\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
\begin{equation}\label{yeq}
 ab \le \int_0^a f(t)dt + \int_0^b f^{-1}(y)dy
\end{equation}
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
\begin{equation}\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),
\end{equation}
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
$\rho(t):=\sup\{s\in\mathbb{T}: s<t\}$, the right jump operator $\sigma$
given by $\sigma(t)=\inf\{s\in\mathbb{T}: s>t\}$, 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
\begin{equation}\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)
\end{equation}
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
\begin{equation}\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,
\end{equation}
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
\begin{equation}\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.
\end{equation}
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
\begin{equation}\label{differncefs}
 F(a,b) - F(a,f(a)) = \int_{f(a)}^{b}[f^{-1}(y)-a]\widetilde{\nabla} y.
\end{equation}
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 $b<f^\rho(a)$ and $b>f(a)$.

(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
\begin{equation}\label{em4}
 F(a,b)+F(\alpha,\beta) \ge - (\alpha-a)(\beta-b),
\end{equation}
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
\begin{equation}\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}
\end{equation}
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
\begin{equation}\label{eq28}
 0 \le F(a,b) \le - \big(f^{-1}(b)-a\big)\big(f^\rho(a)-b\big);
\end{equation}
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
\begin{equation}\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}),
\end{equation}
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 $a_i<a_{i+1}$. Let
$f_1:[\rho(a_1),a_{2}]_\mathbb{T}\to\mathbb{R}$ and
$f_i:[a_i,a_{i+1}]_\mathbb{T}\to\mathbb{R}$ be continuous
strictly monotone functions for $i=2,\dots,m$. Assume
$f:\mathbb{T}\to\mathbb{R}$ is continuous, where
$$
f(t):=f_i(t) \quad\text{and}\quad f^{-1}(y):=f_i^{-1}(y)
$$
for $t\in[a_i,a_{i+1}]_\mathbb{T}$ and
$y\in[\min\{f(a_i),f(a_{i+1})\},\max\{f(a_i),f(a_{i+1})\}]_{f(\mathbb{T})}$,
respectively, and $i=1,2,\dots,m$, that is to say
$f_i(a_{i+1})=f_{i+1}(a_{i+1})$ for $i=1,2,\dots,m-1$. Set
\begin{gather*}
  I_f:=\int_{a_1}^{a_{m+1}} f(t)\Delta t + \int_{b_1}^{b_{m+1}}
 f^{-1}(y)\widetilde{\nabla} y - a_{m+1}b_{m+1}
 + a_1b_1, \quad\text{and} \\
  K_{i}:=-\big(f^{-1}(b_{i})-a_{i}\big)\big(f^\rho(a_{i})-b_{i}\big),
\quad i\in\{1,m+1\},
\end{gather*}
where $f^{-1}(b_1)\in[a_1,a_2]_\mathbb{T}$ and
$f^{-1}(b_{m+1})\in[a_m,a_{m+1}]_\mathbb{T}$. Then we have the following.
\begin{itemize}
\item[(i)] If $f_1$ and $f_m$ are both strictly increasing, then
$$
-K_{1}\le I_f \le K_{m+1}.
$$
The inequalities are reversed if $f_1$ and $f_m$ are both strictly
decreasing.
\item[(ii)] If $f_1$ is strictly increasing and $f_m$ is strictly
decreasing, then
$$
K_{m+1}-K_{1} \le I_f \le 0.
$$
The inequalities are reversed if $f_1$ is strictly decreasing and
$f_m$ is strictly increasing.
\end{itemize}

In all cases, the equalities hold if and only if
$b_1\in\{f^\rho(a_1),f(a_1)\}$ and
$b_{m+1}\in\{f^\rho(a_{m+1}),f(a_{m+1})\}$.
\end{corollary}

