# On Bishop–Phelps and Krein–Milman Properties

## Abstract

**:**

## 1. Introduction

**Property**

**1**

**.**Let X be a real topological vector space. We say that X enjoys the Krein–Milman property provided that every bounded, closed, convex subset of X has an extreme point.

**Property**

**2**

**.**Let X be a real topological vector space. We say that X enjoys the strong Krein–Milman property provided that every bounded, closed, convex subset of X is the closed convex hull of its extreme points.

## 2. Materials and Methods

**Remark**

**1.**

- •
- The inherited ${\mathcal{G}}_{1}$-uniform convergence linear topology of E from F is precisely the ${\mathcal{G}}_{1}$-uniform convergence linear topology of E in view of the fact that ${\mathcal{U}}_{E}(G,U)={\mathcal{U}}_{F}(G,U)\cap E$ for all $G\in {\mathcal{G}}_{1}$ and all $U\in {\mathcal{N}}_{0}\left(M\right)$.
- •
- If ${\mathcal{G}}_{1}\subseteq {\mathcal{G}}_{2}$, then the uniform convergence linear topology on F generated by ${\mathcal{G}}_{1}$ is clearly coarser than the uniform convergence linear topology generated by ${\mathcal{G}}_{2}$.
- •
- If for every ${G}_{2}\in {\mathcal{G}}_{2}$ there exists ${G}_{1}\in {\mathcal{G}}_{1}$ such that ${G}_{2}\subseteq {G}_{1}$, then the uniform convergence linear topology on F generated by ${\mathcal{G}}_{1}$ is finer than the uniform convergence linear topology generated by ${\mathcal{G}}_{2}$.

**Lemma**

**1.**

**Proof.**

## 3. Results

**Property**

**3**

**.**A topological vector space X is said to have the Bishop–Phelps property if for every bounded, closed, convex subset B of X, the set $\mathsf{SA}\left(B\right):=\{f\in {X}^{*}:supf\left(B\right)\phantom{\rule{4.pt}{0ex}}\mathit{is}\phantom{\rule{4.pt}{0ex}}\mathit{attained}\phantom{\rule{4.pt}{0ex}}\mathit{on}\phantom{\rule{4.pt}{0ex}}B\}$, of all functionals of ${X}^{*}$ that attain their supremum on B, is dense in ${X}^{*}$ for the ${\mathcal{BCC}}_{X}$-topology.

**Theorem**

**1.**

**Proof.**

**Lemma**

**2.**

**Proof.**

**Lemma**

**3.**

**Proof.**

**Theorem**

**2.**

**Proof.**

**Lemma**

**4.**

**Proof.**

**Lemma**

**5.**

**Proof.**

**Theorem**

**3.**

**Proof.**

**Corollary**

**1.**

**Proof.**

**Lemma**

**6.**

**Proof.**

**Theorem**

**4.**

**Proof.**

**Proposition 1.**

- 1.
- $T\left({\mathsf{B}}_{{c}_{00}}\right)$ is absolutely convex but free of extreme points.
- 2.
- $T\left({\mathsf{B}}_{{c}_{00}}\right)$ is closed in $\mathrm{span}\left\{{e}_{i}:i\in I\right\}$.
- 3.
- If ${\left({e}_{{i}_{n}}\right)}_{n\in \mathbb{N}}$ is bounded, then $T\left({\mathsf{B}}_{{c}_{00}}\right)$ is bounded, hence T is continuous.

**Proof.**

**Lemma**

**7.**

**Proof.**

**Theorem**

**5.**

**Proof.**

**Lemma**

**8.**

**Proof.**

**Corollary**

**2.**

**Proof.**

**Lemma**

**9.**

**Proof.**

**Definition**

**1**

**.**Let X be a topological vector space X. An absolutely convex prism of X is defined as a set of the form $\mathrm{co}(M\cup -M)$ for M, a bounded, closed, and convex subset of X.

