# Barreled extended locally convex spaces and Uniform boundedness principle

Akshay Kumar and Varun Jindal  Akshay Kumar: Department of Mathematics,
Malaviya National Institute of Technology Jaipur, Jaipur-302017, Rajasthan,
India<EMAIL_ADDRESS>Varun Jindal: Department of Mathematics, Malaviya
National Institute of Technology Jaipur, Jaipur-302017, Rajasthan, India
<EMAIL_ADDRESS>
###### Abstract.

For an extended locally convex space $(X,\tau)$, in [8], the authors studied
the finest locally convex topology (flc topology) $\tau_{F}$ on $X$ coarser
than $\tau$. One can often prove facts about $(X,\tau)$ by applying classical
locally convex space theory on $(X,\tau_{F})$. This paper employs the flc
topology to analyze barreled extended locally convex spaces and establish the
uniform boundedness principle in the extended setting. One of the key results
of this paper is the relationship between the barreledness of an extended
locally convex space $(X,\tau)$ and the barreledness of the associated finest
locally convex space $(X,\tau_{F})$. This is achieved by examining the lower
semi-continuous seminorms on these spaces.

###### Key words and phrases:

Extended locally convex space, extended normed space, finest locally convex
topology, barreled spaces, uniform boundedness principle, equicontinuous
family

###### 2020 Mathematics Subject Classification:

Primary 46A03, 46A08; Secondary 46A17, 46A20, 54C40

##  1\. Introduction

Beer [1] initiated the study of extended normed linear spaces. Beer and
Vanderwerff further developed these spaces in [4, 5]. Motivated by the work of
Beer and Vanderwerff, Salas and Tapia-García introduced the concept of an
extended locally convex space (elcs) in [11]. The structure of these new
spaces differs significantly from that of classical locally convex spaces. In
[8], authors studied the finest locally convex topology (flc topology, for
short) $\tau_{F}$ for an extended locally convex space $(X,\tau)$ that is
still coarser than $\tau$. It is shown that both $(X,\tau)$ and $(X,\tau_{F})$
have the same collection of continuous seminorms. In particular, both have the
same dual. We call the space $(X,\tau_{F})$ the finest locally convex space
associated with the elcs $(X,\tau)$.

In this paper, we define and investigate barreled extended locally convex
spaces. The associated finest locally convex space plays a significant role in
this investigation. To find the relationship between the barreledness of an
elcs $(X,\tau)$ and the barreledness of the corresponding finest space
$(X,\tau_{F})$, we first examine lower semi-continuous seminorms on these
spaces. Finally, by using the relationship between the barreledness of
$(X,\tau)$ and $(X,\tau_{F})$, we prove the uniform boundedness principle for
the elcs $(X,\tau)$.

The entire work of this paper is presented in five sections. Section 2
provides preliminary definitions and features of extended locally convex
spaces. In the third section, we study barreled extended locally convex
spaces. We begin our investigation by characterizing the barreled subspaces of
an elcs. Then as a main result, we show that an elcs $(X,\tau)$ is barreled if
and only if the finest space $(X,\tau_{F})$ is barreled. We use this result to
prove some permanence properties of a barreled elcs.

In the fourth section, we look at the barreledness of the space $C(X)$ endowed
with the topology $\tau_{\mathcal{B}}^{s}$ (or $\tau_{\mathcal{B}}$) of strong
uniform convergence (or uniform convergence) on the elements of a bornology
$\mathcal{B}$. We show that the bornology $\mathcal{K}$ of all relatively
compact sets is coarser than $\mathcal{B}$ if $(C(X),\tau_{\mathcal{B}}^{s})$
is barreled. In the final section, we examine equicontinuous families of
continuous linear operators between two extended locally convex spaces, and
prove the uniform boundedness principle in the context of an elcs.

##  2\. Preliminaries

Throughout the paper, we assume that the underlying field for a vector space
is $\mathbb{K}$ which is either $\mathbb{R}$ or $\mathbb{C}$. We also adopt
the following conventions for the $\infty$: $\infty.0=0.\infty=0$;
$\infty+\alpha=\alpha+\infty=\infty$ for every $\alpha\in\mathbb{R}$;
$\infty.\alpha=\alpha.\infty=\infty$ for $\alpha>0$;
$\inf\\{\emptyset\\}=\infty$.

An extended seminorm $\rho:X\to[0,\infty]$ on a vector space $X$ is a function
which satisfies the following properties.

  * (1)

$\rho(\alpha x)=|\alpha|\rho(x)$ for each scalar $\alpha$ and $x\in X$;

  * (2)

$\rho(x+y)\leq\rho(x)+\rho(y)$ for $x,y\in X$.

An extended seminorm $\parallel\cdot\parallel:X\to[0,\infty]$ is said to be an
extended norm if $\parallel x\parallel=0$ implies $x=0$. A vector space $X$
together with an extended norm $\parallel\cdot\parallel$ is called an extended
normed linear space (enls, for short), and it is denoted by
$(X,\parallel\cdot\parallel)$. In this situation,
$(X,\parallel\cdot\parallel)$ is said to be an extended Banach space if it is
complete with respect to the metric $d(x,y)=\min\\{\parallel
x-y\parallel,1\\}$ for $x,y\in X$. It is easy to prove that an enls
$(X,\parallel\cdot\parallel)$ is an extended Banach space if and only if the
finite subspace $(X_{fin},\parallel\cdot\parallel)$ is a Banach space, where
$X_{fin}=\\{x\in X:\parallel x\parallel<\infty\\}$. For more details about
extended normed linear spaces, we refer to [1, 4, 5].

A vector space $X$ endowed with a Hausdorff topology $\tau$ is said to be an
extended locally convex space (elcs, for short) if $\tau$ is induced by a
collection $\mathcal{P}=\\{\rho_{i}:i\in\mathcal{I}\\}$ of extended seminorms
on $X$, that is, $\tau$ is the smallest topology on $X$ under which each
$\rho_{i}$ is continuous. The finite subspace $X_{fin}$ of an elcs $(X,\tau)$
is defined by

$X_{fin}=\\{x\in X:\rho(x)<\infty\text{ for every continuous extended seminorm
}\rho\text{ on }X\\},$

and we define $X_{fin}^{\rho}=\\{x\in X:\rho(x)<\infty\\}$ for any extended
seminorm $\rho$ on $X$.

Suppose $(X,\tau)$ is an elcs and $\tau$ is induced by a family $\mathcal{P}$
of extended seminorms on $X$. Then the following facts are either easy to
verify or given in [11].

  * (1)

There exists a neighborhood base $\mathcal{B}$ of $0$ in $(X,\tau)$ such that
each element of $\mathcal{B}$ is absolutely convex (balanced and convex);

  * (2)

$X_{fin}$ with the subspace topology is a locally convex space;

  * (3)

$X_{fin}=\bigcap_{\rho\in\mathcal{P}}X_{fin}^{\rho};$

  * (4)

if $\rho$ is any continuous extended seminorm on $(X,\tau)$, then
$X_{fin}^{\rho}$ is a clopen subspace of $(X,\tau)$;

  * (5)

a subspace $Y$ is open in $(X,\tau)$ if and only if there exists a continuous
extended seminorm $\rho$ on $(X,\tau)$ such that $X_{fin}^{\rho}\subseteq Y$.

  * (6)

$X_{fin}$ is an open subspace in $(X,\tau)$ if and only if there exists a
continuous extended seminorm $\rho$ on $(X,\tau)$ such that
$X_{fin}=X_{fin}^{\rho}$. In this case, we say $(X,\tau)$ is a fundamental
elcs.

######  Definition 2.1.

([11]) Suppose $U$ is any absolutely convex set in an elcs $(X,\tau)$. Then
the Minkowski functional $\rho_{U}:X\rightarrow[0,\infty]$ for $U$ is defined
as

$\rho_{U}(x)=\inf\\{\lambda>0:x\in\lambda U\\}.$

The following facts are immediate from the above definition.

  1. (1)

The Minkowski functional $\rho_{U}$ for the set $U$ is an extended seminorm on
$X$. In addition, if $U$ is absorbing, then $\rho_{U}$ is a seminorm on $X$.

  2. (2)

$\\{x\in X:\rho_{U}(x)<1\\}\subseteq U\subseteq\\{x\in X:\rho_{U}(x)\leq
1\\}$.

  3. (3)

The Minkowski functional $\rho_{U}$ is continuous on $X$ if and only if $U$ is
a neighborhood of $0$.

