Ostrowski Type Inequalities for Functions whose Derivatives are Convex

S.S. Dragomir and A. Sofo

School of Communications and Informatics,

Victoria University of Technology,

Victoria, Australia .

(email: sever@matilda.vu.edu.au sofo@matilda.vu.edu.au)

Abstract

An Ostrowski type inequality for absolutely continuous functions whose derivatives are convex and applications are given.

I. Introduction

The following result is known in the literature as Ostrowski’s inequality [1]. Theorem 1: Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with the property that |f′ (t)| ≤ M for all t ∈ (a, b). Then

|f(x)− \frac{1}{b − a}∫_{a}^{b}f (t) dt|≤[\frac{1}{4} + \begin{matrix}(x− \frac{a+b}{2}) & 2 \\ (b − a) & 2 \end{matrix} ](b−a) M (1)

for all x ∈ [a, b]. The constant \frac{1}{4} is the best possible in the sense that it cannot be replaced by a smaller constant. A simple proof of this fact can be done by using the identity:

f (x) = \frac{1}{b − a} ∫_{a}^{b} f (t) dt

+ \frac{1}{b − a} ∫_{a}^{b} p (x, t) f′ (t) dt, x ∈ [a, b] , (2)

where

p (x, t) := \{ \begin{matrix}t − a & if & a ≤ t ≤ x \\ t − b & if & x < t ≤ b \end{matrix}

which also holds for absolutely continuous functions f : [a, b] → ℝ. The following Ostrowski type result for absolutely continuous functions holds (see [2], [3] and [4]). Theorem 2: Let f : [a, b] → ℝ be absolutely continuous on [a, b]. Then, for all x ∈ [a, b], we have:

|f (x) − \frac{1}{b − a} ∫_{a}^{b} f (t) dt|  (3)

≤ \{ \begin{matrix}[\frac{1}{4} + (\frac{x− \frac{a+b}{2}}{b−a} )^{2}] (b − a) ∥f′∥_{∞} if f′ ∈ L_{∞} [a, b] ; \\ \frac{1}{(p+1)^{\frac{1}{p}}} & [(\frac{x−a}{b−a})^{p+1} + (\frac{b−x}{b−a})^{p+1}] & \frac{1}{p} & (b − a) & \frac{1}{p} & ∥f′∥ & q \\ if f′ ∈ L_{q} [a, b] , \frac{1}{p} + \frac{1}{q} = 1, p > 1; \\ [\frac{1}{2} + |\frac{x− \frac{a+b}{2}}{b−a} |] ∥f′∥_{1} ; \end{matrix}

where ∥·∥_{r} (r ∈ [1, ∞]) are the usual Lebesgue norms on L_{r} [a, b], i.e.,

∥g∥_{∞} := ess sup_{t∈[a,b]} |g (t)|

and

∥g∥_{r} := (∫_{a}^{b} |g (t)|^{r} dt)^{\frac{1}{r}} , r ∈ [1, ∞).

The constants \frac{1}{4}, _{\frac{1}{(p+1)^{\frac{1}{p}}}} and \frac{1}{2} respectively are sharp in the

sense presented in Theorem 1. The above inequalities can also be obtained from the Fink result in [5] on choosing n = 1 and performing some appropriate computations. If one drops the condition of absolute continuity and assumes that f is Hölder continuous, then one may state the following result (see [6]): Theorem 3: Let f : [a, b] → ℝ be of r − H−Hölder type, i.e.,

|f (x) − f (y)| ≤ H |x − y|^{r} , for all x, y ∈ [a, b] , (4)

where r ∈ (0, 1] and H > 0 are fixed. Then for all x ∈ [a, b] we have the inequality:

|f (x) − \frac{1}{b − a} ∫_{a}^{b} f (t) dt|

≤ \frac{H}{r + 1} [(\frac{b − x}{b − a})^{r+1} + (\frac{x − a}{b − a})^{r+1}] (b − a)^{r} . (5)

