text:
\begin{document}
\title[INEQUALITIES FOR $m-$CONVEX FUNCTIONS]{NEW OSTROWSKI TYPE
INEQUALITIES FOR $m-$CONVEX FUNCTIONS AND APPLICATIONS}
\author{HAVVA KAVURMACI$^{\blacktriangle ,\bigstar }$}
\address{$^{\blacktriangle }$ATATURK UNIVERSITY, K.K. EDUCATION FACULTY,
DEPARTMENT OF MATHEMATICS, 25240, CAMPUS, ERZURUM, TURKEY}
\email{havva.kvrmc@yahoo.com}
\thanks{$^{\bigstar }$Corresponding Author}
\author{M. EM\.{I}N \"{O}ZDEM\.{I}R$^{\spadesuit }$}
\address{$^{\spadesuit }$GRADUATE SCHOOL OF NATURAL AND APPLIED SCIENCE, A
\u{G}RI \.{I}BRAH\.{I}M \c{C}E\c{C}EN UNIVERSITY, A\u{G}RI, TURKEY}
\email{emos@atauni.edu.tr}
\author{MERVE AVCI$^{\blacktriangle }$}
\email{merveavci@ymail.com}

\begin{abstract}
In this paper, we establish new inequalities of Ostrowski type for functions
whose derivatives in absolute value are $m-$convex. We also give some
applications to special means of positive real numbers. Finally, we obtain
some error estimates for the midpoint formula.
\end{abstract}

\keywords{$m-$convex function, Starshaped function, Convex function,
Ostrowski inequality, Hermite-Hadamard inequality, H\"{o}lder inequality,
Power Mean inequality, Special means, The Midpoint formula, Lipschitzian
mapping.}
\subjclass[2000]{ 26A51, 26D10, 26D15}
\maketitle

\section{INTRODUCTION}

Let $f:I\subset \lbrack 0,\infty )\rightarrow

\mathbb{R}

$ be a differentiable mapping on $I^{\circ }$, the interior of the interval $
I$, such that $f^{\prime }\in L\left( \left[ a,b\right] \right) $ where $a$ $
,$ $b\in I$ with $a<b$ . If $\left\vert f^{\prime }\left( x\right)
\right\vert \leq M$ , then the following inequality holds (see \cite{ADDC}):

\begin{equation*}
\left\vert f(x)-\frac{1}{b-a}\int_{a}^{b}f(u)du\right\vert \leq \frac{M}{b-a}
\left[ \frac{\left( x-a\right) ^{2}+\left( b-x\right) ^{2}}{2}\right] .
\end{equation*}

This inequality is well known in the literature as the \textit{Ostrowski
inequality.}\textbf{\ }For some results which generalize, improve, and
extend the above inequality, see \cite{ADDC},\cite{BCDPS},\cite{CDR},\cite
{DS1} and \cite{DW}, the references therein.

In \cite{GT1}, G. Toader defined $m-$convexity, an intermediate between the
usual convexity and starshaped property, as the following:

\begin{definition}
\label{d.1.1} The function $f:[0,b]\rightarrow

\mathbb{R}

,$ $b>0$ , is said to be $m-$convex, where $m\in \lbrack 0,1]$, if we have
\begin{equation*}
f(tx+m(1-t)y)\leq tf(x)+m(1-t)f(y)
\end{equation*}
for all $x,y\in \left[ 0,b\right] $ and $t\in \left[ 0,1\right] $.
\end{definition}

Denote by $K_{m}(b)$ the set of the $m-$convex functions on $\left[ 0,b
\right] $ for which $f(0)\leq 0$.

\begin{definition}
\label{d.1.2} The function $f:[0,b]\rightarrow

\mathbb{R}

,$ $b>0$ is said to be starshaped if for every $x\in \left[ 0,b\right] $ and
$t\in \lbrack 0,1]$ we have:
\begin{equation*}
f(tx)\leq tf(x).
\end{equation*}
\end{definition}

For $m=1$, we recapture the concept of convex functions defined on $[0,b]$
and $m=0$ we get the concept of starshaped functions on $[0,b]$.

The following theorem contains the Hermite-Hadamard type integral inequality
(see \cite{DCK}).

\begin{theorem}
\label{t.1.1} Let $f:I\subset

\mathbb{R}

\rightarrow

\mathbb{R}

$ be an M-Lipschitzian mapping on $I$ and $a,b\in I$ with $a<b.$ Then we
have the inequality:
\end{theorem}

\begin{equation}
\left\vert f\left( \frac{a+b}{2}\right) -\frac{1}{b-a}\int_{a}^{b}f(x)dx
\right\vert \leq M\frac{(b-a)}{4}.  \label{h.1.1}
\end{equation}

In \cite{SOS} E. Set, M.E. \"{O}zdemir, M.Z. Sar\i kaya established the
following theorem.

\begin{theorem}
\label{t.1.2} Let $f:I^{\circ }\subset \left[ 0,b^{\ast }\right] \rightarrow

\mathbb{R}

,$ $b^{\ast }>0,$ be a differentiable mapping on $I^{\circ }$, $a,b\in $ $
I^{\circ }$ with $a<b.$ If $\left\vert f^{\prime }\right\vert ^{q}$ is $m-$
convex on $[a,b]$, $q>1$ and $m\in (0,1],$ then the following inequality
holds:
\end{theorem}

\begin{equation}
\left\vert f\left( \frac{a+b}{2}\right) -\frac{1}{b-a}\int_{a}^{b}f(x)dx
\right\vert \leq \left( b-a\right) \left( \frac{3^{1-\left( \frac{1}{q}
\right) }}{8}\right) \left( \left\vert f^{\prime }(a)\right\vert +m^{\frac{1
}{q}}\left\vert f^{\prime }(\frac{b}{m})\right\vert \right) .  \label{h.1.2}
\end{equation}
where $\frac{b}{m}<b^{\ast }.$

In \cite{UK} U. Kirmaci proved the following theorem.

\begin{theorem}
\label{t.1.3} Let $f:I^{\circ }\subset

\mathbb{R}

\rightarrow

\mathbb{R}

$ be a differentiable mapping on $I^{\circ }$, $a,b\in I^{\circ }$ with $a<b$
. If the mapping $\left\vert f^{\prime }\right\vert $ is convex on $[a,b]$,
then we have
\end{theorem}

\begin{equation}
\left\vert f\left( \frac{a+b}{2}\right) -\frac{1}{b-a}\int_{a}^{b}f(x)dx
\right\vert \leq \frac{b-a}{8}\left( \left\vert f^{\prime }(a)\right\vert
+\left\vert f^{\prime }(b)\right\vert \right) .  \label{h.1.3}
\end{equation}

In \cite{DG} S.S. Dragomir and G. Toader proved the following
Hermite-Hadamard type inequality for $m-$convex functions.

\begin{theorem}
\label{t.1.4} Let $f:[0,\infty )\rightarrow

\mathbb{R}

$ be an $m-$convex function with $m\in (0,1]$. If $\ 0\leq a<b<\infty $ and $
f\in L^{1}([a,b])$ then
\end{theorem}

\begin{equation}
\frac{1}{b-a}\int_{a}^{b}f(x)dx\leq \min \left\{ \frac{f(a)+mf(\frac{b}{m})}{
2},\frac{f(b)+mf(\frac{a}{m})}{2}\right\} .  \label{h.1.4}
\end{equation}

Some generalizations of this result can be found in \cite{BPR}.

In \cite{BOP} M.K. Bakula, M.E. \"{O}zdemir and J. Pe\v{c}ari\'{c} proved
the following theorems.

\begin{theorem}
\label{t.1.5} Let $I$ be an open real interval such that $[0,\infty )\subset
I$. Let $f:I\rightarrow

\mathbb{R}

$ be a differentiable function on $I$ such that $f^{\prime }\in L([a,b])$,
where $0\leq a<b<\infty $ . If $\left\vert f^{\prime }\right\vert ^{q}$ is $
m-$convex on $[a,b]$ for some fixed $m\in (0,1]$ and $q\in \lbrack 1,\infty
) $ , then
\end{theorem}

\begin{equation}
\left\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a}\int_{a}^{b}f(x)dx\right\vert
\leq \frac{b-a}{4}\left( \mu _{1}^{\frac{1}{q}}+\mu _{2}^{\frac{1}{q}
}\right) ,  \label{h.1.5}
\end{equation}
where

\begin{eqnarray*}
\mu _{1} &=&\min \left\{ \frac{\left\vert f^{\prime }(a)\right\vert
^{q}+m\left\vert f^{\prime }(\frac{a+b}{2m})\right\vert ^{q}}{2},\frac{
\left\vert f^{\prime }(\frac{a+b}{2})\right\vert ^{q}+m\left\vert f^{\prime
}(\frac{a}{m})\right\vert ^{q}}{2}\right\} , \\
\mu _{2} &=&\min \left\{ \frac{\left\vert f^{\prime }(b)\right\vert
^{q}+m\left\vert f^{\prime }(\frac{a+b}{2m})\right\vert ^{q}}{2},\frac{
\left\vert f^{\prime }(\frac{a+b}{2})\right\vert ^{q}+m\left\vert f^{\prime
}(\frac{b}{m})\right\vert ^{q}}{2}\right\} .
\end{eqnarray*}

\begin{theorem}
\label{t.1.6} Let $I$ be an open real interval such that $[0,\infty )\subset
I$. Let $f:I\rightarrow

\mathbb{R}

$ be a differentiable function on $I$ such that $f^{\prime }\in L([a,b])$,
where $0\leq a<b<\infty $ . If $\left\vert f^{\prime }\right\vert ^{q}$ is $
m-$convex on $[a,b]$ for some fixed $m\in (0,1]$ and $q\in \lbrack 1,\infty
) $ , then
\begin{eqnarray}
&&\left\vert f\left( \frac{a+b}{2}\right) -\frac{1}{b-a}\int_{a}^{b}f(x)dx
\right\vert  \label{h.1.6} \\
&\leq &\frac{b-a}{4}\min \left\{ \left( \frac{\left\vert f^{\prime
}(a)\right\vert ^{q}+m\left\vert f^{\prime }(\frac{b}{m})\right\vert ^{q}}{2}
\right) ^{\frac{1}{q}},\left( \frac{m\left\vert f^{\prime }(\frac{a}{m}
)\right\vert ^{q}+\left\vert f^{\prime }(b)\right\vert ^{q}}{2}\right) ^{
\frac{1}{q}}\right\} .  \notag
\end{eqnarray}
\end{theorem}

The main purpose of this paper is to establish new Ostrowski type
inequalities for functions whose derivatives in absolute value are $m-$
convex. Using these results we give some applications to special means of
positive real numbers and we obtain some error estimates for the midpoint
formula.

\section{THE RESULTS}

In \cite{AD}, in order to prove some inequalities related to Ostrowski
inequality, M. Alomari and M. Darus used the following lemma with the
constant $(b-a),$ but we changed it with the constant $(a-b)$ to obtain an
equality in Lemma \ref{l.2.1}.

\begin{lemma}
\label{l.2.1} Let $f:I\subset

\mathbb{R}

\rightarrow

\mathbb{R}

$ be a differentiable mapping on $I^{\circ }$ where $a,b\in I$ with $a<b.$
If $f^{\prime }\in L\left( [a,b]\right) $, then the following equality holds:
\end{lemma}

\begin{equation}
f(x)-\frac{1}{b-a}\int_{a}^{b}f(u)du=(a-b)\int_{0}^{1}p(t)f^{\prime
}(ta+(1-t)b)dt  \label{m.2.1}
\end{equation}
for each $t\in \left[ 0,1\right] $ , where

\begin{equation*}
p(t)=\left\{
\begin{array}{ccc}
t & , & t\in \left[ 0,\frac{b-x}{b-a}\right] \\
&  &  \\
t-1 & , & t\in \left( \frac{b-x}{b-a},1\right]
\end{array}
\right. ,
\end{equation*}
for all $x\in \left[ a,b\right] .$

\begin{theorem}
\label{t.2.1} Let $I$ be an open real interval such that $[0,\infty )\subset
I$. Let $f:I\rightarrow

\mathbb{R}

$ be a differentiable function on $I$ such that $f^{\prime }\in L([a,b])$,
where $0\leq a<b<\infty $ . If $\left\vert f^{\prime }\right\vert $ is $m-$
convex on $[a,b]$ for some fixed $m\in (0,1]$, then the following inequality
holds:
\end{theorem}

\begin{align}
& \left\vert f(x)-\frac{1}{b-a}\int_{a}^{b}f(u)du\right\vert  \label{m.2.2}
\\
\leq (b-a)& \min \left\{
\begin{array}{c}
\left[ \frac{1}{6}-\frac{1}{2}\left( \frac{b-x}{b-a}\right) ^{2}+\frac{2}{3}
\left( \frac{b-x}{b-a}\right) ^{3}\right] \left\vert f^{\prime
}(a)\right\vert \\
+m\left[ \frac{1}{2}\left( \frac{b-x}{b-a}\right) ^{2}-\frac{1}{3}\left(
\frac{b-x}{b-a}\right) ^{3}+\frac{1}{3}\left( \frac{x-a}{b-a}\right) ^{3}
\right] \left\vert f^{\prime }\left( \frac{b}{m}\right) \right\vert
\end{array}
,\right.  \notag \\
& \left.
\begin{array}{c}
\left[ \frac{1}{6}-\frac{1}{2}\left( \frac{b-x}{b-a}\right) ^{2}+\frac{2}{3}
\left( \frac{b-x}{b-a}\right) ^{3}\right] \left\vert f^{\prime
}(b)\right\vert \\
+m\left[ \frac{1}{2}\left( \frac{b-x}{b-a}\right) ^{2}-\frac{1}{3}\left(
\frac{b-x}{b-a}\right) ^{3}+\frac{1}{3}\left( \frac{x-a}{b-a}\right) ^{3}
\right] \left\vert f^{\prime }\left( \frac{a}{m}\right) \right\vert
\end{array}
\right\} .  \notag
\end{align}
for each $x\in \left[ a,b\right] .$

