Ostrowski and generalized trapezoid type inequalities on time scales

Let $f:[a,b]\to \mathbb{R}$ be continuous on $[a,b]$ and differentiable in $(a,b)$ and its derivative $f^{\prime }:(a,b)\to \mathbb{R}$ is bounded in $(a,b)$ . Then for any $x\in [a,b]$ , we have $$\left|f(x)-\frac{1}{b-a}{\int }_a^{b}f(t)\mathit{dt}\right|⩽\left(\frac{1}{4}+\frac{{\left(x-\frac{a+b}{2}\right)}^2}{{(b-a)}^2}\right)(b-a)\parallel f^{\prime }{\parallel }_{\infty },$$ where $\parallel f^{\prime }{\parallel }_{\infty }≔\underset{x\in (a,b)}{\sup }|f^{\prime }(x)|<\infty$ . The inequality is sharp in the sense that the constant $\frac{1}{4}$ cannot be replaced by a smaller one.

Let $f:[a,b]\to \mathbb{R}$ be continuous on $[a,b]$ and differentiable in $(a,b)$ and its derivative $f^{\prime }:(a,b)\to \mathbb{R}$ is bounded in $(a,b)$ . Then for any $x\in [a,b]$ , we have

Assume that the absolutely continuous function $g:\left[a,b\right]\to \mathbb{R}$ satisfies the condition $-\infty <m⩽g^{\prime }\left(t\right)⩽M<\infty$ for a.e $t\in [a,b]$ . Then we have

Assume that the absolutely continuous function $g:\left[a,b\right]\to \mathbb{R}$ satisfies the condition $-\infty <m⩽g^{\prime }\left(t\right)⩽M<\infty$ for a.e $t\in [a,b]$ . Then we have

Let $I\subset \mathbb{R}$ be an open interval and $a,b\in I,a<b$ . If $f:I\to \mathbb{R}$ is a differentiable function such that $m⩽f^{\prime }(t)⩽M$ , for all $t\in [a,b]$ , for some constants $m,M\in \mathbb{R}$ , then we have $$\begin{array}{c}\hfill |\lambda \frac{f(a)+f(b)}{2}+(1-\lambda )f(x)-\frac{m+M}{2}(1-\lambda )\left(x-\frac{a+b}{2}\right)-\frac{1}{b-a}{\int }_a^{b}f(t)\mathit{dt}|\\ \hfill ⩽\frac{M-m}{2(b-a)}\left[\frac{{(b-a)}^2}{4}({\lambda }^2+{(1-\lambda )}^2)+{\left(x-\frac{a+b}{2}\right)}^2\right],\end{array}$$ where $a+\lambda ((b-a)/2)⩽x⩽b-\lambda ((b-a)/2)$ .

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

A time scale $\mathbb{T}$ is an arbitrary nonempty closed subset of the real numbers.

For each $t\in \mathbb{T}$ , the forward jump operator $\sigma :\mathbb{T}\to \mathbb{T}$ by $\sigma (t)=\inf \left\{s\in \mathbb{T}:s>t\right\}$ and the backward jump operator $\rho :\mathbb{T}\to \mathbb{T}$ is defined by $\rho (t)=\sup \left\{s\in \mathbb{T}:s<t\right\}$ .

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. Points that are right-scattered and left-scattered at the same time are called isolated. 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 mapping $\mu :\mathbb{T}\to \left[0,\infty \right)$ defined by $\mu (t)=\sigma (t)-t$ is called the graininess function. The set ${\mathbb{T}}^{\kappa }$ is defined as follows: if $\mathbb{T}$ has a left-scattered maximum m, then ${\mathbb{T}}^{\kappa }=\mathbb{T}-\left\{m\right\}$ ; otherwise, ${\mathbb{T}}^{\kappa }=\mathbb{T}$ .

Let $f:\mathbb{T}\to \mathbb{R}$ and $t\in {\mathbb{T}}^{\kappa }$ . Then we define $f^{\mathrm{\Delta }}(t)$ to be the number (provided it exists) with the property that for any given $\epsilon >0$ there exists a neighborhood U of t such that $$\left|f(\sigma (t))-f(s)-f^{\mathrm{\Delta }}(t)\left[\sigma (t)-s\right]\right|⩽\epsilon \left|\sigma (t)-s\right|,\forall s\in U\text{.}$$ We call $f^{\mathrm{\Delta }}(t)$ the delta derivative of f at t.

