Open Access

On improvements of the Rozanova's inequality

Journal of Inequalities and Applications20112011:33

https://doi.org/10.1186/1029-242X-2011-33

Received: 14 March 2011

Accepted: 18 August 2011

Published: 18 August 2011

Abstract

In the present paper, we establish some new Rozanova's type integral inequalities involving higher-order partial derivatives. The results in special cases yield some of the interrelated results on Rozanova's inequality and provide new estimates on inequalities of this type.

MS (2000) Subject Classifiication: 26D15.

Keywords

Opial's inequality Hölder's inequality Rozanova's inequality

1 Introduction

In the year 1960, Opial [1] established the following integral inequality:

Theorem A Suppose f C1[0, h] satisfies f(0) = f(h) = 0 and f(x) > 0 for all x (0, h). Then
0 h f ( x ) f ( x ) d x h 4 0 h ( f ( x ) ) 2 d x .
(1.1)

The first Opial's type inequality was established by Willett [2] as follows:

Theorem B Let x(t) be absolutely continuous in [0, a], and x(0) = 0. Then
0 a | x ( t ) x ( t ) | d t a 2 0 a | x ( t ) | 2 d t .
(1.2)

A non-trivial generalization of Theorem B was established by Hua [3] as follows:

Theorem C Let x(t) be absolutely continuous in [0, a], and x(0) = 0. Futher, let l be a positive integer. Then
0 a | x ( t ) x ( t ) | d t a l l + 1 0 a | x ( t ) | l + 1 d t .
(1.3)

A sharper inequality was established by Godunova [4] as follows:

Theorem D Let f(t) be convex and increasing functions on [0, ∞) with f(0) = 0. Further, let x(t) be absolutely continuous on [0, τ], and x(α) = 0. Then, following inequality holds
α τ f ( | x ( t ) | ) | x ( t ) | d t f α τ | x ( t ) | d t .
(1.4)

Rozanova [5] proved an extension of inequality (1.4) is embodied in the following:

Theorem F Let f(t) and g(t) be convex and increasing functions on [0, ∞) with f(0) = 0, and let p(t) ≥ 0, p'(t) > 0, t [α, a] with p(α) = 0. Further, let x(t) be absolutely continuous on [α, a), and x(α) = 0. Then, following inequality holds
α a p ( t ) g | x ( t ) | p ( t ) f p ( t ) g | x ( t ) | p ( t ) d t f α a p ( t ) g | x ( t ) | p ( t ) d t .
(1.5)

The inequality (1.5) will be called as Rozanova's inequality in the paper.

Opial's inequality and its generalizations, extensions and discretizations play a fundamental role in establishing the existence and uniqueness of initial and boundary value problems for ordinary and partial differential equations as well as difference equations [613]. For Opial-type integral inequalities involving high-order partial derivatives, see [14, 15]. For an extensive survey on these inequalities, see [16].

The first aim of the present paper is to establish the following Opial-type inequality involving higher-order partial derivatives, which is an extension of the Rozanova's inequality (1.5).

Theorem 1.1 Let f and g be convex and increasing functions on [0, ∞) with f(0) = 0, and let p(s, t) ≥ 0, D 1 D 2 p ( s , t ) = 2 s t p ( s , t ) , D1D2p(s, t) > 0, s [α, a], t [β, b] with p(s, β) = p(α, t) = p(α, β) = 0 and D1D2p(s, t) |t = τ= 0. Further, let x(s, t) be absolutely continuous on [α, a) × [β, b], and x(s, β) = x(α, t) = x(α, β) = 0. Then following inequality holds
α a β b D 1 D 2 p ( s , t ) g | D 1 D 2 x ( s , t ) | D 1 D 2 p ( s , t ) t f p ( s , t ) g | x ( s , t ) | p ( s , t ) d s d t f α a β b D 1 D 2 p ( s , t ) g | D 1 D 2 x ( s , t ) | D 1 D 2 p ( s , t ) d s d t .
(1.6)

We also prove the following Rozanova-type inequality involving higher-order partial derivatives.

