Timeline for Finding $W^{1,\infty}$ solutions to an integral equation by fixed point
Current License: CC BY-SA 4.0
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Oct 26, 2022 at 17:31 | history | edited | username | CC BY-SA 4.0 |
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Oct 26, 2022 at 16:11 | history | edited | username | CC BY-SA 4.0 |
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Oct 26, 2022 at 16:09 | comment | added | username | @GiorgioMetafune You are right, I have expanded that part (to explain what is enough). | |
Oct 26, 2022 at 9:15 | history | edited | username | CC BY-SA 4.0 |
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Oct 26, 2022 at 4:44 | history | edited | username | CC BY-SA 4.0 |
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Oct 25, 2022 at 21:41 | comment | added | Giorgio Metafune | I do not see why $F(C(D))$ is contained in $W^{1,\infty}$ | |
Oct 25, 2022 at 18:59 | history | edited | username | CC BY-SA 4.0 |
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Oct 25, 2022 at 18:58 | comment | added | username | Q1 : yes, just write $u_{n+1} = Fu_n + g$ and use banach fixed point to show the sequence $u_n$ converges for example. Q2. Because that's how you show contractivity (for the Banach Fixed point). Q3. I mean a continuous solution exists and is unique. Now, is this unique solution also $W^{1,\infty}$ ? We also have $u=Fu+g$. so So $u$ (left-hand-side) is equal to $Fu + g$ (right hand side) $Fu$ is in the range of $F$. | |
Oct 24, 2022 at 22:40 | comment | added | user139844 | Thanks! I have a few questions: (1) Is it possible to replace the construction $u = \sum_{k} F^kg$ with an application of Banach fixed point? (2) Why do you compute the operator norm $\sup_{u \neq 0} \| Fu\|_\infty/\|u\|_\infty$? (3) What do you mean when you say that $W^{1,\infty}$ follows from the range of $F$? | |
Oct 24, 2022 at 21:21 | history | edited | username | CC BY-SA 4.0 |
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Oct 24, 2022 at 21:14 | history | edited | username | CC BY-SA 4.0 |
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Oct 24, 2022 at 21:06 | history | edited | username | CC BY-SA 4.0 |
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Oct 24, 2022 at 21:01 | history | edited | username | CC BY-SA 4.0 |
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Oct 24, 2022 at 20:52 | history | answered | username | CC BY-SA 4.0 |