\documentstyle[12pt]{article}

\textwidth=17.0cm \textheight=630pt \evensidemargin=1mm \oddsidemargin=1mm

\newtheorem{theorem}{Theorem}
\renewcommand{\refname}{Bibliography}

\title{Estimations of derivatives of $n$-convex functions}
\author{V.A.Malyshev}
\date{}

\begin{document}
\maketitle

\begin{abstract}
ø{\small\it $\!\!\!\!\!\!\!\!\!\!\!\!$ Estimations of the kind
$$
 ø\begin{array}{lll}
 \|u'-u_0'\|_{C[a+\delta,b-\delta]}&\le&
 A\|u-u_0\|_{C[a,b]}\delta^{-1}+B\|u_0^{(n)}\|_{C[a,b]}\delta^ {n-1},\medskip\\
 \|u^{(k)}-u_0^{(k)}\|_{C[a+\delta,b-\delta]}&\le&C\|u - u_0\|_{C[a,b]}^{(n-k)/n},\medskip\\
 \|u-u_0\|_{C[a+\delta,b-\delta]}&\le&C\|u-u_0\|_{L_1[a,b]}^{n/(n+1)}
\end{array}
$$
are obtained for $n$-convex functions.}
\end{abstract}



\noindent {Introduction}
\medskip

\noindent {1. Estimation of the first derivative}

\noindent {2. Estimation of higher derivatives}

\noindent {3. Estimation of $C$-norm by $L$-norm}

\medskip\noindent {Bibliography}


%%%%%%%%%%%%%%%%%%%
\section*{Introduction}


A function $u(x)$ is called $n$-convex on the interval $[a,b]$ if
$$\left|\begin{array}{llll}
 1&1&\cdots&1\\
 x_0&x_1&\cdots&x_n\\
 &&\cdots&\\
x_0^{n-1}&x_1^{n-1}&\cdots&x_n^{n-1}\\ u(x_0)&u(x_1)&\cdots&u(x_n)
\end{array}\right|\ge 0
$$
for all $a < x_0 < x_1 < \ldots < x_n < b$.



\bigskip
\bigskip
\bigskip

\begin{center}
 \begin{picture}(400,20)
 \put(-40,30){\line(1,0){70}}
 \put(-25,15){{\footnotesize{\it 2000 Mathematics Subject Classification.} Primary 26-06, 26A48,
 26A51,26D10.}}
 \put(-25,0){{\footnotesize{\it Key words and phrases.} Convex functions.}}
\end{picture}
\end{center}

\newpage


An $1$-convex function is increasing and a $2$-convex function is convex. If $n\ge 2$,
then an $n$-convex function $u(x)$ is $C^{n-2}$-differentiable. By the Popoviciu theorem,
a $C^{n-2}$-differentiable $u(x)$ is $n$-convex if and only if $u^{(n-2)}(x)$ is
convex~\cite{Kar}. In particular, a $C^{n}$-differentiable $u(x)$ is $n$-convex if and
only if $u^{(n)}(x)$ is positive. The article is devoted to the estimation of derivatives
of $C^{n}$-differentiable $n$-convex functions. Analogous estimations are given in the
paper~\cite{Mal}.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{1. Estimation of the first derivative}

