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.



Banach lattice subspace of $C([0,1])$ not a sublattice
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.


