\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2011 (2011), No. 74, pp. 1--10.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2011 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2011/74\hfil Young's integral inequality with bounds] {Young's integral inequality with upper and lower bounds} \author[D. R. Anderson, S. Noren, B. Perreault \hfil EJDE-2011/74\hfilneg] {Douglas R. Anderson, Steven Noren, Brent Perreault} % in alphabetical order \address{Douglas R. Anderson \newline Concordia College, Department of Mathematics and Computer Science, Moorhead, MN 56562, USA} \email{andersod@cord.edu http://www.cord.edu/faculty/andersod/} \address{Steven Noren \newline Concordia College, Moorhead, MN 56562, USA} \email{srnoren@cord.edu} \address{Brent Perreault \newline Concordia College, Moorhead, MN 56562, USA} \email{bmperrea@cord.edu} \thanks{Submitted February 14, 2011. Published June 15, 2011.} \subjclass[2000]{26D15, 39A12, 34N05} \keywords{Young's inequality; monotone functions; Pochhammer lower factorial; \hfill\break\indent difference equations; time scales} \begin{abstract} Young's integral inequality is reformulated with upper and lower bounds for the remainder. The new inequalities improve Young's integral inequality on all time scales, such that the case where equality holds becomes particularly transparent in this new presentation. The corresponding results for difference equations are given, and several examples are included. We extend these results to piecewise-monotone functions as well. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{remark}[theorem]{Remark} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{introduction} In 1912, Young \cite{young} presented the following highly intuitive integral inequality, namely that any real-valued continuous function $f:[0,\infty)\to[0,\infty)$ satisfying $f(0)=0$ with $f$ strictly increasing on $[0,\infty)$ satisfies $$\label{yeq} ab \le \int_0^a f(t)dt + \int_0^b f^{-1}(y)dy$$ for any $a,b\in[0,\infty)$, with equality if and only if $b=f(a)$. A useful consequence of this theorem is Young's inequality, $$ab \le \frac{a^p}{p}+\frac{b^q}{q}, \quad \frac{1}{p}+\frac{1}{q}=1,$$ with equality if and only if $a^p=b^q$, a fact derived from \eqref{yeq} by taking $f(t)=t^{p-1}$ and $q=\frac{p}{p-1}$. Hardy, Littlewood, and P\'{o}lya included Young's inequality in their classic book \cite{hardy}, but there was no analytic proof until Diaz and Metcalf \cite{diaz} supplied one in 1970. Tolsted \cite{tolsted} showed how to derive Cauchy, H\"{o}lder, and Minkowski inequalities in a straightforward way from \eqref{yeq}. For other applications and extensions of Young's inequality, see Cerone \cite{gini} and Mitrinovi\'{c}, Pe\v{c}ari\'{c}, and Fink \cite{fink}. For the purposes of this paper we recall some results that consider upper bounds for the integrals in \eqref{yeq}. Merkle \cite{merkle} established the inequality $$\int_0^a f(t)dt + \int_0^b f^{-1}(y)dy \le \max\{af(a),bf^{-1}(b)\},$$ which has been improved and reformulated recently by Minguzzi \cite{em} to the inequality $$\label{emorig} 0 \le \int_{\alpha_1}^a f(t)dt + \int_{\beta_1}^b f^{-1}(y)dy - ab + \alpha_1\beta_1 \le -\big(f^{-1}(b)-a\big)\big(f(a)-b\big),$$ where the hypotheses of Young's integral inequality hold, except that $f(\alpha_1)=\beta_1$ has replaced $f(0)=0$. One might wonder if there is a discrete version of \eqref{yeq} in the form of a summation inequality, or more generally a time-scale version of \eqref{yeq}, where a time scale, introduced by Hilger \cite{hilger}, is any nonempty closed set of real numbers. Wong, Yeh, Yu, and Hong \cite{wyyh} presented a version of Young's inequality on time scales $\mathbb{T}$ in the following form. Using the standard notation \cite{bp1} of the left jump operator $\rho$ given by $\rho(t):=\sup\{s\in\mathbb{T}: st\}$, the compositions $f\circ\rho$ and $f\circ\sigma$ denoted by $f^\rho$ and $f^\sigma$, respectively, the graininess functions defined by $\mu(t)=\sigma(t)-t$ and $\nu(t)=t-\rho(t)$, and the delta and nabla derivatives of $f$ at $t\in\mathbb{T}$, denoted $f^\Delta(t)$ and $f^\nabla(t)$, respectively, (provided they exist) are given by $$f^\Delta(t):=\lim_{s\to t}\frac{f^\sigma(t)-f(s)}{\sigma(t)-s}, \quad f^\nabla(t):=\lim_{s\to t} \frac{f^\rho(t)-f(s)}{\rho(t)-s},$$ we have the following result. \begin{theorem}[Wong, Yeh, Yu, and Hong]\label{amlthm} Let $f$ be right-dense continuous on \\ $[0,c]_\mathbb{T}:=[0,c]\cap\mathbb{T}$ for $c>0$, strictly increasing, with $f(0)=0$. Then for $a\in[0,c]_\mathbb{T}$ and $b\in[0,f(c)]_\mathbb{T}$ the inequality $$ab \le \int_{0}^{a}f^\sigma(t)\Delta t + \int_{0}^{b}(f^{-1})^\sigma(y)\Delta y$$ holds. \end{theorem} If $\mathbb{T}=\mathbb{Z}$ and $f(t)=t$, then Theorem \ref{amlthm} says that $$\label{wongexample} ab \le \sum_{t=0}^{a-1}(t+1) + \sum_{y=0}^{b-1}(y+1) = \frac{1}{2}\big(a(a+1)+b(b+1)\big)$$ holds. Note that an if and only if clause concerning an actual equality is missing in the formulation in Theorem \ref{amlthm}, with equality impossible in the simple example \eqref{wongexample} except for the trivial case $a=0=b$. This omission was rectified in \cite{and} via the following theorem. \begin{theorem}[Anderson]\label{yng2} Let $\mathbb{T}$ be any time scale (unbounded above) with $0\in\mathbb{T}$. Further, suppose that $f:[0,\infty)_{\mathbb{T}}\to\mathbb{R}$ is a real-valued function satisfying \begin{itemize} \item[(i)] $f(0)=0$; \item[(ii)] $f$ is continuous on $[0,\infty)_\mathbb{T}$, right-dense continuous at $0$; \item[(iii)] $f$ is strictly increasing on $[0,\infty)_\mathbb{T}$ such that $f(\mathbb{T})$ is also a time scale. \end{itemize} Then for any $a\in[0,\infty)_\mathbb{T}$ and $b\in[0,\infty)\cap f(\mathbb{T})$, we have $$ab \le \frac{1}{2}\int_0^a [f(t)+f(\sigma(t))]\Delta t + \frac{1}{2}\int_0^b [f^{-1}(y)+f^{-1}(\sigma(y))]\Delta y,$$ with equality if and only if $b=f(a)$. \end{theorem} Motivated by \cite{em}, in this paper we extend \eqref{emorig} to the general time scales setting while, in the process, simplifying and extending Theorem \ref{yng2} as well. As these results on