\begin{theorem}
Let $n \ge 2$. Let $u_{0}(x)$ be $n$-convex and $C^n$-differentiable on the interval
$[a,b]$. There are exist $A$ and $B$ such that any $n$-convex $u(x)$ satisfies the
inequality
$$
\|u'-u_0'\|_{C[a+\delta,b-\delta]}\le
A\|u-u_0\|_{C[a,b]}\delta^{-1}+B\|u_0^{(n)}\|_{C[a,b]}\delta^ {n-1},
$$
where $0<\delta<(b-a)/n$.
\end{theorem}
\noindent{\bf Proof}. Consider the case $n=2$. Let us take a point
$$
a+\delta\le x_0\le b-\delta
$$
and define the functions
$$
ø\begin{array}{lll}
  øv_0(x)&=&u_0(x)-u_0(x_0)-u_0'(x_0)(x-x_0), \medskip\\
  øv(x)  &=&u(x)\;\;  -u_0(x_0)-u_0'(x_0)(x-x_0).
\end{array}
$$
We see that
$$
v_0(x_0)=v_0'(x_0)=0.
$$
Since $v(x)$ is convex, we have
$$
v(x_0)-v(x_0-\delta)\le v'(x_0-0)\delta\le v'(x_0+0)\delta\le v(x_0+\delta)-v(x_0).
$$
Let
$$
ø\varepsilon=\|v-v_0\|_{C[a,b]}=\|u-u_0\|_{C[a,b]}.
$$
Then
$$
v(x_0)-v(x_0-\delta)\ge-2\varepsilon+v_0(x_0)-v_0(x_0-\delta)=-2\varepsilon -
\frac{1}{2}v_0''(x_0-\Theta_-)\delta^2
$$
and
$$
v(x_0+\delta)-v(x_0)\le+2\varepsilon+v_0(x_0+\delta)-v_0(x_0)=+2\varepsilon+
\frac{1}{2}v_0''(x_0+\Theta_+)\delta^2,
$$
where $0<\Theta_-,\Theta_+<\delta$. We see that
$$
ø\begin{array}{lll}
  øu'(x_0-0)-u_0'(x_0)&=&v'(x_0-0), \medskip\\
  øu'(x_0+0)-u_0'(x_0)&=&v'(x_0+0).
\end{array}
$$
Therefore, we have
 ø\begin{eqnarray*}
  øu'(x_0-0)-u_0'(x_0)&\ge&-2\varepsilon\delta^{-1}-\frac{1}{2}\|u_0''\|_{C[a,b]}\delta,\medskip\\
  øu'(x_0+0)-u_0'(x_0)&\le&+2\varepsilon\delta^{-1}+\frac{1}{2}\|u_0''\|_{C[a,b]}\delta.
\end{eqnarray*}
Thus
$$
|u'(x_0)-u_0'(x_0)|\le2\varepsilon\delta^{-1}+\frac{1}{2}\|u_0''\|_{C[a,b]}\delta.
$$
This means that
$$
\|u'-u_0'\|_{C[a+\delta,b-\delta]}\le A\|u-u_0\|_{C[a,b]}\delta^{-1} +
B\|u_0''\|_{C[a,b]} \delta
$$
with $A=2$ and $B=1/2$.

Consider the case $n\ge 3$. Let us take a point
$$
a+\delta\le x_0\le b-\delta.
$$
Consider integers $p,q\ge 1$ satisfying the inequality
$$
a+p\delta\le x_0\le b-q\delta.
$$
Since $(p+q)\delta\le b-a$ and $n\delta\le b-a$, we may assume that $p+q=n$.

