\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 19, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/19\hfil Continuous descent methods]
{Minimizing convex functions by continuous descent methods}

\author[S. Aizicovici, S. Reich, A. J. Zaslavski\hfil EJDE-2010/19\hfilneg]
{Sergiu Aizicovici, Simeon Reich, Alexander J. Zaslavski}
 % in alphabetical order

\address{Sergiu Aizicovici \newline
Department of Mathematics, Ohio University, Athens, OH
45701-2979, USA}
\email{aizicovi@math.ohiou.edu}

\address{Simeon Reich \newline
Department of Mathematics,
The Technion-Israel Institute of Technology,
32000 Haifa, Israel}
\email{sreich@tx.technion.ac.il}

\address{Alexander J. Zaslavski \newline
Department of Mathematics,
The Technion-Israel Institute of Technology,
32000 Haifa, Israel}
\email{ajzasl@tx.technion.ac.il}

\thanks{Submitted June 27, 2009. Published January 28, 2010.}
\subjclass[2000]{37L99, 47J35, 49M99, 54E35, 54E50, 54E52, 90C25}
\keywords{Complete uniform space; convex function; descent method;
\hfill\break\indent initial value problem; minimization problem}

\begin{abstract}
 We study continuous descent methods for minimizing
 convex functions, defined on general Banach spaces, which are
 associated with an appropriate complete metric space of vector
 fields. We show that there exists an everywhere dense open set in
 this space of vector fields such that each of its elements
 generates strongly convergent trajectories.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{proposition}[theorem]{Proposition}

\section{Introduction and preliminaries}

Discrete and continuous descent methods are important topics in
optimization theory and in the study of dynamical systems. See,
for example, the references in this article.
 Given a real-valued continuous convex function
$f$ on a Banach space $X$, we associate with $f$ a complete metric space
of vector fields $V:X \to X$ such that $f^0(x,Vx) \le 0$ for all
$x \in X$. Here $f^0(x,u)$ is the directional derivative of $f$ at $x$
in the direction of $u \in X$. In the discrete case, to each such
vector field there correspond two gradient-like iterative
processes. In the papers \cite{r1,r2,r3} it is shown that for most
of these vector fields, both iterative processes generate
sequences $\{x_n\}_{n=1}^{\infty}$ such that the sequences
$\{f(x_n)\}_{n=1}^{\infty}$ tend to $\inf(f)$ as $n \to \infty$.
Here by ``most'', we mean an everywhere dense $G_{\delta}$ subset
of the space of vector fields. In our recent papers \cite{a1,a2,a3,a4} we
studied the convergence of the trajectories of an analogous
continuous dynamical system governed by such vector fields to the
point where the function $f$ attains its infimum.

In the present paper, we improve upon the results of our previous
papers by showing that in a certain space of Lipschitz vector
fields, equipped with an appropriate complete metric, there exists
an {\it open} everywhere dense set such that each of its elements
generates strongly convergent trajectories.

More precisely, let $(X,\|\cdot\|)$ be a Banach space and let
 $f:X \to R^1$ be a convex continuous function which satisfies the
following conditions:
\begin{itemize}
\item[C(i)] $\lim_{\|x\| \to \infty} f(x)=\infty$;

\item[C(ii)] there is $\bar{x} \in X$ such that $f(\bar{x}) \le f(x)$ for
all $x \in X$;

\item[C(iii)] if $\{x_n\}_{n=1}^{\infty} \subset X$ and $\lim_{n \to
\infty} f(x_n)=f(\bar{x})$, then
$$
\lim_{n \to \infty}\|x_n-\bar{x}\|=0.
$$
\end{itemize}
By C(iii), the point $\bar{x}$, where the minimum of $f$ is
attained, is unique.

