Chen and Wei Journal of Inequalities and Applications (2015) 2015:138 DOI 10.1186/s13660-015-0645-0
RESEARCH
Open Access
A reverse Hölder inequality for α, β-symmetric integral and some related results Guang-Sheng Chen1,2 and Cheng-Dong Wei1,3* *
Correspondence:
[email protected] 1 Key Laboratory for Mixed and Missing Data Statistics of the Education Department of Guangxi Province, Guangxi Teachers Education University, Nanning, Guangxi 530023, China 3 School of Mathematical Sciences, Guangxi Teachers Education University, Nanning, Guangxi 530023, China Full list of author information is available at the end of the article
Abstract In this paper, we establish a reversed Hölder inequality via an α , β -symmetric integral, which is defined as a linear combination of the α -forward and the β -backward integrals, and then we give some generalizations of the α , β -symmetric integral Hölder inequality which is due to Brito da Cruz et al.; some related inequalities are also given. MSC: 26D15; 26E70 Keywords: α , β -symmetric integral; Hölder inequality; Minkowski inequality; Dresher inequality
1 Introduction Let ak ≥ , bk ≥ (k = , , . . . , n), p > , /p + /q = . The classical Hölder inequality [] is stated as follows: n k=
a k bk ≤
n
p ak
/p n
k=
/q q bk
.
(.)
k=
Similarly, the integral version of the Hölder inequality [] is
b a
f (x)g(x) dx ≤ a
b
/p f p (x) dx
b
/q g q (x) dx ,
(.)
a
where f (x) > , g(x) > , p > , /p + /q = , and f (x) and g(x) are continuous real-valued functions on [a, b]. If p = q = , then inequalities (.) and (.) reduce to the well-known Cauchy inequalities [] of the discrete form and the continuous form, respectively. The Hölder inequality and Cauchy inequality play an important role in many areas of pure and applied mathematics. A large number of generalizations, refinements, variations, and applications of these inequalities have been investigated in [–] and references therein. Recently, Brito da Cruz et al. in [] gave a α, β-symmetric integral Hölder inequality as follows. © 2015 Chen and Wei; licensee Springer. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
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Let f , g : R → R and a, b ∈ R with a < b. If |f | and |g| are α, β-symmetric integrable on [a, b], p > with q = p/(p – ). Then
b
f (t)g(t) dα,β t ≤
a
b
f (t)p dα,β t
/p
a
b
g(t)q dα,β t
/q ,
(.)
a
with equality if and only if the functions |f | and |g| are proportional. The aim of this work is to establish a reversed version of the α, β-symmetric integral Hölder inequality and some generalizations of the α, β-symmetric integral Hölder inequality. Moreover, the obtained results will be applied to establish the α, β-symmetric integral reverse Minkowski inequality, Dresher inequality, and their corresponding reverse versions. This paper is organized as follows. In Section , we recall some basic definitions and properties of α, β-symmetric integral, which can also be found in [, ]; in Section , we establish a α, β-symmetric integral reverse Hölder inequality and give some generalizations of the α, β-symmetric integral Hölder inequality, we apply the obtained results to establish the reverse Minkowski inequality, Dresher inequality, and its reverse form involving α, β-symmetric integral, some extensions of the Minkowski and Dresher inequalities are also given; in Section , we establish some further generalizations and refinements of the α, β-symmetric integral Hölder inequality; in Section , we present a subdividing of the α, β-symmetric integral Hölder inequality.
2 Preliminaries In the section, we recall some basic definitions and properties of α, β-symmetric integral. The α-forward and β-backward differences are defined as follows (see []): α [f ](t) :=
f (t + α) – f (t) , α
∇β [f ](t) :=
f (t) – f (t – β) , β
where α > , β > . Definition . (see []) Assume that I ⊆ R with sup I = +∞ and let a, b ∈ I with a < b. If f : I → R and α > , then the α-forward integral of f is defined by
b
f (t)α t = a
where
∞
f (t)α t – a
+∞
f (t)α t = α
x
∞
f (t)α t, b
+∞
k= f (x + kα),
provided the series converges at x = a and x = b.
The α-forward integral has the following properties. Proposition . (see []) Assume that f , g : I → R are α-forward integrable on [a, b], c ∈ [a, b], k ∈ R, then: a . a f (t)α t = ; b c b . a f (t)α t = a f (t)α t + c f (t)α t, when the integrals exist; b a . a f (t)α t = – b f (t)α t; . kf is α-forward integrable on [a, b] and
b
kf (t)α t = k a
b
f (t)α t; a
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. f + g is α-forward integrable on [a, b] and
b
(f + g)(t)α t = a
b a
b
f (t)α t +
a
. if f ≡ , then
b
g(t)α t; a
f (t)α t = .
