%   Master:  proceed.tex%         dvips *-O-22mm,-8mm -Pljh -z -Ppdf*%   -T173mm,252mm%\documentclass[runningheads,dvips]{llncs}%%\usepackage{amsmath,amsfonts}%\usepackage{equadiff}%\usepackage[colorlinks]{hyperref}%\begin{document}\begin{contribution}\newtheorem{amsl}{Lemma}[section]{\bfseries}{\itshape}                                    \newtheorem{amst}[amsl]{Theorem}{\bfseries}{\itshape}\newtheorem{amsc}[amsl]{Corollary}{\bfseries}{\itshape}\newtheorem{amsd}{Definition}[section]{\bfseries}{\upshape}                                    \newtheorem{amsr}{Remark}[section]{\itshape}{\upshape}                                    \newtheorem{amsp}{Problem}{\bfseries}{\upshape}                                    \renewcommand{\theamsl}{\thesection.\arabic{amsl}}\renewcommand{\theHamsl}{\thesection.\arabic{amsl}}\renewcommand{\theamsd}{\thesection.\arabic{amsd}}\renewcommand{\theHamsd}{\thesection.\arabic{amsd}}\renewcommand{\theamsr}{\thesection.\arabic{amsr}}\renewcommand{\theHamsr}{\thesection.\arabic{amsr}}\ifacrobat\renewcommand{\theequation}{\thesection.\arabic{equation}\rlap{\textcolor{white}{.aga}}}\else\renewcommand{\theequation}{\thesection.\arabic{equation}}\fi

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%\let\epsilon\varepsilon\def\tri{\triangle}\def\hdots{\cdots}\def\e{\varepsilon}\def\sgn{\mbox{$\mathrm{sgn}$\,}}\def\matrix{\begin{array}}\def\endmatrix{\end{array}}\def\Cal{\mathcal}\def\boldGamma{\boldsymbol{\Gamma}}\def\flushpar{\noindent}\def\no{\noindent}\def\bold{\bfseries}\def\dis{\displaystyle}%\def\binom#1#2{{#1 \choose #2}}%\def\qed{\hspace{3mm}\hbox{\vrule height7pt width5pt}}\def\l{\left}\def\r{\right}\def\NN{\mathbb{N}}\def\RR{\mathbb{R}}\title{Compactness Condition for Boundary\\ Value Problems}\titlerunning{Compactness Condition for Boundary Value Problems}% abbreviated title (for running head)\toctitle{Compactness Condition for Boundary Value Problems}%\author{Ravi P. \surname{Agarwal}}%\authorrunning{Ravi P. Agarwal}% abbreviated author list (for running head)% modified list of authors for the TOC (add the affiliations)\tocauthor{Ravi P. Agarwal (National University of Singapore)}\institute{Department of Mathematics\\National University of Singapore\\10 Kent Ridge Crescent, Singapore 119260\\\email{MATRAVIP@leonis.nus.sg}\\[8mm]\mbox{}\hfill{\bfseries Dedicated to Professor Lloyd K. Jackson}}\zalozka{Agarwal}{agarwal}
\rejstrikautoru{Agarval@Agarval Ravi P.}\maketitle\begin{abstract} In existence and uniqueness theory of boundaryvalue problems for ordinary differential equations {\itshape Compactness Condition}plays an important role.  It has been a long standing problem whether other conditions imposed on the differential equations imply this compactness condition. In this lecture we shall survey known results on this problem,including its complete unpublished proof essentially due to L. Jackson and K. Schrader. We shall also discuss some related problems.\end{abstract}\begin{AMSclassification} \subject{34B15}, \subject{34C10}\end{AMSclassification}\begin{keywords}\keyword{Boundary value problem}, \keyword{Kamke convergence theorem}, \keyword{Banach indicatrix theorem}, \keyword{Helly selection theorem}, \keyword{total variation}\end{keywords}\section{Introduction}In this lecture we shall consider the following $n~(\geq 2)$th order nonlineardifferential equation\begin{equation}y^{(n)}=f(x,y,y',\dots,y^{(q)}),~~~~0\leq q\leq n-1,~~\mbox{but fixed}. \label{aga1.1}\end{equation}With respect to \eqref{aga1.1} we shall assume that\vspace{2mm}\no (A)\odkaz{A}~~$f(x,u_0,u_1,\dots,u_q) : (a,b)\times \RR^{q+1}\to \RR$ is continuous.\vspace{1mm}\no (B)\odkaz{B}~~Solutions of initial value problems for \eqref{aga1.1} are unique.\vspace{1mm}\no (C)\odkaz{C}~~Solutions of \eqref{aga1.1} extend to $(a,b).$\vspace{1mm}\no (D$_n$)\odkaz{D}~~For any $a<a_1<a_2<\cdots<a_n<b$ and any solutions $y(x)$ and $z(x)$ of \eqref{aga1.1}, it follows that $y(a_i)=z(a_i),~1\leq i\leq n$ implies $y(x)\equiv z(x),$ i.e., the differential equation \eqref{aga1.1} is $n$-point {\it disconjugate} on $(a,b).$ \vspace{2mm}In the study of boundary value problems for the differential equation \eqref{aga1.1}, one of the Propositions which has attracted several Mathematicians and has ledto substantially new mathematics is {\itshape whether conditions \textup{\eqref{agaA} -- \eqrefd{$_n$}} imply the following {\bfseries compactness condition:}} \vspace{2mm}\no (E)\odkaz{E}~~If $[c,d]$ is a compact subinterval of $(a,b)$ and $\{y_m(x)\}$ is a sequence of solutions of \eqref{aga1.1} which is uniformly bounded, i.e., $|y_m(x)|\leq M$ on $[c,d]$ for some $M>0$ and all $m=1,2,\dots,$ then there is asubsequence $\{y_{m(j)}(x)\}$ such that $\{y_{m(j)}^{(i)}(x)\}$ converges uniformly on $[c,d]$ for each $0\leq i\leq n-1.$ \vspace{2mm}In this lecture we shall survey most of the known results on this Proposition, and touch on some related topics.  \section{Preliminary Results}We shall need the following version of Kamke's convergence theorem.\begin{amst} \textbf{\textup{(\cite[p. 14]{aga5})}} \label{agat2.1}Assume that for the differential equation \eqref{aga1.1} the conditions \eqref{agaA} and \eqref{agaC} are satisfied. Then, if $\{y_m(x)\}$ is a sequence of solutions of \eqref{aga1.1} such that there exists a sequence$\{x_m\}\subset (a,b)$ with $\lim_{m\to \infty}x_m=x_0\in (a,b),~\lim_{m\to\infty}y_m^{(i)}(x_m)=y_i,~0\leq i\leq n-1.$ Then, there is a solution $y(x)$ of the differential equation \eqref{aga1.1} satisfying the initial conditions $y^{(i)}(x_0)=y_i,~0\leq i\leq n-1,$ and a subsequence $\{y_{m(j)}(x)\}$ of $\{y_m(x)\}$ such that $\lim_{j\to\infty}y_{m(j)}^{(i)}(x)=y^{(i)}(x),~0\leq i\leq n-1,$ uniformly on each compact subinterval of $(a,b).$ \end{amst}
\unskip\begin{amsl} \label{agal2.2} Let $y(x)\in C^{(n)}[a_1,a_r],$ satisfying\begin{equation}\begin{array}{l}y(a_i)=y'(a_i)=\cdots = y^{(k_i)}(a_i)=0,~~1\leq i\leqr~(\geq 2)\\~\\\displaystylea<a_1<a_2<\cdots<a_r<b,~~k_i\geq 0,~~\sum_{i=1}^{r}k_i+r=n.\end{array}\label{aga2.1}\end{equation}Then, there exist constants $C_{n,k},~0\leq k\leq n-1,$ such that\begin{equation}|y^{(k)}(x)|\leq C_{n,k}(a_r-a_1)^{n-k}\max_{a_1\leq x\leq a_r}|y^{(n)}(x)|.\label{aga2.2}\end{equation}\end{amsl}The problem of finding the best possible constants $C_{n,k}$ in \eqref{aga2.2} is one of the most outstanding problems in polynomial interpolation theory \cite{aga1,aga2}. \vspace{3mm}Inequalities \eqref{aga2.2} will be used now to prove local existence of solutions of the differential equation \eqref{aga1.1} satisfying the {\itshape $r$-point conjugate boundary conditions}\begin{equation}y(a_i)=A_{1,i},~y'(a_i)=A_{2,i},\dots,y^{(k_i)}(a_i)=A_{k_i+1,i},~~1\leq i\leq r.\label{aga2.3}\end{equation}\vspace{-5mm}