For an absolutely convex subset $U$ of an elcs $X$, we define
$X_{fin}^{U}=\\{x\in X:\rho_{U}(x)<\infty\\}$. It is shown in [11] that if
$\mathcal{B}$ is a neighborhood base of $0$ in an elcs $(X,\tau)$ consisting
of absolutely convex sets, then $\tau$ is induced by the collection
$\\{\rho_{U}:U\in\mathcal{B}\\}$.

If $A$ is any nonempty set in a topological space $(X,\tau)$, then we denote
the closure and interior of $A$ in $(X,\tau)$ by $\text{Cl}_{\tau}(A)$ and
$\text{int}_{\tau}(A)$, respectively. For other terms and definitions, we
refer to [10, 12, 13].

##  3\. Barreled extended locally convex spaces

We define and explore barreled extended locally convex spaces in this section.
On an elcs, we also look into barreled subspaces and lower semicontinuous
seminorms. We demonstrate, among other things, that an elcs $(X,\tau)$ is
barreled if and only if $(X,\tau_{F})$ is barreled if and only if every lower
semi-continuous seminorm on $(X,\tau)$ is continuous. At the end of this
section, we examine some permanence properties of a barreled elcs.

######  Definitions 3.1.

A barrel $B$ in an elcs $(X,\tau)$ is a closed, absolutely convex and
absorbing set in $(X,\tau)$. We say $(X,\tau)$ is barreled if every barrel in
it is a neighborhood of $0$.

We first give an example of a barreled elcs.

######  Example 3.2.

Let $X$ be a vector space together with the discrete extended norm

$\parallel x\parallel_{0,\infty}=\begin{cases}\text{0;}&\quad\text{if
$x=0$}\\\ \text{$\infty$;}&\quad\text{if $x\neq 0.$ }\\\ \end{cases}$

Then every subset of $X$ is open in the enls
$(X,\parallel\cdot\parallel_{0,\infty})$. Therefore every barrel in
$(X,\parallel\cdot\parallel_{0,\infty})$ is a neighborhood of $0$. Hence
$(X,\parallel\cdot\parallel_{0,\infty})$ is barreled.

######  Theorem 3.3.

Suppose $(X,\tau)$ is an elcs which is a Baire space. Then $(X,\tau)$ is
barreled.

###### Proof.

Let $B$ be a barrel in $(X,\tau)$. Then
$\displaystyle{X=\cup_{n\in\mathbb{N}}}nB$ as $B$ is absorbing in $X$. Since
$X$ is a Baire space, $\text{ int}(n_{0}B)\neq\emptyset$ for some
$n_{0}\in\mathbb{N}$. Let $x\in\text{ int }(n_{0}B)$. Then there exists a
neighborhood $U$ of $x$ such that $U\subseteq n_{0}B$. Note that
$\frac{(-x)}{2n_{0}}+\frac{U}{2n_{0}}$ is a neighborhood of $0$ and

$\frac{(-x)}{2n_{0}}+\frac{U}{2n_{0}}\subseteq B.$

Thus $B$ is a neighborhood of $0$. Which completes the proof. ∎

######  Corollary 3.4.

If $(X,\parallel\cdot\parallel)$ is an extended Banach space, then
$(X,\parallel\cdot\parallel)$ is barreled.

######  Corollary 3.5.

Let $(X,d)$ be a metric space. Then the collection $C(X)$ of all real-valued
continuous functions on $X$ endowed with the extended norm $\parallel
f\parallel_{\infty}=\sup_{x\in X}|f(x)|$ for $f\in C(X)$ is a barreled enls.

###### Proof.

It follows from the fact that $(C(X),\parallel\cdot\parallel_{\infty})$ is an
extended Banach space.∎

We say a subspace $Y$ of an elcs $(X,\tau)$ is barreled if $(Y,\tau|_{Y})$ is
barreled. To study the barreled subspaces of an elcs, we need the following
lemma.

######  Lemma 3.6.

Let $(X,\tau)$ be an elcs. Then the following properties hold for $A\subseteq
X$.

  * (1)

If $A$ is convex, then $\text{Cl}_{\tau}(A)$ is convex.

  * (2)

If $A$ is balanced, then $\text{Cl}_{\tau}(A)$ is balanced.

###### Proof.

It is similar to the proofs of Proposition 2.13, p. 42, and Proposition
2.5(a), p. 37 in [10].∎

######  Proposition 3.7.

Let $M$ be a subspace of an elcs $(X,\tau)$. Suppose $\rho$ is a continuous
extended seminorm on $(X,\tau)$. Then for every barrel $B$ in $M\cap
X_{fin}^{\rho}$, there exists a barrel $G$ in $M$ such that $G\cap\left(M\cap
X_{fin}^{\rho}\right)=B$.

###### Proof.

Suppose $Z$ is a subspace of $M$ such that $M=\left(M\cap
X_{fin}^{\rho}\right)\oplus Z$. Let $B$ be a barrel in $M\cap X_{fin}^{\rho}$.
Let $G=\text{Cl}_{\tau|_{M}}(B+Z)$. By Lemma 3.6, $G$ is absolutely convex. It
is easy to see that if $x=x_{1}+x_{2}\in M$ for $x_{1}\in M\cap
X_{fin}^{\rho}$ and $x_{2}\in Z$, then there exists a $c>0$ such that
$c(x_{1}+x_{2})\in B+Z$. Therefore $G$ is absorbing in $M$. Consequently, $G$
is a barrel in $M$. Also, $B\subseteq G\cap\left(M\cap X_{fin}^{\rho}\right)$.
For the reverse inclusion, let $x_{0}\in G\cap\left(M\cap
X_{fin}^{\rho}\right)$. If $U$ is a neighborhood of $x_{0}$ in $M\cap
X_{fin}^{\rho}$, then $U$ is a neighborhood of $x_{0}$ in $M$ as $M\cap
X_{fin}^{\rho}$ is an open subspace of $M$. Therefore there is an $x\in M$
such that $x\in U\cap\left(B+Z\right)$. Consequently, $x\in U\cap B$. Since
$B$ is closed in $M\cap X_{fin}^{\rho}$, $x_{0}\in B$. Hence $G\cap\left(M\cap
X_{fin}^{\rho}\right)=B$. ∎

######  Corollary 3.8.

Suppose $\rho$ is a continuous extended seminorm on an elcs $(X,\tau)$. Then
for every barrel $B$ in $X_{fin}^{\rho}$, there exists a barrel $G$ in $X$
such that $G\cap\left(X_{fin}^{\rho}\right)=B$.

######  Theorem 3.9.

Suppose $M$ is a subspace of an elcs $(X,\tau)$. Then the following statements
are equivalent.

  * (1)

$M$ is a barreled subspace of $X$;

  * (2)

$M\cap X_{fin}^{\rho}$ is a barreled subspace of $X$ for every continuous
extended seminorm $\rho$ on $X$;

  * (3)

$M\cap X_{fin}^{\rho}$ is a barreled subspace of $X$ for some continuous
extended seminorm $\rho$ on $X$.

###### Proof.

$(1)\Rightarrow(2)$. Suppose $\rho$ is a continuous extended seminorm on
$(X,\tau)$ and $B$ is a barrel in $M\cap X_{fin}^{\rho}$. By Proposition 3.7,
there exists a barrel $G$ in $M$ such that $G\cap\left(M\cap
X_{fin}^{\rho}\right)=B$. Since $M$ is a barreled subspace of $X$, $G$ is a
neighborhood of $0$ in $M$. Therefore $B=G\cap\left(M\cap
X_{fin}^{\rho}\right)$ is a neighborhood of $0$ in $M\cap X_{fin}^{\rho}$.
Hence $M\cap X_{fin}^{\rho}$ is a barreled subspace of $X$.

$(2)\Rightarrow(3)$. It is immediate.

$(3)\Rightarrow(1)$. Let $G$ be a barrel in $M$. Then $G\cap\left(M\cap
X_{fin}^{\rho}\right)$ is closed in $M\cap X_{fin}^{\rho}$. Note that
$G\cap\left(M\cap X_{fin}^{\rho}\right)$ is absolutely convex and absorbing in
$M\cap X_{fin}^{\rho}$. Therefore $G\cap\left(M\cap X_{fin}^{\rho}\right)$ is
a barrel in $M\cap X_{fin}^{\rho}$. As $M\cap X_{fin}^{\rho}$ is barreled,
$G\cap\left(M\cap X_{fin}^{\rho}\right)$ is a neighborhood of $0$ in $M\cap
X_{fin}^{\rho}$. Consequently, $G$ is a neighborhood of $0$ in $M$. Which
completes the proof.∎

######  Corollary 3.10.

Suppose $(X,\tau)$ is an elcs. Then the following statements are equivalent.