The constant \frac{1}{r+1} is also sharp in the above sense. Note that if r = 1, i.e., f is Lipschitz continuous, then we get the following version of Ostrowski’s inequality for Lipschitzian functions (with L instead of H) (see [7])

|f (x) − \frac{1}{b − a} ∫_{a}^{b} f (t) dt|

≤ [\frac{1}{4} + (\frac{x − \frac{a+b}{2}}{b − a} )^{2}] (b − a) L. (6)

Here the constant \frac{1}{4} is also best. Moreover, if one drops the condition of the continuity of the function, and assumes that it is of bounded variation, then the following result may be stated (see [8]). Theorem 4: Assume that f : [a, b] → ℝ is of bounded variation

and denote by ⋁_{a}^{b} its total variation. Then

|f (x)− \frac{1}{b − a} ∫_{a}^{b}f (t) dt| ≤ [\frac{1}{2} + |\frac{x − \frac{a+b}{2}}{b − a} |] ⋁_{a}^{b} (f) (7)

for all x ∈ [a, b]. The constant \frac{1}{2} is the best possible. If we assume more about f, i.e., f is monotonically increasing, then the inequality (7) may be improved in the following manner [9] (see also [10]). Theorem 5: Let f : [a, b] → ℝ be monotonic nondecreasing. Then for all x ∈ [a, b], we have the inequality:

|f (x) − \frac{1}{b − a} ∫_{a}^{b} f (t) dt|  (8)

≤ \frac{1}{b − a}\{ [2x − (a + b)] f (x) + ∫_{a}^{b} sgn (t − x) f (t) dt\}

≤ \frac{1}{b − a} \{(x − a) [f (x) − f (a)] + (b − x) [f (b) − f (x)]\}

≤ [\frac{1}{2} + |\frac{x − \frac{a+b}{2}}{b − a} |] [f (b) − f (a)] .

All the inequalities in (8) are sharp and the constant \frac{1}{2} is the best possible. In this paper we point out a different Ostrowski type inequality assuming some special properties for the derivative of the function f on (a, b).

II. The Results

In [11], S.S. Dragomir pointed out the following identity in representing an absolutely continuous function. Lemma 6: Let f : [a, b] → ℝ be an absolutely continuous function on [a, b] . Then for any x ∈ [a, b] , one has the equality:

f (x) = \frac{1}{b − a} ∫_{a}^{b} f (t) dt

+ \frac{1}{b − a} ∫_{a}^{b} (x − t) (∫_{0}^{1} f′ [(1 − λ) x + λt] dλ) dt. (9)

Proof: For any t, x ∈ [a, b] , x ≠ t, one has

\frac{f (x) − f (t)}{x − t} = \frac{1}{x − t} ∫_{t}^{x} f′ (u) du
= ∫_{0}^{1} f′ [(1 − λ) x + λt] dλ,

showing that

f (x) = f (t) + (x − t) ∫_{0}^{1} f′ [(1 − λ) x + λt] dλ (10)

for any t, x ∈ [a, b] .

If we integrate (10) over t on [a, b] and divide by (b − a) , we deduce the desired identity (9). The following result holds. Theorem 7: Let f : [a, b] → ℝ be a differentiable function such that its derivative f′ is convex on (a, b) . Then for any x ∈ (a, b) one has the inequality

\frac{1}{2} · \frac{b − x}{b − a} [\frac{1}{b − x}∫_{x}^{b}f (u) du−f (b) − \frac{1}{2} (b − x) f′(x)] (11)

+ 2\frac{(x − a)}{b − a} [\frac{2}{x − a} ∫ x_{\frac{a+x}{2}} f (u) du − f (\frac{a + x}{2})]

≤ f (x) − \frac{1}{b − a} ∫_{a}^{b} f (t) dt

≤ \frac{1}{2} · \frac{x − a}{b − a}[\frac{1}{x − a} ∫_{a}^{x}f (u) du −f (a) − \frac{1}{2} (x − a) f′ (x)]

+ 2\frac{(b − x)}{b − a} [\frac{2}{b − x} ∫x^{\frac{x+b}{2}} f (u) du − f (\frac{x + b}{2})] .