\begin{proof}
We will only prove the first part of (i),
as the other parts follow in a similar manner from
Theorem \ref{newthm1}. Assume $f_1$ and $f_m$ are both
strictly increasing. By Theorem \ref{newthm1} we have the
inequalities
$$
\big(f^{-1}(b_1)-a_1\big)\big(f^\rho(a_1)-b_1\big)
 \le \int_{a_1}^{a_2} f(t)\Delta t
+ \int_{b_1}^{f(a_2)} f^{-1}(y)\widetilde{\nabla} y - a_2f(a_2)
+ a_1b_1 \le 0,
$$
the equalities
$$
0 = \int_{a_i}^{a_{i+1}} f(t)\Delta t
+ \int_{f(a_i)}^{f(a_{i+1})} f^{-1}(y)\widetilde{\nabla} y
 - a_{i+1}f(a_{i+1}) + a_if(a_i) = 0
$$
for $i=2,\dots,m-1$, and the inequalities
\begin{align*}
 0 &\le \int_{a_m}^{a_{m+1}} f(t)\Delta t + \int_{f(a_m)}^{b_{m+1}} f^{-1}(y)\widetilde{\nabla} y - a_{m+1}b_{m+1} + a_mf(a_m) \\
   &\le \big(a_{m+1}-f^{-1}(b_{m+1})\big)
\big(f^\rho(a_{m+1})-b_{m+1}\big),
\end{align*}
where equalities hold in the first line if and only if
$b_1\in\{f^\rho(a_1),f(a_1)\}$, and equalities hold in the third
if and only if $b_{m+1}\in\{f^\rho(a_{m+1}),f(a_{m+1})\}$. If we
add these expressions together, we obtain $-K_{1}\le I_f \le
K_{m+1}$. This completes the proof.
\end{proof}

\begin{remark} \label{rmk2.7} \rm
Corollary \ref{cor26} for continuous piecewise-monotone functions
is new even for $\mathbb{T}=\mathbb{R}$, as well as for $\mathbb{T}=\mathbb{Z}$ and general
time scales. In the next result, Theorem \ref{pieces}, we extend the original Young result for continuous functions to piecewise-continuous piecewise-monotone functions on $\mathbb{R}$.
\end{remark}


\begin{theorem}\label{pieces}
Let $a_i\in\mathbb{R}$ with $a_i<a_{i+1}$, and let
$f_i:[a_i,a_{i+1}]\to\mathbb{R}$ be a continuous strictly monotone
function for $i=1,2,\dots,m$. Let $f:[a_1,a_{m+1}]\to\mathbb{R}$ be the
piecewise-continuous function given by
$$
f(x):=f_i(x)  \quad\text{and}\quad  f^{-1}(y):=f_i^{-1}(y)
$$
for $x\in(a_i,a_{i+1})$ and
$y\in\big(\min\{f(a_i),f(a_{i+1})\},\max\{f(a_i),f(a_{i+1})\}\big)$,
respectively, for $i=1,\dots,m$, with $f(a_1)=f_1(a_1)$ and
$f(a_{m+1})=f_m(a_{m+1})$. Set
\begin{gather*}
\begin{split}
  F(b_1,b_{m+1})&:=\int_{a_1}^{a_{m+1}} f(x)dx
 + \int_{b_1}^{b_{m+1}} f^{-1}(y)dy - a_{m+1}b_{m+1}\\
 &\quad + a_1b_1+\sum_{i=2}^{m}a_i[f_i(a_i)-f_{i-1}(a_i)],
\end{split} \\
  K_{i}:=-\big(f^{-1}(b_{i})-a_{i}\big)
\big(f(a_{i})-b_{i}\big), \quad i\in\{1,m+1\},
\end{gather*}
where $f^{-1}(b_1)\in[a_1,a_2]$ and $f^{-1}(b_{m+1})\in[a_m,a_{m+1}]$.
Then we have the following.
\begin{itemize}
\item[(i)] If $f_1$ and $f_m$ are both strictly increasing, then
$$
-K_{1}\le F(b_1,b_{m+1}) \le K_{m+1}.
$$
The inequalities are reversed if $f_1$ and $f_m$ are both strictly
decreasing.
\item[(ii)] If $f_1$ is strictly increasing and $f_m$ is strictly
decreasing, then
$$
K_{m+1}-K_{1} \le F(b_1,b_{m+1}) \le 0.
$$
The inequalities are reversed if $f_1$ is strictly decreasing
and $f_m$ is strictly increasing.
\end{itemize}
In all cases, the equalities hold if and only if $b_1=f(a_1)$
and $b_{m+1}=f(a_{m+1})$.
\end{theorem}

\begin{proof}
Apply Theorem \ref{newthm1} on $\mathbb{T}=\mathbb{R}$ to the pieces
$f_i$ on $[a_i,a_{i+1}]$, using the appropriate inequalities
for $f_1$ and $f_m$, and equalities for $f_i$, $i=2,\dots,m-1$.
Then add up these expressions to get the result.
\end{proof}

