Consider a real Banach$^1$ 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.


Is it true that a solid, minihedral cone in an infinite-dimensional, real Banach 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.


(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, $C = 1$. But that's not the problem since regularity actually implies normality in Banach spaces [1, Lemma 5.1 and Theorem 5.1].

(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.

(4) I might have found a solution using a representation theorem due to Kakutani.

Theorem (Kakutani) If $V$ is a real Banach space, partially ordered by a solid, normal, minihedral cone $V_+$, then there exists a compact, Hausdorff space $K$ and a linear homeomorphism $\Phi\colon V \rightarrow C(K)$ such that $\Phi(V_+) = C_+(K)$, where $C_+(K)$ is the cone of nonnegative, continuous functions in $C(K)$.

Proof See [1, Theorem 6.6].

Since regularity implies normality, a solid, minihedral, regular cone in a real Banach space satisfies the hypotheses of the theorem. Since $\Phi$ preserves solidity, minihedrality and regularity, it's enough to prove the result for $C(K)$. Now $C(K)$ is finite-dimensional if, and only if $K$ is finite. I showed that if $K$ is infinite, then we can essentially find a copy of $l^\infty$ in $C(K)$, contradicting the hypothesis that $C(K)$ is regular (Example (1) above). Thus if $C(K)$ is regular, then $K$ must be finite.

I'm still checking all the details though, it's quite an elaborate construction, and depends on this nontrivial representation result of Kakutani. So I'm not too happy about it. Since it's an exercise in the book I thought there must have been a simpler solution. Plus, the exercise is given before Kakutani's theorem. So I imagine the authors thought it could be solved without it?


  1. Completeness is assumed throughout [1], seems to be needed even for basic properties such as characterizations of normality, and is present in all applications I've come across so far. So I'm OK with assuming that $V$ is Banach.


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


2013-03-27: Added $l^p$ example and remark about normality.
2013-04-04: Added completeness to the hypotheses in $V$, observation that regularity implies normality, and the representation theorem of Kakutani.

  • $\begingroup$ The question is asked very well. You give the relevant definitions clearly and concisely, you explain why you are interested, and you outline what you've already tried. I agree that (3) is the natural thing to try, but I also don't see the contradiction there. $\endgroup$ – Nik Weaver Apr 5 '13 at 6:59

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