\begin{proof}
By Lemma \ref{l.2.1}, we have
\begin{eqnarray*}
&&\left\vert f(x)-\frac{1}{b-a}\int_{a}^{b}f(u)du\right\vert \\
&\leq &(b-a)\int_{0}^{\frac{b-x}{b-a}}t\left\vert f^{\prime
}(ta+(1-t)b)\right\vert dt \\
&&+(b-a)\int_{\frac{b-x}{b-a}}^{^{1}}(1-t)\left\vert f^{\prime
}(ta+(1-t)b)\right\vert dt
\end{eqnarray*}
Since $\left\vert f^{\prime }\right\vert $ is $m-$convex on $\left[ a,b
\right] $ we know that for any $t\in \left[ 0,1\right] $
\begin{eqnarray*}
\left\vert f^{\prime }(ta+(1-t)b)\right\vert &=&\left\vert f^{\prime
}(ta+m(1-t)\frac{b}{m})\right\vert \\
&\leq &t\left\vert f^{\prime }(a)\right\vert +m(1-t)\left\vert f^{\prime
}\left( \frac{b}{m}\right) \right\vert ,
\end{eqnarray*}
Hence
\begin{eqnarray*}
&&\left\vert f(x)-\frac{1}{b-a}\int_{a}^{b}f(u)du\right\vert \\
&\leq &(b-a)\int_{0}^{^{\frac{b-x}{b-a}}}t\left[ t\left\vert f^{\prime
}(a)\right\vert +m(1-t)\left\vert f^{\prime }\left( \frac{b}{m}\right)
\right\vert \right] dt \\
&&+(b-a)\int_{\frac{b-x}{b-a}}^{^{1}}(1-t)\left[ t\left\vert f^{\prime
}(a)\right\vert +m(1-t)\left\vert f^{\prime }\left( \frac{b}{m}\right)
\right\vert \right] dt \\
&=&(b-a)\left\{
\begin{array}{c}
\left[ \frac{1}{6}-\frac{1}{2}\left( \frac{b-x}{b-a}\right) ^{2}+\frac{2}{3}
\left( \frac{b-x}{b-a}\right) ^{3}\right] \left\vert f^{\prime
}(a)\right\vert \\
+m\left[ \frac{1}{2}\left( \frac{b-x}{b-a}\right) ^{2}-\frac{1}{3}\left(
\frac{b-x}{b-a}\right) ^{3}+\frac{1}{3}\left( \frac{x-a}{b-a}\right) ^{3}
\right] \left\vert f^{\prime }\left( \frac{b}{m}\right) \right\vert
\end{array}
\right\}
\end{eqnarray*}
where we use the facts that
\begin{eqnarray*}
&&\int_{0}^{^{\frac{b-x}{b-a}}}t\left[ t\left\vert f^{\prime }(a)\right\vert
+m(1-t)\left\vert f^{\prime }\left( \frac{b}{m}\right) \right\vert \right] dt
\\
&=&\frac{1}{3}\left( \frac{b-x}{b-a}\right) ^{3}\left\vert f^{\prime
}(a)\right\vert +m\left[ \frac{1}{2}\left( \frac{b-x}{b-a}\right) ^{2}-\frac{
1}{3}\left( \frac{b-x}{b-a}\right) ^{3}\right] \left\vert f^{\prime }\left(
\frac{b}{m}\right) \right\vert ,
\end{eqnarray*}
and
\begin{eqnarray*}
&&\int_{\frac{b-x}{b-a}}^{^{1}}(1-t)\left[ t\left\vert f^{\prime
}(a)\right\vert +m(1-t)\left\vert f^{\prime }\left( \frac{b}{m}\right)
\right\vert \right] dt \\
&=&\left[ \frac{1}{6}-\frac{1}{2}\left( \frac{b-x}{b-a}\right) ^{2}+\frac{1}{
3}\left( \frac{b-x}{b-a}\right) ^{3}\right] \left\vert f^{\prime
}(a)\right\vert +m\frac{1}{3}\left( \frac{x-a}{b-a}\right) ^{3}\left\vert
f^{\prime }\left( \frac{b}{m}\right) \right\vert ,
\end{eqnarray*}
and analogously
\begin{eqnarray*}
&&\left\vert f(x)-\frac{1}{b-a}\int_{a}^{b}f(u)du\right\vert \\
&\leq &(b-a)\left\{
\begin{array}{c}
\left[ \frac{1}{6}-\frac{1}{2}\left( \frac{b-x}{b-a}\right) ^{2}+\frac{2}{3}
\left( \frac{b-x}{b-a}\right) ^{3}\right] \left\vert f^{\prime
}(b)\right\vert \\
+m\left[ \frac{1}{2}\left( \frac{b-x}{b-a}\right) ^{2}-\frac{1}{3}\left(
\frac{b-x}{b-a}\right) ^{3}+\frac{1}{3}\left( \frac{x-a}{b-a}\right) ^{3}
\right] \left\vert f^{\prime }\left( \frac{a}{m}\right) \right\vert
\end{array}
\right\} .
\end{eqnarray*}
The proof is completed.
\end{proof}

\begin{remark}
\label{r.2.1} Suppose that all the assumptions of Theorem \ref{t.2.1} are
satisfied. If we choose $x=\frac{a+b}{2}$, then we have
\begin{eqnarray*}
&&\left\vert f\left( \frac{a+b}{2}\right) -\frac{1}{b-a}\int_{a}^{b}f(u)du
\right\vert \\
&\leq &\frac{b-a}{8}\min \left\{ \left\vert f^{\prime }(a)\right\vert
+m\left\vert f^{\prime }\left( \frac{b}{m}\right) \right\vert ,\left\vert
f^{\prime }(b)\right\vert +m\left\vert f^{\prime }\left( \frac{a}{m}\right)
\right\vert \right\}
\end{eqnarray*}
which is [(\ref{h.1.6}),q=1].
\end{remark}

\begin{remark}
\label{r.2.2} Suppose that all the assumptions of Theorem \ref{t.2.1} are
satisfied. Then
\end{remark}

$(A)$ If we choose $m=1$ and $x=\frac{a+b}{2}$, we obtain

\begin{equation*}
\left\vert f\left( \frac{a+b}{2}\right) -\frac{1}{b-a}\int_{a}^{b}f(u)du
\right\vert \leq \frac{b-a}{8}\left( \left\vert f^{\prime }(a)\right\vert
+\left\vert f^{\prime }(b)\right\vert \right) ,
\end{equation*}
which is (\ref{h.1.3}).

$(B)$ In $(A)$. Additionally, if we choose $\left\vert f^{\prime
}(x)\right\vert \leq M,$ $M>0$

\begin{equation*}
\left\vert f\left( \frac{a+b}{2}\right) -\frac{1}{b-a}\int_{a}^{b}f(u)du
\right\vert \leq M\frac{(b-a)}{4}
\end{equation*}
which is (\ref{h.1.1}).

\begin{theorem}
\label{t.2.2} Let $I$ be an open real interval such that $[0,\infty )\subset
I$. Let $f:I\rightarrow

\mathbb{R}

$ be a differentiable function on $I$ such that $f^{\prime }\in L([a,b])$,
where $0\leq a<b<\infty $ . If $\left\vert f^{\prime }\right\vert ^\frac{p}{p-1}$ is $
m-$convex on $[a,b]$ for some fixed $m\in (0,1]$ and $p>1,$ $\frac{1}{p}+
\frac{1}{q}=1$ , then the following inequality holds:
\begin{eqnarray}
&&\left\vert f(x)-\frac{1}{b-a}\int_{a}^{b}f(u)du\right\vert  \label{m.2.3}
\\
\leq \frac{1}{\left( p+1\right) ^{\frac{1}{p}}} &&\left\{
\begin{array}{c}
\frac{(b-x)^{2}}{b-a}\left[ \min \left\{ \frac{\left\vert f^{\prime
}(b)\right\vert ^{q}+m\left\vert f^{\prime }\left( \frac{x}{m}\right)
\right\vert ^{q}}{2},\frac{\left\vert f^{\prime }(x)\right\vert
^{q}+m\left\vert f^{\prime }\left( \frac{b}{m}\right) \right\vert ^{q}}{2}
\right\} \right] ^{\frac{1}{q}} \\
+\frac{(x-a)^{2}}{b-a}\left[ \min \left\{ \frac{\left\vert f^{\prime
}(a)\right\vert ^{q}+m\left\vert f^{\prime }\left( \frac{x}{m}\right)
\right\vert ^{q}}{2},\frac{\left\vert f^{\prime }(x)\right\vert
^{q}+m\left\vert f^{\prime }\left( \frac{a}{m}\right) \right\vert ^{q}}{2}
\right\} \right] ^{\frac{1}{q}}
\end{array}
\right\}  \notag
\end{eqnarray}
for each $x\in \left[ a,b\right] .$
\end{theorem}

\begin{proof}
From Lemma \ref{l.2.1} and using the H\"{o}lder inequality, we have
\begin{eqnarray*}
&&\left\vert f(x)-\frac{1}{b-a}\int_{a}^{b}f(u)du\right\vert \\
&\leq &(b-a)\left( \int_{0}^{\frac{b-x}{b-a}}t^{p}dt\right) ^{\frac{1}{p}
}\left( \int_{0}^{\frac{b-x}{b-a}}\left\vert f^{\prime }\left(
ta+(1-t)b\right) \right\vert ^{q}dt\right) ^{\frac{1}{q}} \\
&&+(b-a)\left( \int_{\frac{b-x}{b-a}}^{1}(1-t)^{p}dt\right) ^{\frac{1}{p}
}\left( \int_{\frac{b-x}{b-a}}^{1}\left\vert f^{\prime }\left(
ta+(1-t)b\right) \right\vert ^{q}dt\right) ^{\frac{1}{q}} \\
&\leq &(b-a)\left( \frac{b-x}{b-a}\right) ^{\frac{p+1}{p}}\left( \frac{1}{p+1
}\right) ^{\frac{1}{p}}\left( \frac{b-x}{b-a}\right) ^{\frac{1}{q}} \\
&&\times \left( \min \left\{ \frac{\left\vert f^{\prime }(b)\right\vert
^{q}+m\left\vert f^{\prime }\left( \frac{x}{m}\right) \right\vert ^{q}}{2},
\frac{\left\vert f^{\prime }(x)\right\vert ^{q}+m\left\vert f^{\prime
}\left( \frac{b}{m}\right) \right\vert ^{q}}{2}\right\} \right) ^{\frac{1}{q}
} \\
&&+(b-a)\left( \frac{x-a}{b-a}\right) ^{\frac{p+1}{p}}\left( \frac{1}{p+1}
\right) ^{\frac{1}{p}}\left( \frac{x-a}{b-a}\right) ^{\frac{1}{q}} \\
&&\times \left( \min \left\{ \frac{\left\vert f^{\prime }(a)\right\vert
^{q}+m\left\vert f^{\prime }\left( \frac{x}{m}\right) \right\vert ^{q}}{2},
\frac{\left\vert f^{\prime }(x)\right\vert ^{q}+m\left\vert f^{\prime
}\left( \frac{a}{m}\right) \right\vert ^{q}}{2}\right\} \right) ^{\frac{1}{q}
} \\
&=&\frac{1}{(p+1)^{\frac{1}{p}}}\frac{1}{b-a}\left\{
\begin{array}{c}
(b-x)^{2}\left[ \min \left\{ \frac{\left\vert f^{\prime }(b)\right\vert
^{q}+m\left\vert f^{\prime }\left( \frac{x}{m}\right) \right\vert ^{q}}{2},
\frac{\left\vert f^{\prime }(x)\right\vert ^{q}+m\left\vert f^{\prime
}\left( \frac{b}{m}\right) \right\vert ^{q}}{2}\right\} \right] ^{\frac{1}{q}
} \\
+(x-a)^{2}\left[ \min \left\{ \frac{\left\vert f^{\prime }(a)\right\vert
^{q}+m\left\vert f^{\prime }\left( \frac{x}{m}\right) \right\vert ^{q}}{2},
\frac{\left\vert f^{\prime }(x)\right\vert ^{q}+m\left\vert f^{\prime
}\left( \frac{a}{m}\right) \right\vert ^{q}}{2}\right\} \right] ^{\frac{1}{q}
}
\end{array}
\right\}
\end{eqnarray*}
where we use the facts that
\begin{eqnarray*}
\int_{0}^{\frac{b-x}{b-a}}t^{p}dt &=&\left( \frac{b-x}{b-a}\right) ^{p+1}
\frac{1}{p+1}, \\
\int_{\frac{b-x}{b-a}}^{1}(1-t)^{p}dt &=&\left( \frac{x-a}{b-a}\right) ^{p+1}
\frac{1}{p+1},
\end{eqnarray*}
and by Theorem \ref{t.1.4} we get
\begin{eqnarray*}
&&\frac{b-a}{b-x}\int_{0}^{\frac{b-x}{b-a}}\left\vert f^{\prime
}(ta+(1-t)b)\right\vert ^{q}dt \\
&\leq &\min \left\{ \frac{\left\vert f^{\prime }(b)\right\vert
^{q}+m\left\vert f^{\prime }\left( \frac{x}{m}\right) \right\vert ^{q}}{2},
\frac{\left\vert f^{\prime }(x)\right\vert ^{q}+m\left\vert f^{\prime
}\left( \frac{b}{m}\right) \right\vert ^{q}}{2}\right\} , \\
&&\frac{b-a}{x-a}\int_{\frac{b-x}{b-a}}^{1}\left\vert f^{\prime
}(ta+(1-t)b)\right\vert ^{q}dt \\
&\leq &\min \left\{ \frac{\left\vert f^{\prime }(a)\right\vert
^{q}+m\left\vert f^{\prime }\left( \frac{x}{m}\right) \right\vert ^{q}}{2},
\frac{\left\vert f^{\prime }(x)\right\vert ^{q}+m\left\vert f^{\prime
}\left( \frac{a}{m}\right) \right\vert ^{q}}{2}\right\} .
\end{eqnarray*}
The proof is completed.
\end{proof}

\begin{corollary}
\label{c.2.1} Suppose that all the assumptions of Theorem \ref{t.2.2} are
satisfied, if we choose $\left\vert f^{\prime }(x)\right\vert \leq M$, $M>0$
, then we have
\begin{eqnarray*}
&&\left\vert f(x)-\frac{1}{b-a}\int_{a}^{b}f(u)du\right\vert \\
&\leq &\left( \frac{1}{\left( p+1\right) ^{\frac{1}{p}}}\right) \left( \frac{
1+m}{2}\right) ^{\frac{1}{q}}M\left[ \frac{(b-x)^{2}+(x-a)^{2}}{b-a}\right] .
\end{eqnarray*}
\end{corollary}

\begin{corollary}
\label{c.2.2} Suppose that all the assumptions of Theorem \ref{t.2.2} are
satisfied, if we choose $x=\frac{a+b}{2}$ and $\frac{1}{2}<\left(
\frac{1}{p+1}\right) ^{\frac{1}{p}}<1$, then we have
\end{corollary}

\begin{equation*}
\left\vert f\left( \frac{a+b}{2}\right) -\frac{1}{b-a}\int_{a}^{b}f(u)du
\right\vert \leq \frac{b-a}{4}\left( \mu _{1}^{\frac{1}{q}}+\mu _{2}^{\frac{1
}{q}}\right) ,
\end{equation*}
\textit{where}

\begin{eqnarray*}
\mu _{1} &=&\min \left\{ \frac{\left\vert f^{\prime }(b)\right\vert
^{q}+m\left\vert f^{\prime }(\frac{a+b}{2m})\right\vert ^{q}}{2},\frac{
\left\vert f^{\prime }(\frac{a+b}{2})\right\vert ^{q}+m\left\vert f^{\prime
}\left( \frac{b}{m}\right) \right\vert ^{q}}{2}\right\} , \\
\mu _{2} &=&\min \left\{ \frac{\left\vert f^{\prime }(a)\right\vert
^{q}+m\left\vert f^{\prime }(\frac{a+b}{2m})\right\vert ^{q}}{2},\frac{
\left\vert f^{\prime }(\frac{a+b}{2})\right\vert ^{q}+m\left\vert f^{\prime
}(\frac{a}{m})\right\vert ^{q}}{2}\right\} .
\end{eqnarray*}

\begin{remark}
\label{r.2.3} Corollary \ref{c.2.2} is similar to (\ref{h.1.5}) inequality,
but for the left-hand side of Hermite-Hadamard inequality.
\end{remark}

\begin{remark}
\label{r.2.4} Suppose that all the assumptions of Theorem \ref{t.2.2} are
satisfied. Then in Corollary \ref{c.2.2}
\end{remark}

