The title says it all:
Let $E$ be a Banach lattice, which is isomorphic to a Hilbert space (as normed spaces). Is there an equivalent Hilbert norm on $E$, which still makes it a Banach lattice with respect to the original order structure on $E$?
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Yes. This is a consequence of the fact that semi normalized unconditionally basic sequences in a Hilbert space are Riesz bases. For a theorem quoting proof of what you want, note that that the lattice structure on $E$ is order complete by reflexivity, and hence you have a family $E_a$ of finite dimensional sub lattices of $E$ that are directed by inclusion and whose union is dense. On each $E_a$ put a Hilbert lattice (with respect to the lattice structure on $E_a$ that is inherited from $E$) norm $\|\cdot\|_a$ that dominates the original norm and is no more than $C$ times the original norm. $E$ isomorphically embeds as a sublattice into an ultraproduct of $(E_a,\|\cdot\|_a)$, which is a Hilbert lattice.
answered Dec 16, 2020 at 19:04
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$\begingroup$ If you do not like to take an ultraproduct of Banach spaces, you can instead extend each $\|\cdot\|_a$ to $E$ by making it zero off $E_a$ and pass to an ultralimit of these functions. This limit will be a lattice preHilbert norm on the union of the $E_a$, which is dense in $E$. $\endgroup$ Commented Dec 17, 2020 at 17:17
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$\begingroup$ Sorry, I have to look up the ultra-theory, i'll probably do it over the weekend. $\endgroup$– erzCommented Dec 17, 2020 at 22:53
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$\begingroup$ If $E$ is separable then $(E_a)$ can be an increasing sequence of finite dimensional sub lattices and you can just take a subsequence of $\|\cdot\|_a$ (extended to be zero off E_a) that point wise converges on the union of the $E_a$. $\endgroup$ Commented Dec 18, 2020 at 14:18
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$\begingroup$ I'm afraid I don't even understand the part where you select $C$. I guess, if you start with a discrete lattice, then yes, your argument about the Riesz base in fact produces a lattice isomorphism into $l^2$. But what if it's not discrete? Also, once you choose these norms, they have to increase, or how do we take a pointwise limit? $\endgroup$– erzCommented Dec 21, 2020 at 11:01
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$\begingroup$ The finite dimensional sub lattices are discrete. The normalized atoms are all C equivalent to an orthonormal basis since they are 1-unconditional by your hypothesis. You take the limit of the norms along an ultrafilter (or universal subnet, depending on your preference). $\endgroup$ Commented Dec 21, 2020 at 19:12