For each $x \in X$, let
\begin{equation}
f^0(x,u)=\lim_{t \to 0^{+}}[f(x+tu)-f(x)]/t,\; u \in X. \label{e1.1}
\end{equation}
Let $(X^*,\|\cdot\|)$ be the dual space of $(X,\|\cdot\|)$. For
each $x \in X$ and each $r>0$, set
\begin{equation}
B(x,r)=\{z \in X: \|z-x\| \le r\}, \quad
B(r)=B(0,r).\label{e1.2}
\end{equation}


For each mapping $A:X \to X$ and each $r >0$, put
\begin{equation}
\mathop{\rm Lip}(A,r)=\sup\{\|Ax-Ay\|/\|x-y\|:\, x,y \in B(r)
\text{ and } x
\not =y\}. \label{e1.3}
\end{equation}
Denote by $\mathcal{A}_1$ the set of all mappings
$V:X \to X$ such that the restriction of $V$ to any bounded subset
of $X$ is Lipschitz (namely, Lip$(V,r)<\infty$ for any positive
$r$) and $f^0(x,Vx) \le 0$ for all $x \in X$.

For the set $\mathcal{A}_1,$ we consider the uniformity determined by
the base
\begin{equation}
\begin{aligned}
E(n,\epsilon)=\big\{&(V_1,V_2) \in \mathcal{A}_1\times \mathcal{A}_1:
\mathop{\rm Lip}(V_1-V_2,n) \le \epsilon,\\
& \|V_1x-V_2x\| \le \epsilon \text{ for all } x \in B(n)\big\},
\end{aligned} \label{e1.4}
\end{equation}
where $n,\epsilon>0$.

Clearly, this uniform space $\mathcal{A}_1$ is metrizable and complete.
We equip it with the topology induced by this uniformity.
The following existence result was proved in \cite[Section 3]{a2}.

\begin{proposition} \label{prop1.1}
 Let $x_0 \in X$ and $V \in \mathcal{A}_1$.
Then there exists a unique continuously differentiable mapping
$x:[0,\infty) \to X$ such that
$$
x'(t)=Vx(t),\quad t \in [0,\infty),\; x(0)=x_0.
$$
\end{proposition}
Let $x \in W^{1,1}(0,T;X)$; i.e.,
$$
x(t)=x_0+\int_0^Tu(s)ds,\quad t \in [0,T],
$$
where $T>0$, $x_0 \in X$ and $u \in L^1(0,T;X)$.
Then $x:[0,T] \to X$ is absolutely continuous and
$x'(t)=u(t)$ for a.e. $t \in [0,T]$.

It was shown in \cite{a2} that the function
$(f\circ x)(t):=f(x(t))$, $ t \in [0,T]$, is absolutely continuous. It
follows that for almost every $t \in [0,T]$, both the derivatives
$x'(t)$ and $(f\circ x)'(t)$ exist:
\begin{gather*}
x'(t)=\lim_{h \to 0}h^{-1}[x(t+h)-x(t)],\\
(f\circ x)'(t)=\lim_{h \to 0}h^{-1}[f(x(t+h))-f(x(t))].
\end{gather*}
We now quote \cite[Proposition 3.1]{a1}.

\begin{proposition} \label{prop1.2}
Assume that $t \in [0,T]$ and that both the derivatives $x'(t)$
and $(f\circ x)'(t)$ exist. Then
$$
(f\circ x)'(t)=\lim_{h \to 0}h^{-1}[f(x(t)+hx'(t))-f(x(t))].
$$
\end{proposition}

Our results are proved under two additional assumptions on the
function $f$. First, we assume in addition that
\begin{itemize}
\item[C(iv)] $f$ is Lipschitz on bounded subsets of $X$.
\end{itemize}
The second assumption concerns the existence of a sharp minimum.
Specifically, this hypothesis can be formulated as follows.
\begin{itemize}
\item[C(v)] There is $\Delta >0$ such that
$$
f(x) \ge f(\bar{x}) + \Delta \|x - \bar{x}\| \quad\text{for all } x \in X.
$$
\end{itemize}
Clearly, we may assume without any loss of generality
that $\Delta < 1$.