time scales will include new results in difference equations as an important corollary, we will illustrate our new inequalities using discrete examples with $\mathbb{T}=\mathbb{Z}$. \section{Theorem Formulation} We begin this section by introducing a new and improved version of Theorem \ref{yng2} to facilitate the subsequent results. For any time scale $\mathbb{T}$, we have the following result. \begin{theorem}[Young's Inequality on Time Scales]\label{yng3} Let $\mathbb{T}$ be any time scale with $\alpha_1\in\mathbb{T}$ and $\sup\mathbb{T}=\infty$. Further, suppose that $f:[\alpha_1,\infty)_{\mathbb{T}}\to\mathbb{R}$ is a real-valued function satisfying \begin{itemize} \item[(i)] $f(\alpha_1)=\beta_1$; \item[(ii)] $f$ is continuous on $[\alpha_1,\infty)_\mathbb{T}$, right-dense continuous at $\alpha_1$; \item[(iii)] $f$ is strictly increasing on $[\alpha_1,\infty)_\mathbb{T}$ such that $\widetilde{\mathbb{T}}:=f(\mathbb{T})$ is also a time scale. \end{itemize} Then for any $a\in[\alpha_1,\infty)_\mathbb{T}$ and $b\in[\beta_1,\infty)_{\widetilde{\mathbb{T}}}$, we have $$\label{yeq2} ab \le \int_{\alpha_1}^a f(t)\Delta t + \int_{\beta_1}^b f^{-1}(y)\widetilde{\nabla} y + \alpha_1\beta_1,$$ with equality if and only if $b\in\{f^\rho(a),f(a)\}$ for fixed $a$, or with equality if and only if $a\in\{f^{-1}(b),\sigma(f^{-1}(b))\}$ for fixed $b$. The inequality in \eqref{yeq2} is reversed if $f$ is strictly decreasing. \end{theorem} \begin{proof} The proof is modelled after the one given on $\mathbb{R}$ in \cite{diaz}. Note that $f$ is delta integrable and $f^{-1}$ is nabla integrable by the continuity assumption in (ii). For simplicity, define $$\label{Fab} F(a,b):=\int_{\alpha_1}^a f(t)\Delta t + \int_{\beta_1}^b f^{-1}(y)\widetilde{\nabla} y - ab + \alpha_1\beta_1.$$ Then, the inequality to be shown is just $F(a,b)\ge 0$. (I). We will first show that $$F(a,b)\ge F(a,f(a)) \quad \text{for  a\in[\alpha_1,\infty)_\mathbb{T} and b\in[\beta_1,\infty)_{\widetilde{\mathbb{T}}}},$$ with equality if and only if $b\in\{f^\rho(a),f(a)\}$. For any such $a$ and $b$ we have $$\label{differncefs} F(a,b) - F(a,f(a)) = \int_{f(a)}^{b}[f^{-1}(y)-a]\widetilde{\nabla} y.$$ Clearly if $b=f(a)$ then the integrals are empty, and if $b=f^\rho(a)$ then $$F(a,f^\rho(a)) - F(a,f(a)) = \int^{f(a)}_{f^\rho(a)}[a-f^{-1}(y)] \widetilde{\nabla} y = [f(a)-f^\rho(a)][a-f^{-1}(f(a))] = 0.$$ Otherwise, since $f^{-1}(y)$ is continuous and strictly increasing for $y\in\widetilde{\mathbb{T}}$, the integrals in \eqref{differncefs} are strictly positive for $bf(a)$. (II). We will next show that $F(a,f(a))=F(a,f^\rho(a))=0$. For brevity, put $\varphi(a)=F(a,f(a))$, that is $$\varphi(a):=\int_{\alpha_1}^a f(t)\Delta t + \int_{\beta_1}^{f(a)} f^{-1}(y)\widetilde{\nabla} y - af(a) + \alpha_1\beta_1.