On Compact Operators Between Lattice Normed Spaces

Statistically order compact operators on Riesz spaces

“…Not every statistically order compact operator is order bounded. An example illustrating this fact can be found in [10,Exam.6], where the given operator is sequentially order compact. Thus, by applying Remark 2.2, it is also statistically order compact.…”

Multiplicative order compact operators between vector lattices and $l$-algebras

“…An omo-compact operator need not be sequentially omo-compact. To see this, we consider [11,Ex.7] for the next example.…”

“…Consider a sequence f n in {−1, 1} X . Then f n is order bounded in E and f n has no o-convergent subsequence [11,Ex.7(2)]. Thus, every subsequence does not mo-converge because the f -algebra E has a unit element.…”

“…Indeed, take an order bounded sequence (f n ) in E. Then there exist some scalars λ > 0 such that |f n | ≤ λ for all n. On the other hand, we can write f n = β n + g n with reel numbers β n and functions g n in F . It follows from [11,Ex.6] that (g n k ) is order convergent in F . Then it is clear from [18, Thm.VIII.2.3] that (g n k ) is also m r o-convergent in F .…”