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Let $L$ be a Dedekind-complete Banach lattice. Let $\mathcal{B}$ be the family of nonempty norm-compact subsets of $L$ that are bounded from below. Endow $\mathcal{B}$ with the topology induced by the Hausdorff metric.

Consider the function $\psi\colon \mathcal{B}\to L$ defined by $\psi(A)= \inf{(A)}$ for all $A\in \mathcal{B}$.

I have a number of questions on this, the most important of which is the following:

Q1. What conditions on $L$ guarantee that $\psi$ is a continuous function?

The other questions depend on the answer to Q1, in obvious ways.

Q2. What is an example of a Dedekind-complete Banach lattice for which $\psi$ is not continuous?

Q3 Suppose that $\psi$ is continuous and $E$ is a Banach lattice. Will the same hold for the the space of regular operators $\mathcal{L}^r(E,L)$ (the linear span of positive operators)? How about when $L$ has a strong order dual?

Now suppose that $X\subseteq L$ is a non-empty norm-compact sub-lattice of $L$. Take the restriction of $\psi$ to the space $\mathcal{B}(X)$ of nonempty compact subsets of $X$. My final question is the following:

Q4 What conditions on $X$ guarantee that $\psi$ restricted to $\mathcal{B}(X)$ is continuous?

This latter question Q4 is the same as asking, What conditions guarantee that $X$ as a topological lattice has a neighborhood base of $inf$-sub-semilattices?

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I guess finite dimensionality is needed for Q2. For example, in $\ell_2$ consider the sets $X_n = \{n^{-1/2} e_j : 1 \le j \le n \}$. Then in the Hausdorff metric $X_n \to \{0\}$ but the infimum of each $X_n$ has norm one.

It looks like you need to restrict attention to compact sets that are in a common order interval.

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  • $\begingroup$ Thanks Bill. I figured so much riding to uni. But is it always the case that if $X$ is a compact sublattice, the restriction of $\psi$ to $X$ is continuous. So Q4 is of interest to me. $\endgroup$
    – Rabee Tourky
    Commented Jun 18, 2013 at 6:24
  • $\begingroup$ In $L_2(0,1)$ take '$X_n = \{1_{A_i} : 1 \le i \le 2^n \}$', where `$A_i = [(i-1)/2^n, i/2^n]$'. This gives an order bounded counterexample to Q2 which can be generalized to any order continuous non discrete Banach lattice. But the lattice generated by the $X_n$ is not totally bounded. $\endgroup$
    – Bill Johnson
    Commented Jun 18, 2013 at 6:47
  • $\begingroup$ The only infinite dimensional example that I can think of where $X$ is a sublattice that is compact and where $\psi$ restricted to compact subsets of $X$ is continuous is monotone functions in $L_1(0,1)$. $\endgroup$
    – Rabee Tourky
    Commented Jun 18, 2013 at 6:51
  • $\begingroup$ In the previous comment, I mean order continuous infinite dimensional space. $\endgroup$
    – Rabee Tourky
    Commented Jun 18, 2013 at 8:15

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