If $f,g:\mathbb{T}\to \mathbb{R}$ are differentiable at $t\in {\mathbb{T}}^{\kappa }$ , then the product $\mathit{fg}:\mathbb{T}\to \mathbb{R}$ is differentiable at t and $${\left(\mathit{fg}\right)}^{\mathrm{\Delta }}(t)=f^{\mathrm{\Delta }}(t)g(t)+f(\sigma (t))g^{\mathrm{\Delta }}(t)\text{.}$$

The function $f:\mathbb{T}\to \mathbb{R}$ is said to be rd-continuous on $\mathbb{T}$ provided it is continuous at all right-dense points $t\in \mathbb{T}$ and its left-sided limits exist at all left-dense points $t\in \mathbb{T}$ . The set of all rd-continuous function $f:\mathbb{T}\to \mathbb{R}$ is denoted by $C_{\mathrm{rd}}(\mathbb{T},\mathbb{R})$ . Also, the set of functions $f:\mathbb{T}\to \mathbb{R}$ that are differentiable and whose derivative is rd-continuous is denoted by $C_{\mathrm{rd}}^1(\mathbb{T},\mathbb{R})$ .

Let $F:\mathbb{T}\to \mathbb{R}$ be a function. Then  $F:\mathbb{T}\to \mathbb{R}$ is called the antiderivative of f on $\mathbb{T}$ if it satisfies $F^{\mathrm{\Delta }}(t)=f(t)$ for any $t\in {\mathbb{T}}^{\kappa }$ . In this case, the Cauchy integral is defined by $${\int }_a^{b}f(t)\mathrm{\Delta }t=F(b)-F(a),a,b\in \mathbb{T}\text{.}$$

Let $f,g\in C_{\mathrm{rd}}(\mathbb{T},\mathbb{R}),a,b,c\in \mathbb{T}$ and $\alpha ,\beta \in \mathbb{R}$ . Then

• (1)  ${\int }_a^b\left[\alpha f(t)+\beta g(t)\right]\mathrm{\Delta }t=\alpha {\int }_a^{b}f(t)\mathrm{\Delta }t+\beta {\int }_a^{b}g(t)\mathrm{\Delta }t$ .

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

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

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

If f is $\mathrm{\Delta }-$ integrable on $\left[a,b\right]$ , then so is $\left|f\right|$ , and $$\left|{\int }_a^{b}f(t)\mathrm{\Delta }t\right|⩽{\int }_a^b\left|f(t)\right|\mathrm{\Delta }t\text{.}$$

Let $h_k,g_k:{\mathbb{T}}^2\to \mathbb{R}$ , $k\in {\mathbb{N}}_0$ be defined by $h_0(t,s)≔g_0(t,s)≔1$ , for all $s,t\in \mathbb{T}$ and then recursively by $g_{k+1}\left(t,s\right)={\int }_s^t{g}_k\left(\sigma \left(\tau \right),s\right)\mathrm{\Delta }\tau ,h_{k+1}\left(t,s\right)={\int }_s^t{h}_k\left(\tau ,s\right)\mathrm{\Delta }\tau$ , for all $s,t\in \mathbb{T}$ .