Proposition . (see []) Assume that f : I → R is α-forward integrable on [a, b]. Let g : I → R be a nonnegative α-forward integrable function on [a, b]. Then fg is α-forward integrable on [a, b]. Proposition . (see []) Assume that f : I → R and let |f | be α-forward integrable on [a, b]. If p > , then |f |p is also α-forward integrable on [a, b]. Proposition . (see []) Assume that f , g : I → R are α-forward integrable on [a, b] with b = a + kα for some k ∈ N . We have: b . If f (t) ≥ for all t ∈ {a + kα : k ∈ N }, then a f (t)α t ≥ . b b . If g(t) ≥ f (t) for all t ∈ {a + kα : k ∈ N }, then a g(t)α t ≥ a f (t)α t. Proposition . (Fundamental theorem of the α-forward integral, see []) Assume that x f : I → R is α-forward integrable over I. Let x ∈ I and define F(x) = a f (t)α t. Then b α [F](x) = f (x). Conversely, a α [f ](t)α t = f (b) – f (a). Proposition . (α-Forward integration by parts, see []) Assume that f , g : I → R. Let fg and f α [g] be α-forward integrable on [a, b]. Then
b
f (t)α [g](t)α t = f (t)g(t)|ba –
a
b
α [f ](t)g(t + α)α [g](t)α t. a
Similarly, the β-backward integral is defined by Definition .. Definition . (see []) Assume that I ⊆ R with inf I = –∞ and let a, b ∈ I with a < b. For f : I → R and β > , the β-backward integral of f is defined by
b
–∞
x
–∞ f (t)∇β t
a
f (t)∇β t –
a
where
b
f (t)∇β t = =β
f (t)∇β t, –∞
+∞
k= f (x – kβ),
provided the series converges at x = a and x = b.
The β-backward integral has similar results and properties to the α-forward integral. In particular, the β-backward integral is the inverse operator of ∇β . We recall the α, β-symmetric integral which is defined as a linear combination of the α-forward and the β-backward integrals. Definition . (see []) Assume that f : R → R and a, b ∈ R with a < b. If f is α-forward and β-backward integrable on [a, b], α, β ≥ with α + β > , then we define the α, βsymmetric integral of f from a to b by
b
f (t) dα,β t = a
α α+β
b
f (t)α t + a
β α+β
b
f (t)∇β t. a
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The function f is α, β-symmetric integrable if it is α, β-symmetric integrable for all a, b ∈ R. The α, β-symmetric integral has the following properties. Proposition . (see []) Suppose that f , g : R → R are α, β-symmetric integrable on [a, b]. Let c ∈ [a, b] and k ∈ R. Then: a . a f (t) dα,β t = ; b c b . a f (t) dα,β t = a f (t) dα,β t + c f (t) dα,β t, when the integrals exist; b a . a f (t) dα,β t = – b f (t) dα,β t; . kf is α, β-symmetric integrable on [a, b] and
b
b
kf (t) dα,β t = k
f (t) dα,β t;
a
a
. f + g is α, β-symmetric integrable on [a, b] and
b
(f + g)(t) dα,β t = a
b
b
f (t) dα,β t + a
g(t) dα,β t; a
. fg is α, β-symmetric integrable on [a, b] provided g is a nonnegative function. Proposition . (see []) Suppose that f : R → R and p > . Let |f | be symmetric α, βintegrable on [a, b], then |f |p is also α, β-symmetric integrable on [a, b]. Proposition . (see []) Assume that f , g : R → R are α, β-symmetric integrable functions on [a, b], A = {a + kα : k ∈ N } and B = {b – kβ : k ∈ N }. For b ∈ A and a ∈ B, we have: . if |f (t)| ≤ g(t) for all t ∈ A ∪ B, then
b a
f (t) dα,β t ≤
b
g(t) dα,β t; a
. if f (t) ≥ for all t ∈ A ∪ B, then
b
f (t) dα,β t ≥ ;
a
. if g(t) ≥ f (t) for all t ∈ A ∪ B, then a
b
b
g(t) dα,β t ≥
f (t) dα,β t. a
In Proposition . we assume that a, b ∈ R with b ∈ A = {a + kα : k ∈ N } and a ∈ B = {b – kβ : k ∈ N }, where α, β ∈ R+ , α + β = . Proposition . (Mean value theorem, see []) If f , g : R → R are bounded and α, βsymmetric integrable on [a, b] with g nonnegative. Let m and M be the infimum and the
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supremum, respectively, of the function f . Then there exists a real number K satisfying the inequalities m ≤ K ≤ M such that
b
b
f (t)g(t) dα,β t = K a
g(t) dα,β t. a
3 Main results As before, let a, b ∈ R with b ∈ A = {a + kα : k ∈ N } and a ∈ B = {b – kβ : k ∈ N }, where α, β ∈ R+ , α + β = . Theorem . (Reverse Hölder inequality) Let f , g : R → R and a, b ∈ R with a < b. If |f | and |g| are α, β-symmetric integrable on [a, b], < p < (or p < ) with q = p/(p – ), then
b
f (t)g(t) dα,β t ≥
a
b
f (t)p dα,β t
/p
a
b
g(t)q dα,β t
/q ,
(.)
