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This means that ({bar{r}_{n}}_{n=1}^{+infty}) does not have equi-absolutely continuous norms in (E^{Psi}_{h} (Omega)).
Mainly, we focus on the problem when the unit ball BH:="{f∈H:‖f‖p⩽1} of a Λ(p -space H has equi-absolutely continuous norms in Lp -space
Conversely, if a set (Asubset E^{varphi}(Omega)) is relatively compact, then all the functions in A have equi-absolutely continuous norms and A is relatively compact with respect to local convergence in measure.
We say that the operator (T: L^{Phi}(Omega) rightarrow L^{Psi}(Omega)) is equi-absolutely continuous if for any bounded set (A subset L^{Phi}(Omega)) all functions of the set (T (A)subset L^{Psi}(Omega)) have equi-absolutely continuous norms.
If the functions in a set (Asubset L^{varphi}(Omega)) all have equi-absolutely continuous norms and A is relatively compact with respect to local convergence in measure, then A is relatively compact in (E^{varphi}(Omega)), the subspace of absolutely continuous functions in (L^{varphi}(Omega)).
Similar(55)
Suppose that X does not have an absolutely continuous norm.
If X has no absolute continuous norm, then (R X,a)=1+a ).
Since (Phinotinoverline{delta}_{2}), we see that (l_{Phi}) has no absolutely continuous norm.
A Köthe sequence space X is said to have an absolutely continuous norm if (X_{a}=X).
In what follows, E will always denote a minimal symmetric quasi-Banach space with order continuous norm.
Our main technical tool is a principally new property of compact narrow operators which works for a domain space without an absolutely continuous norm.
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CEO of Professional Science Editing for Scientists @ prosciediting.com