  * (1)

$X$ is barreled;

  * (2)

$X_{fin}^{\rho}$ is a barreled subspace of $X$ for every continuous extended
seminorm $\rho$ on $X$;

  * (3)

$X_{fin}^{\rho}$ is a barreled subspace of $X$ for some continuous extended
seminorm $\rho$ on $X$.

######  Corollary 3.11.

Suppose $(X,\parallel\cdot\parallel)$ is an enls. Then $X$ is barreled if and
only if $(X_{fin},\parallel\cdot\parallel)$ is barreled.

In general, we may not replace $X_{fin}^{\rho}$ by $X_{fin}$ in Theorem 3.9.

######  Example 3.12.

Let $X=c_{00}$ be the collection of all real sequences which are eventually
zero. For $n\in\mathbb{N}$, consider the extended norm
$\rho_{n}:X\rightarrow[0,\infty]$ defined by

$\rho_{n}((x^{m}))=\begin{cases}\text{$\infty$,}&\quad\text{ if $x^{m}\neq 0$
for some $1\leq m\leq n$}\\\
\text{$\displaystyle{\sup_{m\in\mathbb{N}}}|x^{m}|$,}&\quad\text{if $x^{m}=0$
for every $1\leq m\leq n$.}\\\ \end{cases}$

Let $\tau$ be the topology induced by the collection
$\mathcal{P}=\\{\rho_{n}:n\in\mathbb{N}\\}$. Then $(X,\tau)$ is an elcs and
$X_{fin}=\\{0\\}$. Therefore $X_{fin}$ is a barreled subspace of $(X,\tau)$.
Now, suppose $B=\\{(x^{m})\in X:|x^{m}|\leq\frac{1}{m}\text{ for
}m\in\mathbb{N}\\}$. Then $B$ is absolutely convex. If $x=(x^{m})$ is any
element in $X$, then there exists an $m_{0}\in\mathbb{N}$ such that $x^{m}=0$
for $m\geq m_{0}$. Consequently, $c.x\in B$ for
$c=\frac{1}{m_{0}\max\\{|x^{m}|+1:m\in\mathbb{N}\\}}$. Thus $B$ is absorbing
in $X$. Let $(z_{n})$ be a sequence in $B$ such that $z_{n}=(z_{n}^{m})$ for
each $n\in\mathbb{N}$. If $z_{n}\to z=(z^{m})$ in $X$, then for every
$m\in\mathbb{N}$, $z_{n}^{m}\to z^{m}$ in $\mathbb{R}$. Therefore
$|z^{m}|\leq\frac{1}{m}$ for every $m\in\mathbb{N}$. Consequently, $z\in B$.
Note that $(X,\tau)$ is a metrizable elcs as $\tau$ is induced by a countable
family of elcs (see, Lemma 4.15 in [11]). Thus $B$ is closed in $X$. Hence $B$
is a barrel in $X$. If $B$ is a neighborhood of $0$ in $(X,\tau)$, then
$\\{x\in X:\rho_{n_{0}}(x)<\epsilon\\}\subseteq B$ for some
$n_{0}\in\mathbb{N}$ and $\epsilon>0$. Suppose $k\in\mathbb{N}$ such that
$\frac{1}{k}<\min\\{\frac{1}{n_{0}},~{}\frac{\epsilon}{2}\\}$. Take
$y=(y^{m})$ defined by

$y^{m}=\begin{cases}\text{$\frac{\epsilon}{2}$;}&\quad\text{ for $m=k$}\\\
\text{0;}&\quad\text{ otherwise. }\\\ \end{cases}$

Then $y\in\\{x\in X:\rho_{n_{0}}(x)<\epsilon\\}\setminus B$. Hence $(X,\tau)$
is not barreled.

Our next aim is to relate the barreledness of an elcs $(X,\tau)$ with the
barreledness of the finest space $(X,\tau_{F})$. To do that, first we study
lower semi-continuous seminorms on $(X,\tau)$.

######  Definitions 3.13.

([13]) A function $f:(X,\tau)\to\mathbb{R}$ on a topological space $(X,\tau)$
is said to be lower semi-continuous at $x_{0}\in X$ if for every $\epsilon>0$,
$f^{-1}((f(x_{0})-\epsilon,\infty))$ is a neighborhood of $x_{0}$ in
$(X,\tau)$. The function $f$ is said to be lower semi-continuous on $X$ if it
is lower semi-continuous at every point of $X$.

Note that a function $f:(X,\tau)\to\mathbb{R}$ on a topological space
$(X,\tau)$ is lower semi-continuous on $X$ if and only if
$f^{-1}((c,\infty))\in\tau$ for every $c\in\mathbb{R}$ if and only if
$f^{-1}((-\infty,c])$ is closed in $(X,\tau)$ for every $c\in\mathbb{R}$.

In [8], it is shown that if $\tau_{F}$ is the flc topology for an elcs
$(X,\tau)$, then both $(X,\tau)$ and $(X,\tau_{F})$ have the same collection
of continuous seminorms. Our next result shows that this is even true for
lower semi-continuous seminorms.

######  Proposition 3.14.

Let $P:X\rightarrow[0,\infty)$ be a seminorm on an elcs $(X,\tau)$. If
$\tau_{F}$ is the flc topology for $(X,\tau)$, then $P$ is lower semi-
continuous with respect to $\tau$ if and only if $P$ is lower semi-continuous
with respect to $\tau_{F}$.

###### Proof.

Suppose $P$ is lower semi-continuous with respect to $\tau$. Then for
$x_{0}\in X$ and $\epsilon>0$, there exists a continuous extended seminorm
$\rho$ and $c>0$ such that

$P\left(x_{0}+\\{x\in
X:\rho(x)<c\\}\right)\subseteq\left(P(x_{0})-\frac{\epsilon}{2},\infty\right).$

Let $M$ be a subspace of $X$ such that $X=X_{fin}^{\rho}\oplus M$. Consider
the seminorm $\mu:X=X_{fin}^{\rho}\oplus M\rightarrow[0,\infty)$ defined by

$\mu(x)=\rho(x_{f})+P(x_{m}),$

where $x=x_{f}+x_{m}$ for $x_{f}\in X_{fin}^{\rho}$ and $x_{m}\in M$. Since
$\mu(x)\leq\rho(x)$ for every $x\in X$, $\mu$ is a continuous seminorm on
$(X,\tau)$. By Theorem 3.5 in [8], $\mu$ is continuous with respect to
$\tau_{F}$. It is easy to prove that

$P\left(x_{0}+\left\\{x\in
X:\mu(x)<\min\left\\{c,\frac{\epsilon}{2}\right\\}\right\\}\right)\subseteq\left(P(x_{0})-\epsilon,\infty\right).$

Therefore $P$ is lower semi-continuous at $x_{0}$ with respect to $\tau_{F}$.
Consequently, $P$ is lower semi-continuous on $X$ with respect to $\tau_{F}$.∎

######  Theorem 3.15.

Let $(X,\tau)$ be an elcs. If $\tau_{F}$ is the flc topology for $(X,\tau)$,
then the following statements are equivalent.

  * (1)

$(X,\tau)$ is barreled.

  * (2)

$(X,\tau_{F})$ is barreled.

  * (3)

Every lower semi-continuous seminorm on $(X,\tau_{F})$ is continuous.

  * (4)

Every lower semi-continuous seminorm on $(X,\tau)$ is continuous.

###### Proof.

$(1)\Rightarrow(2).$ Suppose $B$ is a barrel in $(X,\tau_{F})$. Then $B$ is a
barrel in $(X,\tau)$ as $\tau_{F}\subseteq\tau$. Since $(X,\tau)$ is barreled,
$B$ is a neighborhood of $0$ in $(X,\tau)$. Then the Minkowski functional
$\rho_{B}$ is a continuous seminorm on $(X,\tau)$. By Theorem 3.5 in [8],
$\rho_{B}$ is continuous with respect to $\tau_{F}$. Hence $B$ is a
neighborhood of $0$ in $(X,\tau_{F})$ as $\rho_{B}^{-1}([0,1))\subseteq B$.

$(2)\Rightarrow(3)$. It follows from Theorem 11.4.3, p. 386 in [9].

$(3)\Rightarrow(4)$. It follows from Proposition 3.14.

