in- equality.

Key words and phrases: Hölder’s inequality, Cauchy-Schwarz inequality.

2000 Mathematics Subject Classification. 26D15.

Let(Ω, µ)be a measure space and

L^p(µ)≡L^p(Ω, µ) :={f : Ω→C;kfk^p <∞}

be a Lebesgue space with the L^p-normkfk_p := R

Ω|f|^pdµ¹_p

for 1 ≤ p < ∞and kfk∞ :=

esssup_x∈Ω|f(x)|. Hölder’s Inequality states that:

If p, q ≥ 1 be such that ¹_p + ¹_q = 1, and if f ∈ L^p(µ) and g ∈ L^q(µ), then f g ∈ L¹(µ)and||f g||₁ ≤ ||f||_p||g||_q.

The special case that p = 1andq = ∞is obvious, and the special case p = q = 2is the Cauchy-Schwarz inequality: ||f g||₁ ≤ ||f||₂||g||₂, which actually holds in all inner-product spaces.

Hölder’s inequality can be easily proved (cf. [1, p. 457], [3, pp. 63-64]) by using the arithmetic-geometric mean inequality (or Young’s inequality) ab ≤ ¹_pa^p + ¹_qb^q, ¹_p + ¹_q = 1 (which follows from Jensen’s inequality, a consequence of the convexity of a function). It is also known that the Cauchy-Schwarz inequality implies Lyapunov’s inequality (cf. [1, p.

462]), and from the latter follows the arithmetic-geometric mean inequality. Thus, in a sense,

ISSN (electronic): 1443-5756

c 2006 Victoria University. All rights reserved.

299-05

2 YUAN-CHUANLI ANDSEN-YENSHAW

the arithmetic-geometric mean inequality, Hölder’s inequality, the Cauchy-Schwarz inequality, and Lyapunov’s inequality are all equivalent [1, p. 457]. In the following, we will see that by using the property of convexity one can also deduce Hölder’s inequality directly from the Cauchy-Schwarz inequality.

It suffices to assumef, g≥0and1< p, q <∞. Iff g = 0a.e.[µ], the inequality is obvious.

Therefore we may assumeg >0onΩandf g 6= 0. Define the function F(t) :=

Z

Ω

f^ptg^q(1−t)dµ= Z

Ω

(g^q)(f^pg^−q)^tdµ, t ∈DF,

with the domainDF consisting of all thoset∈Rfor which the integral exists. Then0,1∈DF

andF(1) =kfk^p_pandF(0) =kgk^q_q.

For every ω ∈ Ω, (g^q)(ω)[(f^pg^−q)(ω)]^t is convex on R. Therefore for every t1, t2 ∈ R, 0< λ <1andω∈Ω,

(g^q)(ω)[(f^pg^−q)(ω)]^{λt1+(1−λ)t2}

≤λ(g^q)(ω)[(f^pg^−q)(ω)]^t1 + (1−λ)(g^q)(ω)[(f^pg^−q)(ω)]^t2. By integration with respect toµ, we obtain that fort₁, t₂ ∈D_F and0< λ <1

F(λt₁ + (1−λ)t₂)≤λF(t₁) + (1−λ)F(t₂), i.e.,F is convex onD_F. HenceD_F is an interval containing[0,1].

It is known (cf. [2, Ch. VII]) that a function h : (a, b) → R is convex if and only if h is continuous and midconvex on (a, b). Hence F is continuous on (0,1). Since f g 6= 0, we must have that F(t) ∈ (0,∞) for all t ∈ [0,1] and so lnF is well-defined on [0,1] and is continuous on(0,1). Lett₁, t₂ ∈ (0,1)be arbitrary. The functionsu = [(g^q)(f^pg^−q)^t1]¹² and v = [(g^q)(f^pg^−q)^t2]¹² belong to L²(µ)becausekuk²₂ = F(t₁) < ∞and kvk²₂ = F(t₂) < ∞.

Hence we can apply the Cauchy-Schwarz inequality touandv and obtain F

1 2t₁+ 1

2t₂

= Z

Ω

(g^q)(f^pg^−q)^12t1+12t2dµ

= Z

Ω

[(g^q)(f^pg^−q)^t1]¹²[(g^q)(f^pg^−q)^t2]¹²dµ

≤ Z

Ω

(g^q)(f^pg^−q)^t1dµ ¹₂ Z

Ω

(g^q)(f^pg^−q)^t2dµ ¹₂

=F(t₁)¹²F(t₂)¹². Then we have

lnF 1

2t1+ 1 2t2

≤ 1

2lnF(t1) + 1

2lnF(t2),

i.e., lnF is midconvex on (0,1). By the above remark we have that lnF is convex on(0,1).

Therefore

lnF 1

pt+ 1

q(1−t)

≤ 1

plnF(t) + 1

qlnF(1−t)

= ln F(t)^1/pF(1−t)^1/q , so that

F 1

pt+1

q(1−t)

≤F(t)^1/pF(1−t)^1/q

J. Inequal. Pure and Appl. Math., 7(2) Art. 62, 2006 http://jipam.vu.edu.au/

HÖLDER’SINEQUALITY 3

for allt∈(0,1). SinceF is continuous on(0,1)and convex on[0,1], we have F

1 p

= lim

t↑1 F 1

pt+1

q(1−t)

≤lim sup

t↑1

F(t)^1/plim sup

t↑1

F(1−t)^1/q

≤F(1)^1/pF(0)^1/q, and so||f g||₁ ≤ ||f||_p||g||_q.

REFERENCES

[1] A.W. MARSHALLANDI. OLKIN, Inequalities: Theory of Majorization and Its Applications, Aca- demic Press, 1979.

[2] A.W. ROBERTSANDD.E. VARBERG, Convex Functions, Pure and Applied Mathematics 57, Aca- demic Press, New York, 1973.

[3] W. RUDIN, Real and Complex Analysis, 3rd Ed., McGraw-Hill, Inc. 1987.

J. Inequal. Pure and Appl. Math., 7(2) Art. 62, 2006 http://jipam.vu.edu.au/