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I guess that if $v$ is $P$-integrable then the answer is positive, and actually the set is compact. Indeed, what you are looking for in this case is the compactness of the Aumann integral of the measu …

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 …

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 …

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 …

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 …

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 …

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 …

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 …

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 …

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 …

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 …

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 $ …

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 …

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 …

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 …