$\left( D\right) $ $\left\vert f^{\prime }\right\vert $ is increasing and $
m=1$ then we have

\begin{equation*}
\left\vert f\left( \frac{a+b}{2}\right) -\frac{1}{b-a}\int_{a}^{b}f(u)du
\right\vert \leq \frac{b-a}{2}\left\vert f^{\prime }(b)\right\vert ,
\end{equation*}

$(E)$ $\left\vert f^{\prime }\right\vert $ is decreasing and $m=1$ then we
have

\begin{equation*}
\left\vert f\left( \frac{a+b}{2}\right) -\frac{1}{b-a}\int_{a}^{b}f(u)du
\right\vert \leq \frac{b-a}{2}\left\vert f^{\prime }(a)\right\vert ,
\end{equation*}

$\left( F\right) $ $\left\vert f^{\prime }(b)\right\vert =\left\vert
f^{\prime }(a)\right\vert =\left\vert f^{\prime }\left( \frac{a+b}{2}\right)
\right\vert $ and $m=1$ then we have

\begin{equation*}
\left\vert f\left( \frac{a+b}{2}\right) -\frac{1}{b-a}\int_{a}^{b}f(u)du
\right\vert \leq \frac{b-a}{2}\left\vert f^{\prime }\left( \frac{a+b}{2}
\right) \right\vert .
\end{equation*}

\begin{theorem}
\label{t.2.3} Let $I$ be an open real interval such that $[0,\infty )\subset
I$. Let $f:I\rightarrow

\mathbb{R}

$ be a differentiable function on $I$ such that $f^{\prime }\in L([a,b])$,
where $0\leq a<b<\infty $. If $\left\vert f^{\prime }\right\vert ^{q}$ is $
m- $convex on $[a,b]$ for some fixed $m\in (0,1]$ and $q\in \lbrack 1,\infty
)$, $x\in \left[ a,b\right] $, then the following inequality holds:
\end{theorem}

\begin{eqnarray}
&&\left\vert f(x)-\frac{1}{b-a}\int_{a}^{b}f(u)du\right\vert  \label{m.2.4}
\\
&\leq &(b-a)\left( \frac{1}{2}\right) ^{1-\frac{1}{q}}  \notag \\
&&\left\{
\begin{array}{c}
\left( \frac{b-x}{b-a}\right) ^{2\left( 1-\frac{1}{q}\right) }\left[ \frac{1
}{3}\left( \frac{b-x}{b-a}\right) ^{3}\left\vert f^{\prime }(a)\right\vert
^{q}+m\frac{(b-x)^{2}(b-3a+2x)}{6(b-a)^{3}}\left\vert f^{\prime }\left(
\frac{b}{m}\right) \right\vert ^{q}\right] ^{\frac{1}{q}} \\
+\left( \frac{x-a}{b-a}\right) ^{2\left( 1-\frac{1}{q}\right) }\left[ \left(
\frac{1}{6}+\frac{(b-x)^{2}(3a-b-2x)}{6(b-a)^{3}}\right) \left\vert
f^{\prime }(a)\right\vert ^{q}+m\frac{1}{3}\left( \frac{x-a}{b-a}\right)
^{3}\left\vert f^{\prime }\left( \frac{b}{m}\right) \right\vert ^{q}\right]
^{\frac{1}{q}}
\end{array}
\right\}  \notag
\end{eqnarray}
\textit{for each }$x\in \left[ a,b\right] .$

\begin{proof}
By Lemma \ref{l.2.1} and using the well known power mean inequality we have
\begin{eqnarray*}
&&\left\vert f(x)-\frac{1}{b-a}\int_{a}^{b}f(u)du\right\vert \\
&\leq &(b-a)\int_{0}^{^{\frac{b-x}{b-a}}}t\left\vert f^{\prime
}(ta+(1-t)b)\right\vert dt \\
&&+(b-a)\int_{\frac{b-x}{b-a}}^{^{1}}(1-t)\left\vert f^{\prime
}(ta+(1-t)b)\right\vert dt \\
&\leq &(b-a)\left( \int_{0}^{^{\frac{b-x}{b-a}}}tdt\right) ^{1-\frac{1}{q}
}\left( \int_{0}^{^{\frac{b-x}{b-a}}}t\left\vert f^{\prime
}(ta+(1-t)b)\right\vert ^{q}dt\right) ^{\frac{1}{q}} \\
&&+(b-a)\left( \int_{\frac{b-x}{b-a}}^{^{1}}(1-t)dt\right) ^{1-\frac{1}{q}
}\left( \int_{\frac{b-x}{b-a}}^{^{1}}(1-t)\left\vert f^{\prime
}(ta+(1-t)b)\right\vert ^{q}dt\right) ^{\frac{1}{q}} \\
&\leq &(b-a)\left( \frac{1}{2}\right) ^{1-\frac{1}{q}} \\
&&\left\{
\begin{array}{c}
\left( \frac{b-x}{b-a}\right) ^{2\left( 1-\frac{1}{q}\right) }\left[ \frac{1
}{3}\left( \frac{b-x}{b-a}\right) ^{3}\left\vert f^{\prime }(a)\right\vert
^{q}+m\frac{(b-x)^{2}(b-3a+2x)}{6(b-a)^{3}}\left\vert f^{\prime }\left(
\frac{b}{m}\right) \right\vert ^{q}\right] ^{\frac{1}{q}} \\
+\left( \frac{x-a}{b-a}\right) ^{2\left( 1-\frac{1}{q}\right) }\left[ \left(
\frac{1}{6}+\frac{(b-x)^{2}(3a-b-2x)}{6(b-a)^{3}}\right) \left\vert
f^{\prime }(a)\right\vert ^{q}+m\frac{1}{3}\left( \frac{x-a}{b-a}\right)
^{3}\left\vert f^{\prime }\left( \frac{b}{m}\right) \right\vert ^{q}\right]
^{\frac{1}{q}}
\end{array}
\right\}
\end{eqnarray*}
where we use the facts that
\begin{equation*}
\int_{0}^{\frac{b-x}{b-a}}tdt=\frac{1}{2}\left( \frac{b-x}{b-a}\right) ^{2},
\end{equation*}

\begin{eqnarray*}
&&\int_{0}^{\frac{b-x}{b-a}}t\left\vert f^{\prime }(ta+(1-t)b)\right\vert
^{q}dt \\
&\leq &\frac{1}{3}\left( \frac{b-x}{b-a}\right) ^{3}\left\vert f^{\prime
}(a)\right\vert ^{q}+m\frac{(b-x)^{2}(b-3a+2x)}{6(b-a)^{3}}\left\vert
f^{\prime }\left( \frac{b}{m}\right) \right\vert ^{q},
\end{eqnarray*}

\begin{equation*}
\int_{\frac{b-x}{b-a}}^{1}(1-t)dt=\frac{1}{2}\left( \frac{x-a}{b-a}\right)
^{2},
\end{equation*}

\begin{eqnarray*}
&&\int_{\frac{b-x}{b-a}}^{1}(1-t)\left\vert f^{\prime
}(ta+(1-t)b)\right\vert ^{q}dt \\
&\leq &\left[ \frac{1}{6}+\frac{(b-x)^{2}(3a-2x-b)}{6(b-a)^{3}}\right]
\left\vert f^{\prime }(a)\right\vert ^{q}+m\frac{1}{3}\left( \frac{x-a}{b-a}
\right) ^{3}\left\vert f^{\prime }\left( \frac{b}{m}\right) \right\vert ^{q}.
\end{eqnarray*}
The proof is completed.
\end{proof}

\begin{remark}
\label{r.2.5} Suppose that all the assumptions of Theorem \ref{t.2.3} are
satisfied. If we choose $x=\frac{a+b}{2},$ we obtain
\end{remark}

\begin{equation*}
\left\vert f\left( \frac{a+b}{2}\right) -\frac{1}{b-a}\int_{a}^{b}f(u)du
\right\vert \leq (b-a)\left( \frac{3^{1-\frac{1}{q}}}{8}\right) \left(
\left\vert f^{\prime }(a)\right\vert +m^{\frac{1}{q}}\left\vert f^{\prime
}\left( \frac{b}{m}\right) \right\vert \right)
\end{equation*}
which is (\ref{h.1.2}).

\section{APPLICATIONS TO SPECIAL MEANS}

Let us recall the following means for two positive numbers.

$\left( AM\right) $ \textit{The Arithmetic mean}

\begin{equation*}
A=A(a,b)=\frac{a+b}{2};\text{ }a,b>0,
\end{equation*}

$\left( p-LM\right) $ \textit{The p-Logarithmic mean}
\begin{equation*}
L_{p}=L_{p}(a,b)=\left\{
\begin{array}{cc}
a & if\text{ }a=b \\
\left[ \frac{b^{p+1}-a^{p+1}}{(p+1)(b-a)}\right] ^{\frac{1}{p}} & if\text{ }
a\neq b
\end{array}
\right. ;\text{ }a,b>0,
\end{equation*}

$\left( IM\right) $ \textit{The Identric mean }
\begin{equation*}
I=I(a,b)=\left\{
\begin{array}{cc}
a & if\text{ }a=b \\
\frac{1}{e}\left( \frac{b^{b}}{^{a^{a}}}\right) ^{\frac{1}{b-a}} & if\text{ }
a\neq b
\end{array}
\right. ;\text{ }a,b>0.
\end{equation*}
The following propositions hold:

\begin{proposition}
\label{p.3.1} Let $a,b\in \left[ 0,\infty \right) $, and $a<b$, $n\geq 2$
with $m\in \left( 0,1\right] $. Then we have
\end{proposition}

\begin{equation*}
\left\vert A^{n}(a,b)-L_{n}^{n}(a,b)\right\vert \leq n\frac{b-a}{8}\min
\left\{ 2A\left( a^{n-1},m\left( \frac{b}{m}\right) ^{^{n-1}}\right)
,2A\left( \left( b\right) ^{n-1},m\left( \frac{a}{m}\right) ^{^{n-1}}\right)
\right\} .
\end{equation*}

\begin{proof}
The proof follows by Remark \ref{r.2.1} on choosing $f:\left[ 0,\infty
\right) \rightarrow \left[ 0,\infty \right) $, $f(x)=x^{n},$ $n\in

\mathbb{Z}

,n\geq 2$ which is $m-$convex on $\left[ 0,\infty \right) .$
\end{proof}

\begin{proposition}
\label{p.3.2} Let $a,b\in \left[ 0,\infty \right) $, and $a<b$, with $m\in
\left( 0,1\right] $. Then we have
\end{proposition}

\begin{equation*}
\left\vert \ln \frac{I(a+1,b+1)}{A(a,b)+1}\right\vert \leq \frac{b-a}{4}
\left( \eta _{1}^{\frac{1}{q}}+\eta _{2}^{\frac{1}{q}}\right) ,
\end{equation*}
where

\begin{eqnarray*}
\eta _{1}^{\frac{1}{q}} &=&\min \left\{ \frac{\left( \frac{1}{b+1}\right)
^{q}+m\left( \frac{2m}{a+b+2m}\right) ^{q}}{2},\frac{\left( \frac{2}{a+b+2}
\right) ^{q}+m\left( \frac{m}{b+m}\right) ^{q}}{2}\right\} , \\
\eta _{2}^{\frac{1}{q}} &=&\min \left\{ \frac{\left( \frac{1}{a+1}\right)
^{q}+m\left( \frac{2m}{a+b+2m}\right) ^{q}}{2},\frac{\left( \frac{2}{a+b+2}
\right) ^{q}+m\left( \frac{m}{a+m}\right) ^{q}}{2}\right\} .
\end{eqnarray*}

\begin{proof}
The proof follows by Corollary \ref{c.2.2} on choosing $f:\left[ 0,\infty
\right) \rightarrow (-\infty ,0]$, $f(x)=-\ln (x+1)$ which is $m-$convex on $
[0,\infty )$, $p>1.$
\end{proof}

\section{APPLICATIONS TO THE MIDPOINT FORMULA FOR $1-CONVEX$ FUNCTIONS}

Let $d$ be a division $a=x_{0}<x_{1}<...<x_{n-1}<x_{n}=b$ of the interval $
\left[ a,b\right] $ and consider the quadrature formula

\begin{equation}
\int_{a}^{b}f(x)dx=M(f,d)+E(f,d),  \label{h.4.1}
\end{equation}
where

\begin{equation*}
M(f,d)=\sum_{i=1}^{n-1}\left( x_{i+1}-x_{i}\right) f\left( \frac{
x_{i+1}+x_{i}}{2}\right)
\end{equation*}
is the midpoint formula and $E(f,d)$ denotes the associated approximation
error (see \cite{CPJP}).

