\documentclass[reqno]{amsart}
\usepackage{hyperref}
\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 90, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}
\begin{document}
\title[\hfilneg EJDE-2016/90\hfil Product measurability]
{A general product measurability theorem with applications to
variational inequalities}
\author[K. L. Kuttler, J. Li, M. Shillor \hfil EJDE-2016/90\hfilneg]
{Kenneth L. Kuttler, Ji Li, Meir Shillor}
\address{Kenneth L. Kuttler \newline
Department of Mathematics,
Brigham Young University,
Provo, UT 84602, USA}
\email{klkuttle@math.byu.edu}
\address{Ji Li \newline
Department of Mathematics, Michigan State University,
East Lansing, MI 48824, USA}
\email{liji@math.msu.edu}
\address{ Meir Shillor \newline
Department of Mathematics and Statistics,
Oakland University, Rochester, MI 48309, USA}
\email{shillor@oakland.edu}
\thanks{Submitted January 16, 2016. Published March 31, 2016.}
\subjclass[2010]{35R60, 60H15, 35R45, 35R70, 35S11}
\keywords{Partial differential inclusions; product measurability;
\hfill\break\indent variational inequalities; measurable selection}
\begin{abstract}
This work establishes the existence of measurable weak solutions to
evolution problems with randomness by proving and applying a novel theorem
on product measurability of limits of sequences of functions.
The measurability theorem is used to show that many important
existence theorems within the abstract theory of evolution inclusions or
equations have straightforward generalizations to settings that include random
processes or coefficients. Moreover, the convex set where the solutions are
sought is not fixed but may depend on the random variables.
The importance of adding randomness lies in the fact that real world processes
invariably involve randomness and variability. Thus, this work expands
substantially the range of applications of models with variational
inequalities and differential set-inclusions.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{corollary}[theorem]{Corollary}
\allowdisplaybreaks
\section{Introduction} \label{sec1}
This article concerns the existence of $\mathcal{P}$-measurable
solutions to evolution problems in which some of the input data, such as the
operators, forcing functions or some of the system coefficients, is
random. Such problems, often in the form of evolutionary variational
inequalities, abound in many fields of mathematics, such as optimization and
optimal control, and in applications such as in contact mechanics,
populations dynamics, and many more. The interest in random inputs in
variational inequalities arises from the uncertainty in the identification
of the system inputs or parameters, possibly due to the process of
construction or assembling of the system, and to the imprecise knowledge of
the acting forces or environmental processes that may have stochastic
behavior.
We first establish a very general abstract theorem on the product
measurability of certain limits of sequences of functions. Then, we apply
the result to variational inequalities with monotone, hemicontinuous,
bounded and coercive operators that may or may not be strongly monotone, and
establish the product measurability of the solutions of the variational
inequalities. We note that the abstract result has many other, very diverse,
applications.
The general setting is as follows. We let $(\Omega ,\mathcal{F})$ be
a measurable space with sample space $\Omega $, and a $\sigma $-algebra
$\mathcal{F}$. It is assumed that an evolution problem, in the form of an
inequality or set-inclusion, corresponds to each $\omega \in \Omega $, with a
solution denoted by $u(\cdot ,\omega ) $. When this evolution
problem has a unique solution, it is usually possible to show the existence
of an $\mathcal{F}$-measurable solution, i.e., a solution such that
$(t,\omega ) \to u(t,\omega ) $ is $\mathcal{P}
\equiv \mathcal{B}([0,T] ) \times \mathcal{F}$
measurable, see \cite{kuli2014JDE}. By $\mathcal{P}$ we mean the smallest
$\sigma $ algebra which contains the measurable rectangles $A\times B$
where $A\in \mathcal{B}([0,T] ), B\in \mathcal{F}$. The
case without uniqueness is much more involved and is the focus of this work
as we establish a way to deal with it. No conditions need to be made on the
measurable space and in general, no specific measure is specified,
although we have in mind a probability space.
Our result is very general and we illustrate it by obtaining weak
solutions to a whole class of variational inequalities (see, e.g.,
\cite{KinStamp00,lio69,Mig13}). Our measurability result, stated in
Theorem \ref{zthm1x}, is a substantial improvement of the result we
established in \cite{kuli2014} and \cite{kushi2014}. This version allows
many more applications.
In particular, we apply Theorem \ref{zthm1x} to a very important class of
evolution problems and establish in Theorem \ref{thm9} and Theorem \ref
{thm10} the existence and measurability of the solutions. We refer to
Theorem \ref{zthm1x} as the \emph{measurable selection theorem} because it
establishes the existence of a $\mathcal{P}$-measurable representative in a
set of limits of $\mathcal{P}$-measurable functions.
It is seen from the example that our measurable selection theorem is a
powerful tool and we foresee that it will be used in many problems where
randomness is important, uniqueness is unknown, and measurability of the
solutions is essential. Such cases abound in contact mechanics, especially
when friction is present, see, e.g., \cite{kushi2004, kushi2002, kushi2014}
and the many references in \cite{Mig13, SST04}. We note that our results
may be applied to the results in \cite{mot2000,mot2003}, where
inequality nonconvex problems were studied.
The rest of this article is as follows.