\begin{amst}[\cite{aga1}] \label{agat2.3}Assume that for the differential equation \eqref{aga1.1} the condition \eqref{agaA} is satisfied. Further, assume that\vspace{2mm}\no \textup{(i)} $K_i>0,~0\leq i\leq q$ are given real numbers and let $Q$ be themaximum of $|f(x,u_0,u_1,\dots,u_q)|$ on the compact set $[a_1,a_r]\timesD_0,$ where $$D_0=\{(u_0,u_1,\dots,u_q)~:~|u_i|\leq 2K_i,~0\leq i\leq q\},$$\vspace{2mm}\no \textup{(ii)} $\displaystyle \max_{a_1\leq x\leq a_r}|p^{(i)}(x)|\leq K_i,~0\leq i\leq q,$ where $p(x)$ is the Hermite interpolating polynomial$$p(x)=\sum_{i=1}^{r}\sum_{j=0}^{k_i}\sum_{\ell=0}^{k_i-j}\frac{1}{j!\ell!}\l[\frac{(x-a_i)^{k_i+1}}{\Omega(x)}\r]_{x=a_i}^{(\ell)}\frac{\Omega(x)}{(x-a_i)^{k_i+1-j-\ell}}A_{j+1,i}$$and$$\Omega(x)=\prod_{i=1}^{r}(x-a_i)^{k_i+1},$$\vspace{2mm}\no \textup{(iii)} $\displaystyle (a_r-a_1)\leq \l(\frac{K_i}{QC_{n,i}}\r)^{1/(n-i)},~0\leq i\leq q.$\vspace{2mm}\no Then, the boundary value problem \eqref{aga1.1}, \eqref{aga2.3} has a solution in $D_0.$\end{amst}\unskip\begin{proof}The set$$B[a_1,a_r]=\l\{y(x)\in C^{(q)}[a_1,a_r]~:~\|y^{(i)}\|\leq 2K_i,~0\leq i\leq q\r\},$$where $$\|y^{(i)}\|=\max_{a_1\leq x\leq a_r}|y^{(i)}(x)|$$is a closed convex subset of the Banach space $C^{(q)}[a_1,a_r].$ Consider an operator $T: C^{(q)}[a_1,a_r]\to C^{(n)}[a_1,a_r]$ as follows\begin{equation}(Ty)(x)=p(x)+\int_{a_1}^{a_r}g(x,t)f(t,y(t),y'(t),\dots,y^{(q)}(t))dt,\label{aga2.4}\end{equation}where $g(x,t)$ is the Green's function of the boundary value problem $y^{(n)}=0,$ \eqref{aga2.1}. Obviously, any fixed point of \eqref{aga2.4} is a solution of \eqref{aga1.1}, \eqref{aga2.3}.We note that $(Ty)(x)-p(x)$ satisfies the conditions of Lemma \ref{agal2.2}, and$$(Ty)^{(n)}(x)-p^{(n)}(x)=(Ty)^{(n)}(x)=f(x,y(x),y'(x),\dots,y^{(q)}(x)).$$Thus, for all $y(x)\in B[a_1,a_r],$ $\|(Ty)^{(n)}\|\leq Q,$ and$$\|(Ty)^{(i)}-p^{(i)}\|\leq C_{n,i}Q(a_r-a_1)^{n-i},~~0\leq i\leq q$$which also implies that\begin{equation}\|(Ty)^{(i)}\|\leq \|p^{(i)}\|+ C_{n,i}Q(a_r-a_1)^{n-i}\leq K_i+K_i=2K_i,~~0\leq i\leq q.\label{aga2.5}\end{equation}Thus, the operator $T$ maps $B[a_1,a_r]$ into itself. Further, the inequalities\eqref{aga2.5} imply that the sets $\{(Ty)^{(i)}(x)~:~y(x)\in B[a_1,a_r]\},~0\leq i\leq q$ are uniformly bounded and equicontinuous on $[a_1,a_r].$ Hence, $\overline{TB}[a_1,a_r]$ is compact follows from the Ascoli-Arzela theorem. The Schauder fixed point theorem is applicable and a fixed point of \eqref{aga2.4} in$D_0$ exits.\end{proof}\unskip\begin{amsc} \label{agac2.4} Assume that for the differential equation \eqref{aga1.1} the condition \eqref{agaA} is satisfied. Further, assume that there exist constants$N_i\geq 0,~0\leq i\leq q$ such that $\max_{a_1\leq x\leq a_r}|p^{(i)}(x)|$$\leq N_i,~0\leq i\leq q.$ Then, there exists a $\delta=\delta(N_0,N_1,\dots,N_q)>0$ such that if $a_r-a_1\leq \delta,$ the boundary value problem \eqref{aga1.1},\eqref{aga2.3} has a solution $y(x).$ Furthermore, $$|y^{(i)}(x)|\leq N_i+1,~0\leq i\leq q~~~~\mbox{ on}~~~ [a_1,a_r].$$\end{amsc}\unskip\begin{amst} \label{agat2.5} Assume that for the differential equation \eqref{aga1.1} the condition \eqref{agaA} is satisfied. Further, assume that the conditions (i) and (iii)of Theorem~\upref{agat2.3} are satisfied. Then, for any $g(x)\in C^{(n-1)}[a_1,a_r]$ thedifferential equation \eqref{aga1.1} together with\begin{equation}y^{(j)}(a_i)=g^{(j)}(a_i), ~~~~0\leq j\leq k_i,~~1\leq i\leq r\label{aga2.6}\end{equation}has a solution, if$$\sum_{j=i}^{n-1}M_j(a_r-a_1)^{j-i}\leq K_i,~~~~0\leq i\leq q$$where$$M_j=\max_{a_1\leq x\leq a_r}|g^{(j)}(x)|,~~~~0\leq j\leq n-1.$$\end{amst}\unskip\begin{proof}We need to verify that the condition (ii) of Theorem \ref{agat2.3} issatisfied. For this, in $p(x)$ we take $A_{j+1,i}=g^{(j)}(a_i),~0\leq j\leq k_i,~1\leq i\leq r.$ Then, the function $\phi(x)=g(x)-p(x)$ has $n$ zeros in $[a_1,a_r].$ Thus, from the generalized Rolle's theorem $\phi^{(k)}(x),~1\leq k\leq n-1$ vanishes at least $n-k$ times in $(a_1,a_r).$ Let $x_k\in (a_1,a_r)$ be any zero of $\phi^{(k)}(x),$ then$$\l|p^{(n-1)}(x)\r|=\l|p^{(n-1)}(x_{n-1})\r|=\l|g^{(n-1)}(x_{n-1})\r|\leq \max_{a_1\leq x \leq a_r}\l|g^{(n-1)}(x)\r|=M_{n-1}$$and$$\begin{array}{lll}\displaystyle \l|p^{(n-2)}(x)\r|&\leq & \displaystyle \l|p^{(n-2)}(x_{n-2})\r|+\l|\int_{x_{n-2}}^{x}\l|p^{(n-1)}(t)\r|dt\r|\\~\\&=&\displaystyle \l|g^{(n-2)}(x_{n-2})\r|+\l|\int_{x_{n-2}}^x\l|g^{(n-1)}(x_{n-1})\r|dt\r|\\~\\&\leq & M_{n-2}+M_{n-1}(a_r-a_1).\end{array}$$Using the same argument repeatedly, we obtain$$\l|p^{(i)}(x)\r|\leq \sum_{j=i}^{n-1}M_j(a_r-a_1)^{j-i}.$$\end{proof}\unskip\begin{amsc} \label{agac2.6} Assume that for the differential equation \eqref{aga1.1} the condition \eqref{agaA} is satisfied. Then, for any $g(x)\in C^{(n-1)}[a_1,a_r]$ thereexist constants $\delta>0$, $N_i\geq 0,~0\leq i\leq q,$ all depending on $g(x)$such that the boundary value problem \eqref{aga1.1}, \eqref{aga2.6} has a solution $y(x),$provided $a_r-a_1\leq \delta.$ Furthermore, $|y^{(i)}(x)|\leq N_i+1,~0\leq i\leq q$ on $[a_1,a_r].$\end{amsc}\unskip\begin{amsc} \label{agac2.7} Assume that for the differential equation \eqref{aga1.1} the condition \eqref{agaA} is satisfied. Further, assume that there exist constants$N_i\geq 0,~0\leq i\leq q$ such that $\max_{a_1\leq x\leq a_r}|p^{(i)}(x)|$$\leq N_i,~0\leq i\leq q.$ Then, there exist a $\delta=\delta(N_0,N_1,\dots,N_q)>0,$ and an $\epsilon=\epsilon(a_1,\dots,a_r)$ such that for $a_r-a_1\leq \delta,$ the boundary value problem \eqref{aga1.1},$$y(a_i)=A_{1,i}+\epsilon_{1,i},~y'(a_i)=A_{2,i}+\epsilon_{2,i},\dots,y^{(k_i)}(a_i)=A_{k_i+1,i}+\epsilon_{k_i+1,i},~\,1\leq i\leq r$$has a solution $y_{\epsilon}(x),$ provided $|\epsilon_{j,i}|\leq \epsilon,~ 0\leq j\leq k_i,~1\leq i\leq r.$ Furthermore, $|y^{(i)}_\epsilon(x)|\leq N_i+1,~0\leq i\leq q$ on $[a_1,a_r].$\end{amsc}\section{The Case $n=2$}For the second order differential equation \eqref{aga1.1} only conditions \eqref{agaA} and \eqref{agaC} imply \eqref{agaE}. We shall prove this in the following:\begin{amst}\label{agat3.1} If the differential equation \eqref{aga1.1} is of second order and satisfies conditions \eqref{agaA} and \eqref{agaC}, then \eqref{aga1.1} also satisfies condition \eqref{agaE}.\end{amst}\unskip\begin{proof}If $\{y_m(x)\}$ is a sequence of solutions of \eqref{aga1.1} with$|y_m(x)|\leq M$ on $[c,d]\subset (a,b)$ for some $M>0,$ and each $m\geq 1,$ then for each $m$ there is a $x_m\in (c,d)$ such that $$|y'_m(x_m)|=\frac{|y_m(d)-y_m(c)|}{d-c}\leq \frac{2M}{d-c}.$$ Consequently, $\{x_m\},~\{y_m(x_m)\}$ and$\{y'_m(x_m)\}$ are bounded sequences. By taking subsequences in successionwhich converge, we conclude that there exist values $x_0,~y_0,~y'_0$ such that$x_{m(1)}\to x_0,~y_{m(1)}(x_{m(1)})\to y_0~y'_{m(1)}(x_{m(1)})\to y'_0,$where $\{m(1)\}$ is some subsequence of $\{m\}.$ Thus, by Theorem \ref{agat2.1} thereis a subsequence $\{y_{m(2)}(x)\}$ of $\{y_{m(1)}(x)\}$ and a solution $y(x)$ of \eqref{aga1.1} satisfying $y(x_0)=y_0,~y'(x_0)=y'_0$ such that $\lim_{m\to\infty}y_{m(2)}^{(i)}(x)=y^{(i)}(x),~i=0,1,$ uniformly on $[c,d].$\end{proof}\section{Case $n=3$}If the differential equation \eqref{aga1.1} is of third order, then conditions \eqref{agaA}and \eqref{agaC} are not enough for \eqref{agaE}. In fact, the equation $y'''=-[y']^3$ satisfies\eqref{agaA} and \eqref{agaC} on $\RR^4,$ but the sequence $\{y_m(x)\}$ of solutions of theinitial value problem$$y'''=-[y']^3,~~~~y(0)=y'(0)=0,~~y''(0)=m$$for $m=1,2,\dots,$ is uniformly bounded on $\RR$ and does not contain asubsequence satisfying \eqref{agaE} on any compact subinterval of $\RR.$ Here, we shall show that conditions \eqref{agaA}, \eqref{agaC} and \eqrefd{$_3$} do imply \eqref{agaE}. However, for thisan immediate appeal to Theorem \ref{agat2.1} is not possible and we shall need the following lemmas.\begin{amsl}\label{agal4.1} Assume that the differential equation \eqref{aga1.1} is of third order and satisfies the condition \eqref{agaA}. Then, given any compact subinterval$[c,d]\subset (a,b)$ and any fixed $M>0,$ there is a $\delta(M)>0$ such thatfor any $[a_1,a_2]\subset [c,d]$ with $a_2-a_1\leq \delta(M),$ and any real$\alpha$ with $|\alpha|\leq M,$ \eqref{aga1.1} has solutions satisfying each of the boundary conditions$$y(a_1)=y(a_2)=\alpha,~y'(a_1)=0~~~~\mbox{and}~~~~y(a_1)=y(a_2)=\alpha,~y'(a_2)=0.$$Furthermore, for any such solution $|y'(x)|\leq 1$ and $|y''(x)|\leq 1$ on$[a_1,a_2].