Lemma 18 Xu and Fang, 2016

Suppose that $a,b,t,x\in \mathbb{T},a<b,f:\left[a,b\right]\to \mathbb{R}$ is differentiable, and that g is a function of   $\left[0,1\right]$ into $\left[0,1\right]$ . We then have the inequality $$\begin{array}{cc}\hfill & \left|\frac{1+g\left(1-\lambda \right)-g\left(\lambda \right)}{2}f\left(x\right)+\frac{g\left(\lambda \right)f\left(a\right)+\left(1-g\left(1-\lambda \right)\right)f\left(b\right)}{2}-\frac{1}{b-a}{\int }_a^{b}f(\sigma \left(t\right))\mathrm{\Delta }t\right|\hfill \\ \hfill ⩽& \frac{K}{b-a}\left[h_2\left(a,a+g\left(\lambda \right)\frac{b-a}{2}\right)+h_2\left(x,a+g\left(\lambda \right)\frac{b-a}{2}\right)+h_2\left(x,a+\left(1+g\left(1-\lambda \right)\right)\frac{b-a}{2}\right)\right.\hfill \\ \hfill & \left.h_2\left(b,a+(1+g\left(1-\lambda )\right)\frac{b-a}{2}\right)\right]\hfill \end{array}$$ for all $\lambda \in \left[0,1\right]$ such that $a+g\left(\lambda \right)\frac{b-a}{2}$ and $a+\left(1+g\left(1-\lambda \right)\right)\frac{b-a}{2}$ are in $\mathbb{T}$ , and $x\in \left[a+g\left(\lambda \right)\frac{b-a}{2},a+\left(1+g\left(1-\lambda \right)\right)\frac{b-a}{2}\right]\cap \mathbb{T}$ , where $K=\underset{a<x<b}{\sup }\left|f^{\mathrm{\Delta }}\left(x\right)\right|<\infty$ . This inequality is sharp provided $$\frac{g^2\left(\lambda \right)-2s\left(\lambda \right)}{2}a-\frac{g^2\left(\lambda \right)}{2}b⩾{\int }_{a+g\left(\lambda \right)\frac{b-a}{2}}^{a}t▵t\text{.}$$

Let $a,b,t,x\in \mathbb{T},a<b$ . If $f\in C_{\mathrm{rd}}^1\left(\left[a,b\right],\mathbb{R}\right)$ , then for all $x\in [a,b]$ , we have

Using the integration by parts formula, given in item (iv) of Theorem 15, we get $${\int }_a^b\left(t-x\right)f^{\mathrm{\Delta }}\left(t\right)\mathrm{\Delta }t=\left(b-x\right)f\left(b\right)-\left(a-x\right)f\left(a\right)-{\int }_a^{b}f(\sigma \left(t\right))\mathrm{\Delta }t\text{.}$$ This implies that

If we take $\mathbb{T}=\mathbb{R}$ in Lemma 19, we get Theorem 2.

In the case of $\mathbb{T}=\mathbb{Z}$ in Lemma 19, we have $$\begin{array}{cc}\hfill & \left|\frac{\left(b-x\right)f\left(b\right)+\left(x-a\right)f\left(a\right)}{b-a}-\frac{1}{b-a}{\displaystyle \sum _{t=a}^{b-1}}f(t+1)\right|\hfill \\ \hfill ⩽& K\left[\left(\frac{1}{4}+{\left(\frac{x-\frac{a+b}{2}}{b-a}\right)}^2\right)\left(b-a\right)+\frac{1}{b-a}\left(x-\frac{a+b}{2}\right)\right],\hfill \end{array}$$ where $K=\underset{a<x<b-1}{\sup }\left|\mathrm{\Delta }f\left(x\right)\right|<\infty$ .

By taking $a=0,b=n,f(k)=x_k$ , and $t=i$ in Corollary 21, one gets that for $j\in \{0,1,\cdots ,n-1\}$ $$\begin{array}{cc}\hfill & \left|\frac{\left(n-j\right)x_n+{\mathit{jx}}_0}{n}-\frac{1}{n}{\displaystyle \sum _{i=0}^{n-1}}{x}_{i+1}\right|\hfill \\ \hfill ⩽& K\left[\left(\frac{1}{4}+{\left(\frac{j-\frac{n}{2}}{n}\right)}^2\right)n+\frac{1}{n}\left(j-\frac{n}{2}\right)\right],\hfill \end{array}$$ where $K=\underset{0<j<n-1}{\sup }\left|x_{j+1}-x_j\right|<\infty$ .