Theorem 1.2 Assume that
  1. (i)

    f, g and x(s, t) are as in Theorem 1.1,

     
  2. (ii)

    p(s, t) is increasing on [0, a] × [0, b] with p(s, β) = p(α, t) = p(α, β) = 0,

     
  3. (iii)

    h is concave and increasing on [0, ∞),

     
  4. (iv)

    ϕ(t) is increasing on [0, a] with ϕ(0) = 0,

     
  5. (v)
    For y ( s , t ) = 0 s 0 t D 1 D 2 p ( σ , τ ) g | D 1 D 2 x ( σ , τ ) | D 1 D 2 p ( σ , τ ) d σ d τ ,
    D 1 D 2 f y ( s , t ) D 1 D 2 y ( s , t ) ϕ 1 D 1 D 2 y ( s , t ) c ( a , b ) y ( a , b ) ϕ t y ( a , b ) .
     
Then
0 a 0 b D 1 D 2 f p ( s , t ) g x ( s , t ) p ( s , t ) v D 1 D 2 p ( s , t ) g D 1 D 2 x ( s , t ) D 1 D 2 p ( s , t ) d s d t w 0 a 0 b D 1 D 2 p ( s , t ) g x ( s , t ) D 1 D 2 p ( s , t ) d s d t ,
(1.7)
where
v ( z ) = z h ϕ 1 z , w ( z ) = c ( a , b ) h a ϕ b z ,
and
c ( a , b ) = 0 a 0 b D 1 D 2 f y ( s , t ) D 1 D 2 y ( s , t ) d s d t .

2 Main results and proofs

Theorem 2.1 Let f and g be convex and increasing functions on [0, ∞) with f(0) = 0, and let p(s, t) ≥ 0, D 1 D 2 p ( s , t ) = 2 s t p ( s , t ) , D1D2p(s, t) > 0, s [α, a], t [β, b] with p(s, β) = p(α, t) = p(α, β) = 0 and D1D2p(s, t) |t = τ= 0. Further, let x(s, t) be absolutely continuous on [α, a) × [β, b], and x(s, β) = x(α, t) = x(α, β) = 0. Then, following inequality holds
α a β b D 1 D 2 p ( s , t ) g | D 1 D 2 x ( s , t ) | D 1 D 2 p ( s , t ) t f p ( s , t ) g | x ( s , t ) | p ( s , t ) d s d t f α a β b D 1 D 2 p ( s , t ) g | D 1 D 2 x ( s , t ) | D 1 D 2 p ( s , t ) d s d t .
(2.1)
Proof Let y ( s , t ) = α s β t D 1 D 2 x ( σ , τ ) d σ d τ so that D1D2y(s, t) = |D1D2x(s, t)| and y(s, t) ≥ |x(s, t)|. Thus, from Jensen's integral inequality, we obtain
g x ( s , t ) p ( s , t ) g y ( s , t ) p ( s , t ) g α s β t D 1 D 2 p ( σ , τ ) D 1 D 2 x ( σ , τ ) D 1 D 2 p ( σ , τ ) d σ d τ α s β t D 1 D 2 p ( σ , τ ) d σ d τ 1 p ( s , t ) α s β t D 1 D 2 p ( σ , τ ) g | D 1 D 2 x ( σ , τ ) | D 1 D 2 p ( σ , τ ) d σ d τ .
(2.2)
By using the inequality (2.2), we have
α a β b D 1 D 2 p ( s , t ) g | D 1 D 2 x ( s , t ) | D 1 D 2 p ( s , t ) t f p ( s , t ) g | x ( s , t ) | p ( s , t ) d s d t α a β b D 1 D 2 p ( s , t ) g D 1 D 2 y ( s , t ) D 1 D 2 p ( s , t ) t f α s β t D 1 D 2 p ( σ , τ ) g D 1 D 2 y ( σ , τ ) D 1 D 2 p ( σ , τ ) d σ d τ d s d t .
(2.3)
On the other hand
2 s t f α s β t D 1 D 2 p ( σ , τ ) g D 1 D 2 y ( σ , τ ) D 1 D 2 p ( σ , τ ) d σ d τ = s t f α s β t D 1 D 2 p ( σ , τ ) g D 1 D 2 y ( σ , τ ) D 1 D 2 p ( σ , τ ) d σ d τ α s p σ t ( σ , t ) g D 1 D 2 y ( σ , τ ) D 1 D 2 p ( σ , t ) d σ = 2 s t f α s β t D 1 D 2 p ( σ , τ ) g D 1 D 2 y ( σ , τ ) D 1 D 2 p ( σ , τ ) d σ d τ α s D 1 D 2 p ( σ , t ) g D 1 D 2 y ( σ , τ ) p σ t ( σ , t ) d σ × β t p s τ ( s , τ ) g D 1 D 2 y ( σ , τ ) D 1 D 2 p ( s , τ ) d τ + D 1 D 2 p ( s , t ) g D 1 D 2 y ( s , t ) D 1 D 2 p ( s , t ) × t f α s β t D 1 D 2 p ( σ , τ ) g D 1 D 2 y ( σ , τ ) D 1 D 2 p ( σ , τ ) d σ d τ = D 1 D 2 p ( s , t ) g D 1 D 2 y ( s , t ) D 1 D 2 p ( s , t ) f t α s β t D 1 D 2 p ( σ , τ ) g D 1 D 2 y ( σ , τ ) D 1 D 2 p ( σ , τ ) d σ d τ .
(2.4)
From (2.3) and (2.4), we have
α a β b D 1 D 2 p ( s , t ) g | D 1 D 2 x ( s , t ) | D 1 D 2 p ( s , t ) t f p ( s , t ) g | x ( s , t ) | p ( s , t ) d s d t α a β b 2 s t f α s β t D 1 D 2 p ( σ , τ ) g D 1 D 2 y ( σ , τ ) D 1 D 2 p ( σ , τ ) d σ d τ d s d t = f α a β b D 1 D 2 p ( σ , τ ) g D 1 D 2 y ( σ , τ ) D 1 D 2 p ( σ , τ ) d σ d τ = f α a β b D 1 D 2 p ( s , t ) g | D 1 D 2 x ( s , t ) | D 1 D 2 p ( s , t ) d s d t .

