Extreme points and closure of a set in $l^2(\mathbb{N})$

Question

Consider the subset C of the banach space $l^2(\mathbb{N})$ defined by \begin{equation} C = \left\{ x \in l^2(\mathbb{N}) \,\Bigg|\, \sum_{n = 0}^{\infty} (n+1) x(n) = 1, \, \, x(n) \geq 0 \, \forall n \in \mathbb{N} \right\} \end{equation} It is easy to prove this set is convex. If $x, y \in C$ then $\lambda x + (1-\lambda)y \in C$ for $\lambda \in [0,1]$ since \begin{align} \sum_{n = 0}^{\infty} (n+1)(\lambda x(n) + (1-\lambda)y(n)) &= \lambda \sum_{n = 0}^{\infty} (x+1)x(n) + (1-\lambda) \sum_{n = 0}^{\infty} (n+1)y(n) \\&= \lambda + (1- \lambda) = 1. \end{align} It is given that it's extreme points are the following elements

\begin{equation} \delta_n: \mathbb{N} \to \mathbb{C}: m \mapsto \begin{cases} \frac{1}{n+1} \,\, &\text{for} \,\, m = n\\ 0 \,\, &\text{else}\end{cases} \end{equation}

My question is: how does one prove these are the only extreme points of $C$. And also what is the closure of $C$?

Answer 1

As the closure question was already tackled by Daniel and Floris, let me focus on the extreme point-problem.

Assume that $x\in C$ is an extreme point of $C$ with two or more non-zero entries:

• $x$ has to have at least one entry which is not zero (because $0\not\in C$)
• we already characterized all the extreme points with one non-zero entry (the $\delta_n$)

By assumption we find $N,k\in\mathbb N$ with $x_N,x_k>0$. Due to positivity and the infinite series assumption, the first entry has to be strictly bounded $$0<x_N<\frac1{N+1}\quad\Leftrightarrow\quad 0<\underbrace{x_N(N+1)}_{=:\lambda}<1\,.$$ Now because $(j+1)\delta_j=e_j$ yields the $j$-th standard basis vector we can decompose $x$ as follows $$ \begin{split} x&=x_N(N+1)\delta_N + \big(x-x_N(N+1)\delta_N\big)\\&=x_N(N+1)\delta_N +\big(1-x_N(N+1)\big)\underbrace{\frac{x-x_N(N+1)\delta_N}{1-x_N(N+1)}}_{=:y}\,. \end{split} $$ Nothing fancy happened here, we merely extracted the $N$-th element from $x$. Obviously $\delta_N\in C$ and one readily verifies $y\in C$, the main point here being that $y\neq 0$ because $x$ had more than one entry or, equivalently, $1-x_N(N+1)>0$.

Thus we found $\lambda\in (0,1)$ and $\delta_N,y\in C$ (with $\delta_N\neq y$) such that $$ x=\lambda\delta_N+(1-\lambda)y $$ so by definition $x$ is no extreme point of $C$.

Answer 2

Regarding $\overline C$.

(i). Suppose $x\in \overline C$ and there exists $n_0\in \Bbb N$ and $r>0$ such that $\sum_{n=0}^{n_0}(n+1)x(n)=1+r.$ Let $y\in C$ with $\|y-x\|<r/2(n_0+1)^2.$ So $y(n)>x(n)-r/2(n_0+1)^2$ for $1\le n\le n_0.$ But then $$\sum_{n=0}^{\infty}(n+1)y(n)\ge\sum_{n=0}^{n_0}(n+1)y(n)\ge$$ $$\ge \sum_{n=0}^{n_0}(n+1)(x(n)-r/2(n_0+1)^2)=$$ $$=1+r-\sum_{n=0}^{n_0}(n+1)r/2(n_0+1)^2\ge$$ $$ \ge 1+r-r/2>1$$ contradicting $y\in C.$