The basic measurable selection theorem,
Theorem \ref{zthm1x}, is established in Section \ref{sec2}. The existence of
$\mathcal{P}$-measurable solutions to a variational inequality with a
monotone, hemicontinuous and bounded operator is shown in Theorem \ref
{thm9} in Section \ref{sec3}. Then, in Theorem \ref{thm10}, we allow the
convex set to be a measurable set-valued map of the random variables.
This is a completely novel
and somewhat surprising result never before considered as far as we know,
and our measurable selection result makes it fairly easy to obtain. More
general operators such as maximal monotone operators can be considered also,
but we do not present this here because the necessary existence for an
approximate problem is not readily available.
\section{Measurable selection theorem}
\label{sec2}
We establish in this section our main theorem. It asserts that under very
weak conditions, in particular, the boundedness condition \eqref{21}, there
exists a measurable selection of the set of limit functions of a sequence of
functions in a reflexive and separable Banach space.
We use the usual notations for Sobolev function spaces. In particular,
$\mathcal{V}\equiv L^{p}([0,T] ,V) $ denotes the
space of $p$-integrable functions defined on $[0,T]$ with values in the
space $V$, $C([0,T])$ is the space of continuous functions on $[0,T]$ with
the maximum norm, and we use the abbreviation $\| \cdot \| =\|
\cdot \| _{C([0,T])}$. The numbers $p>1$ and $p'>1$ are
conjugate, i.e., $\frac{1}{p}+\frac{1}{p'}=1$. Also, $\langle
\cdot ,\cdot \rangle _{V}=\langle \cdot ,\cdot \rangle
_{V',V}$ represents the duality pairing between $V$ and $V'$.
In this section $(\Omega ,\mathcal{F}) $ is a measurable
space. We have in mind a probability space but this is not necessary, since no
reference is made to any particular measure. Also, no topological conditions
are necessary on $\Omega $. This is in contrast to the result of \cite{ben73}
in which a complete probability measure was assumed
and $\Omega $ was assumed
to be a completely regular topological space. The theorem here is a
significant improvement over the one we established in
\cite{kuli2014, kushi2014}. Moreover, the steps in the proof differ.
However, for the sake of brevity, we refer to these two papers
for some of the details.
We use the following notation: $C$ represents a generic constant that may
depend on the data but is independent of $n$. The dependence of a constant
on $\omega, t$ or $k$ is denoted explicitly, e.g., if it depends only on
$\omega$ or $k$, we write $C(\omega)$ or $C(k)$, while $C(\omega, t, k)$
depends on all three.
The following is the main result in this work.
\begin{theorem}[The Measurable Selection Theorem]
\label{zthm1x}
Let $V$ be a reflexive separable Banach space with dual $V'$ and let
$p,p'$ be conjugate numbers. Assume that $\{u_n(t,\omega) \}_{n=1}^{\infty }$
is a sequence of $\mathcal{P}\equiv \mathcal{B}([0,T ] ) \times
\mathcal{F}$-measurable functions
with values in $V$ such that for each $\omega \in \Omega$, except for a null
set $N\subset \Omega$, and for each $n$, it satisfies
\begin{equation}
\| u_n(\cdot ,\omega ) \| _{\mathcal{V}}\leq C(\omega ). \label{21}
\end{equation}
Then, there exists a $\mathcal{P}$-measurable function $u(t,\omega )$ such
that for each $\omega \notin N$,
\[
u(\cdot ,\omega ) \in \mathcal{V},
\]
and there is a subsequence $u_{n_{\omega }}$ such that $u_{n_{\omega
}}(\cdot ,\omega ) \to u(\cdot ,\omega ) $
weakly in $\mathcal{V}$ as $n_{\omega }\to \infty $.
\end{theorem}
We note that the bound in \eqref{21} need not be uniform on $\Omega$,
and $u$ is a weak limit of a subsequence that depends on $\omega$.
We prove the theorem in steps, presented as lemmas. First, we need
the space $X=\prod_{k=1}^{\infty }C([0,T]
) $ with the product topology. Then, $X$ is a metric space with the metric
\[
d(\mathbf{f,g}) \equiv \sum_{k=1}^{\infty }2^{-k}\frac{
\| f_{k}-g_{k}\| }{1+\| f_{k}-g_{k}\| },
\]
where $\mathbf{f}=(f_{1},f_{2},\dots ),\mathbf{g}=(g_{1},g_{2},\dots )\in X$
and the norm is the maximum norm on $C([0,T] ) $.
With this metric, $X$ is complete and separable.
The first step follows.
\begin{lemma} \label{xzlemma2x}
Let $\{ \mathbf{f}_n\}_{n=1}^\infty $ be a
sequence in $X$ and suppose that for each component $k$, the sequence
$\{f_{nk}\}_{n=0}^\infty$ is bounded in $C^{0,(1/p')}([0,T] ) $ by $C(k)$.
Then, there exists a subsequence $\{ \mathbf{f}_{n_j}\} $ that converges as
$n_j\to \infty $ to an element $\mathbf{f}\in X$. Thus, $\{
\mathbf{f} _n\} $ is pre-compact in $X$.
\end{lemma}
\begin{proof}
The result follows from Tychonoff's theorem and the compactness of the
embedding of $C^{0,(1/p')}$ in $C([0,1] ) $.