$\end{amsl}\unskip\begin{proof} This is a particular case of Corollary \ref{agac2.4}. \end{proof}\unskip\begin{amsl}[\cite{aga11}] \label{agal4.2}Assume that the differential equation \eqref{aga1.1} is of third order and satisfies the condition \eqref{agaA}. Let $\phi(x),~\psi(x)$ be of class $C^{(2)}$ on $[a_1-\tau,a_1+\tau]\subset (a,b)$ with $\phi(a_1)=\psi(a_1),~\phi'(a_1)=\psi'(a_1)$ and $\phi''(a_1)<\psi''(a_1).$ Then, there is a $\delta,~0<\delta\leq \tau,$ such that all solutions $y(x)$ of \eqref{aga1.1} with the initial conditions $y(a_1)=y_0=\phi(a_1),~y'(a_1)=y_1=\phi'(a_1),$ and $y''(a_1)=y_2=\frac{1}{2}\l[\phi''(a_1)+\psi''(a_1)\r]$ exist on $[a_1-\delta,a_1+\delta]$ and satisfy $\phi(x)<y(x)<\psi(x)$ for $0<|x-a_1|\leq \delta.$\end{amsl}\unskip\begin{proof}Let $8\rho=\psi''(a_1)-\phi''(a_1)$ and choose $\delta_0,~0<\delta_0\leq \tau$ such that $|\phi''(x)-\phi''(a_1)|\leq \rho$ and $|\psi''(x)-\psi''(a_1)|\leq \rho$ for $|x-a_1|\leq \delta_0.$ Let $M>0$be a bound for $f(x,y,y',y'')$ on the compact set $$\l\{(x,y,y',y'') ~:~|x-a_1|~\leq~ \delta_0,~|y-y_0|\leq 1,~|y'-y_1|\leq 1,~|y''-y_2|\leq 1\r\}.$$Then, it follows from the relations $$\begin{array}{l}\displaystyle y(x)=y_0+\int_{a_1}^{x}y'(t)dt,~~~~~~~~y'(x)=y_1+\int_{a_1}^{x}y''(t)dt\\~\\\displaystyle y''(x)=y_2+\int_{a_1}^{x}f(t,y(t),y'(t),y''(t))dt\end{array}$$that all solutions of the stated initial value problem exist on the closed interval $[a_1-\delta_1,a_1+\delta_1],$ where$$\delta_1=\min\l\{\delta_0,\frac{1}{M},\frac{1}{1+|y_2|},\frac{1}{1+|y_1|}\r\}.$$Thus, if $\delta =\min \{\delta_1,\rho/M\},$ it follows that for all solutions $y(x)$ of the initial value problem $|y''(x)-y_2|\leq \rho$ for $|x-a_1|\leq\delta.$ Hence, for all solutions $y(x)$ of the initial value problem$$y''(x)-\phi''(x)\geq 2\rho~~~~\mbox{and}~~~~\psi''(x)-y''(x)\geq 2\rho$$on $[a_1-\delta,a_1+\delta].$ \end{proof}\unskip\begin{amsl}[\cite{aga11}] \label{agal4.3}Assume that the differential equation \eqref{aga1.1} is of third order and satisfies the conditions \eqref{agaA}, \eqref{agaC} and \eqrefd{$_3$}. Then, solutions of two point boundary value problems for \eqref{aga1.1} are unique, i.e., the following condition \eqrefd[2]{$_2$} holds:\vspace{2mm}\no \textup{(D{$_2$})}\odkaz[2]{D}~~ If $a<a_1<a_2<b,$ and $y(x),~z(x)$ are both solutionsof \eqref{aga1.1} satisfying $y(a_1)=z(a_1),~y'(a_1)=z'(a_1),~y(a_2)=z(a_2),$ or $y(a_1)=z(a_1),~y(a_2)=z(a_2),~y'(a_2)=z'(a_2),$ then it follows that $y(x)\equiv z(x)$ on $[a_1,a_2].$\end{amsl}\unskip\begin{proof}We shall consider only the case where $y(a_1)=z(a_1)$,$y'(a_1)=z'(a_1)$, $y(a_2)=z(a_2).$ We first assume that $y''(a_1)\ne z''(a_1),$and to be specific assume that $y''(a_1)>z''(a_1).$ Then, by Lemma \ref{agal4.2} thereis a $\delta>0$ with $a<a_1-\delta<a_1+\delta<a_2$ such that all solutions $w(x)$ of the initial value problem for \eqref{aga1.1} with the initial conditions\begin{equation}w(a_1)=y_0=z(a_1),~~w'(a_1)=y_1=z'(a_1),~~w''(a_1)=y_2=\frac{1}{2}\l[y''(a_1)+z''(a_1)\r]\label{aga4.1}\end{equation}satisfy $z(x)<w(x)<y(x)$ for $0<|x-a_1|\leq \delta.$ Let $\{\epsilon_m\}$ be a monotone decreasing sequence of positive numbers converging to zero, and let$z_m(x)$ be a solution of \eqref{aga1.1} with the initial conditions\begin{equation}z_m(a_1)=y_0,~~z'_m(a_1)=y_1+\epsilon_m,~~z''_m(a_1)=y_2.\label{aga4.2}\end{equation}Then, $\{z_m(x)\}$ contains a subsequence converging uniformly on $[a_1-\delta,a_1+\delta]$ to a solution of \eqref{aga1.1}, \eqref{aga4.1}. Hence, for sufficiently large $m,$there is a solution $z_m(x)$ of \eqref{aga1.1}, \eqref{aga4.2} such that$$z(a_1-\delta)<z_m(a_1-\delta)<y(a_1-\delta)~~~~\mbox{and}~~~~z(a_1+\delta)<z_m(a_1+\delta)<y(a_1+\delta).$$Since $z_m(a_1)=y(a_1)=z(a_1)$ and $z'_m(a_1)=y_1+\epsilon_m>y'(a_1)=z'(a_1),$ it follows that there are $x_1,~x_2$ with $a_1-\delta<x_1<a_1<x_2<a_1+\delta$ such that $z_m(x_1)=z(x_1),$ and $z_m(x_2)=y(x_2).$ Since$y(a_1)=z(a_1)$ at $a_2>a_1+\delta,$ it follows that any extension of $z_m(x)$ intersects either $y(x)$ or $z(x)$ again on $[a_1+\delta,b).$ Since $z_m(x)\not\equiv z(x)$ on $[x_1,a_1]$ and $z_m(x)\not\equiv y(x)$ on $[a_1,x_2],$this contradicts condition \eqrefd{$_3$}.Thus, if $y(x)\not\equiv z(x)$ on $[a_1,a_2],$ then $y^{(i)}(a_1)=z^{(i)}(a_1)$for $i=0,1,2.$ However, if $y^{(i)}(a_1)=z^{(i)}(a_1)$ for $i=0,1,2$ and$y(x)\not\equiv z(x)$ on $[a_1,a_2],$ then for $a<x_3<a_1,$ $u(x)$ and $v(x)$defined by\begin{multline*}u(x)=v(x)=y(x)~~\mbox{on}~~[x_3,a_1],~~u(x)=y(x)~~\mbox{on}~~(a_1,a_2],\\ \mbox{and}~~v(x)=z(x)~~\mbox{on}~~(a_1,a_2]\end{multline*}will be solutions on $[x_3,a_2]$ which again contradicts condition \eqrefd{$_3$}. Thus, we conclude that $y(x)\equiv z(x)$ on $[a_1,a_2].$\end{proof}\unskip\begin{amsl}[\cite{aga12}] \label{agal4.4}Let $y(x)\in C^{(2)}[\alpha,\beta]$ and assume that$|y(x)|\leq M$ on $[\alpha,\beta].$ There is a $K>0$ depending on $M$ and$\beta-\alpha$ such that if, $\max\{|y'(x)|,|y''(x)|\}>K$ for all $\alpha\leqx\leq \beta,$ then $y'(x_0)=0$ for some $x_0$ with $\alpha<x_0<\beta.$\end{amsl}\unskip\begin{proof}Assume that the conclusion is false. We shall determine $N>0$so that the following inequality holds\begin{equation}|y'(x)|+|y''(x)|\geq N+\frac{2M}{\beta-\alpha}+1~~~~\mbox{on}~~[\alpha,\beta].\label{aga4.3}\end{equation}For this, by the Mean value theorem there exists a $x_1\in (\alpha,\beta)$such that$$|y'(x_1)|=\l|\frac{y(\beta)-y(\alpha)}{\beta-\alpha}\r|\leq \frac{2M}{\beta-\alpha}.$$There are now two possible cases, however, since both are similar, we shallconsider only the case$$0<y'(x_1)\leq \frac{2M}{\beta-\alpha}~~~~\mbox{and}~~~~\alpha<x_1\leq \frac{\alpha+\beta}{2}.$$If $y(x_1)=M,$ then $y'(x_1)=0$ and the proof is finished. So, we assume that$y(x_1)\ne M.$ (It is clear that we are assuming $y'(x_1)\ne 0.)$We define $\eta=(\beta-\alpha)/8.$ Now to complete the proof we need to consider the following two subcases:\vspace{2mm}Case (i). Assume that $y''(x_1)\leq 0.$ Then, in order \eqref{aga4.3} holds, it is necessary that $y''(x_1)\leq -N.$ Thus, $y'(x)$ is decreasing on a right neighborhood of $x=x_1.$ In fact, if $0\leq y'(x)\leq (2M/(\beta-\alpha)),$ then it will follow that $y''(x)\leq -N$ on $[x_1,\beta].$ However, then by Taylor's formula, we have$$\begin{array}{lll}y(\beta)&=&\displaystyley(x_1)+(x_1-\beta)y'(x_1)+\frac{(x_1-\beta)^2}{2}y''(\xi),~~~~\xi\in(x_1,\beta)\\~\\&<&\displaystyle M+\frac{2M(\beta-\alpha)}{\beta-\alpha}-\frac{(\beta-\alpha)^2}{4}N\\~\\&\leq& -M\end{array}$$provided $$N\geq \frac{32M}{(\beta-\alpha)^2}=\frac{M}{2\eta^2}.$$But, this implies that $|y(\beta)|>M,$ which is a contradiction. Thus,  there exists a point $x_1\leq x_0<\beta$ such that $y'(x_0)=0.$\vspace{2mm}Case (ii). Assume that $y''(x_1)>0,$ (we shall work across $[\alpha,\beta]$on subintervals of length $\eta.$) Then, from \eqref{aga4.3} it follows that $y''(x_1)>N.$ We assume that $y''(x)\geq N/2$ on $[x_1,x_1+\eta].$ Again, by Taylor'sformula, we have$$y(x_1+\eta)=y(x_1)+\eta y'(x_1)+\frac{1}{2}\eta^2 y''(\xi),~~~~\xi\in(x_1,x_1+\eta).$$As in Case (i), we find $y(x_1+\eta)>-M+N\eta^2/4\geq M,$ provided $N\geq(8M/\eta^2),$ which is a contradiction.From this contradiction, we conclude that there exists a $x_1<x_2<x_1+\eta$such that $y''(x_2)=N/2.$ Since, $y''(x)$ is positive up to $x_2,$ we can assume that $x_2$ is the first point such that $y''(x_2)=N/2.$ Thus,$$y'(x_2)>\frac{1}{2}N+\frac{2M}{\beta-\alpha}~~~~\mbox{on}~~[x_1,x_2).$$Now assume that $$y'(x)\geq \frac{1}{2}N+\frac{2M}{\beta-\alpha}~~~~\mbox{on}~~[x_2,x_2+\eta).$$Then, it follows that$$\begin{array}{lll}y(x_2+\eta)&=&y(x_2)+\eta y'(\xi),~~~\xi\in (x_2,x_2+\eta)\\~\\&>& \displaystyle -M+\frac{1}{2}\eta N+\eta\frac{2M}{\beta-\alpha}\\~\\&=& \displaystyle -M+\frac{1}{2}\eta N+\frac{1}{4}M\\~\\&\geq & M,\end{array}$$provided $N\geq (7M/2\eta).