A new weighted Ostrowski type inequality on arbitrary time scale

Let $f:[a\text{,}b]\to \mathbb{R}$ be a differentiable mapping on $(a\text{,}b)$ with the property that $|f^{\prime }(t)|⩽M$ for all $t\in (a\text{,}b)$ . Then $$\left|f(x)-\frac{1}{b-a}{\int }_a^{b}f(t)\mathit{dt}\right|⩽\left[\frac{1}{4}+\frac{{(x-\frac{a+b}{2})}^2}{{(b-a)}^2}\right](b-a)M\text{,}$$ for all $x\in [a\text{,}b]$ . The constant $\frac{1}{4}$ is the best possible in the sense that it cannot be replaced by a smaller constant.

Let $a\text{,}b\text{,}s\text{,}t\in \mathbb{T}\text{,}a<b$ and $f:[a\text{,}b]\to \mathbb{R}$ be a differentiable. Then

Since the advent of the above result, many Ostrowski and weighted Ostrowski type results on time scales have been published. In order to prove Theorem 2, one needs the so-called Montgomery identity. In the literature, there exist a lot of generalizations of this identity, see for example Karpuz and Özkan (2008), Liu and Tuna (2012), Liu et al. (2014) and Liu et al. (2014). Lately, Liu and Ngô (2009) investigated Theorem 2 by introducing a parameter $\lambda$ . Inspired by the later, Xu and Fang (2016) recently proved the following new generalization of the Montgomery identity.

Suppose that $a\text{,}b\text{,}s\text{,}t\in \mathbb{T}$ , $a<b$ , $f:[a\text{,}b]\to \mathbb{R}$ is differentiable, and $\psi$ is a function of $[0\text{,}1]$ into $[0\text{,}1]$ . Then $$\frac{1+\psi (1-\lambda )-\psi (\lambda )}{2}f(t)+\frac{\psi (\lambda )f(a)+(1-\psi (1-\lambda ))f(b)}{2}=\frac{1}{b-a}{\int }_a^b{f}^{\sigma }(s)\mathrm{\Delta }s+\frac{1}{b-a}{\int }_a^{b}K(s\text{,}t)f^{\mathrm{\Delta }}(s)\mathrm{\Delta }s\text{,}$$ where

Using the above result, Xu and Fang (2016) also proved the following Ostrowski type inequality.

Suppose that $a\text{,}b\text{,}s\text{,}t\in \mathbb{T}$ , $a<b$ , $f:[a\text{,}b]\to \mathbb{R}$ are differentiable, and $\psi$ is a function of $[0\text{,}1]$ into $[0\text{,}1]$ . Then the following inequality holds $$\left|\frac{1+\psi (1-\lambda )-\psi (\lambda )}{2}f(t)+\frac{\psi (\lambda )f(a)+(1-\psi (1-\lambda ))f(b)}{2}-\frac{1}{b-a}{\int }_a^b{f}^{\sigma }(s)\mathrm{\Delta }s\right|⩽\frac{M}{b-a}\left[h_2\left(a\text{,}a+\psi (\lambda )\frac{b-a}{2}\right)+h_2\left(t\text{,}a+\psi (\lambda )\frac{b-a}{2}\right)+h_2\left(t\text{,}a+(1+\psi (1-\lambda ))\frac{b-a}{2}\right)+h_2\left(b\text{,}a+(1+\psi (1-\lambda ))\frac{b-a}{2}\right)\right]\text{,}$$ for all $\lambda \in [0\text{,}1]$ such that $a+\psi (\lambda )\frac{b-a}{2}$ and $a+(1+\psi (1-\lambda ))\frac{b-a}{2}$ are in $\mathbb{T}$ , and $t\in \left[a+\psi (\lambda )\frac{b-a}{2}\text{,}a+(1+\psi (1-\lambda ))\frac{b-a}{2}\right]$ , where $M={\sup }_{a<t<b}|f^{\mathrm{\Delta }}(t)|<\infty$ .