\end{proof}
In the next step we construct a special sequence
$\{ \mathbf{f} _n\} _{n=1}^{\infty }$ in $X$ based on the sequence
$ \{u_n(t,\omega ) \}_{n=1}^{\infty }$. For $m\in \mathbb{N}$ and
$\phi \in V'$, let $l_{m}(t)\equiv \max (0,t-(1/m) ) $ and define
$\psi _{m,\phi }:\mathcal{V}\to C([0,T] ) $ by
\[
(\psi _{m,\phi }u(\cdot ))(t) \equiv \int_{0}^{T}\langle
m\phi \mathcal{X}_{[l_{m}(t),t] }(s) ,u(s) \rangle _{V}ds
=m\int_{l_{m}(t)}^{t}\langle \phi ,u(s) \rangle _{V}ds.
\]
Here, $\mathcal{X}_{[l_{m}(t),t] }(\cdot ) $ is the
characteristic function of the interval $[l_{m}(t),t]$.
Let $\mathcal{D}=\{ \phi _{r}\} _{r=1}^{\infty }$ denote a
countable subset of $V'$. Later, we will describe it more
carefully. For now, all that is important is that it be countable. Then, the
pairs $(m,\phi )$, for $\phi \in \mathcal{D}$ and $m\in \mathbb{N}$ form a
countable set. Let $\{(m_{k},\phi _{r_{k}}) \}_{k=1}^{\infty }$
denote an enumeration of the pairs $(m,\phi ) \in \mathbb{N}
\times \mathcal{D}$. To simplify the notation, we let
\[
(f_{k}(u(\cdot )) )(t) \equiv (\psi _{m_{k},\phi _{r_{k}}}u(\cdot ) )(t)
=m_{k}\int_{l_{m_{k}}(t)}^{t}\langle \phi _{r_{k}},u(s)
\rangle _{V}ds.
\]
We note that for each fixed $\phi \in \mathcal{D}$ there exists a
subsequence such that $m_{k}\to \infty $ and $\phi _{r_{k}}=\phi $.
Thus there are infinitely many indices $k$ such that $\phi _{r_{k}}=\phi $.
We now choose the functions $u_n(\cdot,\omega)$ of Theorem \ref{zthm1x}
as $u(\cdot)$ and so we have constructed a sequence
$\{ \mathbf{f} _n\}_{n=1}^\infty $ in $X$, which depends on $\omega$. We use this
sequence to define the following set-valued map $\Gamma(\omega)$.
It follows from estimate \eqref{21} that for fixed $k$ and $\omega \notin N$
the functions $\{ t\to f_{k}(u_n(\cdot ,\omega ) ) (t) \} _{n=1}^\infty$
are uniformly bounded and equicontinuous since they are bounded in
$C^{0,1/p'}([0,T] ) $. Then, by Lemma \ref{xzlemma2x}, for each $n$ the set
$\{ \mathcal{X}_{N^{C}}(\omega ) \mathbf{f}
(u_j(\cdot ,\omega ) ) \} _{j=n}^{\infty }$
is pre-compact in $X=\prod_{k}C([0,T] ) $. We define
for each $n\in \mathbb{N}$ a set-valued map $\Gamma ^{n}:\Omega \to X $ by
\[
\Gamma ^{n}(\omega ) \equiv \overline{\cup _{j\geq n}\{
\mathcal{X}_{N^{C}}(\omega ) \mathbf{f}(u_j(\cdot
,\omega ) ) \} },
\]
where the closure is taken in $X$. Then $\Gamma ^{n}(\omega ) $
is compact in $X$. From the definition, a function $\mathbf{g}$ is in $
\Gamma ^{n}(\omega ) $ if and only if $d(\mathbf{g},
\mathcal{X}_{N^{C}}(\omega ) \mathbf{f}(w_{l})
) \to 0$ as $l\to \infty $, where each $w_{l}$ is one
of the $u_j(\cdot ,\omega ) $ for $j\geq n$.
The proof of the next two lemmas can be found \cite{kuli2014} or
\cite{kushi2014}.
\begin{lemma}\label{xzlemma4x}
The mapping $\omega \to \Gamma ^{n}(\omega) $ is an
$\mathcal{F}$-measurable set-valued map with values in $X$.
\end{lemma}
\begin{definition}
Let $\Gamma (\omega ) \equiv \cap _{n=1}^{\infty }\Gamma
^{n}(\omega ) $.
\end{definition}
\begin{lemma}\label{xz30decl1f}
$\Gamma $ is a nonempty $\mathcal{F}$-measurable
set-valued mapping with values in compact subsets of $X$ and there exists an
$\mathcal{F}$-measurable selection $\gamma(\omega)\in\Gamma(\omega)$ such
that $\gamma(t,\omega)\equiv (\gamma(\omega))(t)$ is
$\mathcal{P}$-measurable and $(\prod_{k=1}^{ \infty }\mathbb{R})$-valued function.