Suppose that $a,b,t,x\in \mathbb{T},a<b,u:\left[a,b\right]\to \mathbb{R}$ is differentiable, and that g is a function of   $\left[0,1\right]$ into $\left[0,1\right]$ . Then we have $$\begin{array}{cc}\hfill & \left|\frac{1+g\left(1-\lambda \right)-g\left(\lambda \right)}{2}u\left(x\right)-\frac{1+g\left(1-\lambda \right)-g\left(\lambda \right)}{2}\left(x-\frac{a+b}{2}\right)\left(\frac{m+M}{2}\right)\right.\hfill \\ \hfill & +\frac{g\left(\lambda \right)u\left(a\right)+\left(1-g\left(1-\lambda \right)\right)u\left(b\right)}{2}-\left(\frac{b-a}{2}\right)\left(\frac{m+M}{2}\right)\frac{\left(1-g\left(1-\lambda \right)-g\left(\lambda \right)\right)}{2}\hfill \\ \hfill & \left.-\frac{1}{b-a}{\int }_a^{b}u\left(\sigma \left(t\right)\right)\mathrm{\Delta }t-\frac{m+M}{2\left(b-a\right)}\left(g_2\left(a,\frac{a+b}{2}\right)-g_2\left(b,\frac{a+b}{2}\right)\right)\right|\hfill \\ \hfill ⩽& \frac{M-m}{2\left(b-a\right)}\left[h_2\left(a,a+g\left(\lambda \right)\frac{b-a}{2}\right)+h_2\left(x,a+g\left(\lambda \right)\frac{b-a}{2}\right)\right.\hfill \\ \hfill & \left.+h_2\left(x,a+\left(1+g\left(1-\lambda \right)\right)\frac{b-a}{2}\right)+h_2\left(b,a+\left(1+g\left(1-\lambda \right)\right)\frac{b-a}{2}\right)\right],\hfill \end{array}$$ for all $\lambda \in \left[0,1\right]$ such that $a+g\left(\lambda \right)\frac{b-a}{2}$ and $a+\left(1+g\left(1-\lambda \right)\right)\frac{b-a}{2}$ are in $\mathbb{T}$ , and $x\in \left[a+g\left(\lambda \right)\frac{b-a}{2},a+\left(1+g\left(1-\lambda \right)\right)\frac{b-a}{2}\right]\cap \mathbb{T}$ .

In the case when $\mathbb{T}=\mathbb{R}$ in Theorem 23, we get $$\begin{array}{cc}\hfill & \left|\frac{1+g\left(1-\lambda \right)-g\left(\lambda \right)}{2}u\left(x\right)-\frac{1+g\left(1-\lambda \right)-g\left(\lambda \right)}{2}\left(x-\frac{a+b}{2}\right)\left(\frac{m+M}{2}\right)-\frac{1}{b-a}{\int }_a^{b}u\left(t\right)\mathit{dt}\right.\hfill \\ \hfill & \left.+\frac{g\left(\lambda \right)u\left(a\right)+\left(1-g\left(1-\lambda \right)\right)u\left(b\right)}{2}-\left(\frac{b-a}{2}\right)\left(\frac{m+M}{2}\right)\frac{\left(1-g\left(1-\lambda \right)-g\left(\lambda \right)\right)}{2}\right|\hfill \\ \hfill ⩽& \frac{M-m}{2\left(b-a\right)}\left[\frac{g^2\left(\lambda \right){\left(b-a\right)}^2}{8}+\frac{1}{2}{\left(\frac{\left(b-a\right)g\left(\lambda \right)-2\left(x-a\right)}{2}\right)}^2\right.\hfill \\ \hfill & \left.+\frac{{\left(\left(b-a\right)g\left(1-\lambda \right)+a+b-2x\right)}^2}{8}+\frac{{\left(b-a\right)}^2{\left(g\left(1-\lambda \right)-1\right)}^2}{8}\right]\text{.}\hfill \end{array}$$