A time scale $\mathbb{T}$ is an arbitrary nonempty closed subset of $\mathbb{R}$ . The forward jump operator $\sigma :\mathbb{T}\to \mathbb{T}$ and backward jump operator $\rho :\mathbb{T}\to \mathbb{T}$ are defined by $\sigma (t)≔\inf \{s\in \mathbb{T}:s>t\}$ for $t\in \mathbb{T}$ and $\rho (t)≔\sup \{s\in \mathbb{T}:s<t\}$ for $t\in \mathbb{T}$ , respectively. Clearly, we see that $\sigma (t)⩾t$ and $\rho (t)⩽t$ for all $t\in \mathbb{T}$ . If $\sigma (t)>t$ , then we say that $t$ is right-scattered, while if $\rho (t)<t$ , then we say that $t$ is left-scattered. If $\sigma (t)=t$ , then $t$ is called right dense, and if $\rho (t)=t$ then $t$ is called left dense. Points that are both right dense and left dense are called dense. The set ${\mathbb{T}}^k$ is defined as follows: if $\mathbb{T}$ has a left scattered maximum $m$ , then ${\mathbb{T}}^k=\mathbb{T}-m\text{;}$ otherwise, ${\mathbb{T}}^k=\mathbb{T}$ . For $a\text{,}b\in \mathbb{T}$ with $a⩽b$ , we define the interval $[a\text{,}b]$ in $\mathbb{T}$ by $[a\text{,}b]=\{t\in \mathbb{T}:a⩽t⩽b\}$ . Open intervals and half-open intervals are defined in the same manner.

The function $f:\mathbb{T}\to \mathbb{R}$ , is called differentiable at $t\in {\mathbb{T}}^k$ , with delta derivative $f^{\mathrm{\Delta }}(t)\in \mathbb{R}$ , if for any given $∊>0$ there exist a neighborhood $U$ of $t$ such that $$\left|f(\sigma (t))-f(s)-f^{\mathrm{\Delta }}(t)(\sigma (t)-s)\right|⩽∊|\sigma (t)-s|\text{,}\forall s\in U\text{.}$$

If $\mathbb{T}=\mathbb{R}$ , then $f^{\mathrm{\Delta }}(t)=\frac{\mathit{df}(t)}{\mathit{dt}}$ , and if $\mathbb{T}=\mathbb{Z}$ , then $f^{\mathrm{\Delta }}(t)=f(t+1)-f(t)$ .

The function $f:\mathbb{T}\to \mathbb{R}$ is said to be $\mathit{rd}$ -continuous if it is continuous at all dense points $t\in \mathbb{T}$ and its left-sided limits exist at all left dense points $t\in \mathbb{T}$ .

Let $f$ be a $\mathit{rd}$ -continuous function. Then $g:\mathbb{T}\to \mathbb{R}$ is called the antiderivative of $f$ on $\mathbb{T}$ if it is differentiable on $\mathbb{T}$ and satisfies $g^{\mathrm{\Delta }}(t)=f(t)$ for any $t\in {\mathbb{T}}^k$ . In this case, we have $${\int }_a^{b}f(s)\mathrm{\Delta }s=g(b)-g(a)\text{.}$$

If $a\text{,}b\text{,}c\in \mathbb{T}$ with $a<c<b$ , $\alpha \in \mathbb{R}$ and $f\text{,}g$ are $\mathit{rd}$ -continuous, then

• (i) ${\int }_a^b[f(t)+g(t)]\mathrm{\Delta }t={\int }_a^{b}f(t)\mathrm{\Delta }t+{\int }_a^{b}g(t)\mathrm{\Delta }t$ .

• (ii) ${\int }_a^b\alpha f(t)\mathrm{\Delta }t=\alpha {\int }_a^{b}f(t)\mathrm{\Delta }t$

• (iii) ${\int }_a^{b}f(t)\mathrm{\Delta }t=-{\int }_b^{a}f(t)\mathrm{\Delta }t$

• (iv) ${\int }_a^{b}f(t)\mathrm{\Delta }t={\int }_a^{c}f(t)\mathrm{\Delta }t+{\int }_c^{b}f(t)\mathrm{\Delta }t$ .

• (v) $\left|{\int }_a^{b}f(t)\mathrm{\Delta }t\right|⩽{\int }_a^b|f(t)|\mathrm{\Delta }t$ for all $t\in [a\text{,}b]$ .

• (vi) ${\int }_a^{b}f(t)g^{\mathrm{\Delta }}(t)\mathrm{\Delta }t=(\mathit{fg})(b)-(\mathit{fg})(a)-{\int }_a^b{f}^{\mathrm{\Delta }}(t)g(\sigma (t))\mathrm{\Delta }t$ .