\end{lemma}
It follows from the definition of $\Gamma (\omega )$ that, for each
$\omega \notin N$, there exists a subsequence
$u_{n(\omega ) }( \cdot ,\omega ) $ of $u_n(\cdot ,\omega )$ such that for each
component $k$,
\[
\gamma _{k}(t,\omega ) =\lim_{n(\omega ) \to
\infty }\mathcal{X}_{N^{C}}(\omega ) (f_{k}(u_{n(
\omega ) }(\cdot ,\omega ) )) (t)\mbox{ uniformly in} t,
\]
or
\begin{equation}
\gamma _{k}(t,\omega )=\lim_{n(\omega ) \to \infty
}m_{k}\int_{l_{m_{k}}(t)}^{t}\langle \phi _{r_{k}},\mathcal{X}
_{N^{C}}(\omega ) u_{n(\omega ) }(s,\omega
) \rangle _{V}ds. \label{limit}
\end{equation}
We now set $\gamma (t,\omega )\equiv 0$ for $\omega \in N$ and then
\eqref{limit} holds for all $\omega $. We note that it is not clear
whether $\mathcal{X}_{N^{C}}(\omega ) (f_{k}(u_{n(\omega ) }(\cdot ,\omega ) )) (t)$
is $\mathcal{P}$-measurable, however, all that is needed is that the
limit $\gamma (t,\omega )$ is $\mathcal{P}$-measurable.
We have all the ingredients for the proof.
\begin{proof}[Proof of Theorem \ref{zthm1x}]
By assumption, there exists a further subsequence, still
denoted by $n(\omega )$, such that, the weak limit
\[
\lim_{n(\omega ) \to \infty }\mathcal{X}_{N^{C}}(
\omega ) u_{n(\omega ) }(\cdot ,\omega )
=v(\cdot ,\omega )
\]
exists in $\mathcal{V}$. Then,
\begin{align*}
&m_{k}\int_{l_{m_{k}}(t)}^{t}\langle \phi _{r_{k}},v(s,\omega
) \rangle _{V}ds \\
&= \lim_{n(\omega ) \to \infty
}m_{k}\int_{l_{m_{k}}(t)}^{t}\langle \phi _{r_{k}},\mathcal{X}
_{N^{C}}(\omega ) u_{n(\omega ) }(s,\omega
) \rangle _{V}ds \\
&= \gamma _{k}(t,\omega )
\end{align*}
is product measurable. Letting $\phi \in \mathcal{D}$ be given,
there exists a subsequence, denoted
by $k$, such that $m_{k}\to \infty $ and $\phi _{r_{k}}=\phi $.
Recall $(m_{k},\phi _{r_{k}}) $ denoted an enumeration of the
pairs $(m,\phi ) \in \mathbb{N}\times \mathcal{D}$. For a given
$\phi \in \mathcal{D}$ denote this sequence by $m_{\phi }$. Thus, we have
measurability of
\[
(t,\omega ) \to m_{\phi }\int_{l_{m_{\phi}}(t)}^{t}\langle \phi ,v(s,\omega )
\rangle _{V}ds,
\]
for each $\phi \in \mathcal{D}$.
Now we describe the countable set $\mathcal{D}$. Iterate the following.
Let $\phi _{1}\neq 0$. Let $\mathcal{F}$ denote linearly independent subsets
of $V'$ which contain $\phi _{1}$ such that the elements are
further apart than $1/5$. Let $\mathcal{C}$ denote a maximal chain.
Thus $\cup \mathcal{C}$ is also in $\mathcal{F}$.
If $W:=\overline{\operatorname{span}\cup \mathcal{C}}$ fails to be all of $V'$,
then there would exist $\psi \notin W$ such that the distance of $\psi $
to the closed subspace $W$ is at least $1/5$.
Now $\mathcal{C},\cup \{\mathcal{C}\cup \{ \psi \} \}$
would violate maximality of $\mathcal{C}$. Hence $W=V'$. Now it
follows that $\mathcal{C}$ must be countable since otherwise, $V'$
would fail to be separable. Then $\mathcal{D}$ is defined as $\cup \mathcal{C
}$. Let $M$ be the set of rational linear combinations of elements of $\mathcal{D}$,
hence $M$ is dense in $V'$. Note that linear combinations of the $\phi _i$ are
uniquely determined because none is a linear combination of the others. Now,
we define a linear mapping on $M$ which makes sense for
$(t,\omega ) $ on a certain set.
\begin{definition} \rm
Let $E$ be the set of points $(t,\omega ) $ such that the following
limit exists for each $\phi \in \mathcal{D}$
\[
\Lambda (t,\omega ) \phi :=\lim_{m_{\phi }\to \infty
}m_{\phi }\int_{l_{m_{\phi }}(t) }^{t}\langle \phi ,v(
s,\omega ) \rangle ds.
\]
Extend this mapping linearly. That is, for $\psi \in M,\psi
:=\sum_ia_i\phi _i$,
\[
\Lambda (t,\omega ) \psi :=\sum_ia_i\Lambda (t,\omega
) \phi _i
=\sum_ia_i\Big(\lim_{m_{\phi _i}\to \infty
}m_{\phi _i}\int_{l_{m_{\phi _i}}(t) }^{t}\langle \phi
_i,v(s,\omega ) \rangle ds\Big).
