Timeline for When are Weighted $\mathcal{L}^p$-Spaces Topologically Isomorphic?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Jan 7, 2018 at 17:55 | vote | accept | JMJ | ||
| Jan 7, 2018 at 3:42 | answer | added | R W | timeline score: 2 | |
| Jan 7, 2018 at 2:50 | comment | added | JMJ | @RW I believe that's a fair restatement. | |
| Jan 7, 2018 at 2:49 | comment | added | R W | OK - is the following sufficient for your needs? If $X$ is not very big (say, separable, metriziable, complete) - then the $L^p$ spaces associated to all purely non-atomic $\sigma$-finite measures on $X$ are isometrically isomorphic. | |
| Jan 7, 2018 at 2:32 | comment | added | JMJ | @RW I would like an isomorphism which is also a homeomorphism. This way both the algebraic and topological characters of the two spaces will agree. By Banach space isomorphism, I wanted to indicate nothing more than a linear bijection between the spaces, regarded as Banach spaces. | |
| Jan 7, 2018 at 2:28 | comment | added | R W | Yes. Another question: are you asking about topological isomorphism (i.e., isomorphism of these $L^p$ spaces as topological spaces) or about Banach space isomorphism? | |
| Jan 7, 2018 at 2:13 | history | edited | JMJ | CC BY-SA 3.0 |
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| Jan 7, 2018 at 2:12 | comment | added | JMJ | @RW OK. I believe changing the condition to measurable functions will solve this issue. Do you agree? What I was trying to communicate is the $f$ in this space are in some sense "nice enough". | |
| Jan 7, 2018 at 2:08 | comment | added | R W | Yes - if you only consider continuous functions your $L^p$ spaces won't be complete (unless $X$ is pretty exotic). | |
| Jan 7, 2018 at 2:06 | comment | added | JMJ | @RW I may have a terminology error. Please help me understand: why would continuous functions not form a Banach space? Is it a completeness issue? | |
| Jan 7, 2018 at 2:00 | comment | added | R W | How can you claim that these spaces are Banach if the functions $f$ are assumed continuous? | |
| Jan 7, 2018 at 0:53 | comment | added | JMJ | @LSpice okay. Then yes, I agree this is a trivial case. Non-negativity is the key condition. We may not assume $W(x) \neq 0$. | |
| Jan 6, 2018 at 21:36 | history | edited | JMJ | CC BY-SA 3.0 |
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| Jan 6, 2018 at 21:32 | comment | added | LSpice | Note that @BillJohnson specified that he was talking about strictly positive weight functions. | |
| Jan 6, 2018 at 21:18 | comment | added | JMJ | @BillJohnson Yes I meant non-negative. I am very much not an expert in this area, so the basic result eluded me. For the record, what would the operator be? Using the multiplier $W_1/W_2$ (or its inverse) would lead to problems in cases where $W(x) = 0$ for some $x$. | |
| Jan 6, 2018 at 21:15 | comment | added | Bill Johnson | By "positive definite" do you mean non negative? If the weight functions are both strictly positive and finite, then the two weighted $L_p$ spaces are isometrically isomorphic via a multiplication operator. This is easy and basic, so maybe you mean something else? | |
| Jan 6, 2018 at 21:04 | history | edited | JMJ | CC BY-SA 3.0 |
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| Jan 6, 2018 at 20:58 | history | asked | JMJ | CC BY-SA 3.0 |