\begin{theorem}\label{legendrethm}
For some delta differentiable function $g$, let $f=g^\Delta$ satisfy
the hypotheses of Theorem \ref{newthm1}, where
$f^{-1}=(g^\Delta)^{-1}=(g^*)^{\widetilde{\nabla}}$ on
$\widetilde{\mathbb{T}}=f(\mathbb{T})$. [If $\mathbb{T}=\mathbb{R}$ the function $g^*$
is known as the Legendre transform of $g$.] If we
pick $\widehat{a}\widehat{b}=g(\widehat{a})+g^*(\widehat{b})$,
then by Theorem \ref{newthm1} we have
\begin{align*}
 \big((g^*)^{\widetilde{\nabla}}(\widehat{b})-\widehat{a}\big)
\big(g^\nabla(\widehat{a})-\widehat{b}\big)
 \le g(a) + g^*(b) - ab
 \le -\big((g^*)^{\widetilde{\nabla}}(b)-a\big)\big(g^\nabla(a)-b\big),
\end{align*}
where the equalities hold if and only if
$\widehat{b}\in\{g^\nabla(\widehat{a}),g^\Delta(\widehat{a})\}$
and $b\in\{g^\nabla(a),g^\Delta(a)\}$.
\end{theorem}

In the following theorem we reconsider Theorem \ref{newthm1} above.
This allows us to get a Young-type integral inequality without
having to find $f^{-1}$.

\begin{theorem}\label{newthm2}
Let the hypotheses of Theorem \ref{newthm1} hold. Then for
any $a,\alpha,\widehat{a},\widehat{\alpha}\in[\alpha_1,\alpha_2]_\mathbb{T}$
we have
\begin{equation}\label{yeq3}
\begin{split}
 (\widehat{\alpha}-\widehat{a})
\big(f^\rho(\widehat{a})-f(\widehat{\alpha})\big)
 &\le \int_{\widehat{a}}^a f(t)\Delta t - \int_{\widehat{\alpha}}^{\alpha} f(t)\Delta t + (\alpha-a)f(\alpha) + (\widehat{a}-\widehat{\alpha})f(\widehat{\alpha})   \\
 &\le -(\alpha-a)\big(f^\rho(a)-f(\alpha)\big),
\end{split}
\end{equation}
where the equalities hold if and only if
$\widehat{\alpha}\in\{\rho(\widehat{a}),\widehat{a}\}$ and
$\alpha\in\{\rho(a),a\}$.
\end{theorem}

\begin{proof}
By Theorem \ref{newthm1} with $\widehat{a}=\widehat{\alpha}$, $\widehat{b}=f(\widehat{\alpha})$, $a=\alpha$ and $b=f(\alpha)$ we have
\begin{equation}\label{31pf3}
 \int_{f(\widehat{\alpha})}^{f(\alpha)} f^{-1}(y)\widetilde{\nabla} y
= \alpha f(\alpha) - \widehat{\alpha}f(\widehat{\alpha})
 -  \int_{\widehat{\alpha}}^{\alpha} f(t)\Delta t
\end{equation}
for any $\alpha,\widehat{\alpha}\in[\alpha_1,\alpha_2]_\mathbb{T}$.
Since $\alpha,\widehat{\alpha}\in[\alpha_1,\alpha_2]_\mathbb{T}$
 are arbitrary, we substitute \eqref{31pf3} into \eqref{em3}
 to obtain \eqref{yeq3}.
\end{proof}

\begin{example} \label{exmp2.11} \rm
Consider the generalized polynomial functions $h_n(t,s)$ on time
scales defined recursively in the following
way \cite[Section 1.6]{bp1}:
$$
h_0(t,s)\equiv 1, \quad h_{k+1}(t,s)=\int_s^t h_k(\tau,s)\Delta\tau,
 \quad t,s\in\mathbb{T}, \quad k\in\mathbb{N}_0.
$$
If we take $f(t)=h_n(t,\widehat{\alpha})$ for any $n\in\mathbb{N}$,
then by Theorem \ref{newthm2} we have
\begin{equation}\label{exz2}
\begin{split}
 (\widehat{\alpha}-\widehat{a})h_n(\rho(\widehat{a}),\widehat{\alpha})
 &\le h_{n+1}(a,\widehat{a}) - h_{n+1}(\alpha,\widehat{\alpha}) + (\alpha- a)h_n(\alpha,\widehat{\alpha})   \\
 &\le -(\alpha-a)[h_n(\rho(a),\widehat{\alpha})
-h_n(\alpha,\widehat{\alpha})],
\end{split}
\end{equation}
for any $a,\widehat{a},\alpha,\widehat{\alpha}\in[\alpha_1,\alpha_2]_\mathbb{T}$,
where the equalities hold if and only if
$\widehat{\alpha}\in\{\rho(\widehat{a}),\widehat{a}\}$ and
$\alpha\in\{\rho(a),a\}$.
\end{example}