**Theorem**

**6.**

**Proof.**

## 4. Nontrivial Examples

**Example**

**1.**

**Example**

**2.**

**Example**

**3.**

**Lemma**

**10.**

**Proof.**

**Lemma**

**11.**

**Proof.**

**Theorem**

**7.**

**Proof.**

## 5. Discussion

## 6. Conclusions

## Funding

## Institutional Review Board Statement

## Informed Consent Statement

## Data Availability Statement

## Conflicts of Interest

## Appendix A. Extremal Structure

**Property**

**A1**

**.**Let X be a vector space. Let $E\subseteq F\subseteq X$. Then, E is said to enjoy the extremal condition with respect to F whenever the following property is satisfied:

**Example**

**A1**

**.**Let A be a non-empty subset of a vector space X and consider $f\in {X}^{*}$. The supporting hyperplane relative to f in A,

## Appendix B. Bishop–Phelps and Krein–Milman Theorems

**Theorem**

**A1**

**.**Let C be a non-empty compact convex subset of a Hausdorff locally convex topological vector space. Then, C is the closed convex hull of its set of extreme points.

**Theorem**

**A2**

**.**If C is a non-empty bounded closed convex subset of a Banach space X, then the support functionals for C are dense in ${X}^{*}$.

**Theorem**

**A3**

**.**Let X be a Banach space. For every $\epsilon \in (0,1)$, there exists $\delta \left(\epsilon \right):={\epsilon}^{2}/4$ such that for all ${x}^{*}\in {\mathsf{S}}_{{X}^{*}}$ and all $x\in {\mathsf{S}}_{X}$ with ${x}^{*}\left(x\right)>1-\delta \left(\epsilon \right)$ there exist $u\in {\mathsf{S}}_{X}$ and ${y}^{*}\in {\mathsf{S}}_{{X}^{*}}$ with ${y}^{*}\left(u\right)=1$, $\parallel u-x\parallel <\epsilon $ and $\parallel {x}^{*}-{y}^{*}\parallel <\epsilon $.

## Appendix C. Finest Locally Convex Vector Topology

**Theorem A4.**

- 1.
- For every $U\in \mathcal{B}$ there exists $V\in \mathcal{B}$ such that $V+V\subseteq U$.
- 2.
- For every $U\in \mathcal{B}$ there exists $V\in \mathcal{B}$ such that $-V\subseteq U$.
- 3.
- For every $U\in \mathcal{B}$ there exist $V\in \mathcal{B}$ and $W\in {\mathcal{N}}_{0}\left(R\right)$ such that $WV\subseteq U$.
- 4.
- For every $U\in \mathcal{B}$ and every $r\in R$, there exists $V\in \mathcal{B}$ such that $rV\subseteq U$.
- 5.
- For every $U\in \mathcal{B}$ and every $m\in M$, there exists $W\in {\mathcal{N}}_{0}\left(R\right)$ such that $Wm\subseteq U$.

**Remark A1.**

- 1.
- Every neighborhood of 0 is absorbing in view of Theorem A4(5).
- 2.
- If $\mathcal{B}$ is a base of neighborhoods of 0, then ${\mathcal{B}}_{0}:=\left\{{\mathsf{B}}_{\mathbb{K}}U:U\in \mathcal{B}\right\}$ is also base of neighborhoods of 0 and all elements of ${\mathcal{B}}_{0}$ are balanced.

**Remark**

**A2.**

**Remark**

**A3.**

**Lemma A1.**

- 1.
- $\mathrm{int}\left(A\right)\subseteq \mathrm{inter}\left(A\right)\subseteq A$.
- 2.
- If A is open, then $A=\mathrm{int}\left(A\right)=\mathrm{inter}\left(A\right)$.
- 3.
- If A is convex and $\mathrm{int}\left(A\right)\ne \u2300$, then $\mathrm{int}\left(A\right)=\mathrm{inter}\left(A\right)$.
- 4.
- If A is convex, $\mathrm{int}\left(A\right)\ne \u2300$ and $A=\mathrm{inter}\left(A\right)$, then A is open.

