# R infinite space endowed with the infinite norm and fixed point theorem of Schauder

Let $$\mathbb{R}^{\infty}$$ be the vector space of infinite sequences $$(\alpha_1,\alpha_2,\alpha_3,\dots)$$ of real numbers as a natural generalization of $$\mathbb{R}^k$$, $$k\in\mathbb{N}^*$$. Let $$\boldsymbol{C}$$ be the subset of $$\mathbb{R}^{\infty}$$ defined by, $$\boldsymbol{C} := \left\{\boldsymbol{p} = (p_1, p_2, p_3, \dots) \in \mathbb{R}^{\infty}\big/ \forall ~k, p_k \in [0, 1] \text{ and } \sum_{k = 1}^{\infty} p_{k} \leq 1\right\}.$$ I make these statements. Tell me if I am wrong.

1. $$\boldsymbol{C}$$ is a compact and convex nonempty subset of the infinite dimensional space $$\mathbb{R}^{\infty}$$ for the the infinite norm defined by, $$||\boldsymbol{p}||_{\infty} = \max_k p_{k}.$$
2. Schauder's fixed point Theorem (generalization of Brouwer's fixed point Theorem to an infinite dimensional space) can be applied: any continuous mapping from $$\boldsymbol{C}$$ to itself has a fixed point.
• It's not compact. Set $p^{(n)} = (\delta_{in})_{i \in \mathbb{N}}$. You will not be able to find a convergent subsequence. – Peter Morfe Jan 25 at 23:26
• Thank you @PeterMorfe. I don't get it. What is the sequence you set? It seems that you have a sequence of sequence. – Ari.stat Jan 25 at 23:30
• If $\sum_{i = 1}^{\infty}\delta_{in} = 1$ for all $n$ then the this is also true for the limit $(\bar{\delta}_i)$: $\sum_{i = 1}^{\infty}\bar{\delta}_{i} = 1$. – Ari.stat Jan 25 at 23:36

Define a sequence $$(p^{(n)})_{n \in \mathbb{N}}$$ in $$C$$ by $$p^{(n)} = (\delta_{in})_{i \in \mathbb{N}}$$. In other words, $$\begin{equation*} p^{(n)}_{i} = \left\{ \begin{array}{r l} 1, & \text{if} \, \, i = n, \\ 0, & \text{otherwise.} \end{array} \right. \end{equation*}$$

$$(p^{(n)})_{n \in \mathbb{N}}$$ does not have any accumulation points in $$C$$ with the given norm topology. The idea is this: $$\|p^{(n)} - p^{(k)}\|_{\infty} = 1$$ for all $$n,k \in \mathbb{N}$$ so no subsequence is Cauchy.

Edit: Given $$m \in \mathbb{N}$$, if we define $$C^{m}$$ by $$\begin{equation*} C^{m}_{*} = \{x \in \mathbb{R}^{\infty} \, \mid \, \sum_{k = 1}^{\infty} |x_{k}| \leq 1, \, \, |x_{k}| \leq \frac{m}{k} \}, \end{equation*}$$ then $$C^{m}_{*}$$ is compact in $$\mathbb{R}^{\infty}$$ with the norm $$\|\cdot\|_{\infty}$$ given above. Note that the set $$C^{m}$$ defined by Ari.stat below is simply $$C^{m} = C^{m}_{*} \cap [0,1]^{\infty}$$. Since $$[0,1]^{\infty}$$ is a closed subset of $$\mathbb{R}^{\infty}$$ with respect to $$\|\cdot\|_{\infty}$$, $$C^{m}$$ is a closed subset of $$C^{m}_{*}$$ and, thus, is itself compact.

To see this, first, notice that if $$x \in C^{m}_{*}$$, then $$\|x\|_{\infty} \leq \sum_{k = 1}^{\infty} |x_{k}| \leq 1$$.

To see that $$C^{m}_{*}$$ is compact, I will prove that any sequence in $$C^{m}$$ has a subsequence that converges. Suppose that $$(x^{(n)})_{n \in \mathbb{N}} \subseteq C^{m}_{*}$$. Since $$|x^{(n)}_{k}| \leq 1$$ independently of $$n$$ and $$k$$, we can employ a diagonalization argument to find a subsequence $$(n_{j})_{j \in \mathbb{N}} \subseteq \mathbb{N}$$ and a sequence $$x \in \mathbb{R}^{\infty}$$ so that, for each $$k \in \mathbb{N}$$, $$\lim_{j \to \infty} x_{k}^{(n_{j})} = x_{k}$$. In fact, we can show that $$\lim_{j \to \infty} \|x^{(n_{j})} - x\|_{\infty} = 0$$ must hold.

Indeed, given $$\epsilon > 0$$, if we choose $$K_{0} \in \mathbb{N}$$ such that $$\frac{m}{k} \leq \epsilon/2$$ for each $$k \geq K_{0}$$, then the inequality $$\max\{|x^{(n)}_{k}|, |x_{k}|\} \leq \frac{m}{k}$$ implies $$\begin{equation*} \|x^{(n_{j})} - x\|_{\infty} \leq \max \{ |x^{(n_{j})}_{1} - x_{1}|, \dots, |x^{(n_{j})}_{K_{0}} - x_{K_{0}}|, \epsilon \}. \end{equation*}$$ Therefore, by pointwise convergence, $$\limsup_{j \to \infty} \|x^{(n_{j})} - x\|_{\infty} \leq \epsilon$$. We conclude by sending $$\epsilon \to 0^{+}$$.

• Thank you. So what is the contradiction here with the compactness? I understand that $(p^{(n)})_{n\in\mathbb{N}}$ does not have a limit in $C$. But anyway, $(p^{(n)})_{n\in\mathbb{N}}$ does not have a limit in $\mathbb{R}^{\infty}$. – Ari.stat Jan 26 at 0:22
• If $C$ were compact, then any infinite subset would have to have at least one accumulation point. (This is true whenever we are working with compact sets in metric spaces.) – Peter Morfe Jan 26 at 0:46
• Woww I got your point. Very useful. What if I defined the space as $$\boldsymbol{C^m} := \left\{\boldsymbol{p} = (p_1, p_2, p_3, \dots) \in \mathbb{R}^{\infty}\big/ \forall ~k, p_k \in [0, 1]; ~ p_k \leq \frac{m}{k}; \text{ and } \sum_{k = 1}^{\infty} p_{k} \leq 1\right\}?$$ I do not want to change my post because your answer is useful. Can you prove in another answer if the statements are true for some $m > 0$ with this new space? – Ari.stat Jan 26 at 13:52
• @Ari.stat, I edited the answer to address the question in your last comment. – Peter Morfe Jan 30 at 19:36