Let $h_k:{\mathbb{T}}^2\to \mathbb{T}$ , $k\in \mathbb{N}$ be functions that are recursively defined as $$h_0(t\text{,}s)=1$$ and $$h_{k+1}(t\text{,}s)={\int }_s^t{h}_k(\tau \text{,}s)\mathrm{\Delta }\tau \text{,}\mathit{for}\mathit{all}s\text{,}t\in \mathbb{T}\text{.}$$

(a) Using the fact that for all $s\text{,}t\in \mathbb{T}$ , $h_1(t\text{,}s)=t-s$ , we get that $$h_2(t\text{,}s)={\int }_s^t(\tau -s)\mathrm{\Delta }\tau \text{.}$$

(b) When $\mathbb{T}=\mathbb{R}$ , then for all $s\text{,}t\in \mathbb{T}$ , $$h_k(t\text{,}s)=\frac{{(t-s)}^k}{k!}\text{.}$$

(c) When $\mathbb{T}=\mathbb{Z}$ , then for all $s\text{,}t\in \mathbb{T}$ , $$h_k(t\text{,}s)=\left(\begin{array}{c}\hfill t-s\hfill \\ \hfill k\hfill \end{array}\right)={\displaystyle \prod _{i=1}^k}\frac{t-s+1-i}{i}\text{.}$$

(d) When $\mathbb{T}=q^{{\mathbb{N}}_0}$ with $q>1$ , then for all $s\text{,}t\in \mathbb{T}$ , $$h_k(t\text{,}s)=\frac{{(t-s)}_q^k}{[k]!}\mathrm{for}k\in {\mathbb{N}}_0\text{,}$$ where ${[k]}_q≔\frac{q^k-1}{q-1}$ for $q\in \mathbb{R}⧹\{1\}$ and $k\in {\mathbb{N}}_0$ , $[k]!≔{\prod }_{j=1}^k{[j]}_q$ for $k\in {\mathbb{N}}_0$ , $${(t-s)}_q^k≔{\displaystyle \prod _{j=0}^{k-1}}(t-q^{j}s)\mathrm{for}k\in {\mathbb{N}}_0\text{.}$$

Lemma 12 (A weighted generalized Montgomery Identity)

Let $\nu :[a\text{,}b]\to [0\text{,}\infty )$ be $\mathit{rd}-$ continuous and positive and $w:[a\text{,}b]\to \mathbb{R}$ be differentiable such that $w^{\mathrm{\Delta }}(t)=\nu (t)$ on $[a\text{,}b]$ . Suppose also that $a\text{,}b\text{,}s\text{,}t\in \mathbb{T}$ , $a<b$ , $f:[a\text{,}b]\to \mathbb{R}$ is differentiable, and $\psi$ is a function of $[0\text{,}1]$ into $[0\text{,}1]$ . Then we have the following equation

Using item (vi) of Theorem 9, we obtain

Adding Eqs. (5) and (6), and using item (iv) of Theorem 9, gives

Hence, Eq. (3) follows.  □

If $w(t)=t$ , then Lemma 12 reduces to Theorem 3.

For the case when $\mathbb{T}=\mathbb{R}$ in Lemma 12, we get

If we consider $\psi (\lambda )=\lambda$ in Corollary 14, then the equation becomes $$\left[(1-\lambda )f(t)+\lambda \frac{f(a)+f(b)}{2}\right]{\int }_a^b\nu (t)\mathit{dt}={\int }_a^{b}K(s\text{,}t)f^{\prime }(s)\mathit{ds}+{\int }_a^b\nu (s)f(s)\mathit{ds}\text{,}$$ where $\nu (t)=w^{\prime }(t)$ on $[a\text{,}b]$ and

For the case when $\mathbb{T}=\mathbb{Z}\text{,}a=0\text{,}b=n\text{,}f(k)=x_k\text{,}s=j$ and $t=i$ , Lemma 12 becomes

Let $\mathbb{T}=q^{{\mathbb{N}}_0}$ , with $q>1$ , $a=q^m$ and $b=q^n$ with $m<n$ . For this case, $\sigma (t)=\mathit{qt}$ and $f^{\mathrm{\Delta }}(t)=D_{q}f(t)≔\frac{f(\mathit{qt})-f(t)}{(q-1)t}$ . Using this information, Lemma 12 becomes