Here, we obtain some error estimates for the midpoint formula.

\begin{proposition}
\label{p.4.1} Let $I$ be an open real interval such that $[0,\infty )\subset
I$. Let $f:I\rightarrow

\mathbb{R}

$ be a differentiable function on $I$ such that $f^{\prime }\in L([a,b])$, where $0\leq a<b<\infty $ . If $\left\vert f^{\prime }\right\vert ^{q}$ is $
1-$convex on $[a,b]$ for some fixed $m\in (0,1]$ and $p>1,$ $\frac{1}{p}+
\frac{1}{q}=1$, then in (\ref{h.4.1}), for every division $d$ of $\left[ a,b
\right] $, the midpoint error satisfies
\end{proposition}

\begin{equation*}
\left\vert E(f,d)\right\vert \leq \frac{1}{4}\sum_{i=0}^{n-1}\left(
x_{i+1}-x_{i}\right) ^{2}\left( \mu _{1}^{\frac{1}{q}}+\mu _{2}^{\frac{1}{q}
}\right) ,
\end{equation*}
where

\begin{eqnarray*}
\mu _{1} &=&\min \left\{ \frac{\left\vert f^{\prime }(x_{i})\right\vert
^{q}+\left\vert f^{\prime }(\frac{x_{i}+x_{i+1}}{2})\right\vert ^{q}}{2},
\frac{\left\vert f^{\prime }(\frac{x_{i}+x_{i+1}}{2})\right\vert
^{q}+\left\vert f^{\prime }(x_{i})\right\vert ^{q}}{2}\right\} =\frac{
\left\vert f^{\prime }(\frac{x_{i}+x_{i+1}}{2})\right\vert ^{q}+\left\vert
f^{\prime }(x_{i})\right\vert ^{q}}{2}, \\
\mu _{2} &=&\min \left\{ \frac{\left\vert f^{\prime }(x_{i+1})\right\vert
^{q}+\left\vert f^{\prime }(\frac{x_{i}+x_{i+1}}{2})\right\vert ^{q}}{2},
\frac{\left\vert f^{\prime }(\frac{x_{i}+x_{i+1}}{2})\right\vert
^{q}+\left\vert f^{\prime }(x_{i+1})\right\vert ^{q}}{2}\right\} =\frac{
\left\vert f^{\prime }(\frac{x_{i}+x_{i+1}}{2})\right\vert ^{q}+\left\vert
f^{\prime }(x_{i+1})\right\vert ^{q}}{2}.
\end{eqnarray*}

\begin{proof}
On applying Corollary \ref{c.2.2} with $m=1$  on the subinterval $\left[
x_{i},x_{i+1}\right] $ $(i=0,1,2,...,n-1)$ of the division, we have

\begin{equation*}
\left\vert f\left( \frac{x_{i+1}+x_{i}}{2}\right) -\frac{1}{x_{i+1}-x_{i}}
\int_{x_{i}}^{x_{i+1}}f(x)dx\right\vert \leq \frac{x_{i+1}-x_{i}}{4}\left(
\mu _{1}^{\frac{1}{q}}+\mu _{2}^{\frac{1}{q}}\right) ,
\end{equation*}
where
\begin{eqnarray*}
\mu _{1} &=&\frac{\left\vert f^{\prime }(\frac{x_{i}+x_{i+1}}{2})\right\vert
^{q}+\left\vert f^{\prime }(x_{i})\right\vert ^{q}}{2} \\
\mu _{2} &=&\frac{\left\vert f^{\prime }(\frac{x_{i}+x_{i+1}}{2})\right\vert
^{q}+\left\vert f^{\prime }(x_{i+1})\right\vert ^{q}}{2}.
\end{eqnarray*}
Hence, in (\ref{h.4.1}) we have

\begin{eqnarray*}
&&\left\vert \int_{a}^{b}f(x)dx-M(f,d)\right\vert \\
&=&\left\vert \sum_{i=0}^{n-1}\left[ \int_{x_{i}}^{x_{i+1}}f(x)dx-\left(
x_{i+1}-x_{i}\right) f\left( \frac{x_{i+1}-x_{i}}{2}\right) \right]
\right\vert \\
&\leq &\sum_{i=0}^{n-1}\left\vert \int_{x_{i}}^{x_{i+1}}f(x)dx-\left(
x_{i+1}-x_{i}\right) f\left( \frac{x_{i+1}-x_{i}}{2}\right) \right\vert \\
&\leq &\frac{1}{4}\sum_{i=0}^{n-1}\left( x_{i+1}-x_{i}\right) ^{2}\left( \mu
_{1}^{\frac{1}{q}}+\mu _{2}^{\frac{1}{q}}\right) ,
\end{eqnarray*}
which completes the proof.
\end{proof}

\begin{proposition}
\label{p.4.2} Let $I$ be an open real interval such that $[0,\infty )\subset
I$. Let $f:I\rightarrow

\mathbb{R}

$ be a differentiable function on $I$ such that $f^{\prime }\in L([a,b])$,
where $0\leq a<b<\infty $. If $\left\vert f^{\prime }\right\vert ^{q}$ is $
1- $convex on $[a,b]$ for some fixed $m\in (0,1]$ and $q\in \lbrack 1,\infty
)$, $x\in \left[ a,b\right] $, then in (\ref{h.4.1}), for every division $d$
of $\left[ a,b\right] $, the midpoint error satisfies
\end{proposition}

\begin{equation*}
\left\vert E(f,d)\right\vert \leq \left( \frac{3^{1-\frac{1}{q}}}{8}\right)
\sum_{i=0}^{n-1}\left( x_{i+1}-x_{i}\right) ^{2}\left( \left\vert f^{\prime
}(x_{i})\right\vert +\left\vert f^{\prime }\left( x_{i+1}\right) \right\vert
\right) .
\end{equation*}

\begin{proof}
The proof is similar to that of Proposition \ref{p.4.1} and using Remark \ref
{r.2.5} with $m=1$.
\end{proof}

\bigskip


text:
\section{Fundamental Limits for Linear Models: Proofs for Section \ref{sec:linear}}

\subsection{Auxiliary Results}
\begin{lem}[Boundedness of $\tau$ in \eqref{eq:eq_main}]\label{lem:tau_bound_lin}
Let $\Lm(\cdot)$ be a non-linear, convex and twice differentiable function, $\la>0$ and $\delta>0$ and the pair $(\alpha,\tau)$ be a solution to \eqref{eq:eq_main0} where $\alpha>0$. Then,
$
0 < \tau < \frac{1}{\la\delta}.
$
\end{lem}
\begin{proof}
Using Stein's lemma (aka Gaussian integration by parts) we find that
\begin{align*}
\E\Big[ G\cdot \envdx{\Lm}{\al\,G+Z}{\tau}  \Big]&=\alpha\,\E\Big[ \envddx{\Lm}{\al\,G+Z}{\tau}  \Big]
\end{align*}
Therefore the equation in the LHS in \eqref{eq:eq_main0} is equivalent to
\begin{align}\label{eq:steinthird}
\tau\delta\,\E\Big[ \envddx{\Lm}{\al\,G+Z}{\tau}  \Big] = 1-\la\tau\delta.
\end{align}
Next we prove that under the assumptions of the lemma, $\E\Big[ \envddx{\Lm}{\ourx}{\tau}  \Big]$ is positive. First using properties of Morea-envelopes in \eqref{eq:secondderivative_mor}, we have
\begin{align}
\E\Big[ &\envddx{\Lm}{\al\,G+Z}{\tau}  \Big] = \E\left[ \frac{\Lm''(\prox{\Lm}{\al\,G+Z}{\tau})}{1+\tau \Lm''(\prox{\Lm}{\alpha G + Z}{\tau})}\right]\ge0.\label{eq:mor_der_thirdeq}
\end{align}
In particular, we see that equality is achieved in \eqref{eq:mor_der_thirdeq} is achieved if and only if
\begin{align*}
\forall x\in \R: \quad \envddx{\Lm}{x}{\tau} =0.
\end{align*}
Or equivalently,
\begin{align}\label{eq:mor_lin}
\exists\,c_{_1},c_{_2} \in \R : \text{s.t.} \;\; \forall x\in \R : \env{\Lm}{x}{\tau} = c_{_1}x+c_{_2}.
\end{align}
Finally, using Proposition \ref{propo:inverse} to ``invert" the Moreau envelope function, we find that the loss function $\Lm(\cdot)$ satisfying \eqref{eq:mor_lin} is such that
\begin{align*}
\forall x\in\R:\quad\Lm(x) = -\env{-c_{_1}I-c_{_2}}{x}{\tau}= c_{_1}x + \frac{\tau c_{_1}^2}{2}+c_{_2},
\end{align*}
where $I(\cdot)$ is the identity function i.e. $I(t) = t,$  $\forall t\in \R$. But according to the assumptions of the lemma, $\Lm$ is a \emph{non-linear} convex function. Thus, it must hold that $\E\left[ \envddx{\Lm}{\al\,G+Z}{\tau}  \right]>0$. Using this and the assumptions on $\la$ and $\delta$, the advertised claim follows directly from \eqref{eq:steinthird}.
\end{proof}

\subsection{Proof of Theorem \ref{thm:lowerbound_reg}}\label{sec:proof_bin_lowerbound}

Fix a convex loss function $\Lm$ and regularization parameter $\la\geq 0$. Let $(\alpha>0,\tau>0)$ be the unique solution to
\begin{subequations}\label{eq:eq_main}
\begin{align}
 \delta \tau^2\cdot \E \Big[\Big(\envdx{\Lm}{\al\,G+Z}{\tau}\Big)^2\,\Big]&=\alpha^2- \la^2\delta^2 \tau^2, \label{eq:eq_alpha}\\
 \delta \tau \cdot\E\Big[G\cdot\envdx{\Lm}{\al\,G+Z}{\tau}\Big]&=\alpha\,(1-\la\delta\tau) \label{eq:eq_tau}.
\end{align}
\end{subequations}
For convenience, let us define the function $\Psi: \R_{\ge0} \times [0,1)\rightarrow \R$:
\bea\label{eq:psi_reg1}
\quad\Psi(a,x):=\frac{(a^2-x^2\,\delta^2)\,\Ic(V_a)}{(1-x\,\delta)^2}.
\eea
Then,  $\alpha_{\star}>0$ as in \eqref{eq:alphaopt_thm} is equivalently expressed as
\begin{align}\label{eq:psi_reg}
\alpha_{\star}:= \min_{\substack{0\le x<1/\delta}} \left\{a \geq 0: \; \Psi(a,x)= \frac{1}{\delta} \right\}.
\end{align}

Before everything, let us show that $\alpha_\star$ is well-defined, i.e., that the feasible set of the minimization in \eqref{eq:psi_reg} is non-empty for all $\delta>0$ and random variables $Z$ {satisfying Assumption \ref{ass:noise}}.  Specifically, we will show that there exists $a\geq 0$ such that $\Psi\left(a,\frac{a}{(1+a) \delta}\right)=1/\delta$. It suffices to prove that the range of the function $\Psit(a):=\Psi\left(a,\frac{a}{(1+a)\delta}\right)$ is $(0,\infty)$. Clearly, the function $\Psit$ is continuous in $\R_{\ge0}$. Moreover, it can be checked that $\Psit(a)=(a^2+2a)\Psi_0(a)$ where $\Psi_0(a):=a^2\Ic(V_a)$. By Lemma \ref{lem:Fisher_lim},  $\lim_{a\rightarrow 0}\Psi_0(a) = 0$  and $\lim_{a\rightarrow +\infty}\Psi_0(a) = 1$.  Hence, we find that $\lim_{a\rightarrow0}\Psit(a) = 0$ and $\lim_{a\rightarrow+\infty}\Psit(a) = +\infty$, as desired.

We are now ready to prove the main claim of the theorem, i.e.,
\bea\label{eq:desired21}
\alpha \geq \alpha_\star.
\eea
Denote by $\phi_\alpha$ the density of the Gaussian random variable $\alpha G$. We start with the following calculation:
\begin{align}
\E&\bigg[ G\cdot \envdx{\Lm}{V_\alpha }{\tau}  \bigg]  = -\alpha\iint\envdx{\Lm}{ u+z}{\tau}  \phi^\prime_{\alpha}(u)p_{Z}(z)\mathrm{d}u\mathrm{d}z  \nn\\
&= -\alpha\iint\envdx{\Lm}{ v}{\tau}  \phi^\prime_{\alpha}(u)p_{Z}(v-u)\mathrm{d}u\mathrm{d}v  \nn \\
&=  -\alpha\int\envdx{\Lm}{ v}{\tau}  p^\prime_{_{V}}(v)\mathrm{d}v =  -\alpha\,\E\Big[\envdx{\Lm}{V _\alpha}{\tau}\cdot\ksi_{{V_{_{\alpha}}}}(V_\alpha)\Big], \label{eq:equivalence}
\end{align}
where for a random variable $V$, we denote its score function with $\ksi_V(v) := p'_V(v)/p_V(v)$ for $v\in\R$.
Using \eqref{eq:equivalence} and $\alpha>0$,  \eqref{eq:eq_tau} can be equivalently written as following,
\bea\label{eq:ksi}
1-\la\,\delta\,\tau = -\delta\tau\cdot\E\Big[\envdx{\Lm}{V_\alpha}{\tau}\cdot\ksi_{{V_{_{\alpha}}}}(V_\alpha)\Big].
\eea
Next, by applying Cauchy-Shwarz inequality, recalling $\E[(\ksi_{{V_{_{\alpha}}}}(V_\al))^2]=\Ic\left(V_\al\right)$ and using \eqref{eq:eq_alpha}, we have that
\begin{align*}
\left(\E\Big[\envdx{\Lm}{ V_\al}{\tau}\cdot\ksi_{{V_{_{\alpha}}}}(V_\al)\Big]\right)^2&\le \E \Big[\Big(\envdx{\Lm}{V_\al}{\tau}\Big)^2\,\Big]\cdot\Ic\left(V_\al\right) = \frac{(\alpha^2-\la^2\,\delta^2\,\tau^2)\,\Ic(V_\al)}{\delta \tau^2},
\end{align*}
where we have also used the fact that $\tau>0$.
To continue, we use \eqref{eq:ksi} to rewrite the LHS above and deduce that:
\bea
\left(\frac{1-\la\,\delta\,\tau}{\delta\tau}\right)^2\le \frac{(\alpha^2-\la^2\,\delta^2\,\tau^2)\,\Ic(V_\al)}{\delta \tau^2}.
\eea
By simplifying the resulting expressions we have proved that $(\alpha,\tau)$ satisfy the following inequality:

\bea\label{eq:inequality}
\frac{(\alpha^2-\la^2\,\delta^2\,\tau^2)\,\Ic(V_\al)}{(1-\la\,\delta\,\tau)^2} \ge \frac{1}{\delta}.
\eea

In the remaining, we use \eqref{eq:inequality} to prove \eqref{eq:desired21}. For the sake of contradiction to  \eqref{eq:desired21}, assume that there exists a valid triplet $(\alpha,\la,\tau)$ such that $\alpha< \alpha_\star$. Recall by inequality \eqref{eq:inequality} that $\alpha$ satisfies:
\begin{align}\label{eq:psi_ineq}
\Psi\Big({\alpha}\,, \,{\la\,\tau}\Big) \ge \frac{1}{\delta}.
\end{align}
We show first that \eqref{eq:psi_ineq} holds with strict inequality. To see this, suppose that $\Psi({\alpha}, \la\,\tau) =1/\delta$.
From Lemma \ref{lem:tau_bound_lin}, it also holds that $\la\,\tau\in(0,1/\delta)$. Hence, the pair $(\alpha,\la\,\tau)$ is a feasible point in the minimization in \eqref{eq:psi_reg}. Combining this with optimality of $\alpha_\star$ lead to the conclusion that $\alpha_\star\geq \alpha$, which contradicts our assumption $\alpha< \alpha_\star$.