$(4)\Rightarrow(1)$. Suppose $B$ is a barrel in $(X,\tau)$. Then the Minkowski
functional $\rho_{B}$ is a seminorm on $X$. For any $c>0$, let
$(x_{\lambda})_{\lambda\in\Lambda}$ be a net in
$\rho_{B}^{-1}\left((-\infty,c]\right)$ such that $x_{\lambda}\to x$ in
$(X,\tau)$. Then $\left(\frac{1}{c+\frac{1}{n}}\right)x_{\lambda}\in B$ for
every $n\in\mathbb{N}$ and $\lambda\in\Lambda$. Note that for every
$n\in\mathbb{N}$,
$\left(\frac{1}{c+\frac{1}{n}}\right)x_{\lambda}\to\left(\frac{1}{c+\frac{1}{n}}\right)x$.
Since $B$ is closed in $(X,\tau)$, $\left(\frac{1}{c+\frac{1}{n}}\right)x\in
B$ for every $n\in\mathbb{N}$. Which implies that $\rho_{B}(x)\leq
c+\frac{1}{n}$ for every $n\in\mathbb{N}$. Consequently, $\rho_{B}(x)\leq c$.
Therefore $\rho_{B}^{-1}\left((-\infty,c]\right)$ is closed in $(X,\tau)$.
Hence $\rho_{B}$ is lower semi-continuous on $(X,\tau)$. By our assumption,
$\rho_{B}$ is a continuous seminorm on $(X,\tau)$. Thus $B$ is a neighborhood
of $0$ in $(X,\tau)$. ∎

######  Corollary 3.16.

Let $(X,\tau)$ be a barreled elcs. If $\tau_{F}$ is the flc topology for
$(X,\tau)$, then $(X,\tau_{F})$ is a Mackey space.

###### Proof.

Suppose $(X,\tau)$ is barreled. By Theorem 3.15, $(X,\tau_{F})$ is a barreled
locally convex space. Hence $(X,\tau_{F})$ is a Mackey space as every barreled
locally convex space is a Mackey space (see, Corollary 4.9, p. 98 in [10]).∎

######  Corollary 3.17.

Let $(X,\parallel\cdot\parallel)$ be an extended Banach space. If $\tau_{F}$
is the flc topology for $(X,\parallel\cdot\parallel)$, then $(X,\tau_{F})$ is
a Mackey space.

###### Proof.

It follows from Corollary 3.4 and Corollary 3.16. ∎

######  Example 3.18.

Suppose $X=Y=c_{0}$ is the collection of all real sequences converging to $0$.
For $n\in\mathbb{N}$, consider the extended norm
$\rho_{n}:X\rightarrow[0,\infty]$ defined by

$\rho_{n}((x^{i}))=\begin{cases}\text{$\infty$,}&\quad\text{ if $x^{i}\neq 0$
for some $1\leq i\leq n$}\\\
\text{$\displaystyle{\sup_{i\in\mathbb{N}}}|x^{i}|$,}&\quad\text{if $x^{i}=0$
for every $1\leq i\leq n$.}\\\ \end{cases}$

Let $\tau$ be the topology induced by the collection
$\mathcal{P}=\\{\rho_{n}:n\in\mathbb{N}\\}$. Then $(X,\tau)$ is an elcs and
$X_{fin}=\\{0\\}$. By Example 5.12 in [8], the flc topology $\tau_{F}$ for
$(X,\tau)$ is induced by the norm
$\parallel(x^{i})\parallel_{\infty}=\sup_{i\in\mathbb{N}}|x^{i}|$. Note that
$(X,\tau_{F})$ is a Banach space. So by Theorem 3.15, $(X,\tau)$ is a barreled
elcs.

Now, we use Theorem 3.15 to prove several permanence properties related to
barreledness of an elcs.

######  Theorem 3.19.

Let $M$ be a dense subspace of an elcs $(X,\tau)$. If $M$ is a barreled
subspace of $(X,\tau)$, then $(X,\tau)$ is also barreled.

###### Proof.

Suppose $\tau_{F}$ is the flc topology for $(X,\tau)$. Then by Theorem 4.1 in
[8], the flc topology for $(M,\tau|_{M})$ is equal to $\tau_{F}|_{M}$.
Therefore by Theorem 3.15, $(M,\tau_{F}|_{M})$ is barreled. Since $M$ is dense
in $(X,\tau)$, $M$ is a dense subspace of $(X,\tau_{F})$ as
$\tau_{F}\subseteq\tau$. By Theorem 11.2.1, p. 408 in [9], $(X,\tau_{F})$ is
barreled. Hence by Theorem 3.15, $(X,\tau)$ is barreled.∎

######  Theorem 3.20.

Let $M$ be a subspace of a barreled elcs $(X,\tau)$ with countable co-
dimension. Then $M$ is barreled.

###### Proof.

By Theorem 3.15, $M$ is a subspace of the barreled locally convex space
$(X,\tau_{F})$ with countable co-dimension. Therefore by Theorem 11.2.9, p.
412 in [9], $M$ is a barreled subspace of $(X,\tau_{F})$. It is shown in
Theorem 4.3 of [8] that the flc topology for $(M,\tau|_{M})$ is equal to
$\tau_{F}|_{M}$. Hence by Theorem 3.15, $(M,\tau|_{M})$ is barreled. ∎

Our next result shows that if $\tau_{F}$ is the flc topology for an elcs
$(X,\tau)$, then both $(X,\tau)$ and $(X,\tau_{F})$ have the same closed
subspaces.

######  Theorem 3.21.

Let $(X,\tau)$ be an elcs with the flc topology $\tau_{F}$. Suppose $Y$ is a
subspace of $X$. Then $Y$ is closed in $(X,\tau)$ if and only if it is closed
in $(X,\tau_{F})$.

###### Proof.

If $Y$ is closed in $(X,\tau_{F})$, then $Y$ is closed in $(X,\tau)$ as
$\tau_{F}\subseteq\tau$. Conversely, suppose $Y$ is closed in $(X,\tau)$ and
$x_{0}\notin Y$. Let $Z=Y\oplus\text{ span }(x_{0})$. Consider the linear
functional $f$ on $Z$ defined by $f(y+\alpha x_{0})=\alpha$ for every $y\in Y$
and $\alpha\in\mathbb{K}$. If $f$ is not continuous on $(Z,\tau|_{Z})$, then
there exists a net $(z_{\lambda})_{\lambda\in\Lambda}$ in $Z$ such that
$z_{\lambda}\to z$ in $(Z,\tau|_{Z})$ and $f(z_{\lambda})\nrightarrow f(z)$.
Let for every $\lambda$, $z_{\lambda}=y_{\lambda}+\alpha_{\lambda}x_{0}$ and
$z=y+\alpha x_{0}$. Then $\alpha_{\lambda}\nrightarrow\alpha$. Therefore there
exists a $\delta>0$ such that for each $\lambda\in\Lambda$, there is a
$\lambda^{\prime}\geq\lambda$ satisfying
$|\alpha_{\lambda^{\prime}}-\alpha|>\delta$. Since $Y$ is closed in $(X,\tau)$
and $x_{0}\notin Y$, there exist a continuous extended seminorm $\rho$ on $X$
and $0<\epsilon<\delta$ such that $\rho(x_{0}-y)>\epsilon$ for every $y\in Y$.
Also, we can find a $\lambda_{0}\in\Lambda$ such that
$\rho(y_{\lambda_{0}}+\alpha_{\lambda_{0}}x_{0}-y-\alpha x_{0})<\epsilon^{2}$
and $|\alpha_{\lambda_{0}}-\alpha|>\delta$. Since
$\frac{y_{\lambda_{0}}-y}{\alpha-\alpha_{\lambda_{0}}}\in Y$, we have

$\epsilon<\rho\left(\frac{y_{\lambda_{0}}-y}{\alpha-\alpha_{\lambda_{0}}}-x_{0}\right)\leq\frac{\rho\left(y_{\lambda_{0}}-y-(\alpha-\alpha_{\lambda_{0}})x_{0}\right)}{|\alpha_{\lambda_{0}}-\alpha|}<\frac{(\epsilon)^{2}}{\delta}<\epsilon.$

We arrive at a contradiction. So $f$ is continuous on $(Z,\tau|_{Z})$. By
Corollary 4.2 in [8], there exists a $g\in X^{*}$ such that $g|_{Z}=f$. Since
both $(X,\tau)$ and $(X,\tau_{F})$ have the same dual, $g^{-1}(0)$ is closed
in $(X,\tau_{F})$. Therefore $Y$ is closed in $(X,\tau_{F})$ as $Y\subseteq
g^{-1}(0)$ and $g(x_{0})=1$.∎

Recall that if $M$ is a closed subspace of an elcs $(X,\tau)$, then the
quotient space $X/M=\\{x+M:x\in X\\}$ with the quotient topology $\pi(\tau)$
is an elcs, where $\pi(x)=x+M$ for $x\in X$ denotes the quotient map.
Similarly, if $\\{(X_{i},\tau_{i}):i\in\mathcal{I}\\}$ is a collection of
extended locally convex spaces, then the product space
$\Pi_{i\in\mathcal{I}}X_{i}$ with the product topology
$\Pi_{i\in\mathcal{I}}\tau_{i}$ is an elcs. For more details about quotient
and product space of extended locally convex spaces, we refer to [8, 11].