Let $\nu :[a\text{,}b]\to [0\text{,}\infty )$ be $\mathit{rd}$ -continuous and positive and $w:[a\text{,}b]\to \mathbb{R}$ be differentiable such that $w^{\mathrm{\Delta }}(t)=\nu (t)$ on $[a\text{,}b]$ . Suppose also that $a\text{,}b\text{,}s\text{,}t\in \mathbb{T}$ , $a<b$ , $f:[a\text{,}b]\to \mathbb{R}$ is differentiable, and $\psi$ is a function of $[0\text{,}1]$ into $[0\text{,}1]$ . Then we have the following inequality

The proof easily follows by applying the absolute value on both sides of Eq. (3) in Lemma 12 and then using item (v) of Theorem 9. □

Setting $w(t)=t$ in Theorem 18 reduces to Theorem 4 where the equation $${\int }_a^b|K(s\text{,}t)|\mathrm{\Delta }s=h_2\left(a\text{,}a+\psi (\lambda )\frac{b-a}{2}\right)+h_2\left(t\text{,}a+\psi (\lambda )\frac{b-a}{2}\right)+h_2\left(t\text{,}a+(1+\psi (1-\lambda ))\frac{b-a}{2}\right)+h_2\left(b\text{,}a+(1+\psi (1-\lambda ))\frac{b-a}{2}\right)\text{.}$$ holds for all $\lambda \in [0\text{,}1]$ such that $a+\psi (\lambda )\frac{b-a}{2}$ and $a+(1+\psi (1-\lambda ))\frac{b-a}{2}$ are in $\mathbb{T}$ , and $t\in \left[a+\psi (\lambda )\frac{b-a}{2}\text{,}a+(1+\psi (1-\lambda ))\frac{b-a}{2}\right]$ .

Let $a\text{,}b\text{,}s\text{,}t\in \mathbb{T}\text{,}a<b$ , $f:[a\text{,}b]\to \mathbb{R}$ is differentiable, and $\psi$ is a function of $[0\text{,}1]$ into $[0\text{,}1]$ . Then we have the following inequality

For the case when $\mathbb{T}=\mathbb{R}$ in Theorem 18, we get

For the case when $\mathbb{T}=\mathbb{Z}$ , $a=0\text{,}b=n\text{,}f(k)=x_k\text{,}s=j$ and $t=i$ , Theorem 18 amounts to

Let $\mathbb{T}=q^{{\mathbb{N}}_0}$ , with $q>1$ , $a=q^m$ and $b=q^n$ with $m<n$ . Then we have

For $\psi (\lambda )={\lambda }^2$ , the inequality in Theorem 18 boils down to

1. Putting $\lambda =0$ in Eq. (24) gives $$|f(t){\int }_a^b\nu (t)\mathrm{\Delta }t-{\int }_a^b\nu (s)f(\sigma (s))\mathrm{\Delta }s|⩽M{\int }_a^b|K(s\text{,}t)|\mathrm{\Delta }s\text{,}$$ where $$K(s\text{,}t)=\left\{\begin{array}{c}w(s)-w(a)\text{,}s\in [a\text{,}t)\text{,}\hfill \\ w(s)-w(b)\text{,}s\in [t\text{,}b]\text{.}\hfill \end{array}\right.$$

2. For $\lambda =1/2$ in Eq. (24) we obtain the following inequality $$\left|\left[\frac{f(t)}{2}+\frac{f(a)+3f(b)}{8}\right]{\int }_a^b\nu (t)\mathrm{\Delta }t-{\int }_a^b\nu (s)f(\sigma (s))\mathrm{\Delta }s\right|⩽M{\int }_a^b|K(s\text{,}t)|\mathrm{\Delta }s\text{,}$$ where $$K(s\text{,}t)=\left\{\begin{array}{c}w(s)-\frac{7w(a)+w(b)}{8}\text{,}s\in [a\text{,}t)\text{,}\hfill \\ w(s)-\frac{3w(a)+5w(b)}{8}\text{,}s\in [t\text{,}b]\text{.}\hfill \end{array}\right.$$

3. If we take $\lambda =1$ in Eq. (24), we obtain $$\left|\frac{f(a)+f(b)}{2}{\int }_a^b\nu (t)\mathrm{\Delta }t-{\int }_a^b\nu (s)f(\sigma (s))\mathrm{\Delta }s\right|⩽M{\int }_a^b|K(s\text{,}t)|\mathrm{\Delta }s\text{,}$$ where $$K(s\text{,}t)=w(s)-\frac{w(a)+w(b)}{2}s\in [a\text{,}b]\text{.}$$