Proof: For any x ∈ (a, b) , one has the identity (see Lemma 6)

f (x) − \frac{1}{b − a} ∫_{a}^{b} f (t) dt

= \frac{1}{b − a} [∫_{a}^{x} (x − t) (∫_{0}^{1} f′ [(1 − λ) x + λt] dλ) dt

+ ∫_{x}^{b} (x − t) (∫_{0}^{1} f′ [(1 − λ) x + λt] dλ) dt] . (12)

Since f′ is convex, then by the Hermite-Hadamard inequality for f′ we have

f′ (\frac{x + t}{2}) ≤ ∫_{0}^{1} f′ [(1 − λ) x + λt] dλ  (13)

≤ \frac{f′ (x) + f′ (t)}{2}

for any t, x ∈ (a, b) .

Assume a ≤ t ≤ x. Then by (13) we get

∫_{a}^{x} (x − t) f′ (\frac{x + t}{2}) dt  (14)

≤ ∫_{a}^{x} (x − t) (∫_{0}^{1} f′ [(1 − λ) x + λt] dλ) dt

≤ ∫_{a}^{x} [\frac{f′ (x) + f′ (t)}{2}] (x − t) dt.

Assume x ≤ t ≤ b. Then by (13) we also get

∫_{x}^{b} (x − t) [\frac{f′ (x) + f′ (t)}{2}] dt  (15)

≤ ∫_{x}^{b} (x − t) (∫_{0}^{1} f′ [(1 − λ) x + λt] dλ) dt

≤ ∫_{x}^{b} (x − t) f′ (\frac{x + t}{2}) dt.

Summing (14) with (15), dividing with b−a and using the identity (12) we deduce

\frac{1}{b − a} [∫_{a}^{x} (x − t) f′ (\frac{x + t}{2}) dt  (16)

+ ∫_{x}^{b} (x − t) [\frac{f′ (x) + f′ (t)}{2}] dt]

≤ f (x) − \frac{1}{b − a} ∫_{a}^{b} f (t) dt

≤ \frac{1}{b − a} [∫_{a}^{x} [\frac{f′ (x) + f′ (t)}{2}] (x − t) dt

+ ∫_{x}^{b} (x − t) f′ (\frac{x + t}{2}) dt] .

Since

∫_{a}^{x} (x − t) f′ (\frac{x + t}{2}) dt

= 2 ∫_{a}^{x} f (\frac{x + t}{2}) dt − 2f (\frac{x + a}{2}) (x − a)

= 4 ∫ x_{\frac{a+x}{2}} f (u) du − 2f (\frac{a + x}{2}) (x − a) ,

∫_{x}^{b} (x − t) [\frac{f′ (x) + f′ (t)}{2}] dt

= \frac{1}{2} ∫_{x}^{b} f (t) dt − \frac{1}{2} (b − x) f (b) − \frac{1}{4} (b − x)^{2} f′ (x) ,

∫_{a}^{x} [\frac{f′ (x) + f′ (t)}{2}] (x − t) dt

= \frac{1}{2} ∫_{a}^{x} f (t) dt − \frac{1}{2} (x − a) f (a) + \frac{1}{4} (x − a)^{2} f′ (x) ,

and

∫_{x}^{b} (x − t) f′ (\frac{x + t}{2}) dt

= 2 ∫_{x}^{b} f (\frac{x + t}{2}) dt − 2f (\frac{x + b}{2}) (b − x)

= 4 ∫x^{\frac{x+b}{2}} f (u) du − 2f (\frac{x + b}{2}) (b − x) ,

then by (16) we deduce the desired result. The following corollary is natural to be considered. Corollary 8: Assume that f : [a, b] → ℝ is as in Theorem 7. Then

