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\documentclass{rae}
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\received{June 19, 1997}
%\coverauthor{H\'ector H. Cuenya and Felipe Z\'o}
\title{MONOTONE NORMS ON $C(\Omega)$ AND MULTIPLICATIVE FACTORS}
%\covertitle{Monotone Norms on $C(\Omega)$ and Multiplicative Factors}
\markboth{H\'ector H. Cuenya and Felipe Z\'o}{Monotone Norms on
$C(\Omega)$}
\MathReviews{Primary 46H05,46J10}
\keywords{monotone norms, submultiplicative norms, multiplicative
factors, algebra of continuous functions.}
\firstpagenumber{215}
\author{H\'ector H. Cuenya\thanks{This work, was supported by CONICET,
CONICOR, Univ. Nac. de R\'{\i}o Cuarto and Univ. Nac. de San Luis.},
Departamento de Matem\'atica, Fac. de Cs. Exactas F. Q. y Naturales,
Universidad Nacional de R\'{\i}o Cuarto, Campus Universitario, 5800
R\'{\i}o Cuarto, Argentina. e-mail: {\tt hcuenya@unrccc.edu.ar}\and
Felipe Z\'o, Instituto de Matem\'atica Aplicada San Luis, Universidad
Nacional de San Luis, Conicet, 5700 San Luis, Argentina. e-mail: {\tt
fzo@unsl.edu.ar}}
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma} {\theoremstyle{definition}
\newtheorem*{example}{Example}
\newtheorem*{remark}{Remark}}
\DeclareMathOperator{\supp}{supp}
\begin{document}
\maketitle
\begin{abstract} Let $C(\Omega)$ be the algebra of continuous
complex-valued functions on a topological space $\Omega$ and let $\rho$
be a function norm on $C(\Omega).$ We give necessary and sufficient
conditions on the set
$A_{\rho}=\{f\in C(\Omega):\rho(f)<\infty\}$ to be an algebra. Also, we
prove that every complete function norm is quasi-submultiplicative
provided $A_{\rho}$ is an algebra and we give a characterization of
the best multiplicative factor of $\rho.$ Finally we characterize
the infinity norm and we prove that every quasi-submultiplicative
function norm on $C(\Omega)$ is equivalent to the infinity norm.
\end{abstract}
\section{Introduction}
Let $\Omega$ be a topological space and let $C(\Omega)$ be the algebra
of continuous complex-valued functions. In a similar way as it was
introduced in [3] we are going to consider a
{\it function norm} $\rho$ on $C(\Omega)$, i.e., a function
$\rho:C(\Omega)\rightarrow [0,\infty]$ which satisfies the usual
properties of a norm, including the monotonicity condition
$$f, g \in C(\Omega),|f|\le |g|\Rightarrow \rho(f)\le\rho(g).$$ It
follows immediately from the definition that $\rho(|f|)=\rho(f)$ for all
$f\in C(\Omega).$
Throughout this paper $A_{\rho}$ will denote the following subspace of
$C(\Omega),$
$$ A_{\rho}=\{f\in C(\Omega):\rho(f)<\infty \}, $$ and ${\cal B}(\Omega)$
will denote the space of bounded complex-valued functions on $\Omega.$
We are interested in giving necessary and sufficient conditions on
$C(\Omega)$ for existence of a function norm such that
$A_{\rho}$ is an algebra. This problem was studied for other spaces
in [1], and later in [2] and it is related to the existence of
{\it submultiplicative norms}, see [4], [5] and [6].