Therefore we consider only the case where \eqref{eq:psi_ineq} holds with strict inequality i.e., $\Psi({\alpha},\,\la\tau)>1/\delta$.

To proceed, note that $\Psi(0,x)\le0$ for all $x\in[0,1).$ Thus, by continuity of the function $a\mapsto\Psi(a,x)$ for fixed $x\in[0,1/\delta)$:
\begin{align}\label{eq:alphatilde}
\exists\,\widetilde{\alpha} : \;\text{s.t.} \;\;0\le\widetilde{\alpha}<{\alpha},\;\;\text{and}\;\; \Psi\,\Big(\widetilde{\alpha}\,,\,{\la}{\,\tau}\Big) = \frac{1}{\delta}.
\end{align}
By recalling our assumption that ${\alpha}<\alpha_{\star}$, we can deduce that \eqref{eq:alphatilde} in fact holds for $\widetilde{\alpha}<\alpha_{\star}$. However, this is in contradiction with the optimality of $\alpha_{\star}$ defined in \eqref{eq:psi_reg}. This shows that for all achievable $\alpha$ it must hold that $\alpha\ge\alpha_{\star}$. This proves the claim in \eqref{eq:desired21} and completes the proof of the theorem.

\subsection{Proof of Lemma \ref{thm:opt_reg}}
To prove the claim of the lemma, it suffices to show that the proposed loss function and regularization parameter, satisfy the system of equations in \eqref{eq:eq_main} with $\alpha = \alpha_{\star}$. For this purpose we show that $(\Lm, \la, \alpha, \tau) = (\Lm_{\star}, \la_{\star}, \alpha_{\star}, 1)$ satisfy \eqref{eq:eq_main}.

  First, we recognize that for the candidate optimal loss function in Lemma \ref{thm:opt_bin} we have $\forall v\in \R$ that
\bea \label{eq:envdx_opt}
\envdx{\Lm_{\star}}{v}{1} = -\frac{\alpha_{\star}^2 - \la_{\star}^2\,\delta^2}{1-\la_{\star}\delta}\cdot \ksi_{{V_{\star}}}(v).
\eea
Thus by replacing the proposed parameters in \eqref{eq:eq_alpha} we have :
\begin{align*}
\delta\,\E \left[\Big(\envdx{\Lm_{\star}}{V_{\star}}{1}\Big)^2\,\right] &= \delta \left(\frac{\alpha_{\star}^2-\la_{\star}^2\,\delta^2}{1- \la_{\star}\,\delta}\right)^2\Ic\left(V_{\star}\right) = \alpha_{\star}^2-\la_{\star}^2\delta^2,
\end{align*}
where for the last line we used the definitions of $\alpha_{\star}$ and $\la_{\star}$ in the statement of the lemma. This proves the claim for \eqref{eq:eq_alpha}. To show that Equation \eqref{eq:eq_tau} is satisfied we use its equivalent expression in \eqref{eq:ksi} and also replace \eqref{eq:envdx_opt} in \eqref{eq:ksi}. Specifically, this shows that
\begin{align*}
\delta\,\E\bigg[ G\cdot \envdx{\Lm_{\star}}{V_{\star}}{1}  \bigg] &=-\delta\,\alpha_{\star}\,\E\bigg[\envdx{\Lm_{\star}}{ V_{\star}}{1}\cdot\ksi{_{V_\star}}(V_{\star})\bigg] \\
&= \frac{ \delta\,\alpha_{\star}\,(\alpha_{\star}^2-\la_{\star}^2\,\delta^2)\cdot\Ic(V_{\star})}{1-\la_{\star}\,\delta} = \alpha_{\star}(1-\la_{\star}\,\delta),
\end{align*}
from which we conclude that Equation \eqref{eq:eq_tau} is satisfied. This completes the proof of the lemma.

\subsection{Proof of Lemma \ref{cor:LS_reg}}
By letting $\Lm(t) = t^2$ we find that $\env{\Lm}{x}{\tau} = \frac{x^2}{2\tau+1}$ for all $x\in \R$ and $\tau\in\R_{>0}$. Using this in Equations \eqref{eq:eq_main} and a after a few algebraic simplifications we arrive at the following closed-form expression for $\alpha_{{\ell_{_2},\la}}^2$ for all $\la\ge0$ and random variables $Z$ with finite second moment,
\bea\label{eq:alphaLSreg}
\alpha_{{\ell_{_2},\la}}^2 = \frac{1}{2}\left(1-\E[Z^2]-\delta\right)  + \frac{\E[Z^2](\la+2\delta+2)+2(\delta-1)^2+\la(\delta+1)}{2\sqrt{(\la+2\delta-2)^2+8\la}}.
\eea
Next, by using direct differentiation to optimize this over $\la\ge0$, we derive $\la_{\rm opt} = 2\E[Z^2]$ and the resulting expression for $\alpha_{{\ell{_2},\la_{\rm opt}}}^2$ in the statement of the lemma.

\subsection{Proof of Corollary \ref{cor:lowerbound}}\label{sec:proofofcor_lin}
As mentioned in the main body of the paper, the difficulty in deriving a closed-form expression for $\alpha_{\star}$ in \eqref{eq:alphaopt_thm} is due to the fact that in general $\Ic(V_a) = \Ic(a G+Z)$ may not be expressible in closed-form with respect to $a$. The core idea behind this corollary is using Stam's inequality (see Proposition \ref{propo:Fisher}) to bound $\Ic(V_a)$ in terms of $\Ic(a G) = a^{-2}$ and $\Ic(Z)$.

Specifically, applying \eqref{eq:Fisher_Stam} to the random variables $a\,G$ and $Z$ we find that:
\bea\label{eq:stam}
\Ic(V_a) = \Ic(a\, G+Z)\le\frac{\Ic(Z)}{1+a^2\Ic(Z)}.
\eea
Substituting the RHS above in place of $\Ic(V_a)$ in  the definition of $\alpha_\star$ in \eqref{eq:alphaopt_thm}, let us define
 $\widehat{\al}$ as follows:
\bea\label{eq:alphahat}
\widehat{\al} := \min_{0\le x<1/\delta}\left\{a\,\ge0:\frac{(a^2-x^2\,\delta^2)\,\Ic(Z)}{(1-x\,\delta)^2(1+a^2\Ic(Z))}\ge\frac{1}{\delta}\right\}.
\eea

The remaining of proof has two main steps. First, we show that
\bea\label{eq:Cor21_step1}
\alpha_\star^2\ge\widehat{\al}^2.
\eea
Second, we solve the minimization in \eqref{eq:alphahat} to yield a closed-form expression for $\widehat{\al}$.

Towards proving \eqref{eq:Cor21_step1}, note from the definition of $\alpha_\star$ and inequality \eqref{eq:stam} that there exists $x_\star\in[0,1/\delta)$ such that

\bea\nn
\frac{1}{\delta} = \frac{(\al_\star^2-x_\star^2\,\delta^2)\,\Ic(V_{\star})}{(1-x_\star\,\delta)^2} \leq  \frac{(\al_\star^2-x_\star^2\,\delta^2)\,\Ic(Z)}{(1-x_\star\,\delta)^2(1+\al_\star^2\,\Ic(Z))}.
\eea
Thus, the pair $(\al_\star,x_\star)$ is feasible in \eqref{eq:alphahat}. This and optimality of $\widehat{\al}$ in \eqref{eq:alphahat} lead to \eqref{eq:Cor21_step1}, as desired.

The next step is finding a closed-form expression for $\widehat{\alpha}$. Based on \eqref{eq:alphahat} and few algebraic simplifications we have :
\bea
\widehat{\al}^2 &= \min_{0\le \,x\,<1/\delta} \left\{ a^2 : a^2\,\Ic(Z)\cdot(\delta-(1-x\,\delta)^2) \ge(1-x\,\delta)^2 +\delta^3 x^2\Ic(Z)\right\}\nn\\[5pt]
&= \min_{\max\{0, \frac{1-\sqrt{\delta}}{\delta}\}\le \,x \,< 1/\delta}  \left\{a^2 : a^2 \ge \frac{(1-x\delta)^2+\delta^3\,x^2\,\Ic(Z)}{\Ic(Z)\cdot(\delta-(1-x\delta)^2)}\right\}\nn\\[5pt]
&= \min_{\max\{0, \frac{1-\sqrt{\delta}}{\delta}\}\le \,x \,< 1/\delta} \, \left\{\frac{(1-x\delta)^2+\delta^3\,x^2\,\Ic(Z)}{\Ic(Z)\cdot(\delta-(1-x\delta)^2)}\right\}.\label{eq:alphahat2}
\eea

The last equality above is true because the fraction in the constraint in the second line is independent of $a$. Next, by minimizing with respect to the variable $x$ in \eqref{eq:alphahat2}, we reach $\widehat{\alpha}^2 = h_\delta(1/\Ic(Z))$.

Finally, we know from Proposition \ref{propo:Fisher}(f) that equality in \eqref{eq:stam} is achieved if and only if the noise is Gaussian i.e. $Z\sim\mathcal{N}(0,\zeta^2)$ for some $\zeta>0$. Thus, if this is indeed the case, then $\al_\star=\widehat{\al}$ and the lower bound is achieved with replacing the Fisher information of $Z$ i.e, $\Ic(Z) =\zeta^{-2}$. This completes the proof of the corollary.

\subsection{Proof of Equation \eqref{eq:omega_LB}}\label{sec:proof_omega_LB}
First, we prove the bound $\omega_{_\delta} \geq \left(\Ic(Z)\, \E[Z^2]\right)^{-1}.$ Fix $\delta>0$ and consider the function $\widetilde{h}_{\delta}(x):=h_{\delta}(x)/x$ for $x\geq 0$. Direct differentiation and some algebra steps suffice to show that $\widetilde{h}_{\delta}(x)$ is decreasing. Using this and the fact that $1/\Ic(Z) \le \E[Z^2]$ (cf. Proposition \ref{propo:Fisher} (c)), we conclude with the desired.

Next, we prove the lower bound $\omega_\delta\geq 1-\delta$. Fix any $\delta>0$. First, it is straightforward to compute that
$
h_{\delta}(0) = \max\{1-\delta,0\} \geq 1-\delta.
$
Also, simple algebra shows that $h_{\delta}(x)\leq 1, x\geq 0$. From these two and the increasing nature of $h_{\delta}(x)$ we conclude that  $1-\delta \le h_{\delta}(x) \le 1$, for all $x \geq 0$. The desired lower bound follows immediately by applying these bounds to the definition of $\omega_\delta$.

config: arxiv


TITLE: Division by rational (decimal) number meaning
QUESTION [2 upvotes]: When I say, that I exchanged 42 CZK  into 1,5 euro. Why do I get the rate for one euro by dividing?
1) How do you explain this division in words. Like when you say when doing integer division, that you divide 10 cookies to 2 people, so one get 5. That type of style.
2) Does division always gives back an average rate per 1 unit of somethings?
3) the same for mean average, I bought 100 units of something for 2500 unit of any currency. 2500/100 = 25 per one (25/1).
How do you explain this in words too, please. Something like you want to distribute 2500 into 100 equal parts and the answer is the units per one?
Thanks, All I care is the problems being said in words, meaning of division rational number by rational number, in this case in its decimal form

REPLY [1 votes]: Short answer:
Multiplication is repeated addition of the same value, so if your are adding some value to itself over and over again, multiplication by the number of times this value occurs in the sum will give the final result. This is where the concept of multiplication comes from.
Division undoes multiplication. This is where it comes from. So if you know what the sum is, and the number of times the value was included in the summation, then the value that was being added to itself can be obtained by division.
(1) If I may be allowed to skip over the added complications of fractional numbers, this is what is happening in your example. Each Euro is worth some number $c$ of CZK. The value of 1.5 Euros is $1.5 \times c$. But we know that $1.5$ Euros is worth the same as $42 CZK$, so we get $ 1.5 \times c = 42$.
We rewrite this as a division:
$$c = \frac {42}{1.5} = 28$$
to see that the value of each Euro is $28$ CPK.
(2) that is the definition of "average": The average of a collection of values is the number that if every one of values in the collection were changed to this number, then the sum of all the values would not change. From this definition, it follows that
$$\text{Count}\times\text{Average} = \text{Sum}$$ and therefore that $$\text{Average} = \frac{\text{Sum}}{\text{Count}}$$
(3) "mean" and "average" are the same thing. They are just two different words for the same concept.
Long answer:
Suppose you buy 8 widgets that cost 2 Euro each. How much does it cost? Well, you add them up:
$$2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 = 16$$
That's rather tiring, having to repeatedly add the same thing. So we introduce a second operation: multiplication. Since there are $8$ widgets altogether, we define $8 \times 2 = 16$. A little development of rules for multiplying, and we have a useful tool that makes a lot of common calculations much easier. But remember that at its base, multiplication is just repeated addition of the same number.
But when you have multiplication, sometimes you end up needing to ask the opposite question: "I bought some wingdings at a price of $1.5$ Euros each, and the total charge was $30$ Euros. How many wingdings did I purchase? (Fortunately, I am buying in a place without sales tax.) If I call the total number of wingdings purchased $N$, then the total cost will be
$$\underbrace{1.5 + 1.5 + ... + 1.5}_{N\text{ times}} = N \times 1.5$$ Since I paid $30$ Euros total, it must be that $$N \times 1.5 = 30$$ So I need some means of undoing that multiplication by $1.5$, so I can figure out what $N$ is. For this reason, we invented division. With additional development of rules and processes, we find convenient means of calculating that $$\frac {30}{1.5} = 20$$ which means the same thing as saying $20 \times 1.5 = 30$. So I bought $20$ wingdings. No other number when multiplied by $1.8$ Euros would result a total cost of $30$ Euros.
But in these two examples, everything costs exactly the same amount. Suppose next I buy a combination of 10 widgets, wingdings, and whachamacallits, which cost 2.5 Euros each. The total cost now is 20 Euros. How many did I buy of each? Unfortunately, this question cannot be answered fully. There are several different combinations that would result in 20 Euros:

All widgets: $10 \times 2 = 20$.
8 widgets, 1 wingding, 1 whachamacallit: $8 \times 2 + 1 \times 1.5 + 1 \times 2.5 = 20$.