\frac{1}{4} [\frac{2}{b − a} ∫ b_{\frac{a+b}{2}}f (u) du −f (b)−\frac{1}{4} (b−a) f′ (\frac{a + b}{2})] (17)

+ \frac{4}{b − a} ∫ \frac{a+b}{2} ; \frac{3a+b}{4} f (u) du − f (\frac{3a + b}{4})

≤ f (\frac{a + b}{2}) − \frac{1}{b − a} ∫_{a}^{b} f (t) dt

≤ \frac{1}{4} [\frac{2}{b − a} ∫a^{\frac{a+b}{2}} f (u) du

− f (a) + \frac{1}{4} (b − a) f′ (\frac{a + b}{2})]

+ \frac{4}{b − a} ∫ \frac{a+3b}{4} ; \frac{a+b}{2} f (u) du − f (\frac{a + 3b}{4}) .

The proof follows from (11) at the midpoint x = \frac{a+b}{2}.

Corollary 9: Assume that f : [a, b] → ℝ is as in Theorem 7. Then

\frac{8}{5 (b − a)} ∫ b_{\frac{a+b}{2}} f (u) du − \frac{1}{5}f (b)  (19)

− \frac{1}{10} (b − a) f′ (a) − \frac{4}{5}f (\frac{a + b}{2})

≤ \frac{4}{5} [\frac{f (a) + f (b)}{2}] − \frac{1}{b − a} ∫_{a}^{b} f (u) du

≤ \frac{8}{5 (b − a)} ∫a^{\frac{a+b}{2}} f (u) du − \frac{1}{5}f (a)

+ \frac{1}{10} (b − a) f′ (b) − \frac{4}{5}f (\frac{a + b}{2}) .

Proof: At x = a, we have from (11)

\frac{1}{2} [\frac{1}{b − a} ∫_{a}^{b}f (u) du −f (b)− \frac{1}{2} (b−a) f′ (a)]  (20)

≤ f (a) − \frac{1}{b − a} ∫_{a}^{b} f (u) du

≤ 2 [\frac{2}{b − a} ∫a^{\frac{a+b}{2}} f (u) du −f (\frac{a + b}{2})] .

At x = b,

2 [\frac{2}{b − a} ∫ b_{\frac{a+b}{2}} f (u) du − f (\frac{a + b}{2})]  (21)

≤ f (b) − \frac{1}{b − a} ∫_{a}^{b} f (u) du

≤ \frac{1}{2} [\frac{1}{b − a} ∫_{a}^{b} f (u) du − f (a) + \frac{1}{2} (b − a) f′ (b)] .

Manipulating (20) and (21) we arrive at (19).

III. Inequalities for Special Means

In the examples that follow below, the following definitions of special means will be utilized. • Arithmetic mean:

A = A (a, b) = \frac{a + b}{2}; a, b > 0.

• p – Logarithmic mean:

L_{p} (a, b) = \{ \begin{matrix}a, & if a = b \\ [\frac{b^{p+1} − a^{p+1}}{(p + 1) (b − a)}]^{\frac{1}{p}} , & if a ≠ b \end{matrix} .

• Logarithmic mean:

L (a, b) = \{ \begin{matrix}a, & if a = b \\ \frac{b − a}{ln b − ln a}, & if a ≠ b \end{matrix} .

1) Consider a function f with domain [a, b] ⊂ (0, ∞) , f (x) = x^{p} and p ∈ ℝ, p ≥ 2 which is differentiable and its first derivative is convex. From Corollary 8,

\frac{1}{b − a} ∫_{a}^{b} x^{p}dx = L_{p}^{p} (a, b) ,

notice that

\frac{3a + b}{4} = \frac{A (a, b) + a}{2},
\frac{a + 3b}{4} = \frac{A (a, b) + b}{2}

and from (17)

\frac{1}{2 (p + 1) (b − a)}[ b^{p+1} + 7A^{p+1} (a, b)

−8 (\frac{A (a, b) + a}{2})^{p+1}] − (\frac{A (a, b) + a}{2})^{p}

− \frac{p (b − a)}{16}A^{p−1} (a, b) − \frac{b^{p}}{4}

≤ A^{p} (a, b) − L_{p}^{p} (a, b)

≤ \frac{1}{2 (p + 1) (b − a)}[ − a^{p+1} − 7A^{p+1} (a, b)

+ 8 (\frac{A (a, b) + b}{2})^{p+1}] − (\frac{A (a, b) + b}{2})^{p}

+ \frac{p (b − a)}{16}A^{p−1} (a, b) − \frac{a^{p}}{4} .

