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suggest using some other $L$ in place of $C([0,1])$
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Fred Dashiell
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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.

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 $|h| \notin V$.

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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Fred Dashiell
  • 1.7k
  • 9
  • 22

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

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 $|h| \notin V$.