$$ First, assume $a$ is a right-scattered point. Then \begin{align*} \varphi^\sigma(a)-\varphi(a) &= \int_a^{\sigma(a)} f(t)\Delta t + \int_{f(a)}^{f^\sigma(a)} f^{-1}(y)\widetilde{\nabla} y - \sigma(a)f^\sigma(a) + af(a) \\ &= [\sigma(a)-a]f(a) + [f^\sigma(a)-f(a)]f^{-1}(f^\sigma(a)) - \sigma(a)f^\sigma(a) + af(a)\\ & = 0. \end{align*} Therefore, if $a$ is a right-scattered point, then $\varphi^\Delta(a)=0$. Next, assume $a$ is a right-dense point. Let $\{a_n\}_{n\in\mathbb{N}}\subset[a,\infty)_\mathbb{T}$ be a decreasing sequence converging to $a$. Then \begin{align*} \varphi(a_n)-\varphi(a) &= \int_{a}^{a_n} f(t)\Delta t + \int_{f(a)}^{f(a_n)} f^{-1}(y) \widetilde{\nabla} y - a_nf(a_n) + af(a) \\ &\ge (a_n-a)f(a)+[f(a_n)-f(a)]a - a_nf(a_n) + af(a) \\ & = (a_n-a)[f(a)-f(a_n)], \end{align*} since the functions $f$ and $f^{-1}$ are strictly increasing. Similarly, \begin{align*} \varphi(a_n)-\varphi(a) &\le (a_n-a)f(a_n)+[f(a_n)-f(a)]a_n - a_nf(a_n) + af(a) \\ & = (a_n-a)[f(a_n)-f(a)]. \end{align*} Therefore, $$0=\lim_{n\to\infty}[f(a)-f(a_n)] \le \lim_{n\to\infty}\frac{\varphi(a_n)-\varphi(a)}{a_n-a} \le \lim_{n\to\infty}[f(a_n)-f(a)]=0.$$ It follows that $\varphi^\Delta(a)$ exists, and $\varphi^\Delta(a)=0$ for right-dense $a$ as well. In other words, in either case, $\varphi^\Delta(a)=0$ for $a\in[\alpha_1,\infty)_\mathbb{T}$. As $\varphi(\alpha_1)=0$, by the uniqueness theorem for initial value problems we have that $\varphi(a)=0$ for all $a\in[\alpha_1,\infty)_\mathbb{T}$. From earlier we know that $F(a,f^\rho(a))=F(a,f(a))$. Thus as an overall result, we have that $$F(a,b)\ge F(a,f(a))=0,$$ with equality if and only if $b=f(a)$ or $b=f^\rho(a)$, as claimed. The case with $a\in\{f^{-1}(b),\sigma(f^{-1}(b))\}$ for fixed $b$ is similar and thus omitted. If $f$ is strictly decreasing, it is straightforward to see that the inequality in \eqref{yeq2} is reversed; the details are left to the reader. \end{proof} We now focus on establishing an upper bound for Young's integral. Before we state and prove our main theorem, we need the auxiliary result given in the following lemma; this lemma generalizes \cite[Lemma 2.1]{em} to time scales. \begin{lemma}\label{lemma22} Let $f$ satisfy the hypotheses of Theorem \ref{yng3}, and let $F(a,b)$ be given as in \eqref{Fab}. For every $a,\alpha\in\mathbb{T}$ and $b,\beta\in\widetilde{\mathbb{T}}$ we have $$\label{em4} F(a,b)+F(\alpha,\beta) \ge - (\alpha-a)(\beta-b),$$ where equality holds if and only if $\alpha\in\{f^{-1}(b),\sigma(f^{-1}(b))\}$ and $\beta\in\{f^\rho(a),f(a)\}$. \end{lemma} \begin{proof} Fix $a\in\mathbb{T}$ and $b\in\widetilde{\mathbb{T}}$. By Young's integral inequality on time scales (Theorem \ref{yng3}) we have \begin{gather}\label{em5} \int_{\alpha_1}^a f(t)\Delta t + \int_{\beta_1}^{\beta} f^{-1}(y)\widetilde{\nabla} y + \alpha_1\beta_1 \ge a\beta, \quad\text{and} \\\label{em6} \int_{\alpha_1}^{\alpha} f(t)\Delta t + \int_{\beta_1}^{b} f^{-1}(y)\widetilde{\nabla} y + \alpha_1\beta_1 \ge \alpha b, \end{gather} with equality if and only if $\beta\in\{f^\rho(a),f(a)\}$ and $\alpha\in\{f^{-1}(b),\sigma(f^{-1}(b))\}$, respectively. By rearranging it follows that \begin{align*} & \Big(\int_{\alpha_1}^a f(t)\Delta t + \int_{\beta_1}^b f^{-1}(y)\widetilde{\nabla} y - ab + \alpha_1\beta_1\Big)\\ &+ \Big(\int_{\alpha_1}^{\alpha} f(t)\Delta