\]
Thus $(t,\omega ) \to $ $\Lambda (t,\omega )\psi $ is product measurable,
being the sum of limits of product measurable
functions. Let $G$ denote those $(t,\omega ) $ in $E$ such that
there exists a constant $C(t,\omega ) $ such that for all $\psi \in M$,
\[
| \Lambda (t,\omega ) \psi | \leq C( t,\omega ) \| \psi \|.
\]
\end{definition}
We note that $E$ is a product measurable set because it was shown
above that $(t,\omega ) \to m_{\phi }\int_{l_{m_{\phi }}(t)
}^{t}\langle \phi ,v(s,\omega ) \rangle ds$ is
product measurable and in general, the set of points where a sequence of
measurable functions converges is a measurable set.
\begin{lemma}
$G$ is product measurable.
\end{lemma}
\begin{proof}
This follows from the formula
\[
E\cap G^{C}=\cap _n\cup _{\psi \in M}\{ (t,\omega )
:| \Lambda (t,\omega ) \psi | >n\|
\psi \| \}
\]
which is clearly product measurable because $(t,\omega )
\to \Lambda (t,\omega ) \psi $ is. Thus, since $E$ is
measurable, it follows that $E\cap G=G$ is too.
\end{proof}
For $(t,\omega ) \in G,\Lambda (t,\omega ) $ has a
unique extension to all of $V$, the dual space of $V'$, still
denoted as $\Lambda (t,\omega ) $. By the Riesz representation
theorem, for $(t,\omega ) \in G,$there exists $u(t,\omega
) \in V$,
\[
\Lambda (t,\omega ) \psi =\langle \psi ,u(t,\omega) \rangle _{V}.
\]
Thus, $(t,\omega ) \to \mathcal{X}_{G}(t,\omega
) u(t,\omega ) $ is product measurable by the Pettis
theorem. It has been shown that $G$ is measurable and we let $u=0$ off $G$.
Next, we will show using the fundamental theorem of calculus that for each
$\omega$ the set $\{ t:(t,\omega ) \in G\} $ is all but a set of
Lebesgue measure zero.
We now fix $\omega $. By the fundamental theorem of calculus,
\[
\lim_{m\to \infty }m\int_{l_{m}(t) }^{t}v(s,\omega) ds=v(t,\omega ) \quad\text{in }V,
\]
for a.e. $t$, say for all $t\notin N(\omega ) \subseteq [0,T
] $. Of course we do not know that $\omega \to v(
t,\omega ) $ is measurable. However, the existence of this limit for
$t\notin N(\omega ) $ implies that for every $\phi \in V^{\prime} $,
\[
\lim_{m\to \infty }\big| m\int_{l_{m}(t)
}^{t}\langle \phi ,v(s,\omega ) \rangle ds\big|
\leq C(t,\omega ) \| \phi \|,
\]
for some $C(t,\omega ) $. Here $m$ does not depend on $\phi $.
Thus, in particular, this holds for a subsequence and so for each
$t\notin N(\omega ) ,(t,\omega ) \in G$ because for each
$\phi \in \mathcal{D}$,
\[
\lim_{m_{\phi }\to \infty }m_{\phi }\int_{l_{m_{\phi }}(
t) }^{t}\langle \phi ,v(s,\omega ) \rangle ds
\text{ exists and satisfies the above inequality.}
\]
Hence, for all $\psi \in M$,
\[
\Lambda (t,\omega ) \psi =\langle \psi ,u(t,\omega
) \rangle _{V} ,
\]
where $u$ is product measurable.
Also, for $t\notin N(\omega ) $ and $\phi \in \mathcal{D}$,
\[
\langle \phi ,u(t,\omega ) \rangle _{V}
=\Lambda (t,\omega ) \phi \equiv \lim_{m_{\phi }\to
\infty }m_{\phi }\int_{l_{m_{\phi }}(t) }^{t}\langle \phi
,v(s,\omega ) \rangle ds=\langle \phi ,v(
t,\omega ) \rangle _{V} ,
\]
therefore, for all $\phi \in M$,
\[
\langle \phi ,u(t,\omega ) \rangle _{V}
=\langle \phi ,v(t,\omega ) \rangle _{V},
\]
and hence $u(t,\omega ) =v(t,\omega ) $. Thus, for
each $\omega $, the product measurable function $u$ satisfies
$u( t,\omega ) =v(t,\omega ) $ for a.e. $t$.
Hence $u(\cdot ,\omega ) =v(\cdot ,\omega ) $ in $\mathcal{V}$.
This completes the proof of Theorem 1.
\end{proof}
We show next how condition \eqref{21} can be verified by taking expectation
when we deal with a probability space. Thus, we let
$(\Omega,\mathcal{F},P) $ be a probability space, where $P$ is the
probability function.
\begin{proposition}
Let $\{ u_n(\cdot,\omega)\} _{n=1}^{\infty }$ be a sequence of
functions in $L^{p}(\Omega ,\mathcal{V}) $ such that
\begin{equation} \label{eqn5}
\sup_n\int_{\Omega }\| u_n\| _{\mathcal{V}
}^{p}dP=C<\infty.
\end{equation}
Then, there is a set $N$ of measure zero and a subsequence $\{
u_{n,n}(\cdot,\omega)\} _{n=1}^{\infty }$ such that for all $\omega
\notin N$,
\[
\sup_n\| u_{n,n}(\cdot ,\omega ) \| _{
\mathcal{V}}\leq C(\omega ) <\infty.