**Proof.**

- Fix an arbitrary $a\in \mathrm{int}\left(A\right)$. Let $x\in X\backslash \left\{a\right\}$. Let V be a balanced and absorbing neighborhood of 0 such that $a+V\subseteq A$. Since V is absorbing, we can find ${\epsilon}_{x}>0$ such that ${\epsilon}_{x}{\mathsf{B}}_{\mathbb{K}}(x-a)\subseteq V$. Then, $a+{\epsilon}_{x}{\mathsf{B}}_{\mathbb{K}}(x-a)\subseteq a+V\subseteq A,$ which means that $a\in \mathrm{inter}\left(A\right)$.
- By (1), $A=\mathrm{int}\left(A\right)\subseteq \mathrm{inter}\left(A\right)\subseteq A$.
- By (1), we know that $\mathrm{int}\left(A\right)\subseteq \mathrm{inter}\left(A\right)$. Fix $a\in \mathrm{int}\left(A\right)$. Let $b\in \mathrm{inter}\left(A\right)\backslash \left\{a\right\}$. There exists ${\epsilon}_{a}>0$ such that $b+{\epsilon}_{a}{\mathsf{B}}_{\mathbb{K}}(a-b)\subseteq A$. In particular,$$b=\frac{{\epsilon}_{a}}{1+{\epsilon}_{a}}a+\frac{1}{1+{\epsilon}_{a}}\left(b-{\epsilon}_{a}(a-b)\right)\in \left[a,b-{\epsilon}_{a}(a-b)\right)\subseteq \mathrm{int}\left(A\right)$$
- By (4), $A=\mathrm{inter}\left(A\right)=\mathrm{int}\left(A\right)$.

**Theorem A5.**

- 1.
- ${\mathcal{B}}_{X}$ is a base of neighborhoods of 0 for a locally convex vector topology ${\tau}_{X}$ on X.
- 2.
- If τ is a locally convex vector topology on X, then $\tau \subseteq {\tau}_{X}$.
- 3.
- If $A\subseteq X$ is convex and absorbing, then A is a neighborhood of 0 in ${\tau}_{X}$.
- 4.
- If $A\subseteq X$ is convex and $\mathrm{inter}\left(A\right)\ne \u2300$, then $\mathrm{inter}\left(A\right)$ coincides with the interior of A in ${\tau}_{X}$.

**Proof.**

- We will check that ${\mathcal{B}}_{X}$ satisfies all five conditions given in Theorem A4:
- For every $U\in {\mathcal{B}}_{X}$ there exists $V\in {\mathcal{B}}_{X}$ such that $V+V\subseteq U$. Indeed, since U is convex, it suffices to consider $V:=\frac{1}{2}U\in {\mathcal{B}}_{X}$.
- For every $U\in {\mathcal{B}}_{X}$ there exists $V\in {\mathcal{B}}_{X}$ such that $-V\subseteq U$. Indeed, since U is balanced, it suffices to consider $V:=U\in {\mathcal{B}}_{X}$.
- For every $U\in {\mathcal{B}}_{X}$ there exist $V\in {\mathcal{B}}_{X}$ and $W\in {\mathcal{N}}_{0}\left(\mathbb{K}\right)$ such that $WV\subseteq U$. Indeed, since U is balanced, it suffices to consider $V:=U\in {\mathcal{B}}_{X}$ and $W:={\mathsf{B}}_{\mathbb{K}}\in {\mathcal{N}}_{0}\left(\mathbb{K}\right)$.
- For every $U\in {\mathcal{B}}_{X}$ and every $\lambda \in \mathbb{K}$ there exists $V\in {\mathcal{B}}_{X}$ such that $rV\subseteq U$. Indeed, it suffices to take $V:={\lambda}^{-1}U\in {\mathcal{B}}_{X}$ if $\lambda \ne 0$, and $V:=U\in {\mathcal{B}}_{X}$ if $\lambda =0$.
- For every $U\in {\mathcal{B}}_{X}$ and every $x\in X$, there exists $W\in {\mathcal{N}}_{0}\left(\mathbb{K}\right)$ such that $Wx\subseteq U$. Indeed, since U is absorbing, then $0\in \mathrm{inter}\left(A\right)$, so there exists ${\epsilon}_{x}>0$ such that ${\epsilon}_{x}{\mathsf{B}}_{\mathbb{K}}x\subseteq U$, hence it only suffices to take $W:={\epsilon}_{x}{\mathsf{B}}_{\mathbb{K}}\in {\mathcal{N}}_{0}\left(\mathbb{K}\right)$.

