Journal of Inequalities in Pure and
Applied Mathematics
A PROOF OF HÖLDER’S INEQUALITY USING THE
CAUCHY-SCHWARZ INEQUALITY
volume 7, issue 2, article 62,
2006.
YUAN-CHUAN LI AND SEN-YEN SHAW
Department of Applied Mathematics
National Chung-Hsing University
Taichung, 402 Taiwan
EMail: ycli@amath.nchu.edu.tw
Graduate School of Engineering
Lunghwa University of Science and Technology
Taoyuan, 333 Taiwan
EMail: shaw@math.ncu.edu.tw
Received 14 October, 2005;
accepted 15 November, 2005.
Communicated by: H.M. Srivastava
Abstract
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Victoria University
ISSN (electronic): 1443-5756
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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
Lp (µ) ≡ Lp (Ω, µ) := {f : Ω → C; kf kp < ∞}
1
R
be a Lebesgue space with the Lp -norm kf kp := Ω |f |p dµ p for 1 ≤ p < ∞
and kf k∞ := ess supx∈Ω |f (x)|. Hölder’s Inequality states that:
If p, q ≥ 1 be such that p1 + 1q = 1, and if f ∈ Lp (µ) and g ∈ Lq (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. 6364]) by using the arithmetic-geometric mean inequality (or Young’s inequality) ab ≤ p1 ap + 1q bq , p1 + 1q = 1 (which follows from Jensen’s inequality, a
consequence of the convexity of a function). It is also known that the CauchySchwarz 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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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 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 assume f, g ≥ 0 and 1 < p, q < ∞. If f 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
Z
Z
pt q(1−t)
F (t) :=
f g
dµ = (g q )(f p g −q )t dµ, t ∈ DF ,
Ω
Ω
with the domain DF consisting of all those t ∈ R for which the integral exists.
Then 0, 1 ∈ DF and F (1) = kf kpp and F (0) = kgkqq .
For every ω ∈ Ω, (g q )(ω)[(f p g −q )(ω)]t is convex on R. Therefore for every
t1 , t2 ∈ R, 0 < λ < 1 and ω ∈ Ω,
q
p −q
λt1 +(1−λ)t2
(g )(ω)[(f g )(ω)]
≤ λ(g q )(ω)[(f p g −q )(ω)]t1 + (1 − λ)(g q )(ω)[(f p g −q )(ω)]t2 .
A Proof of Hölder’s Inequality
using the Cauchy-Schwarz
Inequality
Yuan-Chuan Li and Sen-Yen Shaw
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By integration with respect to µ, we obtain that for t1 , t2 ∈ DF and 0 < λ < 1
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F (λt1 + (1 − λ)t2 ) ≤ λF (t1 ) + (1 − λ)F (t2 ),
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i.e., F is convex on DF . Hence DF 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
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J. Ineq. Pure and Appl. Math. 7(2) Art. 62, 2006
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(0, 1). Since f g 6= 0, we must have that F (t) ∈ (0, ∞) for all t ∈ [0, 1] and so
ln F is well-defined on [0, 1] and is continuous on (0, 1). Let t1 , t2 ∈ (0, 1) be
1
1
arbitrary. The functions u = [(g q )(f p g −q )t1 ] 2 and v = [(g q )(f p g −q )t2 ] 2 belong
to L2 (µ) because kuk22 = F (t1 ) < ∞ and kvk22 = F (t2 ) < ∞. Hence we can
apply the Cauchy-Schwarz inequality to u and v and obtain
Z
1
1
1
1
F
t1 + t2 = (g q )(f p g −q ) 2 t1 + 2 t2 dµ
2
2
ZΩ
1
1
= [(g q )(f p g −q )t1 ] 2 [(g q )(f p g −q )t2 ] 2 dµ
Ω
Z
≤
12 Z
21
q
p −q t2
(g )(f g ) dµ
(g )(f g ) dµ
p −q t1
q
Ω
A Proof of Hölder’s Inequality
using the Cauchy-Schwarz
Inequality
Yuan-Chuan Li and Sen-Yen Shaw
Ω
1
2
1
2
= F (t1 ) F (t2 ) .
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Then we have
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ln F
1
1
t1 + t2
2
2
≤
1
1
ln F (t1 ) + ln F (t2 ),
2
2
i.e., ln F is midconvex on (0, 1). By the above remark we have that ln F is
convex on (0, 1). Therefore
1
1
1
1
ln F
t + (1 − t) ≤ ln F (t) + ln F (1 − t)
p
q
p
q
= ln F (t)1/p F (1 − t)1/q ,
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so that
F
1
1
t + (1 − t) ≤ F (t)1/p F (1 − t)1/q
p
q
for all t ∈ (0, 1). Since F is continuous on (0, 1) and convex on [0, 1], we have
1
1
1
F
= lim F
t + (1 − t)
t↑1
p
p
q
≤ lim sup F (t)1/p lim sup F (1 − t)1/q
t↑1
≤ F (1)
and so ||f g||1 ≤ ||f ||p ||g||q .
t↑1
1/p
F (0)
1/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 Applied Mathematics 57, Academic Press, New York, 1973.
[3] W. RUDIN, Real and Complex Analysis, 3rd Ed., McGraw-Hill, Inc. 1987.
A Proof of Hölder’s Inequality
using the Cauchy-Schwarz
Inequality
Yuan-Chuan Li and Sen-Yen Shaw
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