Urysohn Lemmas in Topological Vector Spaces

TL;DR

This paper extends the classical Urysohn lemma to topological vector spaces, introducing two variants with quasi-convex continuous functions, broadening the lemma's applicability.

Contribution

It presents two new versions of the Urysohn lemma tailored for topological vector spaces with quasi-convex functions, a novel extension of classical results.

Findings

Two variants of Urysohn lemma for topological vector spaces.

Construction of quasi-convex continuous functions in these lemmas.

Extension of classical Urysohn lemma to a broader context.

Abstract

Two variations of classical Urysohn lemma for subsets of topological vector spaces are obtained in this article. The continuous functions constructed in these lemmas are of quasi-convex type.

Equations

$$A_0\subseteq \underset{i=1}{\overset{n}{\cup }}(x_i+U_{x_i})\cap A$$

Full text

Introduction

The classical technique for the proof of Urysohn’s lemma (See: [1]) is applied to derive metrization theorem (See: [2]) for topological vector spaces. It is possible to find the continuous functions of quasi-convex type, when the same technique is applied on certain subsets of a topological vector spaces. The Urysohn lemma is derived on normal spaces or on locally compact spaces. So, two concepts of convex normal and convex regular subsets of a topological vector spaces are introduced; their properties are studied; and two Urysohn lemmas are derived in this article.

Main Results

Definition 1

A subset $A$ of a topological vector space is said to be locally convex if every point has a local base in $A$ consisting of convex subsets of $A$.

Definition 2

Let $A$ be a locally convex subset of a topological vector space $X$. A set $A$ is said to be convex regular, if for a given point $x$ and a given open convex set $B$(open in $A$) such that $x\in B$, there is an open convex subset $C$ of $A$ such that $x\in C\subseteq cl\ C\subseteq B$.

Definition 3

Let $A$ be a locally convex subset of a topological vector space $X$. A set $A$ is said to be convex normal, if for a given non empty closed set $A_{0}$ and a given open convex set $B$ such that $A_{0}\subseteq B$, there is an open convex subset $C$ of $A$ such that $A_{0}\subseteq C\subseteq cl\ C\subseteq B$.

Proposition 4

Every compact convex set in a locally convex topological vector space is convex normal.

Proof: Let $A_{0}$ be a closed convex subset of a compact convex set $A$ in a locally convex topological vector space $X$. Let $B$ be an open convex set in $A$ containing $A_{0}$. To each $x\in A_{0}$, find an open convex neighbourhood $U_{x}$ of 0 in $X$ such that $(x+U_{x})\mathop{\cap}A\subseteq(x+cl\ (U_{x}))\mathop{\cap}A\subseteq B$. Since $A_{0}$ is compact, find a finite subset $\{x_{1},x_{2}\cdots x_{n}\}$ of $A_{0}$ such that

[TABLE]

The convex hull $C$ of right hand side of (1) is open in $A$ and it is contained in $B$. Note that, any element in $C$ is of the form $\lambda_{1}y_{1}+\lambda_{2}y_{2}\cdots\lambda_{n}y_{n}$ with $y_{i}\in x_{i}+U_{x_{i}},\ 0\leq\lambda\leq 1$ and $\sum\limits_{i=1}^{n}\lambda_{i}=1$. For each $y\in cl\ C$, there is net $\left(\lambda_{1\delta}y_{1\delta}+\lambda_{2\delta}y_{2\delta}\cdots\lambda_{n\delta}y_{n\delta}\right)_{\delta\in\mathcal{D}}$ in $C$ which converges to $y$. We can find a sub net $\left(\lambda_{1\alpha}y_{1\alpha}+\lambda_{2\alpha}y_{2\alpha}\cdots\lambda_{n\alpha}y_{n\alpha}\right)_{\alpha\in\mathcal{G}}$ of this net such that $\lambda_{1\alpha}\to\lambda_{1},\ y_{1\alpha}\to y_{1},\cdots\lambda_{n\alpha}\to\lambda_{n},y_{n\alpha}\to y_{n}$(say). Since $0\leq\lambda\leq 1$, $\sum\limits_{i=1}^{n}\lambda_{i}=1$, $y_{i}\in x_{i}+cl\ (U_{x_{i}})$ and the convex hull of $\mathop{\cup}\limits_{i=1}^{n}(x_{i}+cl\ (U_{x_{i}}))\mathop{\cap}A$ is contained in $B$, we conclude that $y=\sum\limits_{i=1}^{n}\lambda_{i}y_{i}\in B$. Thus $cl\ C\subseteq B$. This proves the result.

Definition 5

A function $f$ from a convex subset $A$ of real vector space into the real line is said to be

1. (i)

convex if $f(\lambda x+(1-\lambda)y)\leq\lambda f(x)+(1-\lambda)f(y)$, for every $x,y\in A$, for every $\lambda\in[0,1]$. 2. (ii)

quasi convex if $\{x\in A:f(x)<r\}$(or equivalently $\{x\in A:f(x)\leq r\}$) is a convex set,for every real number $r$.

