Timeline for Is $A$ is small on bounded functions, is there a large subdomain on which $A$ is small?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Aug 5, 2020 at 6:32 | comment | added | H A Helfgott | Yes, I stated the question wrongly. Just fixed it. Thanks! | |
| Aug 5, 2020 at 6:32 | history | edited | H A Helfgott | CC BY-SA 4.0 |
added 13 characters in body
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| Aug 5, 2020 at 6:17 | comment | added | Jochen Wengenroth | I must misunderstand something. For $t=K/|f|_\infty$ one has $|tf|_\infty\le K$ and hence $$\langle f, A(f)\rangle= t^{-2} \langle tf, A(tf)\rangle \le t^{-2}\alpha |tf|_2=\alpha|f|_2.$$ | |
| Aug 4, 2020 at 18:03 | comment | added | H A Helfgott | I'm not sure how homogeneity would be enough - $A$ could have eigenfunctions with enormous peaks. Sure, define $A(f|_X)$ that way if you wish -- it doesn't matter, since you are going to take the inner product with $f|_X$. | |
| Aug 4, 2020 at 17:23 | comment | added | Jochen Wengenroth | Why doesn't this hold just by homoeneity for $Y=\emptyset $? And how do you define $A (f|_X) $? Do you put the restriction as $0$ outside $X $? | |
| Aug 4, 2020 at 11:40 | history | edited | H A Helfgott | CC BY-SA 4.0 |
edited title
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| Aug 4, 2020 at 9:28 | history | asked | H A Helfgott | CC BY-SA 4.0 |