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This is probably easy, but I did not see it in standard texts. Describe a closed subspace $V$ of $C([0,1])$ such that $V$ is a Banach lattice (in the pointwise ordering), but $V$ is not a sublattice of $C([0,1])$. Note that, by virtue of the formula $2(f\vee g) = f+g+|f-g|$, such a $V$ must contain an element $h$ such that $|h| \notin V$.

Maybe there is a simple example using some other Banach lattice $L$ in place of $C([0,1])$, and exhibiting a closed subspace $V$ of $L$ which is a Banach lattice in the ordering inherited from $L$, but not a sublattice.

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up vote 5 down vote accepted

Yeah, it's easy, take $V$ to be the set of linear functions $ax + b$. Any $f$ and $g$ in $V$ have a least upper bound which is the line from $\max(f(0),g(0))$ to $\max(f(1),g(1))$. The sup norm of $f \vee -f$ equals the sup norm of $f$, so it's a Banach lattice.

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Thanks. Nice example. –  Fred Dashiell Jan 16 '13 at 14:29
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Aside from examples there is a nice and very useful characterizations due to Mayajima http://hmj2.math.sci.hokudai.ac.jp/359/ of lattice subspaces that need not be sublattices. It is of considerable interest in theoretical economics because it allows hedging in finance with incomplete markets. They are described in my late friends' Yuri Abramovich and Roko Aliprantis book: Section 5.4 of An invitation to operator theory.

A vector subspace $X$ of a vector lattice $E$ is a vector lattice-subspace if $X$ equipped with the order induced by $E$ is a vector lattice. The space is a vector sublattice if $x\vee y$ and $x\wedge y$ are in $X$ for every $x,y\in X$.

A sublattice is a lattice-subspace but the converse need not be true. Every two dimensional subspace $X\subseteq E$ such that $X_+=X\cap E_+$ is also two dimensional is a lattice-subspace. To see this notice that in the two dimensional case $(X_+ + x) \cap (X_+ + y)$ is always a translation of the cone $X_+ + z$ and $z$ is the supremum of $\{x,y\}$. Furthermore, in $E=L_p$ (and $C([0,1])$) if such a space $X$ contains the constant function 1 and a strictly monotone function it is not a sublattice. To see this notice that the smallest lattice $Y$ containing $X$ will not be two dimensional because in $L_p$ the closure of $Y$ is the $p$-integrable functions that are measurable in the $\sigma$-algebra generated by the functions in $X$.

Theorem: Let $X$ be a subspace of a vector lattice $E$ and let $Y$ be the smallest vector sublattice of $E$ containing $X$. The vector space $X$ is a lattice-subspace if and only if there is a positive linear projection of $Y$ onto $X$.

This projection is called the Mayajima projection for obvious reasons.

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