...

5 wingdings, 5 whachamacallits: $5 \times 1.5 + 5 * \times 2.5 = 20$.

Well then, if I can't figure out how many I have of each, is there anything useful I can say? Suppose instead of all costing different amounts, everything had cost the same. What would this amount be, given that $10$ items costs $20$ Euros? We divide to see that if everything cost the same, it would be $2$ Euros.
While this example is contrived, there are a number of cases where there really isn't a good way to control how many items come at one value instead of another. For example, manufacturing variation causes the weight of supposedly identical products to vary from item to item. So you cannot say exactly how much each widget weighs. But, if you weigh 1000 widgets, sum up the weight and divide by 1000 to figure out how much they would weigh if they really were the same weight, then it is likely that the 1000 widgets represents a good sample of the values that might occur from the manufacturing variation. So if you get in another order of 2000 widgets, you can now predict that the weight of this 2000 will be close to twice the weight of the first 1000 even though they don't all weigh the same. Now you don't have to weigh every widget in the future to have a good idea how much any large number of them will weigh.
For this reason, we define the concept of "average" or "mean":

The average of a collection of values is the number that, if all of the values in the collection were changed to this number, then the sum of the collection would be same.

Now if all of the items in the collection were change to this same average, then the sum of the new changed collection would be the number of items times the average. Since the average is chosen to leave the sum unchanged, we get this formula, we must have that adding the average value to itself the same number of time as there are values in the collection (i.e., multiplying the average by the collection count) gives the original sum of the collection:
$$\text{Count}\times\text{Average} = \text{Sum}$$
Rewriting it using division gives the familiar formula:
$$\text{Average} = \frac{\text{Sum}}{\text{Count}}$$


TITLE: Coefficient of power series when $p(x) = \sum b_nx^n$ converges for $|x| \le 1$ and $p(x) = 0$ for $|x| \lt \delta$.
QUESTION [0 upvotes]: Suppose that the power series $p(x) = \sum b_nx^n$ converges for $|x| \le 1$. Suppose that for some $\delta \gt 0 , p(x) = 0$ for $|x| \lt \delta$. Show that $b_n = 0$ for all $n \ge 1$.

REPLY [0 votes]: For $|x|<1$ $p(x)$ is analytic. But also $p(x)=0$ for $|x|<\delta $. But
then $p(x)$ must vanish for all $|x|<1$. Hence $b_{n}=0$.


TITLE: Logistic differential question, Please check my work.
QUESTION [0 upvotes]: Consider the logistic differential equation $\frac{dy}{dx} = \frac{y}{4}\left(12 -y \right)$. Let $y = f\left(t\right)$ be the particular solution to the differential equation with $f\left(0\right) = 6$. Find the particular solution $y=f\left(t\right)$ with $f\left(0\right) = 6$.

$\int \frac{y}{4} \left(12 -y \right) \; \mathrm{d}t$
$\rightarrow \dfrac{-t y \left(12 - y\right)}{4} + C$ given $f\left(0\right) = 6$
Therefor $C = 6$ and leading to
$y = \dfrac{-t y \left(12 - y\right)}{4} + 6$.
This is nothing like it says in the answer sheet.
$6 = \dfrac{12e^{c}}{e^c + 1}$ with $C = 1$
Where did I make a mistake?

REPLY [2 votes]: It helps to write the equation in differential form:
$$\frac{dy}{y (12-y)} = \frac{1}{4} dx$$
Now integrate both sides:
$$\int \frac{dy}{y (12-y)} = \frac{1}{4} x + C$$
So the problem boils down to evaluating that integral over $y$.  In case you are stuck:
$$\frac{1}{y (12-y)} = \frac{1}{12} \left( \frac{1}{y} + \frac{1}{12-y}\right)$$

REPLY [2 votes]: From $${dy\over dx}={y\over4}(12-y)$$ you should get $$\int{1\over{y\over4}(12-y)}\,dy=\int\,dx$$


TITLE: Prove an ideal is maximal
QUESTION [0 upvotes]: Question
Prove the ideal $\mathrm I=\{f \in \mathrm R| f(2)=0 \}$ of $\mathrm R=\{f(x) | f: \Bbb R \to \Bbb R $ is continue} is maximal.
DO NOT use the $1$st isomorphism theorem.

I tried.

(1) $\mathrm I \neq \mathrm R$
(2) Suppose that if  "$\mathrm I \subseteq \mathrm J \subseteq \mathrm R$ where $J$ is ideal of $\mathrm R$".
For all  $f(x) \in \mathrm I$,  $f(x) \in \mathrm J=\{ j(x)h(x)$ | $ h(x) \in \mathrm R\}$
Since $f(2)=0$, $f(x)=j(x)h(x)$ for some $h(x) \in \mathrm R$. (ie. $j(2)h(2)=0$)
If $j(2)=0$, then $j(x) \in \mathrm I$. so $\mathrm I = \mathrm J $.
And we show that if $\mathrm I \neq \mathrm J$, then $\mathrm J = \mathrm R $.
If $f(2)=j(2)h(2)=0$ and $j(2) \neq 0$ then, there exist $g(x) \in \mathrm J$
     s.t $g(2) \neq 0$.
moreover there exist $g(x) \in J$ s.t $g(2)=1$. And $g(x)$ is continues.
So we also take $g(x)=1$.
$g(x)=1 \in \mathrm J$. ( ie. $\mathrm J$ contains unit.)
threrfore $\mathrm J=\mathrm R$. ($\mathrm I \neq \mathrm R$)
we conclude $\mathrm I$ is maximal ideal of $\mathrm R$.

is it all right? please fix it or hint. thanks

REPLY [4 votes]: $(x-2)^2\in I$. If there is $g(x) \in J$ such that $g(2)\neq 0$ then
$$(x-2)^2+(g(x))^2 \in J$$ and this function has no zeros so is invertible.


TITLE: Why my this formula doesn't satisfy the formula of volume of sphere?
QUESTION [1 upvotes]: I've defined the function which returns a volume of sphere as a radius value given.
$f(a):=2\int_{0}^{a}\pi*r^{2}*dr$
I thought that above function is correct but actually it is not correct.
The return value of the above function is $\frac{2\pi a^{3}}{3}$
The $2$ at the leftmost of the function represents of a double of volume of hemisphere.
Can anyone tell my why this function is not working?

REPLY [1 votes]: Given a radius $r$, consider the graph of the semicircle given by $y = \sqrt{r^2 - x^2}$ for $x \in [-r, r]$.
Rotate this graph about the $x$-axis, and you will find the volume of the resulting sphere as an integral.


TITLE: Help me understand how to Find f(n)
QUESTION [1 upvotes]: I have this question:

$f(0) = 0$
$f(1) = 1$
$f(n) = f(n-1) + f(n-2)$

help me understand and find $f(n)$.

REPLY [3 votes]: Adding to @gimusi 's answer, if you divide $f(n+1)$ by $f(n)$, then this approaches a certain value, for all $n\in\mathbb{N}$. We will call this value $v$, then $$\frac{f(n+1)}{f(n)}\operatorname*{\longrightarrow}^{n\to\infty} v,$$ or $$\lim_{n\to\infty}\frac{f(n+1)}{f(n)} = v.$$ Whichever choice of notation you prefer. After going here, we find that $$v^2 - v - 1 = 0.\tag{In the link, $v = \Delta$}$$ Do you know the quadratic formula? If so, you can apply it to this equation and obtain that $$v = \frac{1+\sqrt{5}}{2} = \color{orange}{\verb|The Golden Ratio|}\tag{denoted by $\varphi$}.$$
This is why the sequence you have, known as the Fibonacci Sequence, is very well-known!


TITLE: Density matrix element in Jaynes-Cummings model
QUESTION [0 upvotes]: In the Jaynes-Cummings model, when using the density matrix to describe mixed states for the atom-field system, after some calculations I got to this matrix element:
$$ \rho_{ee}^A = \sum_{n=0}^{\infty} \left\langle n \right| \hat{C}(t) \hat{\rho}^F(0) \hat{C}(t)  \left| n \right\rangle $$
where:
$$ \hat{C}(t) = \cos \left( \lambda t \sqrt{ \hat{a} \hat{a}^{\dagger}}\right)$$
Now, I'm supposed to arrive at:
$$ \rho_{ee}^A = \sum_{n=0}^{\infty} \left\langle n \right|  \hat{\rho}^F(0)   \left| n \right\rangle \cos^2(\lambda t \sqrt{n+1}) \hspace{0,8cm} (1)$$
but dont know how to. I tried:
$$ \cos \left( \lambda t \sqrt{ \hat{a} \hat{a}^{\dagger}}\right) = \sum_{m=0}^{\infty} \frac{(-1)^m (\lambda t )^{2m}}{2m!} (\hat{a} \hat{a}^{\dagger})^m$$
but:
$$ (\hat{a} \hat{a}^{\dagger})^m \left| n \right\rangle = \frac{(n+m)!}{n!} \left| n \right\rangle $$
and I don't know how to get to: $\cos^2(\lambda t \sqrt{n+1})$. How do I do this?

REPLY [1 votes]: $$ \hat{a} \hat{a}^{\dagger}\left| n \right\rangle = (n+1) \left| n \right\rangle $$
let
$$\hat{n}= \hat{a} \hat{a}^{\dagger}$$ so we have;
$$B=\cos \left( \lambda t \sqrt{ \hat{a} \hat{a}^{\dagger}}\right)\left| n \right\rangle  = \cos \left(\lambda t \hat{n}^{1/2}\right)\left| n \right\rangle =\sum_m (\lambda t)^{2m} \hat{n}^{m}/(2m!)\left| n \right\rangle  =\sum_m (\lambda t)^{2m} (n+1)^{m}/(2m!)\\=\cos(\lambda t\sqrt{n+1})$$
you have B^2 in you case so QED.
We really do not need this fuss in general if you have a function of an operator and hit it to arguments eigenvector you just replace the operator in the argument with its eigenvalue. This can be shown with Taylor series for any nice function generally here we did for cos


text:
\begin{document}

\title{Feasibility of a fishery regulation model.\\A pulse quota fishing policy\\with environmental stochasticity.}

\author{R. Castro-Santis \and
F. C\'ordova-Lepe \and W. Chambio}

\maketitle

\begin{abstract}
An enviromental-random effect over a deterministic population mo\-del, a resource ({\it e.g.}, a fish stock) is introduced. It is assumed that the harvest activity is concentrated at a non predetermined sequence of instants, at which the abundance reaches a certain predetermined level, for then to fall abruptly a constant capture quota (pulse harvesting). So that, the abundance is modeled by a stochastic impulsive type differential equation, incorporating an standard Brownian motion in the per capita rate of growth. With this  random effect, the pulse times are images of a random variable, more precisely, they are ``stopping-times" of the stochastic process. The proof of the finite expectation of the next access time, {\it i.e.}, the feasibility of the regulation, is the main result.

\end{abstract}

\section{Introduction}
In the context of fisheries resources, this article examines the feasibility of a management model where some regulatory authority limits the amount captured and access times. The above by fixing a fishing quota and a subsequent period of closure that last up to get a certain threshold biomass and so on. It is a mathematical continuous time model of one species unstructured and non-deterministic. The main novelty is the combination of three rules in the dynamics of abundance: (a) The addition to the growth rate of a white noise with amplitude proportional to the stock. (b) The consideration of an impulsive extracction at certain threshold of the stock size. (c) A constant quota fishing policy.

{\it Concerning (a)}: We have hypothesized a very general unstructured stochastic version model by means of a diffusion process. It consider that the infinitesimal increment ($\mathrm{d}N(t)$) of the abundance ($N(t)$) is the sum of the underlying deterministic dynamics tendencies ($N\,r(N)dt$) plus the magnitud of the stochastic fluctuations at different populations sizes ($\sqrt{v(N(t))}\,\mathrm{d}B(t)$). Among the sources of uncertainty to which a population may be subject, we will work with environmental stochasticity, {\it i.e.}, following \cite{Dennis}, where the infinitesimal variation $v(N)$ can be modeled by $\sigma^{2}N^{2}$. The {\it environmental stochasticity} involves considering randomness resulting from any change that affects the whole population and that does not diminish with its growth.

There is abundant literature on population models with the presence of environmental noise. For models with this type of noise and capture, when the harvest is incorporated additively as a density-dependent rate which is subtracted from the natural growth, an author to consider is C.A. Braumann, see \cite{Braumann01}, \cite{Braumann02}, \cite{Braumann03},  \cite{Braumann04}. In \cite{Braumann01}, a quite general model for the growth of populations subjected to harvesting activities in a random environment is studied. There, conditions for non extinction and for the existence of stationary distributions (where the noise intensity was constant or proportional to the rate of growth) are looked for. In \cite{Braumann02} those results are generalized to density-dependent positive noise intensities of very general form.

{\it Concerning (b)}: In population models that we have cited, which combine harvesting and environmental stochasticity, the population abundance variable is of continuous type. Here we will assume pulse harvesting between closures, which implies piecewise continuous curves for the stock, and other type of control problems.

Due to the development of techniques for locating resources and deployability of arts and fishing effort, comparing with closures, the open intervals are considerably shorter. Therefore, modelling the open season as an instant is realistic enough. For impulsive harvesting see: \cite{Cordova01}, \cite{Cordova02}, \cite{Cordova03}, \cite{Zhang-X}, \cite{Zhang-Y}, \cite{Zhao}. Then, the amount harvested is just like a pulse in the abundance curve. The appearance of run for the resource, is another explanation for an instantaneous extraction of the quota. However, although the stock captured could be previously agreed, the run occurs when it appears a strong competition to be first in the markets.

{\it Concerning (c)}: The management rule assumes a continuous sampling of abundance. The open season begins when the size of the stock reaches a certain value $ K^{+} $ and remains until a catch quota $Q$, $0 \leq Q \leq K^{+}$ is  got it. Then, a new closure starts from an abundance level $K_{-}$, $K_{-}=K^{+}-Q$, that does not compromise the survival of the resource by population uncertainty, {\it i.e.}, above the minimum viable population \cite{Brook}. The quota is split between the different actors engaged in fishing effort, when it is completed, it begins a period of closure that lasts until again the regulator announces a level of abundance $K^{+}$. We named this regulation {\it pulse constant quota fishing policy}.