Definition 6

Let $A$ be a convex subset of a topological vector space $X$. The set $A$ is said to be ‘(quasi) convex completely regular’ if for a given point $x\in A$ and a given open convex set $B\subseteq A$ such that $x\in B$, there is continuous (quasi)convex function $f:A\to[0,1]$ by $f(x)=0$ and $f(A\backslash B)=\{1\}=f(B^{c})$.

Proposition 7

Every convex subset of a locally convex space is convex completely regular, when it contains more than one point.

Proof: Let $A$ be a locally convex subset of a locally convex space. Let $x\in A$ and $B$ be an open convex subset of $A$ such that $x\in B$. Without loss of generality, let us assume that there is a continuous semi norm $p$ on $X$ such that $\inf\{p(x-y):y\in A\backslash B\}\geq 1$. Define a map $f:A\to[0,1]$ by $f(y)=\min\{p(x-y),\ 1\}$. Then $f$ is continuous convex mapping such that $f(x)=0$ and $f(y)=1$, for every $y\in A\backslash B$.

First Urysohn lemma 8

Let $A$ be a convex normal subset of a topological vector space $X$. Let $A_{0}$ be a non empty closed convex subset of $A$ and $B$ be an open convex subset of $A$ containing $A_{0}$. Then there is a continuous quasi convex function $f:A\to[0,1]$ such that $f(A_{0})=\{0\}$ and $f(A\backslash B)=\{1\}$.

Proof: Let $P$ be the set of all rational numbers in the interval [0,1]. Define $U_{1}=B$. Find an open convex subset $U_{0}$ in $A$ such that $A\subseteq U_{0}\subseteq cl(U_{0})\subseteq U_{1}\subseteq B$. Find $(U_{r})_{r\in P}$ of open convex sets such that $`p<q\Rightarrow cl(U_{p})\subseteq U_{q}$’. Define $f:A\to[0,1]$ by $f(x)=\begin{cases}\inf\ \{p\in P:x\in U_{p}\},&\mbox{ if }x\in U_{1}\\ 1&\mbox{ if }x\notin U_{1}\end{cases}$

By the proof of the classical Urysohn lemma (See:[1, Theorem 33.1]), $f$ is a continuous function. Since $\{x\in A:f(x)\leq\epsilon\}$ is convex, for every $\epsilon\geq 0$, $f$ is a quasi convex function.

Lemma 9

Let $A$ be a locally convex locally compact subset subset of a locally convex space $(X,(p_{i})_{i\in I})$. Suppose further that $A$ is complete as a uniform space with the uniformity induced by $(p_{i})_{i\in I}$. Let $A_{0}$ be a nonempty compact convex subset of $A$ and $B$ be an open(in $A$) convex subset of $A$ such that $A_{0}\subseteq B$. Then there is an open convex subset $C$ of $A$ such that $A_{0}\subseteq C\subseteq clC\subseteq B$ and $clC$ is compact.

Proof: To each $x\in A_{0}$, find an open convex neighbourhood $U_{x}$ of 0 in $X$ such that $(x+U_{x})\mathop{\cap}A\subseteq(x+cl(U_{x}))\mathop{\cap}A\subseteq B$, and $(x+cl(U_{x}))\mathop{\cap}A$ is compact. Find a finite subset $\{x_{1},x_{2}\cdots x_{n}\}$ of $A_{0}$ such that

[TABLE]

The convex hull $C$ of right hand side of (2) is open in $A$ and it is contained in $B$. By theorem in [2] and by completeness of $A$, the convex hull of $\mathop{\cup}\limits_{i=1}^{n}(x_{i}+cl(U_{x_{i}}))\mathop{\cap}A$ has a compact closure. Thus $clC$ is contained in a compact subset of $A$. So, as in the proof of proposition 4, we see that $clC\subseteq B$. This completes the proof.

Second Urysohn lemma 10

Let $A$ be a locally convex locally compact subset subset of a locally convex space. Suppose further that $A$ is complete. Let $A_{0}$ be a compact convex subset of $A$, and $B$ be an open convex subset of $A$ such that $A_{0}\subseteq B$. Then, there is a continuous function $f:A\to[0,1]$ with compact support such that $f(A_{0})=\{1\},\ f(A\backslash B)=\{0\}$, and $1-f$ is a quasi convex function.

Proof: Find convex open sets $U_{0}$ and $U_{1}$ such that $cl(U_{0})$ and $cl(U_{1})$ are compact and $A\subseteq U_{0}\subseteq cl(U_{0})\subseteq U_{1}\subseteq cl(U_{1})\subseteq B$. As in the proof of first Urysohn lemma, we can find a function $g:A\to[0,1]$ such that $g(A_{0})=\{0\},\ g(A-U_{1})=\{1\},\ g$ is a quasi convex and $g$ is continuous. The required function $f$ is $1-g$.