a
with equality if and only if the functions |f | and |g| are proportional. Proof We assume that
b
f (t)p dα,β t
/p
a
b
g(t)q dα,β t
/q = ,
a
and let p ξ (t) = f (t)
b
f (τ )p dα,β τ
a
and q γ (t) = g(t)
b
f (τ )q dα,β τ .
a
Since the two functions ξ (t) and γ (t) are symmetric α, β-integrable on [a, b], applying the following reverse Young inequality (see []): xy ≥ xp + yq , p q
x, y > , < p < , /p + /q = ,
with equality holding iff x = y, we have a
b
b
|f (t)|
b
|g(t)|
dα,β t |g(τ )|q dα,β τ )/q b b ξ (t) γ (t) = ξ /p (t)γ /q (t) dα,β t ≥ + dα,β t p p a a b |f (t)|p |g(t)|q b d dα,β t = . t + = b b α,β p q p a q a a |f (τ )| dα,β τ a |g(τ )| dα,β τ (
p /p a |f (τ )| dα,β τ )
(
a
Therefore, we obtain the desired inequality. Combining (.) and (.), we have Corollary ..
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Corollary . Let fj : R → R, pj ∈ R, j = , , . . . , m, integrable on [a, b], then: () For pj > , we have
m b
m fj (t) dα,β t ≤
a
j=
b
fj (t)pj dα,β t
m
j= /pj
= . If |fj | is α, β-symmetric
/pj .
(.)
.
(.)
a
j=
() For < p < , pj < , j = , . . . , m, we have
m b
m fj (t) dα,β t ≥
a
j=
b
fj (t)pj dα,β t
/pj
a
j=
Theorem . (Reverse Minkowski inequality) Let f , g : R → R and a, b, p ∈ R with a < b and < p < (or p < ). If |f | and |g| are α, β-symmetric integrable on [a, b], then
b
f (t) + g(t)p dα,β t
a
b
f (t)p dα,β t
≥
/p
/p
b
g(t)p dα,β t
+
a
/p ,
(.)
a
with equality if and only if the functions |f | and |g| are proportional. Proof Let
b
f (t)p dα,β t,
M= a
b
f (t)p dα,β t
W= a
b
g(t)p dα,β t,
N= /p
a
+
b
g(t)p dα,β t
/p .
a
By the α, β-symmetric integral Hölder inequality [], in view of < p < , we have W =
f (t)p M/p– + g(t)p N /p– dα,β t
b
a
≤
b
f (t) + g(t)p M/p + N /p –p dα,β t
a
=W
–p
b
f (t) + g(t)p dα,β t.
(.)
a
By using (.), we immediately arrive at the Minkowski inequality and the theorem is completely proved. An improvement of inequality (.) and its corresponding reverse form are obtained in Theorem .. Theorem . Assume that f , g : R → R and a, b ∈ R with a < b. Assume that |f | and |g| are α, β-symmetric integrable on [a, b], p > , s, t ∈ R\{}, and s = t.
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() Suppose that p, s, t ∈ R are different, such that s, t > and (s – t)/(p – t) > , then
b
f (x) + g(x)p dα,β x
a
s f (x)s dα,β x +
s s(p–t)/(s–t) g(x)s dα,β x
b
b
t b f (x)t dα,β x +
≤ a
a
× a
b
g(x)t dα,β x
t t(s–p)/(s–t) .
(.)
a
() Suppose that p, s, t ∈ R are different, such that < s, t < and (s – t)/(p – t) < , then
b
f (x) + g(x)p dα,β x
a
≥
s f (x)s dα,β x +
b
a
s s(p–t)/(s–t) g(x)s dα,β x
b
a
t f (x)t dα,β x +
b
× a
b
g(x)t dα,β x
t t(s–p)/(s–t) .