Consider $n$-convex functions
$$
v_0(x)=u_0(x)-\sum_{k=0}^{n-1}\frac{1}{k!}u^{(k)}_0(x_{0})(x-x_{0})^k
$$
and
$$
v(x)=u(x)-\sum_{k=0}^{n-1}\frac{1}{k!}u^{(k)}_0(x_{0})(x-x_{0})^k.
$$
We note that
$$
v_0(x_{0})=v_0'(x_{0})=\ldots=v^{(n-1)}_0(x_{0})=0.
$$
Suppose that $p+n+1$ is even. Let
$$
t_k=\left\{\begin{array}{lll}
  øx_{0}-(p+k)\delta & ø{\rm for} & ø0\le k\le p,  \\
  øx_{0}+h & ø{\rm for} & øk=p+1,\\
  øx_{0}+k\delta & ø{\rm for} & øp+2\le k\le n,  \\
\end{array}
\right.
$$
where $0<h<\delta$. Since $v(x)$ is $n$-convex, we have
$$
\left|\begin{array}{llllll}
 1      &\cdots&1      &1            &\cdots&1\\
 t_0    &\cdots&x_0    &x_{0}+h      &\cdots&t_n\\
 t_0^{2}&\cdots&x_0^{2}&(x_{0}+h)^{2}&\cdots&t_n^{2}\medskip\\
   &\cdots&&&\cdots&\medskip\\
 t_0^{n-1}&\cdots&x_0^{n-1}&(x_{0}+h)^{n-1}&\cdots&t_n^{n-1}\\
 v(t_0)    &\cdots&v(x_0)    &v(x_{0}+h)      &\cdots&v(t_n)\\
\end{array}\right|=
\left|\begin{array}{llll}
      \cdots&1      &1            &\cdots\\
     \cdots&x_0    &x_{0}+h      &\cdots\\
 \cdots&x_0^{2}&x_{0}^{2}+2hx_{0}+h^{2}&\cdots\medskip\\
   \cdots&&&\cdots\medskip\\
 \cdots&x_0^{n-1}&x_{0}+(n-1)hx_{0}^{n-2}+\cdots&\cdots\\
 \cdots&v(x_0)    &v(x_{0})+hv'(x_{0}+\Theta_{h})      &\cdots\\
\end{array}\right|\ge 0,
$$
where $0<\Theta_{h}<h$. We put $h\searrow0$ and obtain
$$
 \left|\begin{array}{lllllll}
 1      &\cdots&1          &0                &1            &\cdots&1\\
 t_0    &\cdots&t_p        &1                &t_{p+2}      &\cdots&t_n\\
 t_0^{2}&\cdots&t_p^{2}    &2x_{0}           &t_{p+2}^{2}  &\cdots&t_n^{2}\medskip\\
        &\cdots&           &                 &             &\cdots&\medskip\\
 t_0^{n-1}&\cdots&t_p^{n-1}&(n-2)x_{0}^{n-2} &t_{p+2}^{n-1} &\cdots&t_n^{n-1}\\
 v(t_0)    &\cdots&v(t_p)  &v'(x_{0})        &v(t_{p+2})  &\cdots&v(t_n)\\
\end{array}\right|\ge 0.
$$
Without loss of generality it can be assumed that $x_{0}=0$. Hence
$$
 \left|\begin{array}{lllllll}
 1          &\cdots&1         &0    &1        &\cdots&1                  \\
 -p\delta   &\cdots&-\delta   &1    &\delta   &\cdots&(q-1)\delta        \\
 (-p)^{2}   &\cdots&(-1)^{2}  &0    &1^{2}    &\cdots&(q-1)^{2}\medskip  \\
            &\cdots&          &     &         &\cdots&\medskip           \\
 (-p)^{n-1} &\cdots&(-1)^{n-1}&0    &1^{n-1}  &\cdots&(q-1)^{n-1}        \\
 v(-p\delta)&\cdots&v(-\delta)&v'(0)&v(\delta)&\cdots&v((q-1)\delta)     \\
\end{array}\right|\ge 0.
$$
Since $p+n+1$ is even, we have
$$
\delta v'(0)\ge\sum_{k=-p}^{q-1}A_{k}v(k\delta)
$$
with some constant
$
øA_{-p},\;\ldots,\;A_{-1},\;A_{0}=0,\;A_{1},\;\ldots,\;A_{q-1}.
$
We note that
$$
øv(x)=u(x)-u_{0}(x)+v_{0}(x)
$$
and
$$
v_0(k\delta)=\frac{k^n}{n!}v^{(n)}_0(\Theta_k)\delta^n=
 \frac{k^n}{n!}u^{(n)}_0(\Theta_k)\delta^n,
$$
where $-p\delta<\Theta_k<(q-1)\delta$. Therefore,
$$
\delta v'(0)\ge\sum_{k=-p}^{q-1}A_{k}(u(k\delta)-u_{0}(k\delta))+
 \sum_{k=-p}^{q-1}A_{k}\frac{k^n}{n!}u^{(n)}_0(\Theta_k)\delta^n.
$$
This proves the inequality
$$
\delta v'(x_{0})\ge-\left(A\|u-u_0\|_{C[a,b]}+B\|u^{(n)}_0\|_{C[a,b]}\delta^{n}\right).
$$
We put
$$
t_k=\left\{\begin{array}{lll}
  øx_{0}-(p-1+k)\delta & ø{\rm for} & ø0\le k\le p-1,  \\
  øx_{0}+h & ø{\rm for} & øk=p,\\
  øx_{0}+k\delta & ø{\rm for} & øp+1\le k\le n  \\
\end{array}
\right.
$$
and analogously obtain the inequality
$$
\delta v'(x_{0})\le A\|u-u_0\|_{C[a,b]}+B\|u^{(n)}_0\|_{C[a,b]}\delta^{n}.
$$
This means that
$$
\|u'-u_0'\|_{C[a+\delta,b-\delta]}\le
A\|u-u_0\|_{C[a,b]}\delta^{-1}+B\|u_0^{(n)}\|_{C[a,b]}\delta^ {n-1},
$$
where $0<\delta<(b-a)/n$.

The case with odd $p+n+1$ is considered analogously. The theorem is proved. $\bullet$
\medskip