We next show that seemingly different
variants of this notion are, in fact, equivalent.

\begin{proposition} \label{prop1.3}
 The following conditions are equivalent.
\begin{itemize}
\item[(i)] There is $\Delta>0$ such that
$$
f(x) \ge f(\bar{x})+\Delta\|x-\bar{x}\|\quad \text{for all } x \in X;
$$

\item[(ii)] there exist $\Delta>0$ and $r>0$ such that
$$
f(x) \ge f(\bar{x})+\Delta \|x-\bar{x}\|\quad \text{for all } x \in X
\text{ satisfying } f(x)\le f(\bar{x})+r;
$$

\item[(iii)] there exist $\Delta >0$ and $r>0$ such that
$$
f(x) \ge f(\bar{x})+\Delta\|x-\bar{x}\|\quad \text{for all } x \in
B(\bar{x},r).
$$
\end{itemize}
\end{proposition}

\begin{proof}
 Clearly, (i) implies (ii). By the continuity of $f$,
(ii) implies (iii).

It remains to show that (iii) implies (i).
To this end, assume that (iii) holds. There are then $\Delta>0$
and  $r>0$ such that
\begin{equation}
f(x) \ge f(\bar{x})+\Delta\|x-\bar{x}\| \quad \text{for all } x\in
B(\bar{x},r).\label{e1.5}
\end{equation}
Assume that
\begin{equation}
x \in X \setminus   B(\bar{x},r). \label{e1.6}
\end{equation}
Put
\begin{equation}
z=\bar{x}+r\|x-\bar{x}\|^{-1}(x-\bar{x}).\label{e1.7}
\end{equation}
By \eqref{e1.7},
\begin{equation}
z \in B(\bar{x},r). \label{e1.8}
\end{equation}
By  \eqref{e1.5}, \eqref{e1.7} and \eqref{e1.8},
\begin{equation}
f(z) \ge f(\bar{x})+\Delta\|z-\bar{x}\|
=f(\bar{x})+\Delta r.\label{e1.9}
\end{equation}
Since $f$ is convex, it follows from \eqref{e1.7} and \eqref{e1.6}
that
\begin{align*}
f(z)&=f(r\|x-\bar{x}\|^{-1}x+(1-r\|x-\bar{x}\|^{-1})\bar{x}) \\
 &\le r\|x-\bar{x}\|^{-1}f(x)+(1-r\|x-\bar{x}\|^{-1})f(\bar{x}).
\end{align*}
When combined with \eqref{e1.9}, this implies that
\begin{gather*}
f(\bar{x})+\Delta r\le r\|x-\bar{x}\|^{-1}f(x)+(1-r\|x-\bar{x}\|^{-1})f(\bar{x}), \\
r\|x-\bar{x}\|^{-1}f(\bar{x})+\Delta r \le r\|x-\bar{x}\|^{-1}f(x),
\end{gather*}
whence
\begin{equation}
f(x)\ge f(\bar{x})+\Delta\|x-\bar{x}\|. \label{e1.10}
\end{equation}
Thus \eqref{e1.10} holds for all $x \in X$ satisfying \eqref{e1.6}.
Together with \eqref{e1.5} this implies that (i) holds.
Proposition 1.3 is proved.
\end{proof}

\begin{proposition} \label{prop1.4}
 For each $h \in X$, $f^0(\bar{x},h)\ge \Delta \|h\|$.
\end{proposition}

The above propositon is a direct consequence of \eqref{e1.1} and C(v).

\begin{proposition} \label{prop1.5}
For any $V \in \mathcal{A}_1$, $V\bar{x}=0$.
\end{proposition}

\begin{proof}
Let $V \in \mathcal{A}_1$. By
Proposition 1.1, there is a continuously differentiable mapping
$x:[0,\infty) \to X$ such that
\begin{equation}
x(0)=\bar{x},\quad x'(t)=Vx(t),\quad t \in [0,\infty). \label{e1.11}
\end{equation}
In view of \eqref{e1.11}, the inclusion $V \in \mathcal{A}_1$ and
Proposition 1.2, the composition $f\circ x$ is a decreasing function.
By C(ii) and C(iii), $x(t)=\bar{x}$ for all $t \ge 0$. When combined with
\eqref{e1.11}, this implies that for all $t \ge 0$,
$$
0=x'(t)=V\bar{x}.
$$
The proof is complete.
\end{proof}