A norm $\rho:C(\Omega)\rightarrow [0,\infty]$ will be called
$\sigma$-{\it subadditive} if for all sequences of functions $f_{n}\in
C(\Omega), f_{n}\ge 0$ and $\sum_{1}^{\infty}f_{n}\in C(\Omega)$ it
follows that
$$\rho\Bigl(\sum_{1}^{\infty}f_{n}\Bigr)\le \sum_{1}^{\infty}\rho(f_{n}).$$ We
will say that a norm $ \rho$ on $C(\Omega)$ is {\it complete} if
$(A_{\rho},\rho)$ is complete. It is not difficult to
see that every complete function norm on $C(\Omega)$ is
$\sigma$-subadditive. See for example [3]. We will say that a norm
$\rho$ is {\it quasi}-{\it submultiplicative} if there exists a constant $K>0$
such that
\begin{equation}\label{1}
\rho(fg) \le K\rho(f)\rho(g),%\tag 1
\end{equation}
for all $f, g \in C(\Omega).$
In this case we will say that $K$ is a
{\it multiplicative factor} of $\rho.$ The infimum of all
multiplicative factors of $\rho$ it is called the {\it best
multiplicative factor of} $\rho.$ Obviously given a
quasi-submultiplicative
function norm $\rho$, with $A_\rho\ne \{0\},$ its best multiplicative
factor $M$ is again a multiplicative factor of $\rho,$
in particular $M>0.$
One of our main results states that if $\rho$ is a $\sigma$-subadditive
function norm on $C(\Omega)$ and $A_{\rho}$ is an algebra,
$A_{\rho}$ can not contains an unbounded function. The following
example shows that there exist $\sigma$-subadditive function norms on
$C(\Omega)$ where the subspace $A_{\rho}$ admits unbounded functions.
Let $\Omega$ be the interval $(0,1)$ and
$$\rho(f)=\Bigl(\int_{\Omega}f^{2}\,dx\Bigr)^{1/2}=\|f\|_{2},$$ where $ dx $ stands
for Lebesgue measure. Clearly, $\rho$ is a
$\sigma$-subadditive function norm on $C((0,1))$. However there exist
unbounded square integrable continuous functions on $(0,1).$
We will show that a function norm $\rho$ is quasi-submultiplicative
provided it is a complete norm function and $A_\rho$ is an algebra. For
the function norm $\rho$ we are interested in obtaining an alternative,
expression for the best multiplicative factor of
$\rho$ which is easier to handle. Along the way we will give a characterization of the best
multiplicative factor analogous to the one in [2] in the case
of function norms defined on measurable spaces. More explicitly, we
prove for any non trivial ($A_\rho\ne \{0\}$) quasi-submultiplicative
norm function $\rho$, that its best multiplicative factor is given by
\begin{equation}
M_{\rho}=\text{sup}\{\|f\|_{\infty}:f\in C(\Omega),\rho(f)\le
1\},%\tag 2$$
\end{equation}
where $\|.\|_{\infty}$ denote the infinity norm. Note
that $M_\rho$ is a well defined number in $[0,\infty ]$ for any
function norm $\rho$ and because $M_\rho$ is a finite number it
will characterize quasi-submultiplicative norms.
Finally, we will give a simple characterization of the infinity norm and
we will prove that for every finite complete function norm $\rho$ on
$C(\Omega)$ the multiples $\lambda\rho$ are submultiplicative norms
for
$\lambda\ge M_\rho.$ In general it is easier to decide that a norm is
monotone and complete. Then the previous result gives us a method to
obtain submultiplicative norms.
\section {Results and Proofs}
\begin{theorem}\label{Theorem 1} Let $\rho$ be a function norm on
$C(\Omega).$
\begin{enumerate}
\item[(a)]If $\rho$ is $\sigma$-subadditive, then $ A_{\rho}$ is
an algebra if and only if $A_{\rho} \subset {\cal B}(\Omega),$
\item[(b)] If $A_{\rho}=C(\Omega)$, then $C(\Omega)\subset {\cal
B}(\Omega).$
\item[(c)]If $\rho $ is quasi-submultiplicative, then $A_{\rho}\subset
{\cal B}(\Omega).$
\end{enumerate}
\end{theorem}
\noindent{{\sc Proof.} }
(a) We assume that $A_{\rho}$ is
an algebra. Suppose that there exists an unbounded function $f\in
A_{\rho}.$ Since
$\rho(|f|)=\rho(f)$ we assume that $f\ge 0.$ The next argument is
similar to the one used in Theorem 1 of [4]. Thus we get a sequence of
elements $t_n\in \Omega$ such that
$f(t_{n+1})>f(t_{n})+3$ for each $n\in \mathbb N$, $f(t_{1})>2$ and
$\frac{f(t_n)}{n^2}$ tends to infinity. Let $(I_{n})$ be the sequence of
pairwise disjoint closed intervals $I_{n}=[f(t_{n})-1,f(t_{n})+1]$
and for each $n\in \mathbb N$ we choose a continuous function
$g_{n}:\mathbb R \rightarrow \mathbb R$ with $\supp(g_{n})\subset
I_{n}g_{n}\ge 0$ and $\|g_{n}\|_{\infty}= 1$.