Under this management, without stochasticity and with a growth law autonomous of time, models of this type have monotonic and bounded abundance curves (for values of $ K^{+} $ under carrying capacity), which determines closures with finite length. Closed seasons with finite duration, would theoretically could allow bioeconomic sustainability (preservation and exploitation). However,
if we consider the hypothesis of a stochastic component, for example, coming from environmental factors that determine the vital rates (the biotic potential and the environmental resistance) we have that the finiteness of the closures is not assured a priori. The main novelty problem that guided our research questions and the results was to find conditions on the model parameters and mainly on the noise level, for the viability (finite closures) of this fishery management. Notice that in \cite{Braumann04}, in the framework of a continuous model, it is proved that
in case of a constant quota (a fixed amount harvested by unit of time) in random environment, the abundance always goes to extinction.

In
Theorem \ref{theo:main-1} is technical in nature and aims to prove that the model is well formulated. It solves the existence and uniqueness (in the probabilistic sense) of the growth curves of the resource from an initial time and level of biomass. Its Corollary 1 states that the model determines a finite population variance.

The aim of Theorem 2 is similar to that of Theorem 1, but this one proves the existence and uniqueness for all finite future time interval.

Finally, the main result, Theorem \ref{theo:main-2}, shows that the expectation of a next opening time is finite, by establishing a condition that relates the lower value of the per capita growth rate for biomass values under $K^{+}$ with the parameter that indicates the amplitude of noise.

\section{Description of the model and the problem}
In a first subsection we describe the regulatory basic fishery model (deterministic) and we derive the properties of the length of the closed seasons. In a second one, we will introduce stochasticity on the growth rate and formulate, with details, the research question. In both cases, the main assumptions are highlighted. In all what follows we will denote by $N(t)$ the abundance of the resource at some instant $t \in [0,\infty)$.

\subsection{The deterministic model}
With respect to the growth of the stock, without stocasticity, let us consider the hypotesis that follows:

\vspace{2mm}

\noindent{\bf H1}: The {\it per capita rate of growth (deterministic)} is a continuous function $r: [0,\infty) \to (-\infty,+\infty)$, for which there exists a positive abundance level $K$, the carrying capacity, such that
$r(N) >0$ ({\it resp.} equals to $0$, less than $0$) if $N<K$ ({\it resp.} equals to $K$, biger than $K$).\\

\noindent\textbf{Note:}
There exists statistical evidence that populations present a negative correlation between $r(\cdot)$ and $N(\cdot)$, see \cite{Tanner}. An example of $r(\cdot)$ satisfying the above condition is
the {\it Generalized logistic law}: $r(N)=r_{0}(1-(N/K)^{\mu})^{\nu}$, $r_{0}>0$, $\mu,\,\nu \geq 1$. So that, we have a very natural hypothesis.

If a regulator fixes a minimum level of biomass for harvesting $K^{+}$, $K^{+}<K$, and a catch quota $Q$, the base deterministic model is:
\begin{equation}
\label{1}
\left\{
\begin{array}{lclllcl}
N'(t) & = & r(N(t))\,N(t), & N(t)& <& K^{+},\\
N(t^{+})& = & K_{-}=K^{+}-Q,   & N(t)& =& K^{+},\\
(t,N) & \in & [0,\infty) \times [0,K^{+}].
\end{array}
\right.
\end{equation}
where the solutions are piecewise continuous functions with continuity by the left at its discontinuities.

It is straightforward to prove, for any initial value $N(0)=K_{-}<K^{+}$, that the abundance $N(\cdot)$ is a eventually periodic trajectory that reach all the values in $[K_{-},K^{+}]$. It is always strictly increasing except at a sequence of instants where its value is $K^{+}$, which abruptly drops a quantity $Q$ toward its new value $K_{-}$.

Denoting by $\{t_{k}\}_{k\geq 0}$ the consecutives harvest times ({\it i.e.}, $N(t_{k})=K^{+}$) from a first one $t_{0} \geq 0$, then $N(\cdot)$ has to satisfies the integral equation:
\begin{equation}
N(t) = K_{-} \exp\left(\int_{t_{k}}^{t} r(N(s))ds\right), \quad t\in (t_{k},t_{k+1}].
\end{equation}
So that, substituting $t = t_{k+1}$, using $N(t_{k+1})=K^{+}$ and applying the Mean Value Theorem, we have that the length of the closures in the deterministic model is given by the expression: $
T(K_{-},K^{+})=(1/r(N^{*}))\ln(K^{+}/K_{-})$,
some $N^{*}\in (K_{-},K^{+})$.

In order to get some bounds, defining $\alpha=\inf_{N\in [0,K^{+}]}{r(N)}$ and $\beta = \sup_{N\in[0,K^{+}]}{r(N)}$, we have:
\begin{equation}
\label{determinista}
K_{-}e^{\alpha(t-t_{k})} \leq N(t) \leq K_{-}e^{\beta(t-t_{k})}, \quad t\in(t_{k},t-{k+1}],
\end{equation}
and
\begin{equation}
\label{T}
\frac{1}{\beta}\ln\left(\frac{K^{+}}{K_{-}}\right) \leq T(K_{-},K^{+}) \leq \frac{1}{\alpha}\ln\left(\frac{K^{+}}{K_{-}}\right)
\end{equation}
Note that, except for a desface in time, from any time-state initial condition, all abundance curves are equal.
This is a non traditional situation of sustainability (or equilibrium).

\subsection{The stochastic model}
We modify the deterministic model assuming the following hypothesis:

\vspace{2mm}

\noindent{\bf H2}: The {\it per capita rate of growth (stochastic)} is obtained introducing on the first equation of the deterministic system defined by (\ref{1}), a random component, more precisely we will add to the per capita rate of growth $r(\cdot)$, defined on {\bf H1}, a noise of type $\sigma \mathrm{d}B(t)$, where $B(\cdot)$ is the standard Brownian motion, with the condition that follows:
There exists $K_{0}$, $K_{-}<K_{0}<K^{+}$, such that
\begin{equation}
\label{infm}
\inf_{N \in [0,K_{0}]} \{ r(N) \} > \frac{\sigma^{2}}{2}.
\end{equation}

\noindent\textbf{Note:} The inequality (\ref{infm}) is very natural in order to have a population dynamics no ``phagocytosed" by the noise. When $r(\cdot)$ is a decreasing function as in the {\it generalized logistic law}, we assure (\ref{infm}) if $r(K^{+})>\sigma^{2}/2$.
This condition is the natural generalization of that of the stochastic Maltusian case $\mathrm{d}N(t)=rN(t) \mathrm{d}t +\sigma N(t)\mathrm{d}B(t)$, where $r >\sigma^{2}/2$ ({\it resp.} $<$, {\it i.e.}, extinction) implies $N(t) \to \infty$ ({\it resp.} $\to 0$) with probability one, as $t \to \infty$. Remenber that in the logistic case, $\mathrm{d}N(t)=rN(t)(1-N(t)/K) \mathrm{d}t +\sigma N(t)\mathrm{d}B(t)$, we have that the average $(1/T)\int_{[0,T]}N(t)dt$ tends to $K(1-(1/2)\sigma^{2}/r)$ a.s. as $T \to \infty$ when $r > \sigma^{2}/2$, and the population goes to extinction, with probability one, whenever $r < \sigma^{2}/2$.

Notice that, if the resource has its developement with a random biotic potential or present a random environmental pressure, the capture times are also random variables that depend on the dynamics of growth of the resource. Then, from the mathematical point of view, the model is given by a process represented by the stochastic and impulsive differential equation:
\begin{equation}
\label{eq:SDE-main}
\left\{
\begin{array}{rcllcl}
\mathrm{d}N(t) & = &r(N(t))N(t)\mathrm{d}t+\sigma N(t)\mathrm{d}B(t), & N(t) & < & K^{+},\\
N(t^{+}) & = & K_{-}=K^{+}-Q, & N(t) & = & K^{+},\\
(t,N) & \in & [0,\infty) \times [0,K^{+}].
\end{array}
\right.
\end{equation}

Now, given an initial positive stock size $N(0)=N_{0}$ in $(0, K^{+}]$, it appears the following set of questions: Is there a first harvest time, {\it i.e.}, $\tau_{0}>0$ such that $N(\tau_{0})=K^{+}$? Is it possible to define a sequence of sucesive harvest times $\{\tau_{k}\}_{k\geq 0}$? As the resource population has its development affected by random growth factors, $N(\cdot)$ is a random variable and, therefore, the instants $\tau_{k}$, $k \geq 0$, defined recursively by
\begin{equation}\label{eq:stopping-time}
\tau_{k+1}=\inf \{t>\tau_{k}: \,\,N(t)=K^{+}\},
\end{equation}
are random times.
What is the statistical expectation for the length of the closure periods $\{\tau_{k+1}-\tau_{k}\}$? These are some of the questions that motivate our results.

\section{Existence of solutions}

In order to get solutions of the model (\ref{eq:SDE-main}) we will use the scheme that follows:

\begin{itemize}
 \item[T1] To prove the existence of a unique initial stochastic process (without harvest), so that, a first segment of its trajectory can be taken as a solution of the equation (\ref{eq:SDE-main}), on some interval $[0,T]$, $T>0$.

 \item[T2] Immediately after a probable instant of capture the stochastic model requires restarting the process from an abundance $K_{-}$. Then, it is necessary to prove the existence on each interval $[S,T]$, $0<S<T$, of a unique stochastic process that solves the natural growth rule of the equation (\ref{eq:SDE-main}).

 \item[T3] We must ensure the existence of the harvest times, {\it i.e.} to show the finite expectation for the random variable defined by the stopping time of the equation (\ref{eq:stopping-time}). Finally, we will chain the pieces of trajectories for generating a unique extended solution of (\ref{eq:SDE-main}).
\end{itemize}

The classical theorems for stochastic differential equations, need a Lipschitz condition and sublinear growth over drift and diffusion coefficient (see Theorem 5.2.1 in \cite{Oksendal-1998}). These conditions, are too strong to be used directly in most population dynamics problems. For this reason, we use a more appropriate hypotheses. Note that the positivity of the variable $N(\cdot)$ will be established as a direct consecuence of the Theorem \ref{theo:main-2}.\\

\begin{theorem}\label{theo:main-1}
Let  $T$ and $K_{-}$ be positive constants.
If $r(\cdot)$ satisfies {\bf H1} and $B(t)$ is a standard Brownian motion, then there exists a unique process $N(\cdot)$ solving, on the time-state domain $[0,T]\times (0,\infty)$,
the initial value problem:
\begin{equation}\label{eq:main-1}
\mathrm{d}N(t)=r(N(t))N(t)\mathrm{d}t+\sigma N(t)\mathrm{d}B(t),\,\,\,\,\, N(0)= K_{-}.
\end{equation}
\end{theorem}

\noindent{\it Proof}: For any $L >0$, we will define the auxiliary function $f_{L}(x) = r(x)x,$ ({\it resp.} $r(L)L$) if $x \leq L$ ( {\it resp.} $x > L$).
Thus, an approximate integral equation to (\ref{eq:main-1}) is given by:
\begin{equation}
\label{eq:aprox-L}
N_{L}(t) = K_{-}+\int_{0}^{t} f_{L}(N_{L}(s))ds + \sigma N_{L}(s)dB(s), \,\,\, t \in [0,T].
\end{equation}
Clearly, this equation satisfies the Lipschitz and sublinear growth conditions, on the drift and the diffusion coefficient. Therefore, there exists a unique stochastic process, solution of the equation (\ref{eq:aprox-L}).

Using the It\^o formula, for the fuction $\psi(x)=x^{2}$, $x >0$, in the equation (\ref{eq:aprox-L}), we obtain
\begin{equation}\label{eq:apply-Ito}
 N_{L}^{2}(t) = K_{-}^{2} + 2\int_{0}^{t}f_{L}(N_{L}(s))N_{L}(s)\mathrm{d}s+\int_{0}^{t} \sigma^{2}N_{L}^{2}(s)\mathrm{d}s, \,\,t \in [0,T].
 \end{equation}

Applying expectation to the equation (\ref{eq:apply-Ito}), we obtain:
\begin{equation}
\mathbb{E}[N^{2}_{L}(t)] \leq  K_{-}^{2} + \mathcal{B}K^{2} T +\sigma^{2}\int_{0}^{t} \mathbb{E}[N_{L}^{2}(s)] ds,
\end{equation}
where $\mathcal{B}=\max \{r(N): 0<N<K\}$.

Therefore, using the Gronwall inequality to the above inequality, we have $\mathbb{E}[N_{L}^{2}(t)] \leq C_{T}$, with $C_{T}=(K_{-}^2+\mathcal{B}K^{2}T)e^{\sigma^{2}T}$. Then, the stochastic processes $N_L(t)$ are all square integrable and bounded for any independent constant $L$.

Now, we will estimate the values of $N_{L}(t)$ on $[0,T]$. By using equation (\ref{eq:aprox-L}) and the inequality $f_{L}(\cdot) \leq \mathcal{B}$, it is follows:
\begin{equation}
\label{sup-int}
\sup_{t\in[0,T]}N_L(t) \leq K_{-} +\mathcal{B}T +\sup_{t\in[0,T]}\int_{0}^{t}\sigma N_{L}(s)\mathrm{d}B(s).
\end{equation}

In order to find a bound for the integral in (\ref{sup-int}), we use the Doob inequality for martingales and it is obtained
\begin{equation}\label{eq:expectation-N}
 \begin{array}{rcl}
 \displaystyle \mathbb{E}\left[\sup_{t\in[0,T]}\int_{0}^{t}\sigma N_L(s)\mathrm{d}B(s)\right]&\le&\displaystyle\left(\mathbb{E}\left[\sup_{t\in[0,T]}\left(\int_0^t\sigma N_{L(s)}\mathrm{d}B(s)\right)^{2}\right]\right)^{\frac{1}{2}}\\
 {}\\
 {}&\le&\displaystyle2\left(\mathbb{E}\left[\left(\int_{0}^{T}\sigma N_L(t)\mathrm{d}B(t)\right)^2\right]\right)^{\frac{1}{2}}
  {}\\
  {}\\
  {}&=&\displaystyle 2\sigma^{2}\int_{0}^{T}\mathbb{E}\left[N_{L^{2}}(t)\mathrm{d}t\right]^{\frac{1}{2}}=2\sigma^{2}C_{T}.
 \end{array}
\end{equation}

Therefore, there exists a positive constant, such that
$\mathbb{E} [\sup N_{L}((0,T]) ]$ is less or equal than $C'_{T}=N_{0}+\mathcal{B}T+2\sigma^{2}C_{T}$. As this bound is independent of $L$, using the Markov inequality we obtain:
\begin{equation}
\label{eq:Prob-N}
 \mathbb{P}[\sup N_{L}((0,T]) \geq L] \leq C'_{T}/L.
\end{equation}
This inequality allows us define the following set:
\begin{equation}
\Omega_L=\{\omega\in\Omega :\, \sup N_{L}(\omega,(0,T])<L \},
\end{equation}
where $(\Omega,\mathscr{F},\mathscr{F}_t,\mathbb{P})$ is the stochastic basis, where Brownian motion is defined.\\

From the definition of the set $\Omega_L$ and equation (\ref{eq:Prob-N}), it is follows:
\begin{equation}
  \mathbb{P}(\Omega_L)\ge1-C'_{T}/L.
\end{equation}
Then $\lim_{L\to\infty}\mathbb{P}(\Omega_L)=1$ and if $L<L'$, over $\Omega_L$ and $N_L(t)=N_{L'}(t)$, q.c.