(ii). If $x\in l^2(\Bbb N)$ and $\forall n\in \Bbb N\,(x(n)\ge 0)$ and $\sum_{n=1}^{\infty}(n+1)x(n)=1-d<1,$ then for $j,n\in \Bbb N$ let $x_j(n)=x(n)$ if $n\ne j$ and $x_j(j)= x(j)+d/(j+1).$ Then $x_j\in C$ and $0=\lim_{j\to \infty}d/(j+1)= \lim_{j\to \infty}\|x_j-x\|,$ so $x\in \overline C.$

(iii). If $x\in l^2(\Bbb N)$ and $x(n)=s<0$ for some $n$ then $\|y-x\|>|s|$ for all $y\in C,$ so $x\not \in \overline C.$

(iv). From (i),(ii),(iii), $\;x\in \overline C$ iff $[\;\forall n\in \Bbb N\,(x(n)\ge 0)$ and $\sum_{n=0}^{ \infty}(n+1)x(n)\le 1\;].$

Answer 3

Let $x\in C$. Then there exist $\lambda_{n}\in\mathbb{R}$ such that $x(n)=\lambda_{n}\frac{1}{n+1}$ for all $n$. As $x(n)\geq0$ we find $\lambda_{n}=(n+1)x(n)\geq0$ and as $x\in C$ we find $$\sum^{\infty}_{n=0}\lambda_{n}=\sum^{\infty}_{n=0}(n+1)x(n)=1.$$ So $x$ is a convex combination of $\{\frac{1}{n+1}e_{n}:n\in\mathbb{N}\}$ where $e_{n}$ denote the standard basis vectors of $\ell_{2}$. It follows that $x$ is an extreme point of $C$ if and only $\lambda_{n}=0$ for all $n\in\mathbb{N}\setminus\{m\}$ for some $m\in\mathbb{m}$, hence $\frac{1}{n+1}e_{n}$ are the only extreme vectors of $C$.

$\overline{C}$, the closure of $C$ is given by $$\hat{C}=\{x\in\ell_{2}:0\leq\sum^{\infty}_{n=0}(n+1)x(n)\leq 1,x(n)\geq0\text{ for all }n\in\mathbb{N}\}.$$ To show this first let $x\in\hat{C}$ and let $c_{x}=\sum^{\infty}_{n=0}(n+1)x(n)$. Then clearly $x+\frac{1-c_{x}}{1+n}e_{n}\in C$ and $\lim_{n\rightarrow\infty}x+\frac{1-c_{x}}{1+n}e_{n}=x$ wrt the norm, so $\hat{C}\subset\overline{C}$.

Now let $x\in\overline{C}$ and let $(x_{\alpha})$ be a net in $C$ converging to $x$. Evidently $x(n)\geq0$, now suppose $c_{x}>1$. Let $1<c=\min(\frac{1+c_{x}}{2},2)$, let $N\in\mathbb{N}$ such that $\sum^{N}_{n=0}(n+1)x(n)>c$ and let $\alpha'$ be index such that for all $\alpha\geq\alpha'$ such that $|x(n)-x_{\alpha}(n)|(n+1)<\frac{c-1}{N+1}$ for all $n\leq N$. Such an index exists as $x_{\alpha}$ converges pointwise to $x$. We find for all $\alpha\geq\alpha'$ that $$1=\sum^{\infty}_{n=0}(n+1)x_{\alpha}(n)\geq\sum^{N}_{n=0}(n+1)x_{\alpha}(n)=\sum^{N}_{n=0}(n+1)x(n)+\sum^{N}_{n=0}(n+1)x_{\alpha}(n)-x(n)\\ \geq \sum^{N}_{n=0}(n+1)x(n)-\sum^{N}_{n=0}(n+1)|x_{\alpha}(n)-x(n)|>c-\frac{c-1}{2}=\frac{c+1}{2}>1$$ which is a contradiction. So $\overline{C}=\hat{C}$.