\]
\end{proposition}
\begin{proof}
First, we know that there is a set $\hat{N}$ of measure zero such that for $
\omega \notin \hat{N},\| u_n(\cdot ,\omega )
\| _{\mathcal{V}}^{p}<\infty $ for all $n$. Indeed, we just take
the union of the exceptional sets, one for each $n$. We have,
\begin{align*}
&P\Big(\{ \omega :\lim \inf_{n\to \infty }\|
u_n(\cdot ,\omega ) \| _{\mathcal{V}}^{p}\geq
M_{1}\} \Big) \\
&\leq \frac{1}{M_{1}}\int_{\Omega }\lim \inf_{n\to \infty
}\| u_n(\cdot ,\omega ) \| _{\mathcal{V}
}^{p}dP \\
&\leq \frac{1}{M_{1}}\lim \inf_{n\to \infty }\int_{\Omega
}\| u_n(\cdot ,\omega ) \| _{\mathcal{V}
}^{p}dP\leq \frac{C}{M_{1}}.
\end{align*}
Therefore, by choosing a sufficiently large $M_{1}$, we obtain that the set
\[
B_{1}\equiv \{ \omega :\lim \inf_{n\to \infty }\|
u_n(\cdot ,\omega ) \| _{\mathcal{V}}^{p}\geq
M_{1}\}
\]
has measure less than $1/2$. Letting $G_{1}\equiv \Omega \setminus B_{1}$,
it follows that $P(G_{1}) >1/2$ and for $\omega \in G_{1}$,
\[
\lim \inf_{n\to \infty }\| u_n(\cdot ,\omega )
\| _{\mathcal{V}}^{p}M_{1}$,
\[
B_{2}\equiv \big\{ \omega \in B_{1}:\lim \inf_{n\to \infty
}\| u_{1,n}(\cdot ,\omega ) \| _{\mathcal{V}
}^{p}\geq M_{2}\big\}
\]
has measure less than $1/4$. Let $G_{2}$ be such that $B_{2}\cup G_{2}=B_{1}$.
Then, for $\omega \in G_{2}$ it follows that $\lim \inf_{n\to
\infty }\| u_{1,n}(\cdot ,\omega ) \| _{
\mathcal{V}}^{p}1$; $V\subseteq H=H'\subseteq V'$ and each space
is a separable Hilbert space that is dense in the following one; and
$\mathcal{H}\equiv L^{2}([0,T] ,H) $.
Then, $\mathcal{V\subseteq H}=\mathcal{H}'\subseteq \mathcal{V}'$.
As above, we denote by $\langle \cdot ,\cdot
\rangle _{\mathcal{V}}=\langle \cdot ,\cdot \rangle _{
\mathcal{V}',\mathcal{V}}$ the duality pairing between $\mathcal{V}$
and $\mathcal{V}'$, and $\mathcal{P}$ denotes the product $\sigma
$-algebra which is the smallest $\sigma $ algebra that contains the sets of
the form $B\times F$ where $B$ is Borel in $[0,T] $ and
$F\in \mathcal{F}$.
Let $\mathcal{K}$ be a closed convex subset of $\mathcal{V}$ and let,
for the sake of convenience, $0\in \mathcal{K}$. Note that if $\mathcal{K}$ is
closed and convex in $\mathcal{H,}$ then $\mathcal{K}\cap \mathcal{V}$ is
closed and convex in $\mathcal{V}$. Let $A(\cdot ,\omega ) :
\mathcal{V\to V}'$ be monotone, hemicontinuous, bounded and
coercive operator, i.e.,
\begin{equation}
\lim_{\| u\| \to \infty }\frac{\langle A(
u,\omega ) ,u\rangle _{\mathcal{V}}}{\| u\| _{
\mathcal{V}}}=\infty. \label{2jane1s}
\end{equation}
We assume that $\omega \to A(u(\omega ) ,\omega )$
is measurable into $\mathcal{V}'$ whenever $\omega \to
u(\omega ) $ is measurable into $\mathcal{V}$.
Let $P$ be a penalization operator associated with $\mathcal{K}$,
as discussed in \cite{lio69}. Thus,
\[
Pu(\omega ) =F(u(\omega ) -\operatorname{proj}_{\mathcal{K}}u(\omega ) ),
\]
where $F$ is the duality map for $p$ satisfying
$\| Fx\| =\| x\| ^{p-1}$ and $\langle Fx,x\rangle=\| x\| ^{p}$, \cite{lio69}.
As is well known, $P$ is monotone and demicontinuous, since $P(u) =0$ on
$\mathcal{K}$ and is nonzero for $u\notin \mathcal{K}$.
We assume that $(t,\omega ) \to f(t,\omega ) $ is product measurable.
Under these assumptions, for each $n\in \mathbb{N}$,
there exists a solution $u_n$ to the penalized problem,
\begin{equation}
\begin{gathered}
u_n'+A(u_n(\omega ) ,\omega ) +nP(
u_n(\cdot ,\omega ) ) =f(\cdot ,\omega )
\text{ in }\mathcal{V}', \\
u_n(0,\omega ) =0,
\end{gathered} \label{1jane1s}
\end{equation}
such that $(t,\omega ) \to u_n(t,\omega) $ is product measurable.