In the case of $\mathbb{T}=\mathbb{Z}$ in Theorem 23, we have $$\begin{array}{cc}\hfill & \left|\frac{1+g\left(1-\lambda \right)-g\left(\lambda \right)}{2}u\left(x\right)-\frac{1+g\left(1-\lambda \right)-g\left(\lambda \right)}{2}\left(x-\frac{a+b}{2}\right)\left(\frac{m+M}{2}\right)\right.\hfill \\ \hfill & +\frac{g\left(\lambda \right)u\left(a\right)+\left(1-g\left(1-\lambda \right)\right)u\left(b\right)}{2}-\left(\frac{b-a}{2}\right)\left(\frac{m+M}{2}\right)\frac{\left(1-g\left(1-\lambda \right)-g\left(\lambda \right)\right)}{2}\hfill \\ \hfill & \left.-\frac{1}{b-a}{\displaystyle \sum _{t=a}^{b-1}}u\left(t+1\right)+\frac{m+M}{4}\right|\hfill \\ \hfill ⩽& \frac{M-m}{2\left(b-a\right)}\left[\frac{b-a}{8}g\left(\lambda \right)\left(\left(b-a\right)g\left(\lambda \right)+2\right)+\frac{1}{2}{\left(\frac{\left(b-a\right)g\left(\lambda \right)-2\left(x-a\right)}{2}\right)}^2\right.\hfill \\ \hfill & +\frac{\left(b-a\right)g\left(\lambda \right)-2\left(x-a\right)}{4}+\frac{{\left(\left(b-a\right)g\left(1-\lambda \right)+a+b-2x\right)}^2}{8}\hfill \\ \hfill & \left.+\frac{\left(\left(b-a\right)g\left(1-\lambda \right)+a+b-2x\right)}{4}+\frac{\left(b-a\right)\left(g\left(1-\lambda \right)-1\right)}{8}\left(\left(b-a\right)\left(g\left(1-\lambda \right)-1\right)+2\right)\right],\hfill \end{array}$$ for all $\lambda \in \left[0,1\right]$ such that $a+g\left(\lambda \right)\frac{b-a}{2}$ and $a+\left(1+g\left(1-\lambda \right)\right)\frac{b-a}{2}$ are in $\mathbb{Z}$ , and $x\in \left[a+g\left(\lambda \right)\frac{b-a}{2},a+\left(1+g\left(1-\lambda \right)\right)\frac{b-a}{2}\right]\cap \mathbb{Z}$ .

In the case of $g\left(\lambda \right)=\lambda$ in Theorem 23, we get $$\begin{array}{cc}\hfill & \left|\left(1-\lambda \right)u\left(x\right)-\left(1-\lambda \right)\left(x-\frac{a+b}{2}\right)\left(\frac{m+M}{2}\right)+\lambda \frac{u\left(a\right)+u\left(b\right)}{2}\right.\hfill \\ \hfill & \left.-\frac{1}{b-a}{\int }_a^{b}u\left(\sigma \left(t\right)\right)\mathrm{\Delta }t-\frac{m+M}{2\left(b-a\right)}\left(g_2\left(a,\frac{a+b}{2}\right)-g_2\left(b,\frac{a+b}{2}\right)\right)\right|\hfill \\ \hfill ⩽& \frac{M-m}{2\left(b-a\right)}\left[h_2\left(a,a+\lambda \frac{b-a}{2}\right)+h_2\left(x,a+\lambda \frac{b-a}{2}\right)+h_2\left(x,a+\left(2-\lambda \right)\frac{b-a}{2}\right)\right.\hfill \\ \hfill & \left.h_2\left(b,a+\left(2-\lambda \right)\frac{b-a}{2}\right)\right],\hfill \end{array}$$ for all $\lambda \in \left[0,1\right]$ such that $a+\lambda \frac{b-a}{2}$ and $a+\left(2-\lambda \right)\frac{b-a}{2}$ are in $\mathbb{T}$ , and $x\in \left[a+\lambda \frac{b-a}{2},a+\left(2-\lambda \right)\frac{b-a}{2}\right]\cap \mathbb{T}$ .

In the case of $\lambda =0$ in Corollary 26, we get $$\begin{array}{cc}\hfill & \left|u\left(x\right)-\left(x-\frac{a+b}{2}\right)\left(\frac{m+M}{2}\right)-\frac{m+M}{2\left(b-a\right)}\left(g_2\left(a,\frac{a+b}{2}\right)-g_2\left(b,\frac{a+b}{2}\right)\right)\right.\hfill \\ \hfill & \left.-\frac{1}{b-a}{\int }_a^{b}u\left(\sigma \left(t\right)\right)\mathrm{\Delta }t\right|⩽\frac{M-m}{2\left(b-a\right)}\left[h_2\left(x,a\right)+h_2\left(x,b\right)\right],\hfill \end{array}$$ for all $a,b,t,x\in \mathbb{T}$ .

If we take $\mathbb{T}=\mathbb{R}$ in Corollary 26, then we recapture Theorem 5. Also, Proposition 27 boils down to Theorem 3.