$ But, this implies $y(x_2+\eta)>M,$ which isa contradiction.From this contradiction, there exists a $x_2<x_3<x_2+\eta$ such that $$y'(x_3)=\frac{1}{2}N+\frac{2M}{\beta-\alpha},$$and we take $x_3$ to be the first such point, i.e., $$y'(x)>\frac{1}{2}N+\frac{2M}{\beta-\alpha},~~~~\mbox{on}~~[x_2,x_3).$$So, $y'(x)$ is decreasing on $[x_2,x_3).$Thus, $y''(x_3)<-N/2.$ This implies that in a right neighborhood of $x_3,~y''(x)<-N/2$ and $y'(x)$ is decreasing. Assume that$$0<y'(x) \leq \frac{1}{2}N+\frac{2M}{\beta-\alpha}~~~~\mbox{on}~~[x_3,\beta).$$Then, by Taylor's formula$$y(x_3)=y(\beta)+(x_3-\beta)y'(\beta)+\frac{(x_3-\beta)^2}{2}y''(\xi),~~~~\xi\in (x_3,\beta).$$ Since in the above relation the second term in the right side is nonpositive,in view of $(\beta-x_3)\geq (\beta-\alpha)/4,$ it follows that$$y(x_3)<M-\frac{1}{4}(\beta-\alpha)^2 N\leq -M$$provided $N\geq M/\eta^2.$ But, this leads to the contradiction that $|y(x_3)|>M.$From this construction we conclude that $$0<y'(x)\leq \frac{1}{2}N+\frac{2M}{\beta-\alpha}~~~~\mbox{on}~~[x_3,\beta)$$is false. Thus, in conclusion $y'(x_0)=0,$ for some $x_3<x_0<\beta.$\end{proof}\unskip\begin{amst}[\cite{aga12}] \label{agat4.5}If the differential equation \eqref{aga1.1} is of third order and satisfies conditions \eqref{agaA}, \eqref{agaC} and \eqrefd{$_3$}, then \eqref{aga1.1} also satisfies condition \eqref{agaE}.\end{amst}\unskip\begin{proof}Suppose that the result is not true. Then, since $|y_m(x)|\leqM$ on $[c,d]$ for $n\geq 1,$ it follows from Theorem \ref{agat2.1} that $|y'_m(x)|+|y''_m(x)|\to\infty$ uniformly on $[c,d].$ Let $c\leq a_1<a_2<a_3<a_4\leqd$ be such that $a_4-a_1\leq \delta(M),$ where $\delta(M)$ is as defined inLemma \ref{agal4.1}. By Lemma \ref{agal4.4} there is a $K>0$ such that, if $\max\{|y'_m(x)|,|y''_m(x)|\}>K$ for each $x\in [c,d],$ then $y'_m(x)$ has a zero on $(a_1,a_2),$ on $(a_2,a_3)$ and on $(a_3,a_4).$ Furthermore, we can assumethat $K>1.$ Now from the fact that $|y'_m(x)|+|y''_m(x)|\to\infty$ uniformly on $[c,d]$ we can conclude that there is a positive integer $m_0$ such that$\max\{|y'_{m_0}(x)|,|y''_{m_0}(x)|\}>K$ on $[c,d].$ Let $a_1<x_1<a_2<x_2<a_3<x_3<a_4$ be such that $y'_{m_0}(x_i)=0$ for $i=1,2,3.$ Then, $|y''_{m_0}(x_i)|>K>1$ for $i=1,2,3.$ Now we need to consider the following two cases:If $y_{m_0}(x_i)=y_{m_0}(x_j)$ with $x_i<x_j,$ then $y_{m_0}(x)$ is the solution of the differential equation \eqref{aga1.1} together with the two-point boundary conditions $y(x_i)=y(x_j)=y_{m_0}(x_i),~y'(x_i)=0.$ However, since$x_j-x_i<\delta(M),$ it follows from Lemma \ref{agal4.1} that $|y'_{m_0}(x)|\leq 1$ and$|y''_{m_0}(x)|\leq 1$ on $[x_i,x_j],$ which is a contradiction to$|y''_{m_0}(x_i)|>K>1.$ If $y_{m_0}(x_i)\ne y_{m_0}(x_j)$ for $x_i\ne x_j,$ then it suffices to assume that $y_{m_0}(x_1)<y_{m_0}(x_2)<y_{m_0}(x_3).$ In fact, the sameargument applies to the other orderings of the values of $y_{m_0}(x_i),~i=1,2,3.$ If $y''_{m_0}(x_2)>K,$ there is a $t_1,~x_1<t_1<x_2,$ such that$y_{m_0}(t_1)=y_{m_0}(x_2).$ If $y''_{m_0}(x_2)<-K,$ there is a $t_2,~x_2<t_2<x_3,$ such that $y_{m_0}(t_2)=y_{m_0}(x_2).$ In either case Lemma \ref{agal4.1}is again applied to obtain a contradiction. Hence, the sequence $\{y_m(x)\}$ contains a subsequence converging uniformlyon $[c,d]$ along with its first and second order derivative sequences.\end{proof}\section{Weak Compactness Condition}It seems very difficult, if not impossible, to extend the method of Theorem \ref{agat4.5}to equations of higher orders. Here, we shall show that for equation\eqref{aga1.1} of arbitrary order $n,$ conditions \eqref{agaA}, \eqref{agaC} and \eqrefd{$_n$} do imply a weakertype of compactness condition for the solutions of \eqref{aga1.1}. For this we shallneed the following:\begin{amst} \textbf{\textup{(Banach Indicatrix Theorem, \cite[p. 271]{aga9})}} \label{aga5.1}If $h\in C[c,d] \cap BV[c,d],$ then$$V_c^d(h)=\int_{-\infty}^{\infty}N_h(\alpha)d\alpha,$$where$$N_h(\alpha)=\l\{\begin{array}{l}Card\{x\in [c,d]~:~h(x)=\alpha\},~~\mbox{if this set is finite,}\\~\\+\infty,~~\mbox{if the above set is infinite,}\end{array}\r.$$and where the above integral is in the Lebesgue sense.\end{amst}\unskip\begin{amst}[\cite{aga16}] \label{agat5.2}Assume that the differential equation \eqref{aga1.1} satisfies the conditions \eqref{agaA}, \eqref{agaC}, and \eqrefd{$_n$}. Further, assume that $[c,d]$ is a compact subinterval of $(a,b)$ and $\{y_m(x)\}$ is a sequence of solutionsof \eqref{aga1.1} which is uniformly bounded on $[c,d].$ Then, the sequence$\l\{V^d_c(y_m)\r\}$ of total variations of the functions $y_m(x)$ on $[c,d]$ is bounded, i.e., there exists an $N>0$ such that $V_c^d(y_m)\leq N$ for all $m.$\end{amst}\unskip\begin{proof}Assume the assertion is false. Then there is a sequence ofsolutions $\{y_m(x)\}$ of \eqref{aga1.1}, a compact interval $[c,d]\subset (a,b),$  and an $M>0$ such that $|y_m(x)|\leq M$ on $[c,d]$ for all $m,$ but such that$V^d_c(y_m)\to\infty,$ as $m\to\infty.$We claim that $\sum_{i=0}^{n-1}|y^{(i)}_m(x)|\to\infty$ on $[c,d]$ as $m\to\infty,$ i.e., given $R>0,$ there exists a $L>0$ such that $\sum_{i=0}^{n-1}|y^{(i)}_m(x)|>R,$ on $[c,d]$ for all $m\geq L.$ If the claim is false, thenthere exists a $\beta>0$ and a subsequence $\{y_{m(j)}(x)\}$ such that$\sum_{i=0}^{n-1}|y_{m(j)}^{(i)}(x_j)|\leq \beta,$ for all $j\geq 1,$ and where$\{x_j\}\subset [c,d].$ Now by choosing successive subsequences and relabeling, we obtain points $\{x_p\}$ and solutions $\{y_p(x)\}$ such that$\{x_p\}$ and $\{y_p^{(i)}(x_p)\},~0\leq i\leq n-1$ allconverge. Thus, by Theorem \ref{agat2.1} there exists a further subsequence $\{y_{p(j)}(x)\}$ such that $\{y_{p(j)}^{(i)}(x)\}$ converges uniformly on $[c,d],~0\leq i\leq n-1.$ This implies that $\{y_{p(j)}'(x)\}$ is a uniformly bounded sequence on $[c,d].$ Now since each $y'_{p(j)}(x)$ is absolutely continuous,it follows that$$V_c^d(y_{p(j)})=\int_c^d|y'_{p(j)}(x)|dx.$$Hence, the sequence $\{V_c^d(y_{p(j)})\}$ is a bounded sequence, which is a contradiction. Thus, $\sum_{i=0}^{n-1}|y^{(i)}_m(x)|\to\infty$ on $[c,d]$ as $m\to\infty.$ We shall now apply Corollary \ref{agac2.4} with $N_0=M$ and $N_0=0,~1\leq i\leq n-1,$so that there exists a $\delta(M)>0$ such that for any $\alpha$ with $|\alpha|\leq M$ and any points $c\leq a_1<\cdots<a_n\leq d$ with $a_n-a_1\leq\delta,$and with $w(x)\equiv \alpha$ so that $w^{(i)}(x)\equiv 0,~1\leq i\leq n-1,$the boundary value problem for \eqref{aga1.1} satisfying $y(a_i)=\alpha,~1\leq i\leq n$has a solution $y(x)$ with $|y^{(i)}(x)|\leq N_i+1$ on $[a_1,a_n]$ for$0\leq i\leq n-1.$ In particular, the boundary value problem has a solution$y(x)$ with $|y(x)|\leq M+1,$ and $|y^{(i)}(x)|\leq 1,~1\leq i\leq n-1$ on$[a_1,a_n].$ Further, for such a solution, we have $\sum_{i=0}^{n-1}|y^{(i)}(x)|\leq M+n$ on $[a_1,a_n].$Now let $L_0$ be such that $\sum_{i=0}^{n-1}|y_m^{(i)}(x)|>M+n$ on $[c,d]$for all $m\geq L_0.$ It follows that given $m\geq L_0$ and $\alpha,$ with$|\alpha|\leq M$ the solution $y_m(x)$ intersects the line $y=\alpha$ at most$n-1$ times in any closed subinterval of $[c,d]$ of length less than $\delta.$ For if, there are points $c\leq x_1<\cdots<x_n\leq d,~x_n-x_1\leq \delta$such that $y_m(x_i)=\alpha,~1\leq i\leq n$ where $|\alpha|\leq M$ and$m\geq L_0,$ there is also the above mentioned solution $y(x)$ by Corollary \ref{agac2.4}satisfying $y(x_i)=\alpha,~1\leq i\leq n.$ By \eqrefd{$_n$}, $y_m(x)\equiv y(x)$ on$[x_1,x_n].