Such a solution can be obtained by using arguments similar to those
in stochastic partial differential equations.
We also assume, as in \cite{lio69}, the existence of a regularizing
sequence such that if $u\in \mathcal{K}$, then there exists
$u_i\in \mathcal{K}$ such that $u_i'\in \mathcal{V}',u_i(0) =0$ and
\[
\limsup_{i\to \infty }\langle u_i',u_i-u\rangle _{\mathcal{V}}\leq 0.
\]
Then, using standard arguments, one can pass to an appropriate limit and
obtain the first part of the following theorem.
\begin{theorem} \label{thm9}
Suppose $A(\cdot ,\omega ) $ is monotone,
hemicontinuous, bounded and coercive as a map from $\mathcal{V}$ to
$\mathcal{V}'$. Suppose also that when
$\omega \to u(\omega) $ is measurable into $\mathcal{V}$, then
$\omega \to A(u(\omega ) ,\omega ) $ is measurable into
$\mathcal{V}'$. Let $\mathcal{K}$ be a closed and convex subset
of $\mathcal{V}$ containing 0. Let there be a regularizing sequence
$\{ u_i\} $ for each $u\in \mathcal{K}$ satisfying
$u_i(0)=0,u_i'\in \mathcal{V}',u_i\in \mathcal{K}$,
\[
\limsup_{i\to \infty }\langle u_i',u_i-u\rangle \leq 0.
\]
Then, for each $\omega$ there exists a solution to the variational inequality
\[
\langle v',u-v\rangle +\langle A(u(
\cdot ,\omega ) ,\omega ) ,u(\cdot ,\omega )
-v\rangle \leq \langle f(\cdot ,\omega ), u-v\rangle
\]
that is valid for all $v\in \mathcal{K}$, such that $(Bv) '\in
\mathcal{V}',Bv(0) =0$, and $(t,\omega )
\to u(t,\omega )$ is $\mathcal{B}([0,T
] ) \times \mathcal{F}$-measurable.
\end{theorem}
\begin{proof}
It only remains to verify the assertion about measurability. This
follows from Theorem \ref{zthm1x}, since one can obtain an estimate of
the right form for the measurable functions $u_n(\cdot ,\omega) $ and
$u_n^{\ast }(\cdot ,\omega ) $. Then, the above
argument shows that a subsequence has a convergent subsequence which
converges to a solution.
\end{proof}
We note that one can replace the coercivity condition \eqref{2jane1s} with a weaker
one involving $\lambda I+A$ for large enough $\lambda$, for $p\geq 2$,
see \cite{kushi1999} for details.
Next, we describe a surprising generalization of Theorem \ref{thm9} in which
the convex closed set is not fixed, but depends on the random variable, i.e.,
$\mathcal{K=K}(\omega )$. The proof is exactly the same as long
as the penalization operator satisfies that
$\omega \to P(u(\omega ) ,\omega ) =F(u(\omega ) -\operatorname{proj}_{
\mathcal{K}(\omega ) }u(\omega ) ) $ is
measurable into $\mathcal{V}'$ whenever
$\omega \to u( \omega ) $ is measurable into $\mathcal{V}$.
First, we need the following result that describes the conditions on the
set-valued mapping $\omega \to \mathcal{K}(\omega ) $
that make it satisfy this property.
\begin{proposition}
Let $\omega \to \mathcal{K}(\omega ) $ be measurable
into $\mathcal{V}$. Then, $\omega \to \operatorname{proj}_{\mathcal{K}(
\omega ) }u(\omega ) $ is also measurable into
$\mathcal{V} $ whenever $\omega \to u(\omega ) $ is measurable.
\end{proposition}
\begin{proof}
It follows from standard results on measurable multi-functions, see e.g.,
\cite{pap97}, that there is a countable collection
$\{ w_n(\omega ) \} $, $\omega \to w_n(\omega ) $
being measurable and $w_n(\omega ) \in \mathcal{K}(\omega ) $, for each
$\omega $, such that for each $\omega$ we have
$\mathcal{K}(\omega ) =\overline{\cup _nw_n(\omega ) }$.
Let
\[
d_n(\omega ) \equiv \min \{ \| u(\omega) -w_{k}(\omega ) \| ,k\leq n\}.
\]
Let $u_{1}(\omega ) \equiv w_{1}(\omega ) $. Set
$u_{2}(\omega ) =w_{1}(\omega )$
on the set
\[
\{ \omega :\| u(\omega ) -w_{1}(\omega
) \| <\{ \| u(\omega ) -w_{2}(\omega ) \| \} \}
\]
and $u_{2}(\omega ) \equiv w_{2}(\omega )$ off this set.
Thus, $\| u_{2}(\omega ) -u(\omega )
\| =d_{2}$. Let
\begin{gather*}
u_{3}(\omega ) =w_{1}(\omega ) \quad
\text{on } S_{1} \equiv \big\{\omega :\| u(\omega ) -w_{1}(\omega )\|
<\| u(\omega ) -w_j(\omega ) \|,\; j=2,3 \big\},
\\
u_{3}(\omega ) =w_{2}(\omega ) \quad
\text{on } S_{1}\cap \big\{ \omega :\| u(\omega ) -w_{1}(\omega ) \|
<\| u(\omega ) -w_j(\omega ) \|,\; j=3 \big\}, \\
u_{3}(\omega ) = w_{3}(\omega ) \text{ on the
remainder of }\Omega.