Let $h_{n}(t)= g_{n}(f(t)).$ Then $h_{n}\in C(\Omega),\;0\le
h_{n}(t)\le
f(t)$ for all $t\in \Omega$, so $h_{n}\in A_{\rho}$ and $(fh_{n})(t)\ge
(f(t_{n})-1)h_{n}(t)$ for all $t\in \Omega.$ Thus
$0<\rho(h_{n})<\infty.$
Let $g=\sum_{1}^{\infty}\frac{h_n}{n^2\rho(h_n)}.$ Clearly $ g \in
C(\Omega).$ Since $\rho$ is $\sigma$-subadditive, we have that $g\in
A_{\rho}$. Then $fg\in A_{\rho}.$ On the other hand, by the monotonicity we
get
$$\rho(fg)\ge \frac{\rho(fh_n)}{n^2\rho(h_n)}\ge\frac{
f(t_n)-1}{n^2},$$ which is a contradiction.
Assume now $
A_{\rho}\subset {\cal B}(\Omega),\;f$
and $g$ belong to $A_{\rho}.$ Thus we have
$\rho(fg)\le \|f\|_{\infty}\rho(g)<\infty.$ Therefore $fg \in A_{\rho}.$
(b) Suppose that there exists an unbounded function $f\in
C(\Omega)$.
By defining a function $g$ as in (a), we obtain as before a
contradiction.
(c) Suppose that there exists an unbounded function $f\in
A_{\rho}$. We choose $g_n$ and $h_n$ as in (a). Then for some $K>0$ we
get
$f(t_{n}-1)\rho(h_{n})\le \rho(fh_{n})\le K\rho(f)\rho(h_{n})$
and we obtain that $f(t_{n}-1) \le K\rho(f)$, which is a
contradiction.\qed
\begin{remark} We note that part (c) of Theorem 1, is not a consequence of
[4] or [5], because they used that $A_{\rho}=C(\Omega).$ Also, the
example given in the introduction allows us to observe that monotonicity
does not implies quasi-submultiplicative. In fact, $\rho$ is not
quasi-submultiplicative, otherwise by (c) of Theorem 1, the set
$A_{\rho}$ should not admit an unbounded function, which is false. On
the other hand there exist non-monotone quasi-submultiplicative norms.
In order to see this, let $\rho$ be the
Minkowski's functional associated to a bounded balanced convex
absorbing set
$P$ in $\mathbb R^2.$ If in addition $P$ is a closed set in $\mathbb R^2$,
it is not difficult to see that $\rho$ is monotone if and only if $P$ is
symmetric,
i.e., if $(x_{1},x_{2}) \in P$, then
$(\epsilon_{1}x_{1},\epsilon_{2}x_{2})
\in P$ where $\epsilon_{i}=\pm 1, i=1,2.$ Then if we consider a set $P$
nonsymmetric, the norm $ \rho$ is not monotone. However it is
quasi-submultiplicative, since $\rho$ is equivalent to the
submultiplicative norm $\|.\|_{\infty}.$
\end{remark}
In the remainder
of this section we study existence and characterization of
multiplicative factors.
\begin{theorem}\label{Theorem 2}Let $\rho$ be a function norm on
$C(\Omega).$If any of the two conditions holds
\begin{enumerate}
\item[(a)]$\rho$ is a complete norm and $A_{\rho}$ is a subalgebra of
$C(\Omega)$, or
\item[(b)] $A_{\rho}=C(\Omega)$ where $\Omega$ is a $T_{1}$-space
with a dense set of isolated points without accumulation points.