In a similar fashion, from (19) we have

\frac{8}{5 (p + 1) (b − a)} [b^{p+1} + A^{p+1} (a, b)]

− \frac{b^{p}}{5} − \frac{p (b − a) a^{p−1}}{10} − \frac{4A^{p} (a, b)}{5}

≤ \frac{2}{5} [a^{p} + b^{p}] − L_{p}^{p} (a, b)

≤ \frac{8}{5 (p + 1) (b − a)} [A^{p+1} − a^{p}]

− \frac{a^{p}}{5} + \frac{p (b − a) b^{p−1}}{10} − \frac{4A^{p} (a, b)}{5}.

2) Consider the function f (x) = −\frac{1}{x}, x ≠ 0 then f′ is
convex on (a, b) ⊂ (0, ∞) and

\frac{1}{b − a} ∫_{a}^{b} (−\frac{1}{x}) dx = −L^{−1} (a, b) .

From (17) and (19) respectively, we have that

\frac{3A (a, b) + 7b}{4b (a + A(a, b))} − (\frac{b − a}{16}) A^{−2} (a, b)

+ \frac{1}{2 (b − a)} ln [(\frac{(a + A(a, b))}{2})^{8} \frac{1}{bA^{7} (a, b)}]

≤ L^{−1} (a, b) − A^{−1} (a, b)  (22)

≤ \frac{3A (a, b) + 7a}{4a (a + A(a, b))} + (\frac{b − a}{16}) A^{−2} (a, b)

+ \frac{1}{2 (b − a)} ln [aA^{7} (a, b) (\frac{2}{(b + A(a, b))})^{8}] ,

and

\frac{A (a, b)}{5a^{2}b} (4A (a, b) − 3b) + \frac{4}{5}A^{−1} (a, b)

− \frac{8}{5 (b − a)} ln [bA^{−1} (a, b)]

≤ L^{−1} (a, b) − \frac{4A (a, b)}{5ab}

≤ \frac{A (a, b)}{5a^{2}b} (4A (a, b) − 3a) + \frac{4}{5}A^{−1} (a, b)

− \frac{8}{5 (b − a)} ln [\frac{A (a, b)}{a}] .

IV. Two Plots In this section we produce two – three dimensional plots that demonstrate the validity of the inequality (22). Let

m (a, b) = \frac{3A (a, b) + 7b}{4b (a + A (a, b))} − (\frac{b − a}{16}) A^{−2} (a, b)

+ \frac{1}{2 (b − a)} ln [(\frac{a + A (a, b)}{2})^{8} \frac{1}{bA^{7} (a, b)}] ,

h (a, b) = L^{−1} (a, b) − A^{−1} (a, b)

and

M (a, b) = \frac{3A (a, b) + 7a}{4a (a + A (a, b))} + (\frac{b − a}{16}) A^{−2} (a, b)

+ \frac{1}{2 (b − a)} ln [aA^{7} (a, b) (\frac{2}{b + A (a, b)})^{8}];

then Figure 1 represents the plot of

h (a, b) − m (a, b)

and Figure 2 represents the plot of

M (a, b) − h (a, b)

indicating their positivity over the region (a, b) = (0, 1) × (0, 2) .

Figure 1: Plot of h (a, b) − m (a, b).

Figure 2: Plot of M (a, b) − h (a, b).

References

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http://jipam.vu.edu.au/v3n4.html