t + \int_{\beta_1}^{\beta} f^{-1}(y)\widetilde{\nabla} y - \alpha\beta + \alpha_1\beta_1\Big) \\ & = \Big(\int_{\alpha_1}^a f(t)\Delta t + \int_{\beta_1}^{\beta} f^{-1}(y)\widetilde{\nabla} y + \alpha_1\beta_1\Big) \\ &\quad + \Big(\int_{\alpha_1}^{\alpha} f(t)\Delta t + \int_{\beta_1}^{b} f^{-1}(y)\widetilde{\nabla} y + \alpha_1\beta_1\Big) - ab - \alpha\beta \\ &\ge a\beta + \alpha b - ab - \alpha\beta = - (\alpha-a)(\beta-b). \end{align*} Note that equality holds here if and only if it holds in \eqref{em5} and \eqref{em6}, videlicet if and only if $\alpha\in\{f^{-1}(b),\sigma(f^{-1}(b))\}$ and $\beta\in\{f^\rho(a),f(a)\}$. \end{proof} \begin{theorem}\label{newthm1} Let $\mathbb{T}$ be any time scale and $f:[\alpha_1,\alpha_2]_\mathbb{T}\to[\beta_1, \beta_2]_{\widetilde{\mathbb{T}}}$ be a continuous strictly increasing function such that $\widetilde{\mathbb{T}}=f(\mathbb{T})$ is also a time scale. Then for every $a,\widehat{a}\in[\alpha_1,\alpha_2]_\mathbb{T}$ and $b,\widehat{b}\in[\beta_1,\beta_2]_{\widetilde{\mathbb{T}}}$ we have $$\label{em3} \begin{split} \big(f^{-1}(\widehat{b})-\widehat{a}\big) \big(f^\rho(\widehat{a})-\widehat{b}\big) &\le \int_{\widehat{a}}^a f(t)\Delta t + \int_{\widehat{b}}^b f^{-1}(y)\widetilde{\nabla} y - ab + \widehat{a}\widehat{b}\\ &\le -\big(f^{-1}(b)-a\big)\big(f^\rho(a)-b\big), \end{split}$$ where the equalities hold if and only if $\widehat{b}\in\{f^\rho(\widehat{a}),f(\widehat{a})\}$ and $b\in\{f^\rho(a),f(a)\}$. The inequalities are reversed if $f$ is strictly decreasing. \end{theorem} \begin{proof} Considering $F$ as in \eqref{Fab}, and \eqref{em4} with $\alpha=f^{-1}(b)$ and $\beta=f^\rho(a)$ we have the equality $$F(a,b)+F\big(f^{-1}(b),f^\rho(a)\big) = - \big(f^{-1}(b)-a\big)\big(f^\rho(a)-b\big).$$ As $f^{-1}(b)\in[\alpha_1,\alpha_2]_\mathbb{T}$ and $f^\rho(a)\in[\beta_1,\beta_2]_{\widetilde{\mathbb{T}}}$, via Young's integral inequality on time scales (Theorem \ref{yng3} above) we see that $F\big(f^{-1}(b),f^\rho(a)\big) \ge 0$. Consequently we have that $$\label{eq28} 0 \le F(a,b) \le - \big(f^{-1}(b)-a\big)\big(f^\rho(a)-b\big);$$ note that equality holds if and only if $b\in\{f^\rho(a),f(a)\}$. Thus for any $\widehat{a}\in[\alpha_1,\alpha_2]_\mathbb{T}$ and $\widehat{b}\in[\beta_1,\beta_2]_{\widetilde{\mathbb{T}}}$ we have from \eqref{eq28} that $$\label{eq29} 0 \le - \big(f^{-1}(\widehat{b})-\widehat{a}\big) \big(f^\rho(\widehat{a})-\widehat{b}\big) -F(\widehat{a},\widehat{b}),$$ with equality if and only if $\widehat{b}\in\{f^\rho(\widehat{a}),f(\widehat{a})\}$. Combining inequalities \eqref{eq28} and \eqref{eq29} we get \begin{align*} 0 &\le F(a,b) - \big(f^{-1}(\widehat{b})-\widehat{a}\big) \big(f^\rho(\widehat{a})-\widehat{b}\big)-F(\widehat{a},\widehat{b}) \\ &\le - \big(f^{-1}(b)-a\big)\big(f^\rho(a)-b\big) - \big(f^{-1}(\widehat{b})-\widehat{a}\big) \big(f^\rho(\widehat{a})-\widehat{b}\big)-F(\widehat{a},\widehat{b}). \end{align*} This can be rewritten as \eqref{em3}. If $f$ is strictly decreasing the proof is similar and thus omitted. \end{proof} \begin{remark} \label{rmk2.4}\rm