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Aug 22, 2023 at 13:20 comment added Akira Let $T>0$ and $Y := \mathbb R^d$. By axiom of choice, there is a bijection $\varphi$ from $X:=[0, T]$ onto $\cal J$. We define $g:X \times Y \to \mathbb R$ by $g(x, \cdot) := f_{\varphi (x)}$ for all $x \in X$. Here $f_J$ is as you already defined. The range of $G: X \to L^p_{\text{loc}} (Y), x \mapsto g(x, \cdot)$ is not a.e. separable. So statement S1 (in my other thread) is not true if we can pick $\varphi$ such that $g$ is measurable...
Aug 22, 2023 at 13:08 comment added Iosif Pinelis @Akira : That question seems to be of a different kind, and I don't have an answer to it right now, but I will have it in mind.
Aug 22, 2023 at 8:15 comment added Akira I have a related question here, I would like to ask if you can tailor your counter-example to the statement S1 there. Thank you so much for your consideration!
Aug 10, 2023 at 14:10 vote accept Akira
Aug 10, 2023 at 13:53 history edited Iosif Pinelis CC BY-SA 4.0
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Aug 10, 2023 at 13:41 history edited Iosif Pinelis CC BY-SA 4.0
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Aug 10, 2023 at 13:37 comment added Iosif Pinelis Oops! In my mind, I incorrectly reversed the inequality. Now this should be fixed.
Aug 10, 2023 at 13:32 history edited Iosif Pinelis CC BY-SA 4.0
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Aug 10, 2023 at 12:29 comment added Akira @AymanMoussa Agree! It's clear that there is some $C>0$ such that $\|f\|_{\bar{L}^p} \leq C\|f\|_{L^{\infty}}$. I'm not sure if the reverse inequality holds.
Aug 10, 2023 at 12:26 comment added Ayman Moussa Well $L^1(0,1)$ contains $L^\infty(0,1)$ and the first one is separable whereas it's not the case for the other one. There's something to be checked about the topology, not only set inclusion, but I would also guess that because of the supremum the space $\tilde{L}^p$ is not separable.
Aug 10, 2023 at 12:17 history answered Iosif Pinelis CC BY-SA 4.0