JJ II

volume 7, issue 2, article 62, 2006.

Received 14 October, 2005;

accepted 15 November, 2005.

Communicated by:H.M. Srivastava

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Journal of Inequalities in Pure and Applied Mathematics

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

YUAN-CHUAN LI AND SEN-YEN SHAW

Department of Applied Mathematics National Chung-Hsing University Taichung, 402 Taiwan

EMail:[email protected] Graduate School of Engineering

Lunghwa University of Science and Technology Taoyuan, 333 Taiwan

EMail:[email protected]

2000c Victoria University ISSN (electronic): 1443-5756 299-05

A Proof of Hölder’s Inequality using the Cauchy-Schwarz

Inequality

Yuan-Chuan Li and Sen-Yen Shaw

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J. Ineq. Pure and Appl. Math. 7(2) Art. 62, 2006

Abstract

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

2000 Mathematics Subject Classification:26D15.

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

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 theL^p-normkfkp := R

Ω|f|^pdµ¹_p

for 1≤ p <∞ andkfk∞ :=esssup_x∈Ω|f(x)|. Hölder’s Inequality states that:

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

The special case that p = 1 and q = ∞ is obvious, and the special case p = q = 2 is 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 inequal- ity) 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

A Proof of Hölder’s Inequality using the Cauchy-Schwarz

Inequality

Yuan-Chuan Li and Sen-Yen Shaw

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J. Ineq. Pure and Appl. Math. 7(2) Art. 62, 2006

the latter follows the arithmetic-geometric mean inequality. Thus, in a sense, the arithmetic-geometric mean inequality, Hölder’s inequality, the Cauchy-Schwarz inequality, and Lyapunov’s inequality are all equivalent [1, p. 457]. In the fol- lowing, 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 assume f, g ≥ 0and 1 < p, q < ∞. Iff g = 0 a.e. [µ], the inequality is obvious. Therefore we may assume g > 0 on Ω and f 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∈D_F,

with the domainD_F consisting of all thoset ∈Rfor which the integral exists.

Then0,1∈D_F andF(1) =kfk^p_p andF(0) =kgk^q_q.

For everyω ∈Ω,(g^q)(ω)[(f^pg^−q)(ω)]^tis convex onR. 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 onDF. HenceDF is an interval containing[0,1].

It is known (cf. [2, Ch. VII]) that a functionh : (a, b) → R is convex if and only ifhis continuous and midconvex on(a, b). HenceF is continuous on

A Proof of Hölder’s Inequality using the Cauchy-Schwarz

Inequality

Yuan-Chuan Li and Sen-Yen Shaw

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J. Ineq. Pure and Appl. Math. 7(2) Art. 62, 2006

(0,1). Sincef g 6= 0, we must have thatF(t)∈ (0,∞)for allt ∈ [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 andv = [(g^q)(f^pg^−q)^t2]¹² belong toL²(µ)becausekuk²₂ = F(t₁) <∞andkvk²₂ = 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

q lnF(1−t)

= ln F(t)^1/pF(1−t)^1/q ,

A Proof of Hölder’s Inequality using the Cauchy-Schwarz

Inequality

Yuan-Chuan Li and Sen-Yen Shaw

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J. Ineq. Pure and Appl. Math. 7(2) Art. 62, 2006

so that

F 1

pt+1

q(1−t)

≤F(t)^1/pF(1−t)^1/q

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.

A Proof of Hölder’s Inequality using the Cauchy-Schwarz

Inequality

Yuan-Chuan Li and Sen-Yen Shaw

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References

[1] A.W. MARSHALL AND I. OLKIN, Inequalities: Theory of Majorization and Its Applications, Academic Press, 1979.

[2] A.W. ROBERTS AND D.E. VARBERG, Convex Functions, Pure and Ap- plied Mathematics 57, Academic Press, New York, 1973.

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