(.)
a
Proof () From the assumption, we have (s – t)/(p – t) > , and it is obvious that
b
f (x) + g(x)p dα,β x
a
f (x) + g(x)s (p–t)/(s–t) f (x) + g(x)t (s–p)/(s–t) dα,β x.
b
= a
From the Hölder inequality (see []) with indices (s – t)/(p – t) and (s – t)/(s – p), it follows that b f (x) + g(x)p dα,β x a
≤
b
f (x) + g(x)s dα,β x
(p–t)/(s–t)
a
(s–p)/(s–t) f (x) + g(x)t dα,β x .
b
(.)
a
On the other hand, from the Minkowski inequality (see []) for s > and t > , respectively, we obtain
s f (x) + g(x)s dα,β x
b a
≤
s f (x)s dα,β x +
s g(x)s dα,β x
b
a
b
(.)
a
and
b
f (x) + g(x)t dα,β x
a
≤ a
f (x)t dα,β x
b
t
t
g(x)t dα,β x
b
+
t .
a
From (.), (.), and (.), the desired result is obtained.
(.)
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() Based on the assumption, we have (s – t)/(p – t) < and in view of
b
f (x) + g(x)p dα,β x
a
f (x) + g(x)s (p–t)/(s–t) f (x) + g(x)t (s–p)/(s–t) dα,β x,
b
= a
by using inequality (.) with indices (s – t)/(p – t) and (s – t)/(s – p), we have
f (x) + g(x)p dα,β x
b
a
≥
b
f (x) + g(x)s dα,β x
(p–t)/(s–t)
a
(s–p)/(s–t) f (x) + g(x)t dα,β x .
b
(.)
a
On the other hand, thanks to the Minkowski inequality (.) for the cases of < s < and < t < ,
s f (x) + g(x)s dα,β x
b a
≥
s f (x)s dα,β x +
b
a
s g(x)s dα,β x
(.)
t g(x)t dα,β x .
(.)
b
a
and
b
f (x) + g(x)t dα,β x
t
a
≥
b
f (x)t (dx)α
a
t
+
b
a
It follows from (.), (.), and (.) that the desired result is obtained. Remark . () For Theorem ., for p > , letting s = p + ε, t = p – ε, when p, s, t are different, s, t > , and (s – t)/(p – t)/ > , and letting ε → , we obtain the result of []. () For Theorem ., for < p < , letting s = p + ε, t = p – ε, when p, s, t are different, < s, t < , and < (s – t)/(p – t)/ < , and letting ε → , we obtain (.).
From the Minkowski inequality [] and the reverse Minkowski inequality involving α, β-symmetric integral, we can deduce the following generalization. Corollary . Let fj : R → R, j = , , . . . , m. If |fj | is α, β-symmetric integrable on [a, b], then: () For p > , we have
m b a
j=
p /p /p m b fj (x)p dα,β x fj (x) dα,β x ≤ . a j=
(.)
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() For < p < , we have
m b a
j=
p /p /p m b p fj (x) dα,β x ≥ . fj (x) dα,β x a
(.)
j=
Corollary . is an analog of Corollary .. Corollary . Let fj : R → R, j = , , . . . , m. If |fj | is α, β-symmetric integrable on [a, b], then: () For p > , we have
b a
m p m fj (x) dα,β x ≥ j=
j=
b
fj (x)p dα,β x.
(.)
fj (x)p dα,β x.
(.)
a
() For < p < , we have
b a
m p m fj (x) dα,β x ≤ j=
j=
b
a
Proof () For p > , let s = p, r = , by the Jensen inequality [], it follows that f (x) + f (x) + · · · + fm (x) ≥ f (x)p + f (x)p + · · · + fm (x)p /p , from the above inequality, we obtain
f (x) + f (x) + · · · + fm (x) p ≥ f (x)p + f (x)p + · · · + fm (x)p , by integrating the above inequality with respect to x, we obtain the desired result. () For < p < , let s = , r = p, by the Jensen inequality [], we have f (x) + f (x) + · · · + fm (x) ≤ f (x)p + f (x)p + · · · + fm (x)p /p , it follows from the above inequality that
f (x) + f (x) + · · · + fm (x) p ≤ f (x)p + f (x)p + · · · + fm (x)p , by integrating the above inequality with respect to x, the desired result is obtained.
Theorem . (Dresher inequality) Let f , g : R → R and a, b ∈ R with a < b. If |f | and |g| are α, β-symmetric integrable on [a, b], < r < < p, then b
/(p–r) p a |f (t) + g(t)| dα,β t b r a |f (t) + g(t)| dα,β t b
≤
/(p–r) p a |f (t)| dα,β t b r a |f (t)| dα,β t
+
b
/(p–r) p a |g(t)| dα,β t , b rd |g(t)| t α,β a
with equality if and only if the functions |f | and |g| are proportional.
(.)