Therefore, there exists a process $N(t)$ such that $N_L(t) \to N(t)$ q.c. as $L\to\infty$.

Finally, taking limit in equation (\ref{eq:aprox-L}), we obtain the required result. $\diamond$

\begin{corollary}
\label{cor:expectation-N2}
 Under the hypotheses of Theorem \ref{theo:main-1}, we have the following results of boundedness:\\
(a) $\mathbb{E}[N^{2}(t)] < \infty$ \,\,\,\,\,\mbox{and}\,\,\,\,\,
(b) $\displaystyle\mathbb{E}\left[\int_{0}^{t}N^{2}(s)\mathrm{d}s\right] <
   \infty$, for any $t\in [0,T]$.
\end{corollary}

\noindent{\it Proof:}
 Taking limit in the equation (\ref{eq:expectation-N}), we obtain $\mathbb{E}[N^2(t)] \leq  C_{T}$, for any $t \in [0,T]$.
 This is the proof of the item (a). Item (b) is due to item (a). $\diamond$.

\begin{theorem}\label{theo:solution-ST}
Let $S$, $T$, $S<T$, and $N_{0}$ and be positive constants. If $r(\cdot)$ satisfies {\bf H1} and $B(t)$ is a standard Brownian motion. Then, there exists a unique stochastic process $N(t)$, that solves on the time-state domain $[0,T]\times [0,\infty)$, the initial value problem:
\begin{equation}\label{eq:main-2}
\mathrm{d}N(t)=r(N(t))N(t)\mathrm{d}t+\sigma N(t)\mathrm{d}B(t), \,\,\,N(S)=K_{-}
\end{equation}

\end{theorem}

\noindent{\it Proof:}
Making the change of variable $t'=t-S$, the equation (\ref{eq:main-2}) takes the form
 \begin{equation}\label{eq:main-2.1}
 \mathrm{d}N_1(t')=r(N_1(t'))N_1(t')\mathrm{d}t'+\sigma N_1(t')\mathrm{d}B_1(t'), \,\,\, N_1(0)=K_{-},
 \end{equation}
 with $t'\in[0,T-S)$ and
where $B_1(t')=B(t'+S)-B(S)$, is a standard Brownian motion that satisfies the equality $\mathrm{d}B_1(t)=\mathrm{d}B(t)$. Therefore, due to Theorem \ref{theo:main-1}, there exists a unique stochastic process $N_1(t')$, solution of the equation (\ref{eq:main-2.1}) and such that $N(t)=N_1(t+S)$ is solution of the equation (\ref{eq:main-2}). $\diamond$

\begin{theorem}\label{theo:main-2}
Let us asssume conditions {\bf H1} and {\bf H2}. Given the $k$-th harvest time $\tau_{k}$, some $k \geq 0$, and denoting by $\alpha$ and $\beta$ the infimum and supremum respectively of the set $r([0,K^{+}])$, then:
\begin{itemize}
\item[(a)] For any $t \in (\tau_{k}, \tau_{k+1}]$, the abundance trajectory satisfies
\begin{equation}
\label{cotas1}
K_{-} \exp((\alpha-\frac{\sigma^{2}}{2})(t-\tau_{k})+\sigma B(t)) \leq N(t) \leq
K_{-} \exp((\beta-\frac{\sigma^{2}}{2})(t-\tau_{k})+\sigma B(t)).
\end{equation}

\item[(b)]
The variable length of closures $\tau=\inf \{t-\tau_{k}>0:\, N(t) \geq K^{+}\}$ has finite expectation $\mathbb{E}[\tau]$ such that
\begin{equation}
\label{cotas2}
\frac{1}{\beta-\sigma^2/2}\ln\left(\frac{K^{+}}{K_{-}}\right) \leq \mathbb{E}[\tau] \leq
\frac{1}{\alpha-\sigma^2/2}\ln\left(\frac{K^{+}}{K_{-}}\right).
\end{equation}
\end{itemize}
\end{theorem}

\begin{remark}
At the first stopping time the abundance has reached a level $K^{+}$ and inmediately it is reduced a quota $Q$. So the bounds for the abundance and the expectation of the next stopping time are given respectively by the expressions (\ref{cotas1}) and (\ref{cotas2}). These bounds can be compared with the deterministic value ($\sigma=0$) given in (\ref{determinista}) and (\ref{T}). Moreover, if in (\ref{cotas1}) we have $\alpha=\beta$, then we recover the explicit solution of the Maltusian case $N(t)=K_{-}\exp[(r-\sigma^{2}/2)+\sigma B(t)]$, $t \in (t_{k}, t_{k+1}]$, and now (\ref{cotas2}) defines an accurate expectation.
\end{remark}

\noindent{\it Proof:}
Notice that the solutions of the equations
\begin{equation}\label{eq:Na-solution}
 N_{\gamma}(t)= N_{0}+\gamma \int_{0}^{t}  N_{\gamma}(s)\mathrm{d}s + \sigma\int_0^t N_{\gamma}(s)\mathrm{d}B(s),\,\,\,\gamma \in \{\alpha,\beta\},
\end{equation}
are Geometric Brownian Motions. Their explicit form is given by
$N_{\gamma}(t)=N_{0} \exp\{(\gamma-\sigma^{2}/2)t + \sigma B(t)\}$, $\gamma \in \{\alpha, \beta\}$.

From the inequality $\alpha\le r(x)\le\beta$ for any $x\in (0,K^{+}]$ and the equation (\ref{eq:main-1}), we obtain the estimation
$N_\alpha(t)\le N(t)\le N_\beta(t)$, this is, bounds (\ref{cotas1}), for any $t>0$, such that $N(t)\le K^{+}$,
where $N(t)$ is solution of (\ref{eq:main-1}).

Therefore, the random times $\tau:=\inf \{t>0: N(t)\geq K^{+}\}$ and $\tau_{\gamma}:=\inf \{t>0: N_{\gamma}(t)\geq K^{+} \}$, $\gamma \in \{\alpha, \beta\}$,
satisfy the inequality $\tau_{\beta} \leq \tau \leq \tau_{\alpha}$.

From the equation (\ref{eq:Na-solution}) we obtain $\mathbb{E}[\tau_{\alpha}] = \ln(K^{+}/K_{-}) / (\alpha-\sigma^{2}/2)$, where the quantity $\alpha-\sigma^{2}/2$ is positive, due to the hypotheses of Theorem \ref{theo:main-2}. Analogously, we obtain $\displaystyle\mathbb{E}[\tau_{\beta}]=\ln(K^{+}/K_{-}) / (\beta-\sigma^{2}/2)$. Thus it holds that $ \mathbb{E}[\tau_{\beta}] \le \mathbb{E}[\tau] \le \mathbb{E}[\tau_{\alpha}] $, {\it i.e.}, the result given in (\ref{cotas2}). $\diamond$

\vspace{0.5cm}

\noindent{\it Construction of the extended solution}

The above theorems (up here) allow us the existence, uniqueness and other properties of boundedness for solutions of the problem (8) from some initial condition space-time and over a time interval defined by a stopping time, a stochastic variable that we have shown has finite expectation. However, we need a solution of (6) to prolong the process indefinitely.

Then, in order to monitor the dynamics of the stock beyond a first stop time, we define $\tilde{N}(t)=\mathds{1}_{\{t\in]0,\tau]\}}N(t)$ (where $\mathds{1}_{t\in I}$ is $1$ or $0$ according $t \in I$ or not). Notice that $0<\tilde{N}(t)\le K^{+}$ for any $t \in ]0,\tau]$, $\tilde{N}(0^-)=K_{-}$ and $\tilde{N}$ is a solution of (\ref{eq:SDE-main}) for the case $k=1$, taking $\tau_{0}=0$ y $\tau_{1}=\tau$.

\vspace{0.5cm}
Theorem \ref{theo:solution-ST} implies that the equation
$\mathrm{d}N(t)=r(N(t))N(t)\mathrm{d}t+\sigma N(t)\mathrm{d}B(t)$; with $t\in[\tau,T[$ and $N(\tau)=K_{-}$, has unique solution for every choice of $T>\tau$. In fact, $T$ is also a random variable dependent on $\tau$, but this equation career path is well defined, but is dependent on the solution in the interval $[0,\tau]$.

We define the stopping time $\tau_2=\inf\{t'>\tau;\ N_1(t')\le C\}$, then is a new finite expectation stopping time and the process
$$
\tilde{N}(t)=\mathds{1}_{t\in[\tau_{1},\tau_{2}]}N_{1}(t)+\mathds{1}_{\{t=\tau_{2}\}}K_{-},
$$
satisfies the following properties: (a) $0<\tilde{N}(t)<C$, for all $\tau_1\le t<\tau_2$; (b) $\tilde{N}(\tau_2)=K_{-}$ and $\tilde{N}(\cdot)$ solves the equation (\ref{eq:SDE-main}) for the case $k=2$.

Following the reasoning inductively, we have that:
$$
\begin{array}{c}
\displaystyle\quad\tilde{N}(t)=\sum_{k=1}^{\infty} \mathds{ 1}_{\{t\in[\tau_{k-1},\tau_{k}[\}}N_{k}(t),
\end{array}
$$
is the solution of the equation (\ref{eq:SDE-main}). The uniqueness is a consequence of the uniqueness of the solution in each path.

\section{Numerical simulation}
We will consider a biological resource that is governed by a rule type (\ref{eq:SDE-main}), where $r(N)=r_{0}(1-N/K)$, with $r_{0}=1/9$ and $K=9000[ton]$, {\it i.e.}, we have {\bf H1}.

We assume that the threshold for harvesting is $K^{+}=6000[ton]$ and the permitted quota is $Q=5000[ton]$, this is, $K_{-}=1000[ton]$.

Moreover, the stochastic intensity is given by $\sigma=1/3$. Since $r(\cdot)$ is decreasing and $r(0)=r_{0}=1/9$ a value greater than $\sigma^{2}/2=1/18$, we are under the hypotheses {\bf H2}.

\begin{figure}[h]
\includegraphics[width=12cm, height=6cm]{graph-1.eps}
\caption{Trajectory determined by an initial condition $N(0)=1000[ton]$ showing a sequence of closures with finite ``expectation" lengths.}
\label{S1}
\end{figure}

\begin{figure}[h]
\includegraphics[width=12cm, height=6cm]{graph-2.eps}
\caption{Comparison between a deterministic trajectory and one with stochastic effect, from a same initial condition $N(0)=1000[ton]$.}
\label{S2}
\end{figure}

\section{Discussion}
A fishery regulation, formulated with attachment only  to purely deterministic dynamic rules, can have disastrous effects on their practical implementation. In this paper, we considered a deterministic model of growth and capture, see (1), which does not express a major structuring, but that also is very general in its simplicity. In an effort to add realism, we added an environmental random effect on the per capita growth rate, see (6). We note that good and immediate properties of deterministic model, as the certainty of non-extinction of the resource and the finiteness of the closures, leading a type of bio-economic sustainability, are gone. Then, first, the problem became ensure existence of the resource and after, that a date for an upcoming lifting of closure effectively can be defined. The intuition led us to assume that a sufficiently small noise should approximate the dynamic one that presents the deterministic model and against high noise levels, this dynamic would tend to a total loss of predictability. Since we assumed the amount captured as a constant, the above meant orient our work toward finding a condition of compensation between per capita contribution to growth (per unit time) and the noise level. We refer to the hypothesis {\bf H2}, which allows us to affirm that although theoretically infinite closures may occur, this is unlikely, since the expectation of these is finite. Indeed, it has a bound that is even independent of when these closures begin.

We recognize that in the context of model (1) randomness may also occur in other determinants of the dynamics. For example: (a) The size of the quota effectively that was fished comparatively with the permitted one. (b) The regulatory imprecision in defining moments of access to the resource, which suggests a constant evaluation of the size of the resource. Sources (a) and (b) in principle can be considered ``more controllable" by the regulator, for this reason and for simplicity we have preferred to focus on the environmental noise.

Furthermore, since we have assumed that the per capita growth rate is positive when we are below carrying capacity, we have proven that abundance is always positive and there is no risk of extinction, despite the environmental noise. Thus, the hypothesis {\bf H1} covers cases (no noise) of pure compensation models, and also depensatory, see \cite{Clark}.

We must note that in case of critical depensation ({\it i.e.}, strong Allee effect) proof of Theorem does not apply and the possibility of extinction of the resource, even with moderate noise, it is possible. To find conditions and to develop demonstration for control model in this case means an interesting challenge for the future.\\

\noindent\textbf{This article is partially supported by grant FONDECYT 1120218.}


TITLE: Find a point using multilateration or triangulation
QUESTION [1 upvotes]: Suppose we have 3 points in a 3d coordinate system with the following locations.
A=(100,0,0)
B=(0,100,100)
C=(0,0,100)
If there is a 4th point D, where the distances from A, B, and C are known, can I find the coordinates of point D?  If so, how would I find the coordinates?  In my particular application, only one of the two points could be valid and it would be obvious which one.

REPLY [1 votes]: Suppose the given distances are $a,b,c$. Then the point you seek is at the intersection of three spheres: the sphere of radius $a$ centered at $A$, the sphere of radius $b$ centered at $B$, and the sphere of radius $c$ centered at $C$. There will (usually) be two intersection points, but you say it will be obvious which of the two is the desired one.
To intersect three spheres, look at this answer or this page on trilateration