In \cite[Theorem 1.1]{em} the author assumes $f(\widehat{a})=\widehat{b}$ $($see \eqref{emorig} above$)$, so that the inequality in \eqref{em3} is new for $\mathbb{T}=\mathbb{R}$, as well as for $\mathbb{T}=\mathbb{Z}$ and general time scales. \end{remark} \begin{remark} \label{rmk2.5} \rm Due to Lemma \ref{lemma22} and the first line of the proof of Theorem \ref{newthm1}, there are three other inequalities we could write in place of \eqref{em3}, namely \begin{gather*} \begin{split} \big(f^{-1}(\widehat{b})-\widehat{a}\big) \big(f(\widehat{a})-\widehat{b}\big) &\le \int_{\widehat{a}}^a f(t)\Delta t + \int_{\widehat{b}}^b f^{-1}(y)\widetilde{\nabla} y - ab + \widehat{a}\widehat{b}\\ &\le -\big(f^{-1}(b)-a\big)\big(f(a)-b\big), \end{split}\\ \begin{split} \big(\sigma(f^{-1}(\widehat{b}))-\widehat{a}\big) \big(f^\rho(\widehat{a})-\widehat{b}\big) &\le \int_{\widehat{a}}^a f(t)\Delta t + \int_{\widehat{b}}^b f^{-1}(y)\widetilde{\nabla} y - ab + \widehat{a}\widehat{b}\\ &\le -\big(\sigma(f^{-1}(b))-a\big)\big(f^\rho(a)-b\big), \end{split}\\ \begin{split} \big(\sigma(f^{-1}(\widehat{b}))-\widehat{a}\big) \big(f(\widehat{a})-\widehat{b}\big) &\le \int_{\widehat{a}}^a f(t)\Delta t + \int_{\widehat{b}}^b f^{-1}(y)\widetilde{\nabla} y - ab + \widehat{a}\widehat{b}\\ &\le -\big(\sigma(f^{-1}(b))-a\big)\big(f(a)-b\big). \end{split} \end{gather*} If we focus on just the upper bounds, for $a>\widehat{a}$, $b>\widehat{b}$, and $b \ge f(a)$ we have the least of these upper bounds, leading to $$\int_{\widehat{a}}^a f(t)\Delta t + \int_{\widehat{b}}^b f^{-1}(y)\widetilde{\nabla} y - ab + \widehat{a}\widehat{b} \le -\big(f^{-1}(b)-a\big)\big(f(a)-b\big),$$ whereas for $a>\widehat{a}$, $b>\widehat{b}$, and $b \le f^\rho(a)$ we have $$\int_{\widehat{a}}^a f(t)\Delta t + \int_{\widehat{b}}^b f^{-1}(y)\widetilde{\nabla} y - ab + \widehat{a}\widehat{b} \le -\big(\sigma(f^{-1}(b))-a\big)\big(f^\rho(a)-b\big).$$ \end{remark} \begin{corollary}\label{cor26} Pick $a_i\in\mathbb{T}$ with $a_i1$ and any integers $a\ge\alpha\in\mathbb{Z}$ we have $$B^{\alpha} \le \frac{B^a-B^{\alpha}}{(a-\alpha)(B-1)} \le B^{a-1},$$ where the equalities hold if and only if $\alpha\in\{a-1,a\}$. \end{example} \begin{example} \label{exmp3.5} \rm Consider Theorem \ref{legendrethm}. Let $\mathbb{T}=\mathbb{Z}$ and $f(t)=B^t$ for $B>1$. Then $\widetilde{\mathbb{T}}=B^{\mathbb{Z}}$, and we have $g(t)=\frac{B^t-B^{\alpha}}{B-1}$, $f^{-1}(y)=\log_{B}y$, and $g^*(y)=y\log_{B}y-\frac{y}{B-1} +\beta\big(\alpha+\frac{1}{B-1}-\log_{B}\beta\big)$. It is easy to check that $f^{-1}=(g^*)^{\widetilde{\nabla}}$ on $\widetilde{\mathbb{T}}$ and $\alpha\beta=g(\alpha)+g^*(\beta)$. Therefore we have the inequalities \begin{align*} & (\log_{B} \beta-\alpha)(B^{\alpha-1}-\beta)\\ &\le \frac{B^a-B^{\alpha}}{B-1} + b\log_{B}b-\frac{b}{B-1} +\beta\big(\alpha+\frac{1}{B-1}-\log_{B}\beta\big) - ab \\ &\le -(\log_{B}b-a)(B^{a-1}-b), \end{align*} where the equalities hold if and only if $\beta\in\{B^{\alpha-1},B^{\alpha}\}$ and $b\in\{B^{a-1},B^a\}$. \end{example} \begin{example} \label{exmp3.6} \rm