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Proof Based on the α, β-symmetric integral Hölder and Minkowski inequalities [], we have
b
f (t) + g(t)p dα,β t
a
b
/(p–r)
f (t)p dα,β t
≤
/p
|f (t)| dα,β t
a
|f (t)|r dα,β t
p
b a
b
+
/p p/(p–r)
a
a
b
=
g(t)p dα,β t
+
a
b
b
/p
b
f (t)r dα,β t
/p
a
|g(t)|p dα,β x
/p
b
g(t)r dα,β t
/p p/(p–r)
a |g(t)|r dα,β t /(p–r) b /(p–r) b p p a |f (t)| dα,β t a |g(t)| dα,β t ≤ + b b r r a |f (t)| dα,β t a |g(t)| dα,β t b /r b /r r/(p–r) r r f (t) dα,β t g(t) dα,β t × + . a
a
(.)
a
From Theorem ., we get
b
f (t)r dα,β t
/r
+
a
b
g(t)r dα,β t
/r r
a
b
f (t) + g(t)r dα,β t.
≤
(.)
a
From (.) and (.), we get (.). Hence, the theorem is completely proved.
Corollary . Let fj : R → R, < r < < p, j = , , . . . , m. If |fj | is α, β-symmetric integrable on [a, b], then /(p–r) /(p–r) b m m b p p a | j= fj (t)| dα,β t a |fj (t)| dα,β t ≤ . b m b r r j= a | j= fj (t)| dα,β t a |fj (t)| dα,β t
(.)
Theorem . (Reverse Dresher inequality) Let f , g : R → R and a, b ∈ R with a < b. If |f | and |g| are α, β-symmetric integrable on [a, b], p ≤ ≤ r ≤ , then b
/(p–r) p a |f (t) + g(t)| dα,β t b r a |f (t) + g(t)| dα,β t b
/(p–r) p a |f (t)| dα,β t b r a |f (t)| dα,β t
≥
+
b
/(p–r) p a |g(t)| dα,β t , b r a |g(t)| dα,β t
(.)
with equality if and only if the functions |f | and |g| are proportional. Proof Let α ≥ , α ≥ , β > , and β > , and – < λ < , using the following Radon inequality (see []): n p ak p–
k=
bk
( n a k )p < nk= p– , ( k= bk )
xk ≥ , ak > , < p < ,
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we have αλ+ αλ+ (α + α )λ+ + λ ≤ , (β + β )λ βλ β
(.)
with equality if and only if (α) and (β) are proportional. Let
b
α =
/p |f |p dα,β t
a
b
α =
,
,
β =
a
and set λ =
/r |f |r dα,β t
a
/p |g|p dα,β t
b
β = b
,
(.)
/r |g|r dα,β t
,
(.)
a
r . p–r
From (.)-(.), we have
b b αλ+ αλ+ ( a |f |p dα,β t)(λ+)/p ( a |g|p dα,β t)(λ+)/p + λ = b + b βλ β ( a |f |r dα,β t)λ/r ( a |g|r dα,β t)λ/r /(p–r) b p b p |g| dα,β t /(p–r) (α + α )λ+ a |f | dα,β t = b + ab ≤ r r (β + β )λ a |f | dα,β t a |g| dα,β t b b [( a |f |p dα,β t)/p + ( a |g|p dα,β t)/p ]p/(p–r) = b . b [( a |f |r dα,β t)/r + ( a |g|r dα,β t)/r ]r/(p–r)
(.)
r < , we may assume p < < r, and by Theorem . and < r ≤ , we Since – < λ = p–r obtain, respectively,
b
/p |f | dα,β t p
b
/p p |g| dα,β t p
+
a
b
≥
a
|f + g|p dα,β t,
(.)
a
with equality if and only if f and g are proportional, and
b
/r |f | dα,β t r
a
b
/r r |g| dα,β t r
+ a
≤
b
|f + g|r dα,β t,
(.)
a
with equality if and only if |f | and |g| are proportional. From the equality conditions for (.), (.), and (.), it follows that the sign of equality in (.) holds if and only if |f | and |g| are proportional. From (.)-(.), we obtain the reverse Dresher inequality and the theorem is completely proved. Corollary . Let fj : R → R, p ≤ ≤ r < , j = , , . . . , m. If |fj | is α, β-symmetric integrable on [a, b], then /(p–r) /(p–r) b m m b p p a | j= fj (t)| dα,β t a |fj (t)| dα,β t ≥ . b m b r r j= a | j= fj (t)| dα,β t a |fj (t)| dα,β t
4 Some further generalizations of the Hölder inequality
s Theorem . Suppose that pk > , αkj ∈ R (j = , , . . . , m, k = , , . . . , s), k
s α = , f : R → R. If |f | is α, β-symmetric integrable on [a, b], then: kj j j k=
(.)
pk
= ,
Chen and Wei Journal of Inequalities and Applications (2015) 2015:138
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() For pk > , we have
m b a
j=
m /pk s b +pk αkj fj (t) fj (t) dα,β t ≤ dα,β t . a
(.)
j=
k=
() For < ps < , pk < (k = , , . . . , s – ), we have
m b a
j=
m /pk s b +pk αkj fj (t) fj (t) dα,β t ≥ dα,β t . a
(.)
j=
k=
Proof () Let gk (t) =
m
/pk fj
+pk αkj
(t)
.