\end{enumerate}
Then there exists a constant $K$ such that
$\|f\|_{\infty}\le K\rho(f),$ for all $f\in C(\Omega).$
\end{theorem}
\noindent{{\sc Proof.} }Suppose that (a) holds and the theorem is false. Then there
is a nonnegative function sequence $f_n$, with $\|f_n\|_{\infty}=a_{n},
\rho(f_n)=1$ and
$\sum_{1}^{\infty}\frac{1}{(a_n)^\frac{1}{2}}<\infty.$ We can assume
without lost of generality that $a_{1}\ge 4$ and $ a_{n+1}>a_{n}$ for
all
$n\in \mathbb N.$ Let $t_{n}$ a sequence in $\Omega$ be such that
$f_{n}(t_{n})>\frac{a_{n}}{2},$ and set
$J_n=[f_{n}(t_{n})-1,f_{n}(t_{n})+1].$ For each $n\in \mathbb N$, let
$g_{n}$ be a non negative function in $ C(\mathbb R)$ with
$\supp(g_{n})\subset J_{n},
\; g_n(f_{n}(t_{n}))=\frac{a_{n}}{4}=\|g_{n}\|_{\infty}.$ We define a
function $h_{n}\in C(\Omega)$ by $h_{n}(t)=g_{n}(f_{n}(t)).$ Given
$t\in \Omega, $ if $ f_{n}(t)\in J_{n}$, then
$$f_{n}(t)\ge f_{n}(t_{n})-1>\frac{a_{n}}{2}-1\ge\frac{a_{n}}{4}=
\|g_{n}\|_{\infty}\ge g_{n}(f_{n}(t))=h_{n}(t),$$ while
$f_{n}(t)\notin J_{n},$ implies that $h_{n}(t)=0.$
Therefore $h_{n}\le f_{n}$ and hence $\rho(h_{n})\le \rho(f_{n})=1.$
As $\sum_{n=1}^{\infty}
\frac{\rho(h_n)}{\sqrt{a_n}} \le \sum_1^{\infty}
\frac{1}{/\sqrt{a_n}} < \infty,$ the function
$s_k := \sum_{n=1}^k \frac{h_n}{\sqrt{a_n}}$
belongs to $A_{\rho}$ and $(s_k)$ is a Cauchy sequence. Since
$A_{\rho}$ is
a complete space, there exists $s\in A_{\rho}$ such that $\rho (s_k - s)
\to 0.$ Moreover, as in part (b) of the proof of theorem 4.8 in [3],
we have $s\ge s_k,$ for every $k.$ Then
$$\|s\|_{\infty} \ge \|s_k\|_{\infty} \ge
\frac{{\|h_k\|_{\infty}}}{\sqrt{a_k}}
\ge \frac{\sqrt{a_k}}{4} \to \infty.$$
Therefore $s$ is not bounded, contrary to (a) of
Theorem 1.
Now we assume (b) and suppose the theorem is false. Then there exists a
sequence of functions $f_{n}\in A_{\rho}, f_{n}\ge 0$ such that
$\rho(f_{n})=1$ and $\|f_{n}\|_{\infty}\to \infty $ for $n\to \infty.$
Thus we can get a sequence of isolated points $t_{n}$ such that
$f_n(t_{n})\to \infty$ and $t_{n}$ has no accumulation points. If
$\delta_{n}$ is the characteristic function of the unitary set
$\{t_{n}\},$ then the function $\delta_{n}$ is continuous. We let
$h=\sum_{1}^{\infty}f_n(t_{n})\delta_{n}.$ Since the set $\{t_{n}:n\in
\mathbb N\}$ has no accumulation points, it follows that $h\in
C(\Omega).$ On the other hand,
$\|h\|_{\infty}\ge f_n(t_{n})$ for all $n\in \mathbb N.$ Therefore
$h\notin B(\Omega),$ contrary to part (b) of Theorem 1.