This completes the proof.

Remark 2.2 Let x(s, t) reduce to s(t), and with suitable modifications in the proof of Theorem 2.1, then (2.1) becomes inequality (1.5) stated in Section 1.

Remark 2.3 Taking for g(x) = x in (2.1), then (2.1) becomes the following inequality.
α a β b D 1 D 2 x ( s , t ) t ( f ( x ( s , t ) ) ) d s d t f α a β b D 1 D 2 x ( s , t ) d s d t .
(2.5)

Let x(s, t) reduce to s(t), and with suitable modifications, then (2.5) becomes inequality (1.4) stated in Section 1.

Remark 2.4 For f(t) = t l +1, l ≥ 0, the inequality (2.5) reduces to
α a β b x ( s , t ) l t ( x ( s , t ) ) d s d t 1 l + 1 α a β b D 1 D 2 x ( s , t ) d s d t l + 1 .
(2.6)
In the right side of (2.6), by Hölder inequality with indices l + 1 and (l + 1)l, gives
α a β b x ( s , t ) l t ( x ( s , t ) ) d s d t [ ( a - α ) ( b - β ) ] l l + 1 α a β b D 1 D 2 x ( s , t ) l + 1 d s d t .
(2.7)

Let x(s, t) reduce to s(t) and α = β = 0, then (2.7) becomes Hua's inequality (1.3) stated in Section 1.

Theorem 2.5 Assume that
  1. (i)

    f, g and x(s, t) are as in Theorem 2.1,

     
  2. (ii)

    p(s, t) is increasing on [0, a] × [0, b] with p(s, β) = p(α, t) = p(α, β) = 0,

     
  3. (iii)

    h is concave and increasing on [0, ∞),

     
  4. (iv)

    ϕ(t) is increasing on [0, a] with ϕ(0) = 0,

     
  5. (v)
    For y ( s , t ) = 0 s 0 t D 1 D 2 p ( σ , τ ) g | D 1 D 2 x ( σ , τ ) | D 1 D 2 p ( σ , τ ) d σ d τ ,
    D 1 D 2 f y ( s , t ) D 1 D 2 y ( s , t ) ϕ 1 D 1 D 2 y ( s , t ) c ( a , b ) y ( a , b ) ϕ t y ( a , b ) .
    (2.8)
     