(.)
j=
Applying the assumptions that s
s
k pk
= and
s
k= αkj
= , it follows from a direct computation
gk (t) = g (t)g (t) · · · gs (t)
k=
=
m
/p fj
+p αj
(t)
j=
=
m
fj
m
/p fj
+p αj
···
(t)
/p +αj
(t)
m
fj
/p +αj
m
/ps fj
+ps αsj
(t)
j=
(t) · · ·
j=
fj
j=
j=
=
m
/p +/p +···/ps +αj +αj +···+αsj
m
fj
/ps +αsj
(t)
j=
(t) =
j=
m
fj (t).
j=
That is, s
gk (t) =
m
fj (t).
j=
k=
It is obvious that a
m b j=
b s fj (t) dα,β t = gk (t) dα,β t. a
(.)
k=
From the Hölder inequality (.), it follows that a
s b k=
/pk s b pk gk (t) dα,β t gk (t) dα,β t ≤ . a k=
Substituting gk (t) in (.), we have inequality (.) immediately.
(.)
Chen and Wei Journal of Inequalities and Applications (2015) 2015:138
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() The proof of inequality (.) is similar to the proof of inequality (.); by (.), (.), and (.), we obtain a
s b k=
/pk s b gk (t)pk dα,β t gk (t) dα,β t ≥ . a
(.)
k=
Substitution of gk (t) in Eq. (.) gives inequality (.) immediately.
Remark . Let s = m, αkj = –/pk for j = k, and αkk = – /pk . Then the inequalities (.) and (.) are respectively reduced to (.) and (.). Many existing inequalities concerned with the Hölder inequality are special cases of the inequalities (.) and (.). For example, we have the following. Corollary . Under the assumptions of Theorem ., assume that s = m, αkj = –t/pk for j = k, and αkk = t( – /pk ) with t ∈ R, then: () For pk > , one obtains
m b a
j=
m –t /pk m b pk t
fk (t) fj (t) fj (t) dα,β t ≤ dα,β t . a k=
(.)
j=
() For < pm < , pk < (k = , , . . . , m – ), one obtains
m b a
j=
m –t /pk m b pk t
fk (t) fj (t) fj (t) dα,β t ≥ dα,β t . a k=
Theorem . Suppose that pk > , r ∈ R, αkj ∈ R (j = , , . . . , m, k = , , . . . , s),
s k= αkj = , fj : R → R. If |fj | is α, β-symmetric integrable on [a, b], then: () For rpk > , we have
m b a
j=
/rpk m s b +rpk αkj fj (t) fj (t) dα,β t ≤ dα,β t . a k=
(.)
j=
s
k pk
= r,
(.)
j=
() For < rps < , rpk < (k = , , . . . , s – ), we have
m b a
j=
m /rpk s b +rpk αkj fj (t) fj (t) dα,β t ≥ dα,β t . a k=
(.)
j=
Proof () Since rpk > and sk p = r, it follows that sk rp = . Then by (.), we immek k diately find inequality (.).
() From < rps < , rpk < , and sk p = r, it follows that sk rp = . By (.), we imk k mediately get inequality (.). This completes the proof. From Theorem ., we establish Corollary ., which is a generalization of Theorem .. Corollary . Under the assumptions of Theorem ., assume that s = , p = p, p = q, αj = –αj = αj , then:
Chen and Wei Journal of Inequalities and Applications (2015) 2015:138
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() For rp > , we get
fj (t) dα,β t
m b a
j=
m b
fj (t)+rpαj dα,β t
≤ a
/rp
m b
fj (t)–rqαj dα,β t
a
j=
/rq .
(.)
.