\noindent{\bf  Corollary.} There exists $A$ such that any $n$-convex $u(x)$ satisfies the
inequality
$$
\|u'\|_{C[a+\delta,b-\delta]}\le A\|u\|_{C[a,b]},
$$
where $0<\delta<(b-a)/n$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{2. Estimation of higher derivatives}

\begin{theorem}
Let $n\ge 2$. Let $u_0(x)$ be $n$-convex and $C^n$-differentiable on the interval
$[a,b]$. For any $R>0$ and $0<\Delta<(b-a)/2$ there exists $C>0$ such that any $n$-convex
$u(x)$ from the ball
$$
\|u-u_0\|_{C[a,b]}\le R
$$
satisfies the inequality
$$
\|u^{(k)} - u_0^{(k)}\|_{C[a+\Delta,b-\Delta]}\le C\|u - u_0\|_{C[a,b]}^{(n-k)/n},
$$
where  $1\le k\le n-1$.
\end{theorem}
\noindent{\bf Proof.}  Consider the case $k=1$. Let
$$
0<r<\min\left\{\left(\frac{b-a}{n}\right)^n,\Delta^n\right\}.
$$
Assume that
$$
\|u-u_0\|_{C[a,b]}\le r.
$$
We put
$$
\delta=\|u - u_0\|_{C[a,b]}^{1/n}
$$
and see that $\delta<\Delta$ and $\delta<(b-a)/n$. Let
$$
C' = A + B\|u_0^{(n)}\|_{C[a,b]},
$$
where $A$ and $B$ are taken from Theorem 1.

By Theorem 1, we have
\begin{eqnarray*}
ø\|u' - u_0'\|_{C[a+\Delta,b-\Delta]}&\le& \|u' - u_0'\|_{C[a+\delta,b-\delta]}\medskip\\
 &\le& A\|u - u_0\|_{C[a,b]}\delta^{-1}+B \|u_0^{(n)}\|_{C[a,b]}\delta^{n-1}\medskip\\
 &=&\left(A+ B\|u_0^{(n)}\|_{C[a,b]}\right)\|u-u_0\|_{C[a,b]}^{(n-1)/n}\medskip\\
 &=&C'\|u-u_0\|_{C[a,b]}^{(n-1)/n}.
\end{eqnarray*}
Hence the inequality
$$
\|u-u_0\|_{C[a,b]}\le r
$$
implies the inequality
$$
\|u'-u_0'\|_{C[a+\Delta,b-\Delta]}\le C'\|u-u_0\|_{C[a,b]}^{(n-1)/n}.
$$
If $R \le r$, the statement is proves. Let $R > r$. Assume that
$$
r<\|u-u_0\|_{C[a,b]}\le R.
$$
Let
$$
\delta = \frac{1}{2}\min\left\{\frac{b-a}{n},\Delta\right\}
$$
and
$$
C'' = AR^{1/n}\delta^{-1} + B\|u_0^{(n)}\|_{C[a,b]}\delta^{n-1}r ^{(1-n)/n}.
$$
By Theorem 1, we have
\begin{eqnarray*}
\|u'-u_0'\|_{C[a+\Delta,b-\Delta]}&\le&\|u'-u_0'\|_{C[a+\delta,b-\delta]}\medskip\\
 &\le& A\|u-u_0\|_{C[a,b]}\delta^{-1}+ B\|u_0^{(n)}\|_{C[a,b]}\delta^{n-1}\medskip\\
 &=&\left\{A\|u-u_0\|_{C[a,b]}^{1/n}\delta^{-1}+
B\|u_0^{(n)}\|_{C[a,b]}\delta^{n-1}\|u-u_0\|_{C[a,b]}^{(1-n)/n}
\right\}\|u-u_0\|_{C[a,b]}^{(n-1)/n}\medskip\\
 &\le& C''\|u-u_0\|_{C[a,b]}^{(n-1)/n}.
\end{eqnarray*}
Let $C=\max\{C',C''\}$. Then the inequality
$$
\|u-u_0\|_{C[a,b]}\le R
$$
implies the inequality
$$
\|u'-u_0'\|_{C[a+\Delta,b-\Delta]}\le C\|u-u_0\|_{C[a,b]}^{(n-1)/n},
$$
where $C$ depends only on $\Delta,R$ and $u_0$.


Consider the case $2\le k\le n-1$. Since the derivative of an $n$-convex function is
$(n-1)$-convex, we prove by induction.

Let
$$
0 = \Delta_0<\Delta_1<\Delta_2<\ldots<\Delta_{n-2}<\Delta_{n-1}=\Delta.
$$
We put
$$
R_0 = R
$$
and define $C_0$ in such a way that any $n$-convex function $u(x)$ satisfying the
inequality
$$
\|u-u_0\|_{C[a+\Delta_0,b-\Delta_0]}\le R_0
$$
satisfies the inequality
$$
\|u'-u_0'\|_{C[a+\Delta_1,b-\Delta_1]}\le
C_0\|u-u_0\|_{C[a+\Delta_0,b-\Delta_0]}^{(n-1)/n}.
$$
Then we put
$$
R_1=C_0R_0^{(n-1)/n}
$$
and define $C_1$ in such a way that any $(n-1)$-convex function $u(x)$ satisfying the
inequality
$$
\|u-u_0'\|_{C[a+\Delta_1,b-\Delta_1]}\le R_1
$$
satisfies the inequality
$$
\|u' - u_0''\|_{C[a+\Delta_2,b-\Delta_2]}\le C_1 \|u -
u_0'\|_{C[a+\Delta_1,b-\Delta_1]}^{(n-2)/(n-1)}.
$$
Then we put
$$
R_2=C_1R_1^{(n-2)/(n-1)}
$$
and continue the process.

