**Background**

Consider a real, normed vector space $V$. We'll call a subset $V_+ \subseteq V$ a *cone* if

 - $V$ is closed,
 - $\alpha V_+ \subseteq V_+$ for every $\alpha \geqslant 0$ and
 - $V_+ \cap (-V_+) = \{0\}$.

Cones *induce* a partial order $\leqslant$ in $V$: 

$$x \leqslant y \quad \Longleftrightarrow \quad y - x \in V_+.$$

A cone is said to be *solid* if it has nonempty interior $intV_+ \neq \varnothing$. It is said to be *minihedral* if every finite subset $B$ of $V$ has a supremum (least upper bound). It is said to be *regular* if every order-bounded, monotone sequence converges in norm:

$$x_1 \leqslant x_2 \leqslant \cdots \leqslant x_n \leqslant \cdots \leqslant u \quad \Longrightarrow \quad x_n \longrightarrow x_\infty. \quad \quad \quad \quad (*)$$

I'm interested in cone-induced partial orders in the context of Monotone Dynamical Systems. Properties such as the ones listed above often come up in the discussion of asymptotic behavior.


**Question**

Is it true that a solid, minihedral cone in an infinite-dimensional real normed space cannot be regular?

This is listed as Exercise 6.7 on page 61 in [1]. However I've found statements in exercises in this book to be false---in fact, the very first exercise is false---and there are many other mistakes such as theorems without all hypotheses listed. So I'm skeptical.

**Progress**

(1) The cone $l^\infty_+$ of nonnegative, bounded sequences in $l^\infty$ is solid and minihedral, but not regular. Setting

$$x_n := (1, 1, 1, \ldots, 1, 0, 0, 0, \ldots) \quad \text{($n$ $1$'s)}$$

we have a monotone sequence order-bounded from above by $(1, 1, 1, \ldots)$ which does not converge in norm.

(2) Let $1 \leqslant p < \infty$. The cone $l^p_+$ of nonnegative sequences in $l^p$ is minihedral and regular, but not solid. *Minihedral*: given two sequences in $l^p$, just take the maximum of the two coordinatewise. *Regular*: given an order-bounded monotone sequence as in $(*)$, it converges pointwise by monotonicity and so it converges in norm by dominated convergence---$(x_n)$ is sandwiched between $x_1$ and $u$. *Not solid*: for any sequence $(s_n)$ in $l^p$, $s_n \rightarrow 0$. So we can construct sequences in $l^p$ arbitrarily close to $(s_n)$ having negative terms.

*Remark*: The cones in (1) and (2) above are also normal---meaning that there exists a $C \geqslant 0$ such that $0 \leqslant x \leqslant y$ implies $\|x\| \leqslant C\|y\|$. In fact, $C = 1$.

(3) I've tried to prove it by contradiction, assuming that the cone is solid, minihedral and regular in an infinite-dimensional space. I picked a sequence of unit vectors $(v_n)$ such that

$$dist(v_{n+1}, span \{ v_1,\ldots,v_n \} ) \geqslant 1/2.$$

Because the cone is solid, there exists an $u \in V$ such that

$$\{v_n\} \subseteq B_1(0) \subseteq [-u, u] := \{x \in V\,;\ -u \leqslant x \leqslant u\}.$$

By the minihedrality assumption, a monotone sequence $(w_n)_{n \in \mathbb{N}}$ is well-defined by

$$w_n := \sup\{v_1,\ldots,v_n\}.$$

In particular, $(w_n)$ is order-bounded by $u$, so it converges in norm by regularity. I haven't yet been able to see the contradiction here though.



**References**

[1] Krasnosel'skij, Lifshits and Sobolev; *Positive Linear Systems---The Method of Positive Operators*. Heldermann Verlag Berlin, 1989.

**Edits**

2013-03-27: Added $l^p$ example and remark about normality.