\section{Results for difference equations}

In this section we concentrate on the discrete case.
For $\mathbb{T}=\mathbb{Z}$ we have the following new discrete results,
which are corollaries of the theorems above. Recall that
$[\alpha_1,\alpha_2]_\mathbb{Z}=\{\alpha_1,\alpha_1+1,\alpha_1+2,
\dots,\alpha_2-1,\alpha_2\}$. The first two theorems are direct
translations to $\mathbb{T}=\mathbb{Z}$ of Theorem \ref{newthm1} and
Theorem \ref{newthm2}, respectively.

\begin{theorem}\label{newthm1Z}
Let $f:[\alpha_1,\alpha_2]_\mathbb{Z}\to[\beta_1,\beta_2]_{\widetilde{\mathbb{Z}}}$
be a strictly increasing function, where $\widetilde{\mathbb{Z}}=f(\mathbb{Z})$.
Then for every $a,\widehat{a}\in[\alpha_1,\alpha_2]_\mathbb{Z}$ and
$b,\widehat{b}\in[\beta_1,\beta_2]_{\widetilde{\mathbb{Z}}}$ we have
\begin{equation}\label{em3z}
\begin{split}
 \big(f^{-1}(\widehat{b})-\widehat{a}\big)
\big(f(\widehat{a}-1)-\widehat{b}\big)
 &\le \sum_{t=\widehat{a}}^{a-1} f(t)
 + \sum_{y\in(\widehat{b},b]\cap\widetilde{\mathbb{Z}}}
 f^{-1}(y)\widetilde{\nu}(y) - ab
 + \widehat{a}\widehat{b}  \\
 &\le -\big(f^{-1}(b)-a\big)\big(f(a-1)-b\big),
\end{split}
\end{equation}
where the equalities hold if and only if
$\widehat{b}\in\{f(\widehat{a}-1),f(\widehat{a})\}$ and
$b\in\{f(a-1),f(a)\}$.
\end{theorem}

\begin{theorem}\label{newthm2Z}
Let $f:\mathbb{Z}\to\mathbb{R}$ be a strictly increasing function.
Then for any integers $a,\widehat{a},\alpha,\widehat{\alpha}$ we have
\begin{equation}\label{yeq3z}
\begin{split}
(\widehat{\alpha}-\widehat{a})[f(\widehat{a}-1)
 -f(\widehat{\alpha})]
 &\le \sum_{t=\widehat{a}}^{a-1} f(t)
- \sum_{t=\widehat{\alpha}}^{\alpha-1} f(t)
+ (\alpha-a)f(\alpha) + (\widehat{a}
-\widehat{\alpha})f(\widehat{\alpha})   \\
 &\le -(\alpha-a)[f(a-1)-f(\alpha)],
\end{split}
\end{equation}
where the equalities hold if and only if
$\widehat{\alpha}\in\{\widehat{a}-1,\widehat{a}\}$ and
$\alpha\in\{a-1,a\}$.
\end{theorem}

\begin{example} \label{exmp3.3} \rm
Consider the Pochhammer lower factorial function
$$ f_k(t)=t^{\underline{k}}:=t(t-1)\dots(t-k+1), \quad t,k\in\mathbb{Z}, $$
also known as $t$ to the $k$ falling \cite{kp},
or the falling factorial power function \cite{vl}.
It is clear that $f_k$ is strictly increasing on the
integer interval $[k-1,\infty)_\mathbb{Z}$.
By Theorem \ref{newthm2Z} we have
\begin{equation}\label{exz1}
 (a-\alpha)f_k(\alpha) \le \frac{1}{k+1}[f_{k+1}(a)-f_{k+1}(\alpha)]
 \le (a-\alpha)f_k(a-1)
\end{equation}
for $a,\alpha\in\{k-1,k,k+1,\dots\}$, where the equalities hold
if and only if $\alpha\in\{a-1,a\}$.
\end{example}

\begin{example} \label{exmp3.4} \rm
For any real $B>1$ 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}


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\end{document}