######  Theorem 3.22.

Let $(X,\tau)$ be a barreled elcs with the flc topology $\tau_{F}$. Suppose
$M$ is a closed subspace $X$. Then the quotient space $X/M$ with the quotient
topology $\pi(\tau)$ is a barreled elcs.

###### Proof.

By Theorems 3.15 and 3.21, $M$ is a closed subspace of the barreled locally
convex space $(X,\tau_{F})$. Then by Theorem 11.12.3, p. 409 in [9], the
quotient space $(X/M,\pi(\tau_{F}))$ is a barreled locally convex space. It is
shown in Theorem 4.3 of [8] that the flc topology for
$\left(X/M,\pi(\tau)\right)$ is equal to $\pi(\tau_{F})$. Therefore by Theorem
3.15, $\left(X/M,\pi(\tau)\right)$ is barreled. ∎

######  Theorem 3.23.

Suppose $\\{(X_{i},\tau_{i}):i\in\mathcal{I}\\}$ is a collection of barreled
extended locally convex spaces. Then the product space
$\Pi_{i\in\mathcal{I}}X_{i}$ with the product topology
$\Pi_{i\in\mathcal{I}}\tau_{i}$ is a barreled elcs.

###### Proof.

Let $\tau_{F_{i}}$ be the flc topology of the elcs $(X_{i},\tau_{i})$ for
every $i\in\mathcal{I}$. Then by Theorem 3.15, $(X_{i},\tau_{F_{i}})$ is a
barreled locally convex space for every $i\in\mathcal{I}$. Therefore by
Theorem 11.12.4, p. 409 in [9], the product space
$\left(\Pi_{i\in\mathcal{I}}X_{i},\Pi_{i\in\mathcal{I}}\tau_{F_{i}}\right)$ is
a barreled locally convex space. It is shown in Theorem 4.4 of [8] that the
flc topology for
$\left(\Pi_{i\in\mathcal{I}}X_{i},\Pi_{i\in\mathcal{I}}\tau_{i}\right)$ is
equal to $\Pi_{i\in\mathcal{I}}\tau_{F_{i}}$. Hence by Theorem 3.15,
$\left(\Pi_{i\in\mathcal{I}}X_{i},\Pi_{i\in\mathcal{I}}\tau_{i}\right)$ is a
barreled elcs.∎

##  4\. Function Spaces

For a metric space $(X,d)$, let $C(X)$ denote the space of real-valued
continuous functions on $(X,d)$. A collection $\mathcal{B}$ of nonempty sets
in a metric space $(X,d)$ is said to be a bornology if it covers $X$ and is
closed under finite union and taking subsets of its members. A base
$\mathcal{B}_{0}$ is a subfamily of $\mathcal{B}$ which is cofinal in
$\mathcal{B}$ under the set inclusion. If every member of $\mathcal{B}_{0}$ is
closed in $(X,d)$, then $\mathcal{B}$ is said to have a closed base. For
details on metric bornologies, we refer to the monograph [2].

In this section, we study the barreledness of the spaces
$(C(X),\tau_{\mathcal{B}})$ and $(C(X),\tau_{\mathcal{B}}^{s})$, where
$\tau_{\mathcal{B}}$ denotes the topology of uniform convergence on
$\mathcal{B}$, and $\tau_{\mathcal{B}}^{s}$ denotes the topology of strong
uniform convergence on $\mathcal{B}$.

######  Definition 4.1.

Let $\mathcal{B}$ be a bornology on a metric space $(X,d)$. Then the topology
$\tau_{\mathcal{B}}$ of uniform convergence on $\mathcal{B}$ is determined by
a uniformity on $C(X)$ having base consisting of sets of the form

$[B,\epsilon]=\left\\{(f,g):\forall x\in
B,~{}|f(x)-g(x)|<\epsilon\right\\}~{}(B\in\mathcal{B},~{}\epsilon>0).$

######  Definition 4.2.

([3]) Let $\mathcal{B}$ be a bornology on a metric space $(X,d)$. Then the
topology $\tau^{s}_{\mathcal{B}}$ of strong uniform convergence on
$\mathcal{B}$ is determined by a uniformity on $C(X)$ having base consisting
of sets of the form

$[B,\epsilon]^{s}=\left\\{(f,g):\exists~{}\delta>0~{}\forall x\in
B^{\delta},~{}|f(x)-g(x)|<\epsilon\right\\}~{}(B\in\mathcal{B},~{}\epsilon>0),$

where for $B\subseteq X$, $B^{\delta}=\displaystyle{\bigcup_{y\in B}}\\{x\in
X:d(x,y)<\delta\\}$.

The topology $\tau_{\mathcal{B}}$ on $C(X)$ can also be induced by the
collection $\mathcal{P}=\\{\rho_{B}:B\in\mathcal{B}\\}$ of extended seminorms,
where $\rho_{B}(f)=\sup_{x\in B}|f(x)|$ for $f\in C(X)$. Similarly, the
topology $\tau_{\mathcal{B}}^{s}$ on $C(X)$ is induced by the collection
$\mathcal{P}=\\{\rho_{B}^{s}:B\in\mathcal{B}\\}$ of extended seminorms, where
$\rho_{B}^{s}(f)=\inf_{\delta>0}\left\\{\sup_{x\in B^{\delta}}|f(x)|\right\\}$
for $f\in C(X)$. Therefore both $(C(X),\tau_{\mathcal{B}})$ and
$(C(X),\tau_{\mathcal{B}}^{s})$ are extended locally convex spaces. When
$\mathcal{B}$ is the bornology of all finite subsets and we extend our
function space to $\mathbb{R}^{X}$, the set of all real-valued functions on
$(X,d)$, $\tau_{\mathcal{B}}^{s}$ is the weakest topology on $\mathbb{R}^{X}$
for which $C(X)$ is closed ([3], Corollary 6.8). For more details related to
$\tau_{\mathcal{B}}$ and $\tau_{\mathcal{B}}^{s}$, we refer to [3, 6].

######  Theorem 4.3.

Let $\mathcal{B}$ be a bornology with a closed base on a metric space $(X,d)$.
If $(C(X),\tau_{\mathcal{B}}^{s})$ is barreled, then the bornology
$\mathcal{K}$ of all non-empty relatively compact subsets of $X$ is contained
in $\mathcal{B}$.

###### Proof.

Let $K$ be a compact subset of $X$. Consider $V_{K}=\\{f\in C(X):\sup_{x\in
K}|f(x)|\leq 1\\}$. Then $V_{K}$ is an absolutely convex and absorbing subset
of $C(X)$. If $(f_{\lambda})_{\lambda\in\Lambda}$ is a net in $V_{K}$
converging to $f$ in $(C(X),\tau_{\mathcal{B}}^{s})$, then $f_{\lambda}(x)\to
f(x)$ for every $x\in X$. For every $\epsilon>0$ and $x\in K$, there exists an
$\lambda_{o}\in\Lambda$ such that $|f_{\lambda_{0}}(x)-f(x)|<\epsilon$. So
$|f(x)|\leq 1$ as $|f(x)|\leq|f(x)-f_{\lambda}(x)|+|f_{\lambda}(x)|$ for every
$\lambda\in\Lambda$. Consequently, $f\in V_{K}$. Therefore $V_{K}$ is closed
in $(C(X),\tau_{\mathcal{B}}^{s})$. Hence $V_{K}$ is a barrel in
$(C(X),\tau_{\mathcal{B}}^{s})$. Since $(C(X),\tau_{\mathcal{B}}^{s})$ is
barreled, $V_{K}$ is a neighborhood of $0$ in $C(X,\tau_{\mathcal{B}}^{s})$.
Therefore $U=\\{f\in C(X):\rho_{B}^{s}(f)<\epsilon\\}\subseteq V_{K}$ for some
closed set $B\in\mathcal{B}$ and $\epsilon>0$. It is sufficient to show that
$K\subseteq B$. Suppose $x_{0}\in K\setminus B$. Then there exists an
$m\in\mathbb{N}$ such that $d(x_{0},y)>\frac{1}{m}$ for every $y\in B$ as
$B=\bigcap_{n\in\mathbb{N}}\\{x\in X:d(x,y)\leq\frac{1}{n}\text{ for some
}y\in B\\}$. It is easy to see that
$\text{Cl}_{X}(B^{\frac{1}{2m}})\subseteq\\{x\in X:d(x,y)\leq\frac{1}{m}\text{
for some }y\in B\\}$. So by Urysohn’s lemma, there exists a continuous
function $f$ on $X$ such that $f|_{\text{Cl}_{X}(B^{\frac{1}{2m}})}=0$ and
$f(x_{0})=2$. Consequently, $f\in U\setminus V_{K}$. We arrive at a
contradiction. Therefore $K\subseteq B$. Hence $K\in\mathcal{B}$.∎

######  Theorem 4.4.

Let $\mathcal{B}$ be a bornology with a closed base on a metric space $(X,d)$.
If $(C(X),\tau_{\mathcal{B}})$ is barreled, then the bornology $\mathcal{K}$
of all non-empty relatively compact subsets of $X$ is contained in
$\mathcal{B}$.