Let $f(t)=\sin[\pi t/(2k)]$ for $k\in\mathbb{N}$. Then $f$ is strictly increasing on $[-k,k]_{\mathbb{Z}}$, so that for any $a\ge\alpha\in[-k,k]_{\mathbb{Z}}$ we have by Theorem \ref{newthm2Z} that $$\sin[\frac{\alpha\pi}{2k}] \le \frac{1}{2(a-\alpha)} \Big(\cos[\frac{(2\alpha-1)\pi}{4k}] - \cos[\frac{(2a-1)\pi}{4k}]\Big) \csc[\frac{\pi}{4k}] \le \sin[\frac{(a-1)\pi}{2k}],$$ with equalities if and only if $\alpha\in\{a-1,a\}$. \end{example} \begin{example} \label{exmp3.7} \rm Let $f(t)=\binom{t}{k}$ for $t,k\in\mathbb{N}$ with $t\ge k$. Then $f$ is strictly increasing on $[k,\infty)_{\mathbb{Z}}$, whereby for any $a\ge\alpha\in[k,\infty)_{\mathbb{Z}}$ we have $$\binom{\alpha}{k} \le \frac{(a-k)\binom{a}{k}-(\alpha-k) \binom{\alpha}{k}}{(a-\alpha)(k+1)} \le \binom{a-1}{k},$$ with equalities if and only if $\alpha\in\{a-1,a\}$. \end{example} \begin{thebibliography}{99} \bibitem{and} D. R. Anderson; Young's integral inequality on time scales revisited, \emph{J. Inequalities in Pure Appl. Math.}, Volume 8 (2007), Issue 3, Article 64, 5 pages. \bibitem{bp1} M. Bohner and A. Peterson; \emph{Dynamic Equations on Time Scales: An Introduction with Applications}, Birkh\"auser, Boston (2001). \bibitem{gini} P. Cerone; On Young's inequality and its reverse for bounding the Lorenz curve and Gini mean, \emph{J. Math. Inequalities} 3:3 (2009) 369--381. \bibitem{diaz} J. B. Diaz and F. T. Metcalf; An analytic proof of Young's inequality, \emph{American Math. Monthly}, 77 (1970) 603--609. \bibitem{hardy} G. H. Hardy, J. E. Littlewood, and G. P\'{o}lya; \emph{Inequalities}, Cambridge, 1934. \bibitem{hilger} S. Hilger; Analysis on measure chains---a unified approach to continuous and discrete calculus, \emph{Results Math.} 18 (1990), no. 1--2, 18--56. \bibitem{kp} W. G. Kelley and A. C. Peterson; \emph{Difference Equations: An Introduction with Applications}, 2e, Harcourt/Academic Press, San Diego, 2001. \bibitem{vl} V. Lampret; Approximating real Pochhammer products: a comparison with powers, \emph{Cent. Eur. J. Math.}, 7(3) (2009) 493--505. \bibitem{merkle} M. Merkle; A contribution to Young's inequality, \emph{Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz.}, 461--497 (1974), 265--267. \bibitem{em} E. Minguzzi; An equivalent form of Young's inequality with upper bound, \emph{Appl. Anal. Discrete Math.} 2 (2008) 213--216. \bibitem{fink} D. S. Mitrinovi\'{c}, J. E. Pe\v{c}ari\'{c}, and A. M. Fink; \emph{Classical and New Inequalities in Analysis}, Kluwer Academic Publishers, Dordrecht, 1993. \bibitem{tolsted} E. Tolsted; An elementary derivation of the Cauchy, Holder, and Minkowski inequalities from Young's inequality, \emph{Math. Mag.}, 37 (1964), 2--12. \bibitem{wyyh}F. H. Wong, C. C. Yeh, S. L. Yu, and C. H. Hong; Young's inequality and related results on time scales, \emph{Appl. Math. Letters}, 18 (2005) 983--988. \bibitem{young} W. H. Young; On classes of summable functions and their Fourier series, \emph{Proc. Royal Soc.}, Series (A), 87 (1912) 225--229. \end{thebibliography} \end{document}