$ But, then $\sum_{i=0}^{n-1}|y^{(i)}(x)|\leq M+n$ on $[a_1,a_n]$ and $\sum_{i=0}^{n-1}|y_m^{(i)}(x)|>M+n$ on $[a_1,a_n]$ is not possible. Thus,for every $m\geq L_0,~y_m(x)$ intersects each line $y=\alpha,~|\alpha|\leq M$at most $n-1$ times in any closed subinterval of $[c,d]$ of length less than$\delta.$ So, for all $m\geq L_0,$$$N_{y_m}(\alpha)\leq (n-1)\l(\l[\frac{d-c}{\delta}\r]+1\r),~~~~\mbox{if}~~~~|\alpha|\leq M$$and $N_{y_m}(\alpha)=0,$ if $|\alpha|>M.$ Thus, by the Banach Indicatrix Theorem it follows that for $m\geq L_0,$$$\begin{array}{lll}\displaystyle V_c^d(y_m)&=&\displaystyle\int_{-M}^{M}N_{y_m}(\alpha)d\alpha\\~\\&\leq&\displaystyle \int_{-M}^{M}(n-1)\l(\l[\frac{d-c}{\delta}\r]+1\r)d\alpha\\~\\&=&\displaystyle2M(n-1)\l(\l[\frac{d-c}{\delta}\r]+1\r).\end{array}$$But, this contradicts $V_c^d(y_m)\to\infty,$ as $m\to \infty.$ Hence, $\{V_c^d(y_{m})\}$ is a bounded sequence.\end{proof}\unskip\begin{amst} \textbf{\textup{(Helly's Selection (or Choice) Theorem, \cite[p. 398]{aga22})}} \label{agat5.3}If  $\{y_m(x)\}$ is a sequence of functions on $[c,d]$ such that for some $M,~|y_m(x)|\leq M$ on $[c,d]$ for all $m\geq 1,$ and such that $|V_c^d(y_m)|\leq H,$ for all $m\geq 1,$ and some $H>0,$ then there exists a subsequence $\{y_{m(j)}(x)\}$ which converges point-wise on $[c,d].$ Moreover, the limit function is of bounded variation on $[c,d].$\end{amst}\unskip\begin{amsc} \label{agac5.4} Assume that the differential equation \eqref{aga1.1} satisfies the conditions \eqref{agaA}, \eqref{agaC} and \eqrefd{$_n$}. Then, if $[c,d]$ is a compact subinterval of $(a,b)$ and if $\{y_m(x)\}$ is a sequence of solutions of \eqref{aga1.1} which is uniformly bounded on $[c,d],$ there is a subsequence $\{y_{m(j)}(x)\}$which converges point-wise on $[c,d]$ and $z(x)=\lim_{j\to\infty}y_{m(j)}(x)$ is of bounded variation on $[c,d]$ (as a consequence $z(x)$ has finite derivative almost everywhere, and also $\int_c^dz(x)dx$ exists).\end{amsc}\unskip\begin{proof} The result follows from Theorems \ref{agat5.2} and \ref{agat5.3}. \end{proof}\unskip\begin{amsr} \label{agar5.1}In conjunction with Corollary \ref{agac5.4} we remark that Schrader \cite{aga19,aga20} has proven that, if $\{y_m(x)\}$ is a uniformly bounded sequence of functions on a compact interval $[c,d],$ and if the functions $y_m(x)$ satisfy only the uniqueness condition \eqrefd{$_n$} on $[c,d],$ then there is a subsequenceof $\{y_m(x)\}$ which converges point-wise on $[c,d].$\end{amsr}\section{Generalized Solutions} \label{agas6}To prove the Proposition now we shall follow another possible approach. Forthis, if the differential equation \eqref{aga1.1} satisfies \eqref{agaA} -- \eqrefd{$_n$}, then it isstraightforward to show that the compactness condition \eqref{agaE} is equivalent tothe following:\vspace{2mm}\no (E$^*$)\odkaz[h]{E}~~If $\{y_m(x)\}$ is a sequence of solutions of \eqref{aga1.1} which is monotone and bounded on some compact subinterval $[c,d]\subset (a,b),$ then $\lim_{m\to\infty}y_m(x)$ is a solution of \eqref{aga1.1} on $[c,d].$ \vspace{2mm}Thus, to prove the Proposition it suffices to show that the conditions \eqref{agaA} -- \eqrefd{$_n$} imply that the limit of a bounded monotone sequence of solutionsof \eqref{aga1.1} is also a solution.\begin{amsd} \label{agad6.1} A function $\phi(x)$ defined on an interval $J\subset(a,b)$ is said to be a {\itshape generalized solution} of \eqref{aga1.1} on $J$ if for each set of points $a_1<a_2<\cdots<a_n$ contained in $J$ and any solution $y(x)$ of\eqref{aga1.1}, the inequalities $(-1)^{n+i}\l[y(a_i)-\phi(a_i)\r]$ $<0$, $1\leq i\leqn$ imply that $y(x)<\phi(x)$ on $J\cap [a_n,b)$ and $(-1)^{n+1}[y(x)-\phi(x)]<0$ on $J\cap (a,a_1],$ and the inequalities $(-1)^{n+i}\l[y(a_i)-\phi(a_i)\r]>0,~1\leq i\leq n$ imply $y(x)>\phi(x)$ on $J\cap[a_n,b)$ and $(-1)^{n+1}[y(x)-\phi(x)]>0$ on $J\cap (a,a_1].$\end{amsd}\unskip\begin{amst}[\cite{aga13,aga17}] \label{agat6.1}Assume that the differential equation \eqref{aga1.1} satisfies conditions \eqref{agaA} -- \eqrefd{$_n$}, and that $\lim_{m\to\infty}y_m(x)=\phi(x)$ on $J\subset (a,b),$ where $\{y_m(x)\}$ is a sequence of solutions of \eqref{aga1.1}. Then, $\phi(x)$ is a generalized solution of \eqref{aga1.1} on $J.$\end{amst}\unskip\begin{proof}Assume that for $a_1<a_2<\cdots<a_n$ contained in $J$ thereis a solution $y(x)$ of \eqref{aga1.1} such that $(-1)^{n+i}\l[y(a_i)-\phi(a_i)\r]<0$ for $1\leq i\leq n,$ but that also $y(a_0)>\phi(a_0)$ for some $a_0>a_n$ in $J.$ Then, since $\lim_{m\to\infty}y_m(x)=\phi(x),$ there is asolution $y_m(x)$ of \eqref{aga1.1} such that $(-1)^{n+i}\l[y(a_i)-y_m(a_i)\r]<0$for $1\leq i\leq n$ and $y(a_0)>y_m(a_0).$ This contradicts the condition \eqrefd{$_n$}. The remaining inequalities can be proved in a similar way.Thus, the limit of a bounded monotone sequence of solutions $\{y_m(x)\}$ of\eqref{aga1.1} satisfying \eqref{agaA} -- \eqrefd{$_n$} is a generalized solution.\end{proof}\pagebreak
\begin{amsl} \label{agal6.2}Assume that the differential equation \eqref{aga1.1} satisfiescondition \eqref{agaA} and that $\phi(x)\in C^{(n-1)}[c,d],$ where $[c,d]$ is a compact subinterval of $(a,b).$ Assume that $M>0$ is such that $|\phi^{(j)}(x)|\leq M$ on $[c,d]$ for $0\leq j\leq n-1.$ Then, there exists a $\delta>0$such that, for any $c\leq a_1<a_2<\cdots<a_n\leq d$ with $a_n-a_1\leq \delta,$\eqref{aga1.1} has a solution $y(x)$ with $y(a_i)=\phi(a_i),~1\leq i\leq n$ and$|y^{(j)}(x)|\leq 2M$ on $[a_1,a_n]$ for $0\leq j\leq n-1.$ Furthermore, $\delta$ can be chosen in such a way that, for each fixed set $a_1<a_2<\cdots <a_n$ satisfying the above conditions, there is an $\epsilon>0$ such that for any $y_i,~1\leq i\leq n$ with $|y_i-\phi(a_i)|<\epsilon,~1\leq i\leqn,$ \eqref{aga1.1} has a solution $y(x)$ satisfying $y(a_i)=y_i,~1\leq i\leq n,$ and$|y^{(j)}(x)|\leq 3M$ on $[a_1,a_n]$ for $0\leq j\leq n-1.$\end{amsl}\unskip\begin{proof} The proof follows from Corollary \ref{agac2.7}. \end{proof}\unskip\begin{amst}[\cite{aga13,aga17}] \label{agat6.3}Assume that the differential equation \eqref{aga1.1} satisfies conditions \eqref{agaA} and \eqrefd{$_n$}, and that $\lim_{m\to\infty}y_m(x)=\phi(x)$ on $[c,d]\subset (a,b),$ where $\{y_m(x)\}$ is a sequence of solutions of \eqref{aga1.1}. Then, if $\phi(x)\in C^{(n-1)}[c,d],$ $\phi(x)$ is a solution of \eqref{aga1.1} on $[c,d]$ and $\lim_{m\to\infty}y_m^{(j)}(x)=\phi^{(j)}(x)$ uniformly on $[c,d]$ for each $0\leq j\leq n-1.$\end{amst}\unskip\begin{proof}Let $M>0$ be such that $|\phi^{(j)}(x)|\leq M$ on $[c,d]$ for$0\leq j\leq n-1.$ By Lemma \ref{agal6.2} there is a $\delta>0$ such that, if $c\leqa_1<a_2<\cdots<a_n\leq d$ is a fixed set of points with $a_n-a_1\leq \delta,$there is an $\epsilon>0$ with the property that $|y_i-\phi(a_i)|<\epsilon,~1\leq i\leq n$ implies that \eqref{aga1.1} has a solution $y(x)$ satisfying $y(a_i)=y_i,~1\leq i\leq n$ and $|y^{(j)}(x)|\leq 3M$ on $[a_1,a_n]$ for $0\leq j\leq n-1.$ It follows that there is an $N>0$ such that $m\geq N$ implies $|y_m(a_i)-\phi(a_i)|<\epsilon,~1\leq i\leq n.$ Hence, by condition \eqrefd{$_n$} and the choice of $\epsilon,~|y_m^{(j)}(x)|\leq 3M$ on $[a_1,a_n]$ for $0\leq j\leq n-1$ and all $m\geq N.$ From this the conclusion follows.\end{proof}Now let $\phi(x)$ be a real valued function defined on $(c,d).$ At a point$x_0\in (c,d)$ where $\phi(x)$ has a finite right limit $\phi(x_0+0),$ we define$$D^1\phi(x_0+0)=\lim_{x\to x_0^+}\frac{\phi(x)-\phi(x_0+0)}{x-x_0}$$provided the limit exists. The left derivative $D^1\phi(x_0-0)$ is similarlydefined. Likewise, if $\phi(x_0+0)$ and $D^1\phi(x_0+0)$ exist and are finite,we define$$D^2\phi(x_0+0)=\lim_{x\to x_0^+}\l\{\frac{2}{(x-x_0)^2}\l[\phi(x)-\phi(x_0+0)-D^1\phi(x_0+0)(x-x_0)\r]\r\}$$provided the limit exists. In general, if the limits defining $\phi(x_0+0)$and $D^j\phi(x_0+0),~1\leq j\leq k-1$ exist and are finite, we define\begin{multline*}D^k\phi(x_0+0)=\\\lim_{x\to x_0^+}\biggl\{\frac{k!