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With $C=\lVert a\rVert_{p,\omega}'$ there holds $a_j^*\le j^{-1/p}C$ for every $j$. Hence, $$\lVert a\rVert_{p,\omega}\le \sup_nn^{-1+1/p}\sum_{j=1}^nj^{-1/p}C\le \sup_nn^{-1+1/p}\Bigl(1+\int_1^nx^{-1 …

To make Gerald Edgar's answer complete: There always does exist a product-measurable choice. More precisely, if $f\colon\mathbb{R}^n\to L_2(\mathbb{R}^d)$ is measurable then there exists a product mea …

If $f_0\ne0$ or if $f_0=0$ and the $f_1,\dotsc,f_n$ are not linearly independent, then the answer is trivial: In this case there is another functional, say $f_1$, which is in the span of the remaining …

I had the same problem many years ago (if I remember correctly), and the answer was negative. I think that I had found a counterexample in the monograph of Diestel and Uhl. If $E$ has finite dimension …

Diestel & Uhl can only mean the first interpretation you gave for two reasons: For the second interpretation, the term “off a fixed set of measure zero” makes no sense. Even in case $X=\mathbb R$ (wh …

As Michael already observed: It is simple to construct counterexamples if $B$ has a nontrivial kernel $N(B)$ and $A$ maps $N(B)$ into itself, since on $N(B)$ the size of $c$ plays no role. But even if …

No, this is not possible, even if you define $\text{supp}(f)$ only as an equivalence class (up to a null set). For a counterexample take the characteristic function $f$ of a closed set $E$ of positive …

I do not have a counterexample, but a strong feeling that the conjecture is false, based on the following positive proof. If you require slightly more, namely Lebesgue integrability of $t\mapsto f(t,x …

This does certainly not follow from your other hypotheses, as what you want to conclude is not much weaker than the equi-integrability of $\{K(t,\cdot):t\in I\}$ (sometimes also called absolute contin …

Only now I realize the condition that $\zeta$ is nonnegative. (Was it really there in the first formulation of the question?) With this condition, it is possible to get the required a-priori bound req …

It's many years ago that I read it, but I think that some of the most general interpolation type results for Orlicz spaces were contained in O’Neil, Richard, Integral transforms and tensor products on …

The definition makes no sense due to the mixing up of "relatively" (weakly) compact and (not relatively) compact. I guess that what you mean is: $\psi$ is strongly-weakly proper on closed balls (that …

What is needed is an a-priori $L_\infty$ bound for the solution $v_k$. If you know such an a-priori bound, you can modify $\zeta$ outside of this bound, and you can assume without of generality that $ …

Not really an answer, but some remarks: Even if the spectral radius of $K$ is less than $1$, there are counterexamples that the resolvent need not have the required form: This is related to the fact …

It should be straightforward to verify that $\mathcal A$ consists exactly of the lower semi-continuous radial and decreasing functions (which are non-negative and vanish at $\infty$): You already know …