Timeline for Is there a meaningful interpretation of an $L^i$-space?
Current License: CC BY-SA 4.0
13 events
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| Jun 25, 2023 at 1:50 | comment | added | Daniel Asimov | In my opinion it is not at all a good idea to redefine standard notation, even temporarily. Surely there are enough letters in the Roman and Greek alphabets that you could avoid redefining L^p. | |
| Jun 16, 2023 at 23:43 | vote | accept | TheSimpliFire | ||
| Jun 16, 2023 at 23:28 | comment | added | TheSimpliFire | I think there's just a simple notational misunderstanding. I interpreted $\Im(a+bi)=b$ instead of $\Im(a+bi)=bi$ which seems to be what you have written. I've deleted my comment, and it makes sense to me now. Thanks for the extensive answer. | |
| Jun 16, 2023 at 23:23 | history | edited | Dmitri Pavlov | CC BY-SA 4.0 |
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| Jun 16, 2023 at 23:21 | comment | added | Dmitri Pavlov | @openletter.mousetail.nl: $\Im p$ cannot be negative, since it is purely imaginary. Raising a positive real number to a purely imaginary power produces a complex number with absolute value 1. | |
| Jun 16, 2023 at 21:39 | comment | added | Dmitri Pavlov | @FShrike: I added details for the equivalence of two constructions. | |
| Jun 16, 2023 at 21:37 | history | edited | Dmitri Pavlov | CC BY-SA 4.0 |
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| Jun 16, 2023 at 21:16 | comment | added | FShrike | Thank you for the clarifications. I like the new definition in terms of a quotient by pairs $(f;\mu)$ where $|f|^{1/\Re p}$ is $\mu$-integrable. Though I dont understand how your original formulation is equivalent to this new definition, I see no reason why that would be true as we still have a problem with functions not being replaceable with ae bounded functions (and it is not true in general that $g\mu$ is still finite and $fg^p$ is still bounded). I still feel the old version is not well defined (this is not critical, I’m a student after all, just trying to see how it could make sense) | |
| Jun 16, 2023 at 20:29 | comment | added | Dmitri Pavlov | @FShrike: Concerning the first comment, I added a paragraph explaining this. I also added the definition of the multiplication map and the L^p-norm. | |
| Jun 16, 2023 at 20:28 | history | edited | Dmitri Pavlov | CC BY-SA 4.0 |
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| Jun 16, 2023 at 20:13 | comment | added | FShrike | Also what algebra structure / norm structure do you impose on this new $\mathsf{L}^p$ space? You refer to the algebraic tensor products but, for pure imaginary $q$ (say) we don't have a traditional $L^q$ space to refer back to. At first I thought you were using the isomorphism $L^p\cong\mathsf{L}^p$ (isomorphism of sets?) to get some norm/algebra structure on $\mathsf{L}^p$ but I can't see what you're doing for $q=i$ (say), a pure imaginary | |
| Jun 16, 2023 at 20:08 | comment | added | FShrike | In the "observe for any faithful finite measure... $f\mapsto(f,\mu)$ is an isomorphism" I don't believe this is well-defined. For $f$ in the traditional $L^p$ or $L^{1/p}$ space needn't be (equivalent to) a bounded measurable function, which you took to be a hypothesis in the construction of this new $\mathsf{L}^p$ space | |
| Jun 16, 2023 at 19:44 | history | answered | Dmitri Pavlov | CC BY-SA 4.0 |