We shall call a subset $V_+ \subseteq V$ of a Banach space $V$ a cone if

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

Cones induce a partial order on the underlying space: $$ x \leqslant y \quad \Longleftrightarrow \quad y - x \in V_+\,. $$

A cone is said to be solid if its interior $\operatorname{int} V_+$ is nonempty. A cone is said to be normal if there exists a constant $K \geqslant 1$ such that $0 \leqslant x \leqslant y$ implies $\|x\| \leqslant K\|y\|$. (Theorem 2.1.1 on pages 27--28 of Guo/Cho/Zhu (2004) gives several equivalent formulations.)

The following is not difficult to show.

Lemma. Let $V$ be a Banach space partially ordered by a solid, normal cone $V_+ \subseteq V$. Suppose that $(x_\alpha)_{\alpha \in A}$ is a net such that

  • $x_\alpha \longrightarrow x_\infty$ for some $x_\infty \in V$, and
  • $x_\alpha^- := \inf_{\alpha' \geqslant \alpha} x_\alpha$ and $x_\alpha^+ := \sup_{\alpha' \geqslant \alpha} x_\alpha$ are well-defined for each $\alpha \in A$.

Then $$ x_\alpha^- \longrightarrow x_\infty \quad \text{and} \quad x_\alpha^+ \longrightarrow x_\infty\,. $$


To what extent are the hypotheses that the cone be solid and normal necessary?

That the cone be always solid and normal is not necessary, as noted below. However I am still wondering whether normality is ever needed. In other words, is there an example in which the hypotheses of the Lemma are all satisfied, except for the normality of the cone, and for which the conclusion does not hold?


  1. I have an example showing that the conclusion may fail if the cone is not solid. (It's a straightforward construction in $L^1([0,1];[0,1])$.)

  2. The canonical example of a cone which is not normal seems to be the cone $V_+$ of nonnegative functions in $V := C^2([0,1];{\mathbb R})$ with the norm $\|\cdot\|\colon V \rightarrow [0,\infty)$ given by $$ \|f\| := \|f\|_\infty + \|f'\|_\infty\,, \quad f \in V\,. $$ The Lemma seems to still hold in this case.

  3. The cone ${\mathbb R}_{\geqslant 0} \times \{0\}$ is not solid in ${\mathbb R}^2$. But the Lemma still holds here as well.

  • $\begingroup$ Hi @orlandoweber, it's late but did you find something here? I'm working on something similar.. $\endgroup$ – Motaka Jul 18 at 8:33

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