We are finally ready to state our main result. It improves upon
previous results because it guarantees the existence of a good
{\it open} everywhere dense set, and not just a good $G_{\delta}$ dense
set. Its proof is given in Section 3 while Section 2 is devoted
to several auxiliary results.

\begin{theorem} \label{thm1.1}
 Let $M_0>0$. Then there exists an open
everywhere dense $\mathcal{F} \subset \mathcal{A}_1$ such that for each
$U \in \mathcal{F}$, there exists a neighborhood $\mathcal{U}$
of $U$ in $\mathcal{A}_1$ so that the following property holds:

For each $\epsilon >0$, there exists $T_{\epsilon}>0$ such that
for each $W \in \mathcal{U}$ and each $x \in C^1([0,\infty);X)$
satisfying
$$
x'(t)=Wx(t),\quad t \in [0,\infty),\quad \|x(0\|\le M_0,
$$
the inequality $\|x(t)-\bar{x}\|\le \epsilon$ holds  for all
$t \ge T_{\epsilon}$.
\end{theorem}

\begin{corollary} \label{coro1.1}
Let $M_0>0$ and $U \in \mathcal{F}$, where
$\mathcal{F}$ is the set the existence of which is guaranteed
 by Theorem \ref{thm1.1}. Let $x \in C^1([0,\infty);X)$ satisfy
$x'(t)=Ux(t)$, $t \in [0,\infty)$, $\|x(0\|\le M_0$.
Then $\lim_{t \to \infty} x(t)=\bar{x}$.
\end{corollary}


\section{Auxiliary results}

For each $V \in \mathcal{A}_1$ and each $\gamma \in (0,1)$, set
\begin{equation}
V_{\gamma}x=Vx+\gamma(\bar{x}-x),\; x \in X.\label{e2.1}
\end{equation}

\begin{lemma}[{\cite[Lemma 4.1]{a2}}] \label{lem2.1}
 Let $V \in \mathcal{A}_1$ and
$\gamma \in (0,1)$. Then $V_{\gamma} \in \mathcal{A}_1$.
\end{lemma}

\begin{lemma}[{\cite[Lemma 4.2]{a2}}] \label{lem2.2}
Let $V \in \mathcal{A}_1$.
Then $\lim_{\gamma \to 0^{+}}V_{\gamma}=V$.
\end{lemma}

\begin{lemma} \label{lem2.3}
Let $M_0>0$. Then there exists $M>0$ such
that for each $V \in \mathcal{A}_1$ and each $x \in C^1([0,\infty); X)$
which satisfies
\begin{equation}
x'(t)=Vx(t),\quad t \in [0,\infty),\quad \|x(0)\|\le M_0,\label{e2.2}
\end{equation}
the following inequality holds:
\begin{equation}
\|x(t)\|\le M\quad \text{for all } t \in [0,\infty). \label{e2.3}
\end{equation}
\end{lemma}

\begin{proof} By C(iv), there is $M_1>0$ such that
\begin{equation}
f(z) \le M_1\text{ for all } x \in B(0,M_0). \label{e2.4}
\end{equation}
By C(i), there is $M>0$ such that
\begin{equation}
\|z\| \le M\quad \text{for all } z \in X\text{ satisfying } f(z)
\le M_1. \label{e2.5}
\end{equation}
Assume that $V \in \mathcal{A}_1$ and that
$x \in C^1([0,\infty); X)$ satisfies \eqref{e2.2}.
By \eqref{e2.2}, Proposition 1.2 and
the inclusion $V \in \mathcal{A}_1$, the function $f\circ x$ is
decreasing. Together with \eqref{e2.2}, \eqref{e2.4}
 and \eqref{e2.5} this implies that
$f(x(t)) \le f(x(0))\le M_1$ and $\|x(t)\|\le M$ for all $t \ge
0$. The proof is complete.
\end{proof}

\begin{lemma} \label{lem2.4}
Let $V \in \mathcal{A}_1$, $\gamma \in (0,1)$ and
$x \in X$. Then
\begin{equation}
f^0(x, V_{\gamma}x) \le \gamma(f(\bar{x})-f(x))\le -\gamma\Delta
\|x-\bar{x}\|.\label{e2.6}
\end{equation}
\end{lemma}