In the case of $\lambda =\frac{1}{2}$ in Corollary 26, we get $$\begin{array}{cc}\hfill & \left|\frac{1}{2}u\left(x\right)-\frac{1}{2}\left(x-\frac{a+b}{2}\right)\left(\frac{m+M}{2}\right)+\frac{u\left(a\right)+u\left(b\right)}{4}\right.\hfill \\ \hfill & \left.-\frac{1}{b-a}{\int }_a^{b}u\left(\sigma \left(t\right)\right)\mathrm{\Delta }t-\frac{m+M}{2\left(b-a\right)}\left(g_2\left(a,\frac{a+b}{2}\right)-g_2\left(b,\frac{a+b}{2}\right)\right)\right|\hfill \\ \hfill ⩽& \frac{M-m}{2\left(b-a\right)}\left[h_2\left(a,\frac{3a+b}{4}\right)+h_2\left(x,\frac{3a+b}{4}\right)+h_2\left(x,\frac{a+3b}{4}\right)\right.\hfill \\ \hfill & \left.+h_2\left(b,\frac{a+3b}{4}\right)\right],\hfill \end{array}$$ where $\frac{3a+b}{4}$ and $\frac{a+3b}{4}$ are in $\mathbb{T}$ , and $x\in \left[\frac{3a+b}{4},\frac{a+3b}{4}\right]\cap \mathbb{T}$ .

In the case of $\lambda =1$ in Corollary 26, we get $$\begin{array}{cc}\hfill & \left|\frac{u\left(a\right)+u\left(b\right)}{2}-\frac{1}{b-a}{\int }_a^{b}u\left(\sigma \left(t\right)\right)\mathrm{\Delta }t-\frac{m+M}{2\left(b-a\right)}\left(g_2\left(a,\frac{a+b}{2}\right)-g_2\left(b,\frac{a+b}{2}\right)\right)\right|\hfill \\ \hfill ⩽& \frac{M-m}{2\left(b-a\right)}\left[h_2\left(a,\frac{a+b}{2}\right)+2h_2\left(x,\frac{a+b}{2}\right)+h_2\left(b,\frac{a+b}{2}\right)\right]\hfill \end{array}$$ where $\frac{a+b}{2}\in$ $\mathbb{T}$ .

Let $a,b,t,x\in \mathbb{T},a<b$ . If $u\in C_{\mathrm{rd}}^1\left(\left[a,b\right],\mathbb{R}\right)$ , then for all $x\in [a,b]$ , we have $$\begin{array}{cc}\hfill & \left|\frac{\left(b-x\right)u\left(b\right)+\left(x-a\right)u\left(a\right)}{b-a}+\frac{m+M}{2}\left(x-\frac{a+b}{2}\right)\right.\hfill \\ \hfill & \left.-\frac{1}{b-a}{\int }_a^{b}u\left(\sigma \left(t\right)\right)\mathrm{\Delta }t-\frac{m+M}{2\left(b-a\right)}\left(g_2\left(a,\frac{a+b}{2}\right)-g_2\left(b,\frac{a+b}{2}\right)\right)\right|\hfill \\ \hfill ⩽& \frac{M-m}{2\left(b-a\right)}\left[h_2\left(a,x\right)+h\left(b,x\right)\right]\text{.}\hfill \end{array}$$

We observe that Theorem 31 amounts to Theorem 4 if the time scale is the set of real numbers.

Setting $\mathbb{T}=\mathbb{Z}$ in Theorem 31, we obtain $$\begin{array}{cc}\hfill & \left|\frac{\left(b-x\right)u\left(b\right)+\left(x-a\right)u\left(a\right)}{b-a}+\left(\frac{m+M}{2}\right)\left(x-\frac{a+b}{2}\right)\right.\hfill \\ \hfill & \left.-\frac{1}{b-a}{\displaystyle \sum _{t=a}^{b-1}}u\left(t+1\right)+\frac{m+M}{4}\right|\hfill \\ \hfill ⩽& \frac{M-m}{2}\left[\left(\frac{1}{4}+{\left(\frac{x-\frac{a+b}{2}}{b-a}\right)}^2\right)\left(b-a\right)+\frac{1}{b-a}\left(x-\frac{a+b}{2}\right)\right]\text{.}\hfill \end{array}$$