}{(x-x_0)^k}\biggl[\phi(x)-\phi(x_0+0)-\sum_{j=1}^{k-1}\frac{D^j\phi(x_0+0)(x-x_0)^j}{j!}\biggr]\biggr\}\end{multline*}provided the limit exists. The left derivatives $D^j\phi(x_0-0)$ are defined correspondingly. \begin{amst}[\cite{aga13,aga17}] \label{agat6.4}Assume that the differential equation \eqref{aga1.1} satisfies conditions \eqref{agaA} and \eqrefd{$_n$}, and that $\phi(x)$ is a bounded generalized solution of \eqref{aga1.1} on $(c,d)\subset (a,b).$ Then, $\phi(x)$ has right and left limits at each point of $(c,d)$ and $D^1\phi(x_0-0)$ and $D^1\phi(x_0+0)$ exist in the extended reals for all $x_0\in (c,d).$ Furthermore, if at a point $x_0\in (c,d),~D^j \phi(x_0+0)$ exists and is finite for each $1\leq j\leq k-1\leq n-2,$ then the limit defining $D^k\phi(x_0+0)$ exists in the extended reals. The same assertion applies to the left derivative $D^k\phi(x_0-0).$\end{amst}\unskip\begin{proof}Assume that for some $x_0\in (c,d),~\liminf_{x\to x_0^+}\phi(x)<\limsup_{x\to x_0^+}\phi(x)$ and choose a real number $r$ such that$\liminf_{x\to x_0^+}\phi(x)<r<\limsup_{x\to x_0^+}\phi(x).$ Then, there exist sequences $\{t_m\}$ and $\{x_m\}$ in $(c,d)$ such that $\lim t_m=\lim x_m=x_0,~x_0<t_{m+1}<x_m<t_m$ for each $m\geq 1,$ $\lim \phi(t_m)=\limsup_{x\tox_0^+}\phi(x),$ and $\lim \phi(x_m)=\liminf_{x\to x_0^+}\phi(x).$ let $y(x)$be a solution of \eqref{aga1.1} satisfying the initial conditions $y(x_0)=r$ and$y^{(j)}(x_0)=0,~1\leq j\leq n-1.$ This solution exists on $[x_0,x_0+\delta]$for some $\delta>0,$ and since $\lim_{x\to x_0}y(x)=r,$ there is an $N>0$ such that $m\geq N$ implies that $x_0<t_m<x_0+\delta$ and $\phi(t_m)>y(t_m),~\phi(x_m)<y(x_m).$ This contradicts the fact that $\phi(x)$ is a generalizedsolution on $(c,d).$ The existence of $\phi(x_0-0)$ can be proved similarly.Now assume that for some $x_0\in (c,d)$ the limit defining $D^1\phi(x_0+0)$does not exist in the extended reals. Then, choose the real number $r$ such that$$\liminf_{x\to x_0^+}\frac{\phi(x)-\phi(x_0+0)}{x-x_0}<r<\limsup_{x\tox_0^+}\frac{\phi(x)-\phi(x_0+0)}{x-x_0}.$$If $y(x)$ is a solution of \eqref{aga1.1} satisfying the initial conditions $y(x_0)=\phi(x_0+0),~y'(x_0)=r,$ and $y^{(j)}(x_0)=0,~2\leq j\leq n-1,$ againsequences $\{t_m\}$ and $\{x_m\}$ can be chosen so that $\lim t_m=\lim x_m=x_0,~x_0<t_{m+1}<x_m<t_m$ for each $m\geq 1,$ and $\phi(t_m)>y(t_m),~\phi(x_m)<y(x_m)$ for all sufficiently large $m.$ This again contradicts $\phi(x)$being a generalized solution. Thus, $D^1\phi(x_0+0)$ and $D^1\phi(x_0-0)$exist in the extended reals for all $x_0\in (c,d).$Finally, if we assume that for some $x_0\in (c,d),~D^j\phi(x_0+0)$ existsand is finite for each $1\leq j\leq k-1\leq n-2,$ then by considering a solution of \eqref{aga1.1} satisfying the initial conditions $y(x_0)=\phi(x_0+0),~y^{(j)}(x_0)=D^j\phi(x_0+0)$ for $1\leq j\leq k-1,~y^{(k)}(x_0)=r,$ and$y^{(j)}(x_0)=0$ for $k+1\leq j\leq n-1,$ we can as above prove that the limit defining $D^k\phi(x_0+0)$ exists in the extended reals.\end{proof}\unskip\begin{amsc} \label{agac6.5} Assume that the differential equation \eqref{aga1.1} satisfiesconditions \eqref{agaA} and \eqrefd{$_n$}, and that $\phi(x)$ is a bounded generalized solution of \eqref{aga1.1} on $(c,d)\subset (a,b).$ Then, $\phi(x)$ has a finite derivative $\phi'(x)$ almost everywhere on $(c,d).$\end{amsc}\unskip\begin{amst}[\cite{aga13,aga17}] \label{agat6.6}Assume that the differential equation \eqref{aga1.1} satisfies conditions \eqref{agaA} -- \eqrefd{$_n$}. Let $\{y_m(x)\}$ be a sequence of solutions of \eqref{aga1.1} on $(c,d)\subset (a,b)$ such that $\{y_m(x)\}$ is uniformly bounded on $(c,d)$ and $\lim y_m(x)=\phi(x)$ on $(c,d).$ Then, if for some $x_0\in (c,d)$ the derivatives $D^j\phi(x_0+0),~1\leq j\leq n-1$ all exist andare finite, or the derivatives $D^j\phi(x_0-0),~1\leq j\leq n-1$ all exist and are finite, it follows that there is a subsequence $\{y_{m(j)}(x)\}$ such that $\{y_{m(j)}^{(i)}(x)\}$ converges uniformly on each compact subinterval of $(a,b)$ for each $0\leq i\leq n-1.$\end{amst}\unskip\begin{proof}Assume that for some $x_0\in (c,d)$ the derivatives $D^j\phi(x_0+0),~1\leq j\leq n-1$ exist and are finite. Let $p(x)$ be the polynomial$$p(x)=\phi(x_0+0)+\sum_{j=1}^{n-1}\frac{D^j\phi(x_0+0)(x-x_0)^j}{j!}$$then, it follows from the definition of $D^{n-1}\phi(x_0+0)$ that given any$\epsilon>0$ there is a $\delta>0$ such that $x_0+\delta<d,$ and$$|p(x)-\phi(x)|<\frac{\epsilon (x-x_0)^{n-1}}{(n-1)!}$$for $x_0<x\leq x_0+\delta.$ Let $d_0$ be a fixed number satisfying $x_0<d_0<d.$ By Lemma \ref{agal6.2} there is a $\delta_0>0$ such that for $x_0<x_1<x_2<\cdots<x_n\leq d_0$ with $x_i-x_{i-1}=\eta\leq \delta_0$ for each $1\leq i\leq n,$\eqref{aga1.1} has a solution $y(x)$ with $y(x_i)=p(x_i)$ for $1\leq i\leq n$ and $|y^{(j)}(x)|\leq 2M$ on $[x_1,x_n]$ for $0\leq j\leq n-1$ where $|p^{(j)}(x)|\leq M$ on $[x_0,d_0]$ for $0\leq j\leq n-1.$ Furthermore, there is an $\epsilon_0>0$ such that, if $|y_i-p(x_i)|<\epsilon_0$ for $1\leq i\leq n,$ then \eqref{aga1.1} has a solution $y(x)$ with $y(x_i)=y_i$ for $1\leq i\leq n$ and$|y^{(j)}(x)|\leq 3M$ on $[x_1,x_n]$ for $0\leq j\leq n-1.$ It is not difficultto show that with equal spacing $\eta$ between the $x_i's$ a suitable $\epsilon_0$ has the form $\epsilon_0=Mh_n\eta^{n-1},$ where $h_n$ is a fixedconstant depending on $n.$ Now as noted above, if we choose $\epsilon=Mh_n/(2n^{n-1}),$ there is a $\eta,~0<\eta\leq \delta_0$ such that $x_0<x<x_0+n\eta$ implies$$|p(x)-\phi(x)|<\frac{\epsilon (x-x_0)^{n-1}}{(n-1)!}\leq \frac{\epsilon_0}{2(n-1)!}\leq \frac{\epsilon_0}{2}.$$For such a choice of $\eta>0,$ we have $|p(x_i)-\phi(x_i)|\leq \epsilon_0/2$for $1\leq i\leq n$ where $x_i-x_{i-1}=\eta$ for $1\leq i\leq n.$ Consequently,if $N>0$ is such that $m\geq N$ implies $|y_m(x_i)-\phi(x_i)|<\epsilon_0/2$for $1\leq i\leq n,$ then $|p(x_i)-y_m(x_i)|<\epsilon_0$ for $m\geq N$ and $1\leq i \leq n.$ It follows from our construction and condition \eqrefd{$_n$} that $|y_m^{(j)}(x)|\leq 3M$ on $[x_1,x_n]$ for $0\leq j\leq n-1$ and all $m\geq N.$The conclusion of the theorem now follows.\end{proof}Thus, we see that, in order to prove that conditions \eqref{agaA} -- \eqrefd{$_n$} imply thecompactness condition \eqref{agaE}, it is sufficient to prove that, if $\phi(x)$ is thepoint-wise limit of a bounded sequence of solutions of \eqref{aga1.1} on $(c,d)\subset(a,b),$ then there is at least one $x_0\in (c,d)$ at which either $D^j\phi(x_0+0),~1\leq j\leq n-1$ or $D^j\phi(x_0-0),~1\leq j\leq n-1$ are finite.\section{The Case $q=0$}As we have remarked in Section \ref{agas6} for the differential equation \eqref{aga1.1} thecompactness condition \eqref{agaE}, under the assumptions \eqref{agaA} -- \eqrefd{$_n$}, is equivalent to \eqrefe[h]{$^*$}. This observation is used in the following result to establish theProposition for the case $q=0.$\begin{amst}[\cite{aga17}] \label{agat7.1}If the differential equation \eqref{aga1.1} with $q=0$ satisfies conditions \eqref{agaA} -- \eqrefd{$_n$}, then \eqref{aga1.1} with $q=0$ also satisfies condition \eqrefe[h]{$^*$}.\end{amst}\unskip\begin{proof}Let $\{y_m(x)\}$ be a monotone, bounded sequence of solutionsof \eqref{aga1.1} with $q=0$ which converges point-wise to a function $\phi(x)$ on$[c,d]\subset (a,b).$ Let $c=a_1<a_2<\cdots<a_n=d$ and $p_m(x)$ be theunique polynomial of degree $n-1$ such that $p_m(a_i)=y_m(a_i),~1\leq i\leqn$ and $m=1,2,\dots.