\qed
In particular, the hypothesis (b) of Theorem 2 holds on $\Omega$
when
$\Omega=\mathbb Z,$ the set of integers with the discrete topology.
\begin{corollary}\label{Corollary 3}Let $\rho$ be a function norm on $C(\Omega).$
\begin{enumerate}
\item[(a)]If $\rho$ is complete, then $A_{\rho}$ is a subalgebra of
$C(\Omega)$ if and only if $\rho$ is quasi-submultiplicative.
\item[(b)]If $\Omega$ is a $T_{1}$ space with a dense set of
isolated points without accumulation points, such that
$A_{\rho}=C(\Omega)$, then
$\rho$ is quasi-submultiplicative.
\end{enumerate}
\end{corollary}
\noindent{{\sc Proof.} } We only prove (a) since (b) follows by analogous
arguments. Suppose that $A_{\rho}$ is a subalgebra of $C(\Omega).$ Let
$f,g \in A_{\rho}.$
By Theorem 2 there exists a constant $M$ such that
$\|h\|_{\infty}\le M\rho(h)$ for all $h\in C(\Omega).$ It follows that
$\rho(fg)\le \|f\|_{\infty}\rho(g) \le M \rho(f)\rho(g),$ i.e.
$\rho$ is
quasi-submultiplicative. The remaining of the statement is obvious.
\qed
The condition that $\rho$ be complete cannot be substituted by the
weaker condition of $\sigma$-subadditive, though
$A_{\rho}=C(\Omega),$ as the next example shows.
\begin{example}\label{Example} Let $\Omega=[0,1]$ and define a function
norm $\rho$ on
$C(\Omega)$ by $\rho(f)=\int_{\Omega}|f|d\mu=\|f\|_{1}.$ Clearly,
$A_{\rho}= C(\Omega)$ and $\rho$ is $\sigma$-subadditive. We can
construct a sequence $(f_n)\in C(\Omega)$ such that $\frac
{\|f_n\|_{2}}{\|f_n\|_{1}}$ is arbitrarily large. Thus there is no
constant $K$ such that $\rho(f^2)\le K(\rho(f))^{2}$ for all $f\in
C(\Omega).$ Consequently $\rho$ is not quasi-submultiplicative.
\end{example}
Next we will give a characterization
of the best multiplicative factor. If $f\in C(\Omega) $and
$K$ is a nonnegative real number, we consider the following subset
of $\Omega,\;A(f,K)=\{t\in
\Omega:f(t)>K\rho(f)\}.$
\begin{lemma}\label{Lemma 4} Let $\rho$ be a quasi-submultiplicative function norm
on $C(\Omega)$ and let $f\in C(\Omega).$ If $K$ is a
multiplicative factor of $\rho$ and $A(f,K)$ is nonempty, then
there exists a function
$b\in A_{\rho}, b\ne 0$ such that $\rho(bf)=K\rho(b)\rho(f).$
\end{lemma}
\noindent{{\sc Proof.} }We may assume without lost of generality that $f\ge 0.$
Since $A(f,K)\ne \emptyset,$ the function $f\in A_{\rho}$ is
by Theorem 1 a bounded function. Now set
$$r=\inf \{f(t):t\in A(f,K)\}.$$ We have two cases, $r>K\rho(f)$
or $r=K\rho(f).$\newline In the first case, we take a nonnegative
function $g\in C(\mathbb R)$ such that $g(x)=0$ for $x\le
K\rho(f)$ and $\|g\|_{\infty}=g(r)=r.$
Now, we define $b(t)=g(f(t))$ for $ t\in \Omega.$
Clearly $b\in C(\Omega)$ and $b\ne 0.$ If $t\in A(f,K), $ we
have $K\rho (f)
\le (f)(t)$, otherwise we have $b(t)=0.$ Thus
$(bf)(t)\ge K\rho (f)b(t)$ for all $t\in \Omega.$\newline In the
second case there is $t_{0}\in \Omega$ such that $ K\rho(f)<
f(t_{0})\le \|f\|_{\infty}.$ Here we choose a nonnegative function