(.)
j=
() For < rp < , we get
fj (t) dα,β t
m b a
j=
m b
fj (t)+rpαj dα,β t
≥ a
/rp
m b
a
j=
fj (t)–rqαj dα,β t
/rq
j=
Now we present a refinement of inequalities (.) and (.), respectively. Theorem . Under the assumptions of Theorem .: () For rpk > , one has
m b a
j=
m /rpk s b +rpk αkj fj (t) dα,β t ≤ ϕ(c) ≤ dα,β t , fj (t) a
(.)
j=
k=
where c s m fj (t) dα,β t + ϕ(c) ≡ a
j=
k=
m b
fj (t)+rpk αkj dα,β t
c
/rpk
j=
is a nonincreasing function with a ≤ c ≤ b. () For < rps < , rpk < (k = , , . . . , s – ), one has
m b a
j=
/rpk m s b +rpk αkj fj (t) fj (t) dα,β t ≥ ϕ(c) ≥ dα,β t , a j=
k=
where c m s fj (t) dα,β t + ϕ(c) ≡ a
j=
k=
m b
fj (t)+rpk αkj dα,β t
c
j=
is a nondecreasing function with a ≤ c ≤ b. Proof () Let gk (t) =
m j=
/rpk fj
+rpk αkj
(t)
.
/rpk
(.)
Chen and Wei Journal of Inequalities and Applications (2015) 2015:138
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By rearrangement and the assumptions of Theorem ., it follows that m
fj (t) =
j=
s
gk (t).
k=
Then thanks to the Hölder inequality (.), we have a
m b j=
b s fj (t) dα,β t = gk (t) dα,β t a k=
b c s s = gk (t) dα,β t + gk (t) dα,β t a k= c k= /rpk c s s b rpk g g (t) d t + (t) d t ≤ α,β k k α,β a c k=
≤
s
gk (t)rpk dα,β t +
s k=
=
c
s k=
a
k=
=
k=
b
gk (t)rpk dα,β t
/rpk
c b
gk (t)rpk dα,β t
/rpk
a
m b
fj (t)+rpk αkj dα,β t
a
/rpk .
j=
Therefore, we obtain the desired result. () The proof of inequality (.) is similar to the proof of inequality (.).
5 A subdividing of the Hölder inequality Theorem . Let f , g : R → R and a, b ∈ R with a < b. Assume that |f | and |g| are α, βsymmetric integrable on [a, b], and s, t ∈ R, and let p = (s – t)/ – t, p = (s – t)/(s – ). () If s < < t or s > > t, then
b
f (x)g(x) dα,β x ≤
a
b
f (x)sp dα,β x
/p
a
b
f (x)tp dα,β x
×
g(x)tq dα,β x
a
a
b
b
g(x)sq dα,β x
/q
/pq (.)
a
with equality if and only if f (x) and g(x) are proportional. () If s > t > or s < t < ; t > s > or t < s < , then
b
f (x)g(x) dα,β x ≥
a
b
f (x)sp dα,β x
/p
a
b
f (x)tp dα,β x
× a
b
g(x)tq dα,β x
a
b
g(x)sq dα,β x
a
with equality if and only if f (x) and g(x) are proportional.
/q
/pq (.)
Chen and Wei Journal of Inequalities and Applications (2015) 2015:138
Proof () Set p =
s–t , –t
and it follows from s < < t or s > > t that
s–t > . –t
p=
s–t –t
From inequality (.) with indices
Page 16 of 18
s–t , s–
and
it follows that
b
f (x)g(x) dα,β x
a
b
f (x)g(x)s(–t)/(s–t) f (x)g(x)t(s–)/(s–t) dα,β x
= a
b
f (x)g(x)s dα,β x
≤
(–t)/(s–t)
a
b
f (x)g(x)t dα,β x
(s–)/(s–t) ,
with equality if and only if (fg)s and (fg)t are proportional. On the other hand, by the Hölder inequality again, for p = inequalities are obtained:
(.)
a
s–t –t
> , the following two
b
f (x)g(x)s dα,β x
a
b
f (x)s(s–t)/(–t) dα,β x
≤
(–t)/(s–t)
a
(s–)/(s–t) g(x)s(s–t)/(s–) dα,β x ,
b
(.)
a
with equality if and only if f s(s–t)/(–t) and g s(s–t)/(s–) are proportional, and
b
f (x)g(x)t dα,β x
a
b
f (x)t(s–t)/(–t) dα,β x
≤
(–t)/(s–t)
a
b
g(x)t(s–t)/(s–) dα,β x
(s–)/(s–t) ,
(.)
a
with equality if and only if f t(s–t)/(–t) and g t(s–t)/(s–) are proportional. It follows from (.), (.), and (.) that the case () of Theorem . is proved. s–t , and in view of s > t > or s < t < , we have () Let p = –t s–t < , –t
p=
and t > s > or t < s < , we have < have
s–t –t
< , by inequality (.) with indices
s–t –t
and
s–t , s–
we
b
f (x)g(x) dα,β x
a
b
f (x)g(x)s(–t)/(s–t) f (x)g(x)t(s–)/(s–t) dα,β x
= a
≥ a
b
f (x)g(x)s dα,β x
(–t)/(s–t)
b
f (x)g(x)t dα,β x
a
with equality if and only if (fg)s and (fg)t are proportional.