$ Then, $p_m(x)$ converges uniformly to $p(x),$ where$p(x)$ is the unique polynomial of degree $n-1$ such that $p(a_i)=\lim_{m\to\infty}y_m(a_i),~1\leq i\leq n.$  Now, since $y_m(x)$ are uniformly boundedon $[c,d],$ it is clear that $M=\sup\{|f(x,y_m(x))| : c\leq x\leq d,~m\geq 1\}$exists. Further, from the properties of the Green's function $g(x,t),$ it follows that $\partial g/\partial x$ exists and is continuous on $[c,d]\times[c,d],$ and hence $|\partial g/\partial x|\leq K$ for all $x,t\in [c,d].$ Thus, if $x\ne t$ from the integral representation\begin{equation}y_m(x)=p_m(x)+\int_{c}^{d}g(x,t)f(t,y_m(t))dt,\label{aga7.1}\end{equation}which is the same $$\omega_m(x)\equiv y_m(x)-p_m(x)=\int_{c}^{d}g(x,t)f(t,y_m(t))dt$$we find$$\begin{array}{lll}|\omega_m(x)-\omega_m(s)|&\leq&\displaystyle \int_c^d|g(x,t)-g(s,t)||g(t,y_m(t))|dt\\~\\&\leq& MK|x-s|(d-c).\end{array}$$Hence, $\{\omega_m(x)\}$ is uniformly bounded and equicontinuous on $[c,d].$Thus, a subsequence and by monotonicity the whole sequence $\{y_m(x)\}$ converges uniformly to $\phi(x)$ on $[c,d].$ Finally, taking limits through \eqref{aga7.1} yields that $\phi(x)$ is a solution of \eqref{aga1.1} with $q=0$ on $[c,d],$ and hence the condition \eqrefe[h]{$^*$} is satisfied.  \end{proof}\section{The Uniform Convergence}In \cite{aga8} Henderson and Jackson in there closing remarks have mentioned the validity of the Proposition for fourth order differential equations. To prove the Proposition for arbitrary order differential equations we let$P_n$ denote the set of all real-valued polynomials of degree at most $n.$\begin{amsd}[\cite{aga3}] \label{agad8.1}Given $S\subset [c,d],~x_0$ is a {\itshape bilateralaccumulation point} of $S,$ in case $x_0$ is an accumulation point of both$S\cap [c,x_0]$ and $S\cap [x_0,d].$\end{amsd}\unskip\begin{amsd} \label{agad8.2} A function $g(x) : I\to \RR,~I$ an interval, issaid to be {\itshape $n$-convex ($n$-concave),} on $I$ in case for any distinctpoints $x_0,x_1,\dots,x_n$ in $I,$ $$\sum_{i=0}^{n}\frac{g(x_i)}{\omega'(x_i)}\geq 0,~(\leq 0),$$where$$\omega(x)=\prod_{i=0}^{n}(x-x_i),~~~~~\mbox{so that}~~~~\omega'(x_j)=\prod_{i=0,i\ne j}^{n}(x_j-x_i).$$\end{amsd}The following results for $n$-convex functions are well known.\begin{amsl} \label{agal8.1} Suppose $g(x)\in C^{(n)}(I).$ Then, $g(x)$ is $n$-convex on $I,$ if and only if, $g^{(n)}(x)\geq 0$ on $I.$\end{amsl}
\unskip\begin{amsl} \label{agal8.2} The function $g(x)$ is $n$-convex, if and only if, $g(x)\in C^{(n-2)}(I)$ and $g^{(n-2)}(x)$ is convex.\end{amsl}
\unskip\begin{amsr} \label{agar8.1} In Lemmas \ref{agal8.1} and \ref{agal8.2}, `convex' can be replaced by`concave'.\end{amsr}
\unskip\begin{amsl}[\cite{aga3}] \label{agal8.3}Let $g(x)\in C[c,d]$ and assume that, for each $p(x)\in P_n,$ the set $\{x : p(x)=g(x)\}$ does not have a bilateral accumulation point in $(c,d).$ Then, there exists a subinterval $I\subseteq [c,d]$ on which $g(x)$ is either $(n+1)$-convex or $(n+1)$-concave.\end{amsl}
\unskip\begin{amst}[\cite{aga21}] \label{agat8.4}Assume that the differential equation \eqref{aga1.1} satisfies the conditions \eqref{agaA} -- \eqrefd{$_n$}. Then, \eqref{aga1.1} also satisfies condition \eqref{agaE}.\end{amst}
\unskip\begin{proof}Let $\{y_m(x)\}$ be a sequence of solutions of \eqref{aga1.1}which is uniformly bound\-ed on some subinterval $[c,d]\subset (a,b).$ Then, byCorollary \ref{agac5.4} there exists a subsequence $\{y_{m(j)}(x)\}$ and a function$z(x)\!\in\! BV[c,d]$ such that $\lim_{j\to\infty}y_{m(j)}(x)$ $=z(x)$ point-wiseon $[c,d].$ Thus, $z'(x)$ exists a.e. on $[c,d]$ and $\int_c^dz(x)dx$ exists.We set$$Z(x)=\int_c^xz(t)dt.$$Then, $Z\in C[c,d],$ and by Lemma \ref{agal8.3}, either one of the following holds.\vspace{2mm}\no (i)~~$Z(x)$ is $(n+2)$-convex or $(n+2)$-concave on some $[c_1,d_1]\subseteq [c,d],$ or\vspace{2mm}\no (ii)~~ there exists a $p(x)\in P_{n+1}$ such that $\{x : p(x)=Z(x)\}$has a bilateral accumulation point in $(c,d).$\vspace{2mm}Case (i). Let us relabel the sequence $\{y_{m(j)}(x)\}$ as $\{y_m(x)\}.$By Lemma \ref{agal8.2}, $Z(x)\in C^{(n)}[c_1,d_1]$ and $Z^{(n)}(x)$ is convex, (orconcave). Thus, $Z'(x)\in C^{(n-1)}[c_1,d_1]$ and $Z'(x)=z(x)$ a.e. on$[c_1,d_1].$ Thus, as a consequence of Corollaries \ref{agac2.6} and \ref{agac2.7} there existsa $\delta=\delta(Z',d_1-c_1)>0$ such that, for fixed $c_1\leq a_1<a_2<\cdots<a_n\leq d_1,$ with $a_n-a_1\leq \delta,$ there exists an $\epsilon_0>0$such that the boundary value problem for \eqref{aga1.1} satisfying $y(a_j)=Z'(a_j)+\epsilon_j,~1\leq j\leq n,$ where $|\epsilon_j|\leq \epsilon_0,~1\leq j\leqn,$ has a solution $y(x).$ Furthermore, the first $n-1$ derivatives of thissolution are bounded, with bounds depending on $Z'$ and $d_1-c_1.$ We callthese bounds as $N_0+1,\dots, N_{n-1}+1.$Let us now choose points $c_1\leq a_1<a_2<\cdots<a_n\leq d_1,$ with $a_n-a_1\leq \delta$ and such that $Z'(a_j)=z(a_j),~1\leq j\leq n.$ Then, there existsan $M$ such that $|y_m(a_j)-z(a_j)|\leq \epsilon_0,~1\leq j\leq n$ for all$m\geq M.$ Now for $m\geq M$ and $1\leq j\leq n,$ let $\epsilon_{m(j)}=y_m(a_j)-z(a_j).$ Then, for $m\geq M,$$$y_m(a_j)=z(a_j)+\epsilon_{m(j)}=Z'(a_j)+\epsilon_{m(j)},~~~1\leq j\leq n$$and it follows from condition \eqrefd{$_n$} that $y_m(x)$ is the solution referred to above resulting from Corollaries \ref{agac2.6} and \ref{agac2.7}. As a consequence, we have for $m\geq M,$ $|y^{(i)}_m(x)|\leq N_i+1$ on $[a_1,a_n]$ for each $0\leqi\leq n-1.$ Now we can apply Theorem \ref{agat2.1} to obtain a further subsequence $\{y_{m(\ell)}(x)\}$ such that $\{y^{(i)}_{m(\ell)}(x)\}$ converges uniformlyon each compact subinterval of $(a,b)$ for each $0\leq i\leq n-1.$  \vspace{2mm}Case (ii). Assume that $\{x_m\}\downarrow x_0$ is such that $p(x_m)=Z(x_m),$for all $m\geq 1.$ Thus, if on some subinterval $[x_{j+1},x_j],$ $Z'(x)=z(x)\geq p'(x)$ a.e., then we have$$\begin{array}{lll}Z(x_j)-Z(x_{j+1})&=&\displaystyle\int_{x_{j+1}}^{x_j}z(t)dt\\~\\&\geq &\displaystyle\int_{x_{j+1}}^{x_j}p'(t)dt\\~\\&=&p(x_j)-p(x_{j+1})~=~Z(x_j)-Z(x_{j+1}),\end{array}$$so that the inequality is in fact an equality. However, from $z(x)-p'(x)\geq 0,$ a.e. on $[x_{j+1},x_j],$ we have $\int_{x_{j+1}}^{x_j}(z(t)-p'(t))dt=0,$ we conclude that $z(x)-p'(x)=0$ a.e. on $[x_{j+1},x_j].$ In particular,$$Z'(x)=z(x)=p'(x)~~~~\mbox{a.e.}~~~~\mbox{on}~~ [x_{j+1},x_j].$$Similarly, if we assume that on some subinterval $[x_{j+1},x_j],$ $Z'(x)=z(x)\leq p'(x)$ a.e., then we would arrive at $Z'(x)=z(x)=p'(x)$ a.e. on $[x_{j+1},x_j].$Now that $p'(x)\in C^{(n-1)}[c,d],$ by Corollaries \ref{agac2.6} and \ref{agac2.7} there existsa $\delta=\delta(p',d-c)>0$ such that for fixed $c\leq a_1<a_2<\cdots<a_n\leq d,~a_n-a_1\leq \delta,$ there exists an $\epsilon_0>0$ such that theboundary value problem for \eqref{aga1.1} satisfying $y(a_j)=p'(a_j)+\epsilon_j,~1\leq j\leq n,$ where $|\epsilon_j|\leq \epsilon_0,~1\leq j\leq n,$ hasa solution $y(x).$ As in Case (i) this solution has bounds on its first $n-1$derivatives depending only on $p'$ and $d-c;$ again we call these bounds as $N_0+1,\dots,N_{n-1}+1.$Let us choose points $x_{\ell+n}<x_{\ell+n-1}<\cdots<x_{\ell}$ from $\{x_m\}$such that $x_\ell-x_{\ell+n}\leq \delta.$ We need to consider two subcases:\vspace{2mm}Case (a). For some $1\leq r\leq n,$ on $[x_{\ell+r},x_{\ell+r-1}]$ we have$z(x)\geq p'(x)$ a.e., or $z(x)\leq p'(x)$ a.e. From the above arguments we, however, have $z(x)=p'(x)$ a.e. on $[x_{\ell+r},x_{\ell+r-1}].$ We repeatthe arguments of Case (i). Choose points $x_{\ell+r}\leq a_1<\cdots<a_n\leqx_{\ell+r-1}$ such that $z(a_j)=p'(a_j),~1\leq j\leq n.$ Then, there exists an $M$ such that $|y_m(a_j)-z(a_j)|\leq \epsilon_0,~1\leq j\leq n,~m\geq M.