Then
0 a 0 b D 1 D 2 f p ( s , t ) g x ( s , t ) p ( s , t ) v D 1 D 2 p ( s , t ) g D 1 D 2 x ( s , t ) D 1 D 2 p ( s , t ) d s d t w 0 a 0 b D 1 D 2 p ( s , t ) g D 1 D 2 x ( s , t ) D 1 D 2 p ( s , t ) d s d t ,
(2.9)
where
v ( z ) = z h ϕ 1 z ,
(2.10)
w ( z ) = c ( a , b ) h a ϕ b z .
(2.11)
and
c ( a , b ) = 0 a 0 b D 1 D 2 f y ( s , t ) D 1 D 2 y ( s , t ) d s d t .
Proof From (2.2), we easily obtain
p ( s , t ) g x ( s , t ) p ( s , t ) 0 s 0 t D 1 D 2 p ( σ , τ ) g | D 1 D 2 x ( σ , τ ) | D 1 D 2 p ( σ , τ ) d σ d τ = y ( s , t ) .
(2.12)
From (2.8), (2.10-2.12) and Jensen's inequality(for concave function), hence
0 a 0 b D 1 D 2 f p ( s , t ) g x ( s , t ) p ( s , t ) v D 1 D 2 p ( s , t ) g D 1 D 2 x ( s , t ) D 1 D 2 p ( s , t ) d s d t 0 a 0 b D 1 D 2 f y ( s , t ) v D 1 D 2 y ( s , t ) d s d t = 0 a 0 b D 1 D 2 f y ( s , t ) D 1 D 2 y ( s , t ) h ϕ 1 D 1 D 2 y ( s , t ) d s d t = 0 a 0 b D 1 D 2 f y ( s , t ) D 1 D 2 y ( s , t ) h ϕ 1 D 1 D 2 y ( s , t ) d s d t 0 a 0 b D 1 D 2 f y ( s , t ) D 1 D 2 y ( s , t ) d s d t × 0 a 0 b D 1 D 2 f y ( s , t ) D 1 D 2 y ( s , t ) d s d t h 0 a 0 b D 1 D 2 f y ( s , t ) D 1 D 2 y ( s , t ) ϕ 1 D 1 D 2 y ( s , t ) d s d t 0 a 0 b D 1 D 2 f y ( s , t ) D 1 D 2 y ( s , t ) d s d t c ( a , b ) h 0 a 0 b c ( a , b ) y ( a , b ) ϕ t y ( a , b ) d s d t c ( a , b ) c ( a , b ) = h 1 y ( a , b ) 0 a y ( a , b ) ϕ t y ( a , b ) | t = 0 t = b d s c ( a , b ) = h a ϕ b y ( a , b ) c ( a , b ) = w 0 a 0 b D 1 D 2 p ( s , t ) g | D 1 D 2 x ( s , t ) | D 1 D 2 p ( s , t ) d s d t .

This completes the proof.

Remark 2.6 Let x(s, t) reduce to s(t), and with suitable modifications in the proof of Theorem 2.5, then (2.9) becomes the following inequality:
0 a f p ( t ) g x ( t ) p ( t ) v p ( t ) g x ( t ) p ( t ) d t w 0 a p ( t ) g | x ( t ) | p ( t ) d t .
(2.13)

The inequality has been obtained by Rozanova in [17]. For x ( t ) = x 1 ( t ) , x 1 ( t ) > 0 , x 1 ( 0 ) = 0 , x ( a ) = b , g ( t ) = t , f ( t ) = ϕ ( t ) = t 2 and h ( t ) = 1 + t , the inequality (2.13) reduces to Polya's inequality (see [17]).

Remark 2.7 Taking for g(x) = x in (2.9), then (2.9) becomes the following interesting inequality.
0 a 0 b D 1 D 2 f ( x ( s , t ) ) v ( D 1 D 2 x ( s , t ) ) d s d t w 0 a 0 b D 1 D 2 x ( s , t ) d s d t .

Declarations

Acknowledgements

The authors express their deep gratitude to the referees for their many very valuable suggestions and comments. The research of Chang-Jian Zhao was supported by National Natural Science Foundation of China (10971205), and the research of Wing-Sum Cheung was partially supported by a HKU URC grant.

Authors’ Affiliations

(1)
Department of Mathematics, China Jiliang University
(2)
Department of Mathematics, The University of Hong Kong

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© Zhao and Cheung; licensee Springer. 2011

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