\begin{proof}
By \eqref{e2.1}, the properties of the directional
derivative of a convex function and the relation $V \in \mathcal{A}_1$,
\begin{align*}
f^0(x,V_{\gamma}x)
& =f^0(x,Vx+\gamma(\bar{x}-x))\\
& \le f^0(x,Vx)+\gamma f^0(x,\bar{x}-x)\\
&\le \gamma f^0(x,\bar{x}-x)\le \gamma(f(\bar{x})-f(x))\\
&\le -\gamma\Delta\|x-\bar{x}\|.
\end{align*}
The proof is complete.
\end{proof}

\begin{lemma} \label{lem2.5}
Let $V \in \mathcal{A}_1$, $\gamma \in (0,1)$ and
$M>\|\bar{x}\|$. Then there exists a neighborhood $\mathcal{U}$ of
$V_{\gamma}$ in $\mathcal{A}_l$ such that for each $W \in {\mathcal{U}}$
and each $x \in B(0,M)$,
$$
f^0(x,Wx) \le (\gamma/2)(f(\bar{x})-f(x)).
$$
\end{lemma}

\begin{proof} By C(iv), there exists $L_0>1$ such that
\begin{equation}
|f(z_1)-f(z_2)|\le L_0\|z_1-z_2\|\quad \text{for all }z_1,z_2 \in
B(0,M+1). \label{e2.7}
\end{equation}
Choose
\begin{equation}
\delta \in (0,\gamma \bar \Delta (2L_0)^{-1}). \label{e2.8}
\end{equation}
Put
\begin{equation}
\mathcal{U}=\{W \in \mathcal{A}_1: (W,V_{\gamma}) \in
E(M+1,\delta)\}.\label{e2.9}
\end{equation}
Let
\begin{equation}
W \in \mathcal{U} \text{ and } x \in B(0,M).\label{e2.10}
\end{equation}
By Lemma \ref{lem2.4},
\begin{equation}
f^0(x,V_{\gamma}x)\le \gamma(f(\bar{x})-f(x)).\label{e2.11}
\end{equation}
By  the properties of the directional derivative of  a convex
function, \eqref{e2.11}, \eqref{e2.7}, Proposition 1.5, \eqref{e2.10},
the inequality
$M>\|\bar{x}\|$, \eqref{e2.9}, C(v) and \eqref{e2.8},
\begin{align*}
f^0(x,Wx)
&\le f^0(x, V_{\gamma}x) +f^0(x,Wx-V_{\gamma}x) \\
&\le \gamma(f(\bar{x})-f(x))+L_0\|Wx-V_{\gamma}x\| \\
&\le \gamma(f(\bar{x})-f(x))+L_0\|(W-V_{\gamma})x-(W-V_{\gamma})
\bar{x}\|\\
&\le \gamma(f(\bar{x})-f(x))+L_0\delta\|x-\bar{x}\|\\
&\le (f(\bar{x})-f(x))(\gamma-L_0\delta \bar \Delta^{-1})\\
&\le (\gamma/2)(f(\bar{x})-f(x)).
\end{align*}
The proof is complete.
 \end{proof}

\section{Completion of the proof of Theorem \ref{thm1.1}}

Let $M_0>0$. By Lemma \ref{lem2.3}, there exists $M>\|\bar{x}\|$ such that
the following property holds:
\begin{itemize}
\item[(P1)] For each $V \in \mathcal{A}_1$ and each
$x \in C^1([0,\infty);X)$
which satisfies
$$
x'(t)=Vx(t),\quad t \in [0,\infty),\quad \|x(0)\|\le M_0,
$$
the following inequality holds:
$$
\|x(t)\|\le M,\quad t\in [0,\infty).
$$
\end{itemize}

Let $V \in \mathcal{A}_1$ and $\gamma \in (0,1)$.
By Lemma \ref{lem2.5}, there
exists an open neighborhood $\mathcal{U}(V,\gamma)$ of $V_{\gamma}$ in
$\mathcal{A}_1$ such that following property holds:
\begin{itemize}
\item[(P2)] For each $W \in \mathcal{U}(V,\gamma)$ and
each $x \in B(0,M)$,
$$
f^0(x,Wx)\le (\gamma/2)(f(\bar{x})-f(x)).
$$
\end{itemize}
Put
\begin{equation}
\mathcal{F}=\cup\{\mathcal{U}(V,\gamma): V \in \mathcal{A}_1,\;
 \gamma \in (0,1)\}.\label{e3.1}
\end{equation}
 Clearly, $\mathcal{F}$ is an open everywhere dense
subset of $\mathcal{A}_1$.

