# Banach space that is not a Banach lattice

There is a well-known criterium that distinguishes Banach spaces into the following two classes: those Banach spaces that can be made into a Hilbert space $(X, \langle .,. \rangle)$ and those that cannot. A norm $\lVert . \rVert$ is induced by an inner product $\langle ., . \rangle$ iff $\lVert \cdot \rVert$ satisfies the parallelogram identity $\lVert x + y \rVert^2 + \lVert x - y \rVert^2 = 2(\lVert x \rVert^2 + \lVert y \rVert^2)$ in which case one can set $\langle x, y \rangle := \frac{1}{4}(\lVert x + y\rVert^2 - \lVert x - y \rVert^2)$.

Is there also a criterium that similarly distinguishes Banach spaces from Banach lattices? Let us split this question into two questions:

1. Given a Banach space $(X, \lVert . \rVert)$, is there a partial order $\leq$ on $X$ such that $(X, \leq, \lVert . \rVert)$ is a Banach lattice meaning that (i) $(X, \leq)$ is a Riesz space and (ii) $\lVert . \rVert$ is a Riesz norm?

2. If question 1 has a negative answer, under what additional (possibly sufficient and necessary) conditions on the Banach space $(X, \lVert . \rVert$) does question 1 has a positive answer?

Note that we start here with a Banach space structure and search for a compatible lattice structure. In contrast, if we start with a Riesz space structure $(X, \leq)$, then it is known (Fremlin, Measure Theory, 355Xb) that there is at most one norm $\lVert . \rVert$ (up to equivalence) that turns $(X, \leq, \lVert . \rVert)$ into a Banach lattice.

Added: Maybe this is also helpful to make an abstract Riesz space more concrete: any Riesz space $(X, \leq)$ can be represented as a subspace of a quotient of some $\mathbb{R}^X$ (Fremlin, Measure Theory, 352L). In more detail, there exists a set $Y$, a solid linear subspace $V \subseteq \mathbb{R}^Y$ and an injective Riesz homomorphism $\varphi : X \to \mathbb{R}^Y / V$.

• One of the necessary conditions is the l.u.st. property – Norbert Sep 9 '16 at 19:45
• @Norbert Thank you. Here is some more information with further references on the local unconditional structure property (l.u.st. property and the weaker GL-l.u.st. property (Gordon-Lewis)): "Handbook of the Geometry of Banach Spaces", pp. 59. Every Banach lattice has l.u.st. Examples for Banach spaces that do not have (GL-)l.u.st. are compact bounded operators on $l^2$, the disc algebra and James' space. – yada Sep 10 '16 at 6:32

## 1 Answer

This is a classical theorem of James.

Let $X$ be a Banach lattice. Then either $X$ is reflexive or it contains an isomorphic copy of $c_0$ or $\ell_1$.

Now, there are non-reflexive spaces that contain neither $c_0$ nor $\ell_1$ (for example, the James space) so they cannot be turned into Banach lattices.