$For $m\geq M$ and $1\leq j\leq n$ let $\epsilon_{m(j)}=y_m(a_j)-z(a_j).$Then, for $m\geq M,$$$y_m(a_j)=z(a_j)+\epsilon_{m(j)}=p'(a_j)+\epsilon_{m(j)},~~~~1\leq j\leq n$$ and it follows that, from condition \eqrefd{$_n$}, $y_m(x)$ is the solution referred to the above problem arising from the Corollaries \ref{agac2.6} and \ref{agac2.7}. Thus, for all$m\geq M,$$$|y_m^{(i)}(x)|\leq N_i+1~~~~\mbox{on}~~~~[a_1,a_n]$$for each $0\leq i\leq n-1.$ Then, by Theorem \ref{agat2.1} there exists a furthersubsequence $\{y_{m(s)}(x)\}$ such that $\{y_{m(s)}^{(i)}(x)\}$ convergesuniformly on each compact subinterval of $(a,b),$ for each $0\leq i\leq n-1.$\vspace{2mm}Case (b). For each $1\leq r\leq n,$ there exist sets $A_r,~B_r\subset [x_{\ell+r},x_{\ell+r-1}],$ each having positive Lebesgue measure, and$z(x)>p'(x)$ on $A_r,$ and $z(x)<p'(x)$ on $B_r.$ However, since $\lim_{m\to\infty}y_m(x)=z(x),$ and so there exists a $M$ such that for $m\geq M,$$y_m(x)>p'(x),$ for some $A_r,$ and $y_m(x)<p'(x),$ for some $x\in B_r.$By continuity, for all $1\leq r\leq n,$ there exists $a_r\in (x_{\ell+r},x_{\ell+r-1})$ such that $y_m(a_r)=p'(a_r).$ In particular, there are points $x_{\ell+n}\leq \tilde{a}_1<\cdots<\tilde{a}_n\leq x_\ell~(\tilde{a}_n-\tilde{a}_1\leq \delta),$ so that, for some $M'\geq M,$$$y_m(\tilde{a}_j)=p'(\tilde{a}_j)+\epsilon_{m(j)},~~~~1\leq j\leq n$$where $|\epsilon_{m(j)}|\leq \epsilon_0,~1\leq j\leq n$ and all $m\geq M'.$It follows from condition \eqrefd{$_n$} that $y_m(x)$ is the solution referred tobefore Case (a) arising from Corollaries \ref{agac2.6} and \ref{agac2.7}. Thus, for all $k\geq M',$$$|y_m^{(i)}(x)|\leq N_i+1~~~~\mbox{on}~~[\tilde{a}_1,\tilde{a}_n],$$for each $0\leq i\leq n-1.$ Now an application of Theorem \ref{agat2.1} leads to asubsequence $\{y_{m(j)}(x)\}$ such that $\{y_{m(j)}^{(i)}(x)\}$ convergesuniformly on each compact subinterval of $(a,b),$ for each $0\leq i\leq n-1.$\end{proof}\section{Problems and Comments}The establishment of Theorem \ref{agat8.4} implies that in the known results on conjugate boundary value problems the condition \eqref{agaE} is, in fact, superfluous. As an example we state an important result, which was independently proved by Hartman \cite{aga6} and Klassen \cite{aga17} with the additional condition \eqref{agaE}.\pagebreak
\begin{amst} \label{agat9.1} Assume that for the differential equation \eqref{aga1.1} conditions \eqref{agaA} -- \eqrefd{$_n$} are satisfied. Then, each $r$ point boundary value problem, i.e., for any $a<a_1<a_2\cdots<a_r<b$ and any $A_{j+1,i},~0\leq j\leq k_i,~1\leq i\leq r$ the problem \eqref{aga1.1}, \eqref{aga2.3} has a unique solution. \end{amst}\unskip\begin{amsp} \label{agap1}For the third order differential equations in Theorem \ref{agat4.5} we have proved that conditions \eqref{agaA}, \eqref{agaC} and \eqrefd{$_3$} imply condition \eqref{agaE}. Itwill be interesting to extend this result to equations of arbitrary order, i.e., whether it is possible to prove Theorem \ref{agat8.4} without the assumption \eqref{agaB}.\end{amsp}In \cite{aga16} Jackson has indicated that for $n$-point boundary value problems,Klassen has used a result he established in \cite{aga18} to prove the existence of solutions under the assumptions \eqref{agaA}, \eqref{agaC}, \eqrefd{$_n$} and \eqref{agaE}. Thus, if the answer to Problem~\ref{agap1} is affirmative, then for $r=n,$ Theorem \ref{agat9.1} holds without the assumption \eqref{agaB}. \vspace{3mm}Let $2\leq r\leq n$ and let $m_i,~1\leq i\leq r,$ be positive integers such that $\sum_{i=1}^{r}m_i=n.$ Let $s_0=0$ and for $1\leq k\leq r,$ let$s_k=\sum_{i=1}^{k}m_i.$ A boundary value problem for \eqref{aga1.1} with the boundaryconditions \begin{equation}y^{(i)}(a_k)=y_{i,k},~~~~s_{k-1}\leq i\leq s_k-1,~~1\leq k\leq r\label{aga9.1}\end{equation}where $a<a_1<a_2<\cdots<a_r<b$ is called a {\itshape right $(m_1,\dots,m_r)$-focalpoint boundary value problem for} \eqref{aga1.1} {\itshape on $(a,b).$}With respect to the boundary conditions \eqref{aga9.1} we replace the condition \eqrefd{$_n$}by the following:\vspace{2mm}\no (D$_n^{rf}$)\odkaz[rf]{D}~~ For any $a<a_1<a_2<\cdots<a_n<b$ and any solutions $y(x)$ and $z(x)$ of \eqref{aga1.1}, it follows that $y^{(i-1)}(a_i)=z^{(i-1)}(a_i),~1\leq i\leq n$ implies $y(x)\equiv z(x),$ i.e., the differential equation \eqref{aga1.1} is right $(1,1,\dots,1)$ disfocal on $(a,b).$ \vspace{2mm}As an application of Rolle's theorem it follows that condition \eqrefd[rf]{$_n^{rf}$}implies the condition \eqrefd{$_n$}. Thus, in Theorem \ref{agat8.4} condition \eqrefd{$_n$} can bereplaced by \eqrefd[rf]{$_n^{rf}$}. We state this observation in the following result.\begin{amst} \label{agat9.2} Assume that the differential equation \eqref{aga1.1} satisfiesconditions \eqref{agaA} -- \eqref{agaC} and \eqrefd[rf]{$_n^{rf}$}. Then, \eqref{aga1.1} also satisfies condition\eqref{agaE}.\end{amst}Of course, in Theorem \ref{agat8.4}, we can always replace condition \eqrefd{$_n$} by a stronger condition. The point is now whether it is possible to replace condition \eqrefd{$_n$} by some other condition which does not imply \eqrefd{$_n$}. The first `round about' result in this direction is the following:\begin{amst} \label{agat9.3} If the differential equation \eqref{aga1.1} is of third order and satisfies conditions \eqref{agaA}, \eqref{agaC} and \eqrefd[2]{$_2$}, then \eqref{aga1.1} also satisfies condition \eqref{agaE}.\end{amst}\unskip\begin{proof} The proof is similar to that of Theorems \ref{agat4.5}. \end{proof}\unskip\begin{amsp} \label{agap2} A result similar to that of Theorem \ref{agat9.3} for arbitrary order differential equations \eqref{aga1.1} remains undecided, i.e., does conditions \eqref{agaA}, \eqref{agaC} and (D$_r$) imply condition \eqref{agaE}.\end{amsp}For arbitrary order differential equations \eqref{aga1.1} Jackson \cite{aga14} has established that conditions \eqref{agaA}, \eqref{agaB} and \eqrefd{$_n$} imply (D$_r$). A converse  of this result for third order differential equations \eqref{aga1.1} is that the conditions \eqref{agaA}, \eqref{agaC} and \eqrefd[2]{$_2$} imply \eqrefd{$_3$}. Jackson's proof \cite{aga15} of this converse result uses Theorem \ref{agat9.3}, i.e., under the assumptions, condition \eqref{agaE} is implied, and then this fact is used to prove \eqrefd{$_3$}. Thus, if we accept Jackson's converseresult without looking at its proof, then we can argue that conditions\eqref{agaA}, \eqref{agaC} and \eqrefd[2]{$_2$} imply \eqrefd{$_3$}, and therefore Theorem \ref{agat4.5} gives condition \eqref{agaE}.\begin{amsp} \label{agap3} The question for arbitrary order differential equations \eqref{aga1.1} which remains open is whether conditions \eqref{agaA}, \eqref{agaC} and (D$_r$) imply condition \eqrefd{$_n$}. \end{amsp}Finally, we state one more result which is similar in nature to that of Theorem \ref{agat9.3}.\begin{amst}[\cite{aga7}] \label{agat9.4}If the differential equation \eqref{aga1.1} is of third order and satisfies conditions \eqref{agaA}, \eqref{agaC} and \end{amst}\no \textup{(D$_2^{rf}$)}~~ each right $(2,1)$-focal point boundary value problemfor \eqref{aga1.1} on $(a,b)$ has at most one solution,\vspace{2mm}\no then \eqref{aga1.1} also satisfies condition \eqref{agaE}.\begin{amsp} \label{agap4} A result similar to that of Theorem \ref{agat9.4} for arbitrary order differential equations \eqref{aga1.1} is not known.\end{amsp}\bigskipThe author is grateful to Professor Johnny Henderson for his help in thepreparation of this lecture.\begin{thebibliography}{22}\bibitem{aga1} R. 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