$g\in C(\mathbb R)$ with $\supp(g) \subset [r,\|f\|_{\infty}],\;
g(f(t_{0}))\ne 0$ and $\|g\|_{\infty}=r.$ We define a function
$b$ in $C(\Omega)$ by $b(t)=g(f(t)).$ Then if $t\in
A(f,K),$ we have $K\rho (f)0.$ Now
$\rho(fg)\le \|f\|_{\infty}\rho(g)$ for all $f,g\in A_{\rho}$ which
implies that $M\le M_{\rho}.$ We are going to show that $M\ge M_\rho.$
Let $f\ne 0$ be a function in $A_\rho$ and $\epsilon>0.$ Then the set
$A(f,M +\epsilon)$ is empty. In fact if this where not so, by Lemma 4
there exists $ b\in A_{\rho}, b\ne 0$ such that $\rho(bf)=
(M+\epsilon)\rho(b)\rho(f),$ which is contradiction. Hence, we
must have $A(f,M+\epsilon)=\emptyset.$ Therefore
$\|f\|_{\infty}\le(M+\epsilon)\rho(f).$ Thus
$M_{\rho}\le M+\epsilon,$ for every $\epsilon >0. $
\qed
\begin{corollary}\label{Corollary 6} Let $\rho$ be a function norm on
$C(\Omega)$
with $A_{\rho}\ne \{0\}$ and which satisfies the conditions (a) or
(b) of Theorem 2. Then $\rho$ is quasi-submultiplicative and the best
multiplicative factor is given by (2).
\end{corollary}
\noindent{{\sc Proof.} }It follows immediately from Corollary 3 and Theorem 5.
\qed
\begin{corollary}\label{Corollary 7} Let $\rho$ be a norm on $C(\Omega)$ and
$A_{\rho}\ne \{0\}.$ Then $\rho$ is quasi-submultiplicative
(submultiplicative) if and only if
$M_\rho <\infty,\; (M_\rho \le 1).$ Moreover if $M_\rho<\infty $ and
$\lambda >0$, the function norm $\lambda\rho$ is submultiplicative if and
only if $\lambda\ge M_\rho.$
\end{corollary}
\noindent{{\sc Proof.} }If $M_\rho<\infty,\;(M_\rho\le 1)$ the monotonicity of $\rho$
implies that $\rho$ is quasi-submultiplicative
(submultiplicative). The converse statement follows by Theorem 5.
Observe that $\lambda >0$ and $A_\rho\ne \{0\}.$ Then
$M_{\lambda\rho}=\frac {1}{\lambda}M_\rho.$ Thus the proof is completed.
\qed
\begin{theorem}\label{Theorem 8}
\begin{enumerate}
\item[(a)]Let $\rho$ be a
function norm on $C(\Omega).$ If $1\in A_{\rho}$ and
$M_{\rho}$ satisfies
$\rho(1)M_{\rho} =1$, then $\|f\|_{\infty}=M_{\rho}\rho(f),$ for
all $f\in C(\Omega).$
\item[(b)] The infinity norm is the unique submultiplicative function
norm on $C(\Omega)$ such that $\rho(1)=1.$
\item[(c)] Every quasi-submultiplicative function norm $\rho$ on
$C(\Omega)$ such that $1\in A_\rho,$ is equivalent to infinity norm.
\end{enumerate}
\end{theorem}
\noindent{{\sc Proof.} }Since $\|f\|_{\infty}\le M_{\rho}\rho(f)
\le M_{\rho}\rho(1)\|f\|_{\infty},$ we have (a).\newline Now, by Theorem
5 the best multiplicative factor for a quasi-submultiplicative
function norm $\rho$ is given by $M_{\rho}.$ As $\rho$ is
submultiplicative $M_{\rho}\le 1.$ On the other hand, since
$\rho(1)\le M_{\rho}(\rho(1))^{2}$ and $\rho(1)=1$, we have
$M_{\rho}\ge 1.$ Thus, (b) follows from (a).
Finally (c) is a direct consequence of Theorem 5 and of the monotonicity of
$\rho.$
\qed
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