According to Theorem A4, there exists a unique vector topology ${\tau}_{X}$ on X for which ${\mathcal{B}}_{X}$ is a base of neighborhoods of 0. This topology is by construction locally convex. - In accordance with Remark A2, there exists a base $\mathcal{B}$ of neighborhoods of 0, whose elements are convex, balanced, and absorbing for the topology $\tau $. Then, $\mathcal{B}\subseteq {\mathcal{B}}_{X}$, which implies that $\tau \subseteq {\tau}_{X}$.
- Let $\mathcal{L}:=\{B\subseteq A:B\phantom{\rule{4.pt}{0ex}}\mathrm{is}\phantom{\rule{4.pt}{0ex}}\mathrm{absolutely}\phantom{\rule{4.pt}{0ex}}\mathrm{convex}\}$. Clearly $\mathcal{L}$ is non-empty because A is absorbing. Notice that $\mathcal{L}$ can be partially ordered by the inclusion. If ${\left({B}_{i}\right)}_{i\in I}$ is a chain of $\mathcal{L}$, then it is clear that ${\bigcup}_{i\in I}{B}_{i}\in \mathcal{L}$. Using Zorn’s lemma, there exists a maximal element ${B}_{0}$. If ${B}_{0}$ is not absorbing, then there exists $x\ne 0$ and ${\epsilon}_{x}>0$ such that ${\epsilon}_{x}{\mathsf{B}}_{\mathbb{K}}x\subseteq A$ but ${\epsilon}_{x}{\mathsf{B}}_{\mathbb{K}}\u2288{B}_{0}$. Then, $C:=\mathrm{co}\left({\epsilon}_{x}{\mathsf{B}}_{\mathbb{K}}x\cup {B}_{0}\right)\subseteq A$ is absolutely convex and contains ${B}_{0}$ strictly. This is a contradiction, so ${B}_{0}$ is absorbing. As a consequence, ${B}_{0}\in {\mathcal{B}}_{X}$, so A is a neighborhood of 0 in ${\tau}_{X}$.
- Suppose now that A is convex and $\mathrm{inter}\left(A\right)\ne \u2300$. By Lemma A1(1), the interior of A in ${\tau}_{X}$ is contained in $\mathrm{inter}\left(A\right)$. Fix an arbitrary $a\in \mathrm{inter}\left(A\right)$. Now, $0\in \mathrm{inter}(A-a)$, so $A-a$ is absorbing, and it is also convex, therefore, by (3), $A-a$ is a neighborhood of 0 in ${\tau}_{X}$. As a consequence, a is an interior point of A in ${\tau}_{X}$.

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García-Pacheco, F.J.
On Bishop–Phelps and Krein–Milman Properties. *Mathematics* **2023**, *11*, 4473.
https://doi.org/10.3390/math11214473

**AMA Style**

García-Pacheco FJ.
On Bishop–Phelps and Krein–Milman Properties. *Mathematics*. 2023; 11(21):4473.
https://doi.org/10.3390/math11214473

**Chicago/Turabian Style**

García-Pacheco, Francisco Javier.
2023. "On Bishop–Phelps and Krein–Milman Properties" *Mathematics* 11, no. 21: 4473.
https://doi.org/10.3390/math11214473