\end{gather*}
Thus, $\| u_{3}(\omega ) -u(\omega ) \| =d_{3}$.
We continue in this way, at each step $n$ obtaining
$u_n(\omega ) $ such that
\[
\| u_n(\omega ) -u(\omega) \| =d_n(\omega ),
\]
and $u_n(\omega) \in \mathcal{K}(\omega ) $ with $u_n$ measurable.
Thus, in effect, we pick the closest of all the $w_{k}(\omega )
$ for $k\leq n$ as the value of $u_n(\omega ) $ and $u_n$ is
measurable. Then, by the density of $\{w_n(\omega ) \} $ in
$\mathcal{K}(\omega ) $, for each $\omega$ we have that $\{ u_n(\omega ) \} $ is a
minimizing sequence for
\[
\lambda (\omega ) \equiv \inf \{ \| u(\omega) -z\| :z\in \mathcal{K}(\omega ) \}.
\]
Then, it follows that $u_n(\omega ) \to \operatorname{proj}_{
\mathcal{K}(\omega ) }u(\omega ) $ weakly in
$ \mathcal{V}$. Indeed, suppose the sequence fails to converge to
$\operatorname{proj}_{\mathcal{K}(\omega ) }u(\omega ) $. Since
the sequence is minimizing, it is bounded. Thus, there is a subsequence,
still denoted as $u_n(\omega ) $ that converges to
$q(\omega ) \neq \operatorname{proj}_{\mathcal{K}(\omega )}u(\omega ) $.
Then,
\[
\lambda (\omega ) =\lim_{n\to \infty }\| u(
\omega ) -u_n(\omega ) \| \geq \|
u(\omega ) -q(\omega ) \|,
\]
because a convex and lower semicontinuous function is weakly lower
semicontinuous.This implies that $q(\omega ) =\operatorname{proj}
_{\mathcal{K}(\omega ) }(u(\omega ) ) $ because the projection map
is well defined thanks to strict convexity of the norm used, which is a
contradiction. Hence, $\operatorname{proj}_{\mathcal{K}(\omega ) }u(
\omega ) =\lim_{n\to \infty }u_n(\omega ) $ and
so is a measurable function. It follows that $\omega \to P(
u(\omega ) ,\omega ) $ is measurable into $\mathcal{V}$.
\end{proof}
The following resutl is now immediate, which we state as a theorem
because of its importance.
\begin{theorem} \label{thm10}
Suppose that: $A(\cdot ,\omega ) $ is monotone,
hemicontinuous, bounded and coercive as a map from $\mathcal{V}$ to
$\mathcal{V}'$; when $\omega \to u(\omega ) $ is measurable into
$\mathcal{V}$ then $\omega \to A(u(\omega ) ,\omega ) $ is measurable into
$\mathcal{V}'$; $\mathcal{K}(\omega ) $ is a closed and
convex subset of $\mathcal{V}$ containing 0; and $\omega \to \mathcal{
K}(\omega ) $ is a set-valued measurable multifunction. Moreover, suppose
that there is a regularizing sequence $\{ u_i\} $, for each $u\in
\mathcal{K}$, satisfying $u_i(0) =0,u_i'\in \mathcal{
V}',u_i\in \mathcal{K}$,
\[
\limsup_{i\to \infty }\langle u_i',u_i-u\rangle \leq 0.
\]
Then, for each $\omega $, there exists a solution $u(\omega)$ to
the variational inequality
\[
\langle v',u(\cdot ,\omega ) -v\rangle
+\langle A(u(\cdot ,\omega ) ,\omega ) ,u(
\cdot ,\omega ) -v\rangle \leq \langle f(\cdot
,\omega ) ,u(\cdot ,\omega ) -v\rangle
\]
valid for all $v\in \mathcal{K}(\omega ) $ such that $v'\in \mathcal{V}',Bv(0) =0$,
and $(t,\omega ) \to u(t,\omega ) $, is
$\mathcal{B}([0,T] ) \times \mathcal{F}$-measurable.
\end{theorem}
\begin{proof}
The proof is identical to the proof of Theorem \ref{thm9}. We obtain a
measurable solution to \eqref{1jane1s} in which $P$ is replaced with
$P(\cdot ,\omega )$. Then, we follow exactly the same steps
and finally use Theorem \ref{zthm1x} to obtain the measurability of a
solution to the variational inequality.
\end{proof}
The above result, in addition top its importance, is quite interesting.
Indeed, it is not obvious that if $u(\omega ) \in \mathcal{K}(
\omega ) $, for each $\omega $, that $(t,\omega )
\to u(t,\omega ) $ has any kind of product
measurability. This would not be obvious even if $\mathcal{K}$ were
independent of $\omega $. However, the theorem says that there
is a measurable solution to the variational inequality, as long as
$\omega \to \mathcal{K}(\omega ) $ is a set-valued
measurable multifunction.
\subsection*{Acknowledgements}
The authors would like to thank the anonymous referee
for the warm report.
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\end{document}