###### Proof.

It is similar to the proof of Theorem 4.3.∎

######  Corollary 4.5.

Let $\mathcal{B}$ be a bornology with closed base on a metric space $(X,d)$.
Suppose $\tau_{F}$ and $\tau_{F}^{s}$ are the flc topologies for the spaces
$(C(X),\tau_{\mathcal{B}})$ and $(C(X),\tau_{\mathcal{B}}^{s})$, respectively.
If $(C(X),\tau_{\mathcal{B}})$ and $(C(X),\tau_{\mathcal{B}}^{s})$ are
barreled, then the compact open topology $\tau_{\mathcal{K}}$ is coarser than
both $\tau_{F}$ and $\tau_{F}^{s}$.

The next example shows that Theorem 4.3 and Theorem 4.4 may not hold if
$\mathcal{B}$ does not have any closed base in $X$.

######  Example 4.6.

Let $X=\mathbb{R}$ with the usual metric, and let
$\mathcal{B}=\\{A\subseteq\mathbb{R}:A\text{ is a non-empty subset of
}\mathbb{Q}\cup F\text{ for a finite subset }F\text{ of }\mathbb{R}\\}.$ Then
$\mathcal{B}$ forms a bornology on $X$ without any closed base. Note that
$A=[0,1]$ is a compact set in $X$ and $A\notin\mathcal{B}$. Also, it is easy
to see that $\tau_{\mathcal{B}}=\tau_{\mathcal{B}}^{s}=\tau_{u}$, where
$\tau_{u}$ is the topology of uniform convergence on $C(X)$ induced by the
extended norm $\parallel f\parallel_{\infty}=\sup_{x\in X}|f(x)|$ for $f\in
C(X)$. By Corollary 3.5, $(C(X),\tau_{u})$ is barreled.

The next example shows that if $\mathcal{B}$ is a bornology with a closed base
on a metric space $(X,d)$, and both $(C(X),\tau_{\mathcal{B}})$ and
$(C(X),\tau_{\mathcal{B}}^{s})$ are barreled, then the inclusions in Corollary
4.5 may be strict, that is, if $\mathcal{B}$ is a bornology with a closed base
on a metric space $(X,d)$, then the compact open topology $\tau_{\mathcal{K}}$
may not be equal to the flc topologies $\tau_{F}$ and $\tau_{F}^{s}$ for the
spaces $(C(X),\tau_{\mathcal{B}})$ and $(C(X),\tau_{\mathcal{B}}^{s})$,
respectively.

######  Example 4.7.

Let $(X,d)$ be a discrete metric space such that
$X=\\{x_{n}:n\in\mathbb{N}\\}$. If $\mathcal{B}$ is the collection of all non-
empty subsets of $X$, then both $\tau_{\mathcal{B}}$ and
$\tau_{\mathcal{B}}^{s}$ are induced by the extended norm $\parallel
f\parallel_{\infty}=\sup_{x\in X}|f(x)|$ for $f\in C(X)$. By Corollary 3.5,
$(C(X),\parallel\cdot\parallel_{\infty})$ is barreled. Let $\tau_{F}$ be the
flc topology for $(C(X),\parallel\cdot\parallel_{\infty})$. We show that
$\tau_{\mathcal{K}}\neq\tau_{F}$. For $m\in\mathbb{N}$, define a function
$f_{m}$ on $X$ by

$f_{m}(x_{n})=\begin{cases}\text{$1$,}&\quad\text{ if $m=n$ }\\\
\text{0,}&\quad\text{ if $m\neq n$.}\\\ \end{cases}$

Since $(X,d)$ is a discrete space, $f_{m}\in C(X)$ for every $m\in\mathbb{N}$.
Also, it is easy to see that $f_{m}\to 0$ in $(C(X),\tau_{\mathcal{K}})$. Let
$M=\text{span}(\\{f_{m}:m\in\mathbb{N}\\})$. Then
$(M,\parallel\cdot\parallel_{\infty})$ is a normed linear space. Since
$\parallel f_{m}\parallel_{\infty}=1$ for every $m\in\mathbb{N}$,
$f_{m}\nrightarrow 0$ in $(M,\parallel\cdot\parallel_{\infty})$. Thus
$f_{m}\nrightarrow 0$ in $(C(X),\tau_{F})$ as
$\tau_{F}|_{M}=\tau_{\parallel\cdot\parallel_{\infty}}$ on $M$.

##  5\. uniform boundedness principle

In this section, we establish the uniform boundedness principle for a barreled
extended locally convex space. In order to do that we use the barreledness of
the corresponding finest space. We first relate the continuity of a linear
operator between extended locally convex spaces to the continuity of the same
under the corresponding finest spaces. Then we extend this result to a
equicontinuous family of continuous linear operators.

The following result follows from Proposition 4.8 of [11].

######  Proposition 5.1.

Let $(X,\tau),(Y,\sigma)$ be two extended locally convex spaces. Suppose
$T:X\rightarrow Y$ is a linear map. Then $T$ is continuous if and only if for
every continuous extended seminorm $q$ on $Y$, there exists a continuous
extended seminorm $p$ on $X$ such that $q(T(x))\leq p(x)$ for every $x\in X$.

######  Corollary 5.2.

Let $(X,\tau),(Y,\sigma)$ be two extended locally convex spaces. Suppose
$T:X\rightarrow Y$ is a continuous linear map. Then for every continuous
extended seminorm $q$ on $(Y,\sigma)$, there exists a continuous extended
seminorm $p$ on $(X,\tau)$ such that $T^{-1}\left(Y_{fin}^{q}\right)\supseteq
X_{fin}^{p}.$

######  Theorem 5.3.

Let $(X,\tau),(Y,\sigma)$ be two extended locally convex spaces. Suppose
$T:X\rightarrow Y$ is a linear map such that for every continuous extended
seminorm $q$ on $(Y,\sigma)$, there exists a continuous extended seminorm $p$
on $(X,\tau)$ such that

$T^{-1}\left(Y_{fin}^{q}\right)\supseteq X_{fin}^{p}.$

Then the following statements are equivalent.

  * (1)

$T:(X,\tau)\rightarrow(Y,\sigma)$ is continuous.

  * (2)

$T:(X,\tau{{}_{F}})\rightarrow(Y,\sigma{{}_{F}})$ is continuous, where
$\tau_{F}$ and $\sigma_{F}$ are the flc topologies for the spaces $(X,\tau)$
and $(Y,\sigma)$, respectively.

###### Proof.

$(1)\Rightarrow(2)$. Suppose $q$ is a continuous seminorm on $(Y,\sigma_{F})$.
Then $q$ is a continuous with respect to $(Y,\sigma)$. Since
$T:(X,\tau)\rightarrow(Y,\sigma)$ is continuous, by Proposition 5.1, there
exists a continuous extended seminorm $p$ on $(X,\tau)$ such that $q(T(x))\leq
p(x)$ for all $x\in X$. Let $M$ be a subspace of $X$ such that
$X=X_{fin}^{p}\oplus M$. Consider the seminorm $\rho:X\rightarrow[0,\infty)$
defined by

$\rho(x)=p(x_{f})+q(T(x_{m})),$

where $x=x_{f}+x_{m}$ for $x_{f}\in X_{fin}^{p}$ and $x_{m}\in M.$ Since
$\rho(x)\leq p(x)$ for every $x\in X$, $\rho$ is a continuous seminorm on
$(X,\tau)$. By Theorem 3.5 in [8], $\rho$ is continuous on $(X,\tau_{F})$. It
is easy to prove that $q(T(x))\leq\rho(x)$ for $x\in X$. Therefore by Theorem
5.7.3, p. 126 in [9], $T:(X,\tau{{}_{F}})\rightarrow(Y,\sigma{{}_{F}})$ is
continuous.