Let an $n$-convex function $u(x)$ satisfy the inequality
$$
\|u-u_0\|_{C[a,b]}\le R.
$$
We have
$$
\|u' - u_0'\|_{C[a+\Delta_1,b-\Delta_1]}\le C_0\|u - u_0\|_{C[a,b]}^{(n-1)/n}\le R_1.
$$
Then we have
$$
\|u'' - u_0''\|_{C[a+\Delta_2,b-\Delta_2]}\le C_1\|u' -
u_0'\|_{C[a+\Delta_1,b-\Delta_1]}^{(n-2)/(n-1)}\le C_1C_0^{(n-2)/(n-1)}\|u - u_0\|_
{C[a,b]}^{(n-2)/n}\le R_2.
$$
And so on. Therefore we have
$$
\|u^{(k)} - u_0^{(k)}\|_{C[a+\Delta,b-\Delta]}\le C\|u - u_0\|_{C[a,b]}^{(n-k)/n}
$$
with some $C>0$. $\bullet$
\medskip


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{3. Estimation of $C$-norm by $L$-norm}

\begin{theorem}
Let $n\ge 1$. Let $u_0(x)$ be $n$-convex and $C^n$-differentiable on the interval
$[a,b]$. For any $R>0$ and $0<\Delta<(b-a)/2$ there exists $C>0$ such that any $n$-convex
function $u(x)$ from the ball
$$
\|u-u_0\|_{L_1[a,b]}\le R
$$
satisfies the inequality
$$
\|u-u_0\|_{C[a+\Delta,b-\Delta]}\le C\|u-u_0\|_{L_1[a,b]}^{n/(n+1)}.
$$
\end{theorem}
\noindent{\bf Proof.} Consider $(n+1)$-convex function
$$
U_0(x)=\int_a^xu_0(t)\,dt.
$$
By Theorem 2, there exists $C>0$ such that any $(n+1)$-convex function $U(x)$ satisfying
the inequality
$$
\|U-U_0\|_{L_1[a,b]}\le R
$$
satisfies the inequality
$$
\|U'-U_0'\|_{C[a+\Delta,b-\Delta]}\le C\|U-U_0\|_{C[a,b]}^{n/(n+1)}.
$$
Let an $n$-convex $u(x)$ satisfy the inequality
$$
\|u-u_0\|_{L_1[a,b]}\le R.
$$
We put
$$
U(x)=\int_a^xu(t)\,dt.
$$
Then
\begin{eqnarray*}
 \|U-U_0\|_{C[a,b]}
 & = &\sup_{a\le x\le b}\left|\int_a^xu(t)\,dt-\int_a^xu_0(t)\,dt\right|\medskip\\
 &\le&\sup_{a\le x\le b}\int_a^x|u(t)\,dt-u_0(t)|\,dt\medskip\\
 &\le&\|u-u_0\|_{L_1[a,b]}.
\end{eqnarray*}
Hence
$$
\|U-U_0\|_{L_1[a,b]}\le R.
$$
Consequently
\begin{eqnarray*}
 \|u-u_0\|_{C[a+\Delta,b-\Delta]}
 &=&\|U'-U_0'\|_{C[a+\Delta,b-\Delta]}\medskip\\
 &\le&C\|U-U_0\|_{C[a,b]}^{n/(n+1)}\medskip\\
 &\le&C\|u-u_0\|_{L_1[a,b]}^{n/(n+1)}.\medskip\\
\end{eqnarray*}
The theorem is proved. $\bullet$
\medskip





\begin{thebibliography}{9}


\bibitem{Kar} S.Karlin, W.J.Studden. Tchebysheff systems: with applications in analysis
and statistics. Interscience publications. Wiley $\&$ Sons.

\bibitem{Mal} V.A.Malyshev. Nonlimear embedding theorems. St.Peterburg Math. J. Vol. 5
(1994), No. 6



\end{thebibliography}

Vladimir Malyshev

wmal@ryb.adm.yar.ru

\end{document}