Let $U \in \mathcal{F}$. By \eqref{e3.1} there are
$V \in \mathcal{A}_1$ and
$\gamma \in (0,1)$ such that
\begin{equation}
U \in \mathcal{U}(V,\gamma). \label{e3.2}
\end{equation}
Let $\epsilon >0$. By C(ii), there is $\epsilon_1\in (0,\epsilon)$
such that
\begin{equation}
\|x-\bar{x}\|\le \epsilon \quad \text{for all } x \in X
\text{ satisfying } f(x) \le f(\bar{x})+\epsilon_1.\label{e3.3}
\end{equation}
Choose a number
\begin{equation}
T_{\epsilon}>1+4(\epsilon_1\gamma)^{-1}[\sup\{|f(z)|: z\in
B(0,M_0)\}-f(\bar{x})].\label{e3.4}
\end{equation}
Assume that
\begin{equation}
W \in \mathcal{U}(V,\gamma),\quad
 x\in C^1([0,\infty);X)\label{e3.5}
\end{equation}
and that
$$
x'(t)=Wx(t),\quad
t \in [0,\infty),\quad \|x(0)\|\le M_0.
$$
To complete the proof it is sufficient to show that
\begin{equation}
\|x(t)-\bar{x}\|\le \epsilon,\quad t \ge T_{\epsilon}.\label{e3.6}
\end{equation}
By \eqref{e3.3}, it suffices to show that
\begin{equation}
f(x(t))\le f(\bar{x})+\epsilon_1,\quad t\in
[T_{\epsilon},\infty).\label{e3.7}
\end{equation}
Since $(f\circ x)$ is decreasing,
\eqref{e3.7} would follow once we prove that
\begin{equation}
\min\{f(x(t)): t\in [0,T_{\epsilon}]\}\le f(\bar{x})
+\epsilon_1.\label{e3.8}
\end{equation}
Assume the contrary. Then
\begin{equation}
f(x(t))>f(\bar{x})+\epsilon_1,\quad t\in [0,T_{\epsilon}].\label{e3.9}
\end{equation}
By \eqref{e3.5} and (P1), we have
\begin{equation}
\|x(t)\|\le M,\quad t \in [0,\infty).\label{e3.10}
\end{equation}
By \eqref{e3.10}, \eqref{e3.9}, (P2) and \eqref{e3.5}, for all
 $t \in [0,T_{\epsilon}]$,
\begin{equation}
f^0(x(t),Wx(t))\le (\gamma/2)(f(\bar{x})-f(x(t))\le
-\epsilon_1\gamma/2.\label{e3.11}
\end{equation}
By Proposition 1.2, \eqref{e3.5} and \eqref{e3.11},
\[
f(x(T_{\epsilon}))-f(x(0))
=\int_0^{T_0}(f\circ x)'(t)dt
=\int_0^{T_{\epsilon}}f^0(x(t),x'(t))dt
<-T_{\epsilon}\epsilon_1\gamma/2
\]
and ({\it cf.} \eqref{e3.5} and \eqref{e3.9})
$$
T_{\epsilon}\epsilon_1\gamma/2
\le f(x(0))-f(x(T_{\epsilon}))
\le \sup\{f(z): z\in B(0,M_0)\}-f(\bar{x}).
$$
This contradicts \eqref{e3.4}. The contradiction we have reached shows
that \eqref{e3.8} holds and therefore the proof
of Theorem \ref{thm1.1} is complete.


\subsection*{Acknowledgments}
The second author was partially supported by the
Fund for the Promotion of Research at the Technion and by the Technion
President's Research Fund.

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