$(2)\Rightarrow(1)$. Suppose $q$ is a continuous extended seminorm on
$(Y,\sigma)$. Then by our assumptions, there exists a continuous extended
seminorm $p$ on $(X,\tau)$ such that $T^{-1}\left(Y_{fin}^{q}\right)\supseteq
X_{fin}^{p}$. Let $N$ be a subspace of $Y$ such that $Y=Y_{fin}^{q}\oplus N$.
Define a seminorm $\mu$ on $Y$ by $\mu(y)=q(y_{f})$, where $y=y_{f}+y_{n}$ for
$y_{f}\in Y_{fin}^{q}$ and $y_{n}\in N$. Since $\mu(y)\leq q(y)$ for every
$y\in Y$, $\mu$ is a continuous seminorm on $(Y,\sigma)$. By Theorem 3.5 in
[8], $\mu$ is continuous on $(Y,\sigma_{F})$. Since
$T:(X,\tau{{}_{F}})\rightarrow(Y,\sigma_{F})$ is continuous, by Theorem 5.7.3,
p. 126 in [9], there exists a continuous seminorm $\rho$ on $(X,\tau_{F})$
such that $\mu(T(x))\leq\rho(x)$ for every $x\in X$. It is easy to see that
$p+\rho$ is a continuous extended seminorm on $(X,\tau)$. Note that if
$q(T(x))=\infty$, then $p(x)=\infty$. Otherwise,
$q(T(x))=\mu(T(x))\leq\rho(x)$. Therefore

$q(T(x))\leq(p+\rho)(x)\text{ for every }x\in X.$

Hence $T:(X,\tau)\rightarrow(Y,\sigma)$ is continuous. ∎

######  Corollary 5.4.

Let $(X,\tau),(Y,\sigma)$ be two barreled extended locally convex spaces.
Suppose $T:X\rightarrow Y$ is a linear map such that for every continuous
extended seminorm $q$ on $Y$, there exists a continuous extended seminorm $p$
on $X$ such that

$T^{-1}\left(Y_{fin}^{q}\right)\supseteq X_{fin}^{p}.$

Then the following statements are equivalent.

  * (1)

$T:(X,\tau)\rightarrow(Y,\sigma)$ is continuous.

  * (2)

$T:(X,\tau{{}_{w}})\rightarrow(Y,\sigma{{}_{w}})$ is continuous, where
$\tau_{w}$ and $\sigma_{w}$ are the weak topologies for the spaces $(X,\tau)$
and $(Y,\sigma)$, respectively.

###### Proof.

It follows from Corollary 11.3.7, p. 385 in [9], Theorem 3.15, and Theorem
5.3.∎

In general, we may not drop the condition
$T^{-1}\left(Y_{fin}^{q}\right)\supseteq X_{fin}^{p}$ in Theorem 5.3.

######  Example 5.5.

Let $X=Y=\mathbb{R}^{2}$. If $\tau$ is the topology induced by
$\parallel(x,y)\parallel=|x|+|y|$ for $(x,y)\in X$, and $\sigma$ is the
topology induced by the discrete extended norm, then $X_{fin}=\mathbb{R}^{2}$
in $(X,\tau)$ and $Y_{fin}=\\{(0,0)\\}$ in $(Y,\sigma)$. Note that
$\tau_{F}=\sigma_{F}$ as the dimension of $X$ is finite. Then the identity map
$I:(X,\tau_{F})\rightarrow(Y,\sigma_{F})$ is continuous but
$I:(X,\tau)\rightarrow(Y,\sigma)$ is not continuous.

To define a pointwise bounded family of continuous linear operators, we recall
the definition of a bounded set in an elcs and some of its basic properties
from [8, 7].

######  Definition 5.6.

We say a subset $A$ of an elcs $(X,\tau)$ is bounded if for every neighborhood
$U$ of $0$, there exist a finite subset $F$ of $X$ and $r>0$ such that
$A\subseteq F+rU$.

######  Proposition 5.7.

Suppose $(X,\tau)$ is an elcs. Then the following statements hold.

  * (1)

The collection of all bounded sets in an elcs forms a bornology.

  * (2)

$A$ is bounded in $(X,\tau)$ if and only if for every neighborhood $U$ of $0$,
there exist a finite subset $F$ of $A$ and $r>0$ such that $A\subseteq F+rU$.

  * (3)

Suppose $(Y,\sigma)$ is an elcs and $T:(X,\tau)\rightarrow(Y,\sigma)$ is a
continuous linear operator. Then for each bounded set $A$ in $(X,\tau)$,
$T(A)$ is bounded in $(Y,\sigma)$.

  * (4)

No subspace $\left(\text{except }\\{0\\}\right)$ of $X$ is bounded.

  * (5)

If $(x_{n})$ converges to $0$ in $(X,\tau)$, then $\\{x_{n}:n\in\mathbb{N}\\}$
is bounded in $(X,\tau)$.

  * (6)

For a conventional locally convex space $(X,\tau)$, $A\subseteq X$ is bounded
in the sense of Definition 5.6 if and only if it is absorbed by each
neighborhood of $0$ in $(X,\tau)$.

Let $\mathcal{L}(X,Y)$ denote the collection of all continuous linear maps
from an elcs $(X,\tau)$ to an elcs $(Y,\sigma)$. Then
$\mathcal{A}\subseteq\mathcal{L}(X,Y)$ is said to be pointwise bounded if for
every $x\in X$, $\\{Tx:T\in\mathcal{A}\\}$ is bounded in $(Y,\sigma)$.

######  Definition 5.8.

Suppose $(X,\tau)$ and $(Y,\sigma)$ are extended locally convex spaces. Then
$\mathcal{A}\subseteq\mathcal{L}(X,Y)$ is said to be equicontinuous in
$\mathcal{L}(X,Y)$ if for every neighborhood $V$ of $0$ in $(Y,\sigma)$, there
exists a neighborhood $U$ of $0$ in $(X,\tau)$ such that $T(U)\subseteq
V\text{ for }T\in\mathcal{A}$.

The following facts regarding equicontinuity of a subset $A$ of $X^{*}$ are
easy to verify ([7]).

  * (1)

$A$ is equicontinuous on $(X,\tau)$ if and only if $A_{\circ}=\\{x\in
X:|f(x)|\leq 1\text{ for every }f\in A\\}$ is a neighborhood of $0$ in
$(X,\tau)$.

  * (2)

If $A\subseteq B\subseteq X^{*}$ and $B$ is equicontinuous, then $A$ is also
equicontinuous.

  * (3)

If $(X,\parallel\cdot\parallel)$ is an enls, then $A\subseteq X^{*}$ is
equicontinuous if and only if there exists an $M>0$ such that $\parallel
f\parallel_{op}\leq M$ for each $f\in A$, where

$\parallel f\parallel_{op}=\sup\\{|f(x)|:\parallel x\parallel\leq 1\\}.$

In particular, the closed unit ball $B_{X^{*}}=\\{f\in X^{*}:\parallel
f\parallel_{op}\leq 1\\}$ is always equicontinuous on $X$.

######  Remark 5.9.

If $(X,\parallel\cdot\parallel)$ is an enls and $x\in X\setminus X_{fin}$,
then for any $n\in\mathbb{N}$ there exists a linear functional $f_{n}\in
B_{X^{*}}$ such that $f_{n}(x)=n$. So $B_{X^{*}}$ is not pointwise bounded.
Hence in contrast to classical locally convex spaces, an equicontinuous family
on an elcs may not be pointwise bounded.

######  Theorem 5.10.

Suppose $(X,\tau)$ and $(Y,\sigma)$ are extended locally convex spaces. Then
$\mathcal{A}\subseteq\mathcal{L}(X,Y)$ is equicontinuous if and only if for
every continuous extended seminorm $q$ on $Y$, there exists a continuous
extended seminorm $p$ on $X$ such that $q(T(x))\leq p(x)\text{ for every
}T\in\mathcal{A}\text{ and }x\in X$.