(s–)/(s–t) ,
(.)
Chen and Wei Journal of Inequalities and Applications (2015) 2015:138
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On the other hand, from the reverse Hölder inequality again for < p = < , we obtain the following two inequalities:
s–t –t
< or p =
s–t –t
b
f (x)g(x)s dα,β x
a
≥
b
f (x)s(s–t)/(–t) dα,β x
(–t)/(s–t)
a
(s–)/(s–t) g(x)s(s–t)/(s–) dα,β x ,
b
(.)
a
with equality if and only if f s(s–t)/(–t) and g s(s–t)/(s–) are proportional, and
b
f (x)g(x)t dα,β x
a
≥
b
f (x)t(s–t)/(–t) dα,β x
(–t)/(s–t)
a
b
g(x)t(s–t)/(s–) dα,β x
(s–)/(s–t) ,
(.)
a
with equality if and only if f t(s–t)/(–t) and g t(s–t)/(s–) are proportional. From (.), (.), and (.), the proof of the case () of Theorem . is completed.
Competing interests The authors declare that they have no competing interests. Authors’ contributions All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript. Author details 1 Key Laboratory for Mixed and Missing Data Statistics of the Education Department of Guangxi Province, Guangxi Teachers Education University, Nanning, Guangxi 530023, China. 2 Department of Construction and Information Engineering, Guangxi Modern Vocational Technology College, Hechi, Guangxi 547000, China. 3 School of Mathematical Sciences, Guangxi Teachers Education University, Nanning, Guangxi 530023, China. Acknowledgements The authors thank the editor and the referees for their valuable suggestions to improve the quality of this paper. This paper was partially supported by the Key Laboratory for Mixed and Missing Data Statistics of the Education Department of Guangxi Province (No. GXMMSL201404), the Scientific Research Project of Guangxi Education Department (Nos. YB2014560 and KY2015YB468), and the Natural Science Foundation of Guangxi Province (No. 2013JJAA10097). Received: 1 December 2014 Accepted: 30 March 2015 References 1. Beckenbach, EF, Bellman, R: Inequalities. Springer, Berlin (1961) 2. Mitrinovi´c, DS: Analytic Inequalities. Springer, New York (1970) 3. Hardy, G, Littlewood, JE, Pólya, G: Inequalities, 2nd edn. Cambridge University Press, Cambridge (1952) 4. Yang, X: A generalization of Hölder inequality. J. Math. Anal. Appl. 247, 328-330 (2000) 5. Yang, X: Refinement of Hölder inequality and application to Ostrowski inequality. Appl. Math. Comput. 138, 455-461 (2003) 6. Yang, X: A note on Hölder inequality. Appl. Math. Comput. 134, 319-322 (2003) 7. Yang, X: Hölder’s inequality. Appl. Math. Lett. 16, 897-903 (2003) 8. Abramovich, S, Peˇcari´c, JE, Varošanec, S: Sharpening Hölder’s and Popoviciu’s inequalities via functionals. Rocky Mt. J. Math. 34, 793-810 (2004) 9. Abramovich, S, Peˇcari´c, JE, Varošanec, S: Continuous sharpening of Hölder’s and Minkowski’s inequalities. Math. Inequal. Appl. 8(2), 179-190 (2005) 10. Wu, S, Debnath, L: Generalizations of Aczél’s inequality and Popoviciu’s inequality. Indian J. Pure Appl. Math. 36(2), 49-62 (2005) 11. He, WS: Generalization of a sharp Hölder’s inequality and its application. J. Math. Anal. Appl. 332, 741-750 (2007) 12. Wu, S: A new sharpened and generalized version of Hölder’s inequality and its applications. Appl. Math. Comput. 197, 708-714 (2008) 13. Kwon, EG, Bae, EK: On a continuous form of Hölder inequality. J. Math. Anal. Appl. 343, 585-592 (2008) 14. Nikolova, L, Varošanec, S: Refinements of Hölder’s inequality derived from functions ψp,q,λ and φp,q,λ . Ann. Funct. Anal. 2(1), 72-83 (2011) 15. Xu, B, Wang, X-H, Wei, W, Wang, H: On reverse Hilbert-type inequalities. J. Inequal. Appl. 2014, 198 (2014) 16. Yang, B, Chen, Q: A more accurate half-discrete reverse Hilbert-type inequality with a non-homogeneous kernel. J. Inequal. Appl. 2014, 96 (2014)
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