###### Proof.

Suppose $\mathcal{A}\subseteq\mathcal{L}(X,Y)$ is equicontinuous and $q$ is
any continuous extended seminorm on $(Y,\sigma)$. Then there exists an
absolutely convex neighborhood $U$ of $0$ in $(X,\tau)$ such that
$T(U)\subseteq q^{-1}\left([0,1]\right)$ for every $T\in\mathcal{A}$. Let
$p=\rho_{U}$, where $\rho_{U}$ is the Minkowski functional on $X$
corresponding to $U$. Then $\rho_{U}$ is a continuous extended seminorm on
$(X,\tau)$. It is easy to see that if $p(x)<\infty$, then

$\frac{x}{p(x)+\epsilon}\in p^{-1}\left([0,1)\right)\subseteq U\subseteq
T^{-1}\left(q^{-1}\left([0,1]\right)\right)$

for every $\epsilon>0$. Therefore $q(T(x))\leq p(x)$ for every $x\in X$ and
$T\in\mathcal{A}$.

Conversely, suppose $\mathcal{A}\subseteq\mathcal{L}(X,Y)$ has the given
property. If $V$ is a neighborhood of $0$ in $(Y,\sigma)$, then there exists a
continuous extended seminorm $q$ on $(Y,\sigma)$ such that
$q^{-1}([0,1))\subseteq V$. By our assumption, there exists a continuous
extended seminorm $p$ on $(X,\tau)$ such that $q(T(x))\leq p(x)$ for every
$x\in X$ and $T\in\mathcal{A}$. It is easy to see that $U=p^{-1}([0,1))$ is a
neighborhood of $0$ in $(X,\tau)$ and that $T(U)\subseteq V$ for every
$T\in\mathcal{A}$. ∎

######  Corollary 5.11.

Let $(X,\tau)$ and $(Y,\sigma)$ be two extended locally convex spaces. Suppose
$\mathcal{A}\subseteq\mathcal{L}(X,Y)$ is equicontinuous. Then for every
continuous extended seminorm $q$ on $(Y,\sigma)$, there exists a continuous
extended seminorm $p$ on $(X,\tau)$ such that

$\bigcap_{T\in\mathcal{A}}T^{-1}\left(Y_{fin}^{q}\right)\supseteq
X_{fin}^{p}.$

######  Theorem 5.12.

(Uniform Boundedness Principle) Let $(X,\tau)$ be a barreled elcs, and let
$(Y,\sigma)$ be any elcs. Suppose $\mathcal{A}\subseteq\mathcal{L}(X,Y)$ is
pointwise bounded and for every continuous extended seminorm $q$ on
$(Y,\sigma)$, there exists a continuous extended seminorm $p$ on $(X,\tau)$
such that $T^{-1}\left(Y_{fin}^{q}\right)\supseteq X_{fin}^{p}\text{ for all
}T\in\mathcal{A}$. Then $\mathcal{A}$ is an equicontinuous family in
$\mathcal{L}(X,Y)$.

###### Proof.

Let $\mathcal{L}(X^{F},Y^{F})$ be the collection of all continuous linear maps
from $(X,\tau_{F})$ to $(Y,\sigma_{F})$, where $\tau_{F}$ and $\sigma_{F}$ are
the flc topologies for $(X,\tau)$ and $(Y,\sigma)$, respectively. By Theorem
5.3, $\mathcal{A}\subseteq\mathcal{L}(X^{F},Y^{F})$. Therefore $\mathcal{A}$
is a pointwise bounded subset of $\mathcal{L}(X^{F},Y^{F})$ as
$\sigma_{F}\subseteq\sigma$. Since $(X,\tau)$ is barreled, by Theorem 3.15,
$(X,\tau_{F})$ is barreled. Thus by Theorem 11.9.1, p. 400 in [9],
$\mathcal{A}$ is an equicontinuous family in $\mathcal{L}(X^{F},Y^{F})$.
Suppose $q$ is a continuous extended seminorm on $(Y,\sigma)$. Then there
exists a subspace $N$ of $Y$ such that $Y=Y_{fin}^{q}\oplus N$. Consider the
seminorm $q_{1}$ defined by

$q_{1}(y=y_{f}+y_{n})=q(y_{f})$

for $y_{f}\in Y_{fin}^{q}$ and $y_{n}\in N$. Since $q_{1}(y)\leq q(y)$ for
every $y\in Y$, $q_{1}$ is a continuous seminorm on $(Y,\sigma)$. By Theorem
3.5 in [8], $q_{1}$ is a continuous seminorm on $(Y,\sigma_{F})$. Since
$\mathcal{A}$ is an equicontinuous family in $\mathcal{L}(X^{F},Y^{F})$, there
exists a continuous seminorm $p_{1}$ on $(X,\tau_{F})$ such that
$q_{1}(T(x))\leq p_{1}(x)$ for every $T\in\mathcal{A}\text{ and }x\in X$ (see,
Theorem 8.6.3, p. 245 in [9]). Also, by our assumption, there exists a
continuous extended seminorm $p$ on $(X,\tau)$ such that
$T^{-1}\left(Y_{fin}^{q}\right)\supseteq X_{fin}^{p}\text{ for all
}T\in\mathcal{A}$. It is easy to see that $\mu(x)=p(x)+p_{1}(x)$ for $x\in X$
is a continuous extended seminorm on $(X,\tau)$ and that $q(T(x))\leq\mu(x)$
for every $x\in X$. Hence by Theorem 5.10, $\mathcal{A}$ is an equicontinuous
family in $\mathcal{L}(X,Y)$. ∎

######  Corollary 5.13.

Let $(X,\tau)$ be a barreled fundamental elcs, and let $(Y,\sigma)$ be an
elcs. Suppose $\mathcal{A}\subseteq\mathcal{L}(X,Y)$ is pointwise bounded.
Then $\mathcal{A}$ is an equicontinuous family in $\mathcal{L}(X,Y)$.

###### Proof.

Suppose $p$ is a continuous extended seminorm on $(X,\tau)$ such that
$X_{fin}^{p}=X_{fin}$. Note that $X_{fin}$ and $Y_{fin}$ are the connected
components of zero in $(X,\tau)$ and $(Y,\sigma)$, respectively. Therefore
$T(X_{fin})\subseteq Y_{fin}$ for every $T\in\mathcal{A}$. Since
$Y_{fin}\subseteq Y_{fin}^{q}$ for every continuous extended seminorm $q$ on
$(Y,\sigma)$, we have

$X_{fin}^{p}\subseteq T^{-1}\left(Y_{fin}^{q}\right)$

for every continuous extended seminorm $q$ on $(Y,\sigma)$. Hence by Theorem
5.12, $\mathcal{A}$ is an equicontinuous family in $\mathcal{L}(X,Y)$. ∎

######  Corollary 5.14.

Let $(X,\parallel\cdot\parallel)$ be an extended Banach space, and let
$(Y,\sigma)$ be an elcs. Suppose $\mathcal{A}\subseteq\mathcal{L}(X,Y)$ is
pointwise bounded. Then $\mathcal{A}$ is an equicontinuous family in
$\mathcal{L}(X,Y)$. Moreover, there exists an $M>0$ such that $\parallel
T\parallel_{op}\leq M$ for every $T\in\mathcal{A}$.

######  Theorem 5.15.

Suppose $(X,\tau)$ is an elcs with flc topology $\tau_{F}$. Then $(X,\tau)$ is
barreled if and only if every pointwise bounded subset $\mathcal{A}$ of
$X^{*}$ is equicontinuous on $(X,\tau)$.

###### Proof.

Suppose $(X,\tau)$ is barreled. Then by Theorem 5.12, every pointwise bounded
subset $\mathcal{A}$ of $X^{*}$ is equicontinuous on $(X,\tau)$.

Conversely, let $B$ be a barrel in $(X,\tau_{F})$. Then $B^{\circ}=\\{f\in
X^{*}:|f(x)|\leq 1\text{ for }x\in B\\}$ is pointwise bounded. By our
assumptions, $B^{\circ}$ is equicontinuous on $(X,\tau)$. Therefore
$(B^{\circ})_{\circ}=\\{x\in X:|f(x)|\leq 1\text{ for}f\in B^{\circ}\\}$ is a
neighborhood of $0$ in $(X,\tau)$. Since $B$ is a barrel in $(X,\tau_{F})$, by
Bipolar Theorem (Theorem 8.3.8, p. 235 in [9]), $B=(B^{\circ})_{\circ}$. Which
implies that $B$ is a neighborhood of $0$ in $(X,\tau)$. Consequently, the
Minkowski functional $\rho_{B}$ is a continuous seminorm on $(X,\tau)$. By
Theorem 3.4 in [8], $\rho_{B}$ is a continuous on $(X,\tau_{F})$. Therefore
$B$ is a neighborhood of $0$ in $(X,\tau_{F})$. Hence $(X,\tau_{F})$ is
barreled. Consequently, by Theorem 3.15, $(X,\tau)$ is barreled. ∎

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