Timeline for Operators from $L^{\infty}$ to $L^{\infty}$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Jul 5, 2013 at 20:55 | history | edited | Danqing | CC BY-SA 3.0 |
added 33 characters in body
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| Jul 5, 2013 at 17:34 | answer | added | Nik Weaver | timeline score: 1 | |
| Jul 5, 2013 at 13:35 | comment | added | Danqing | We cannot say $\int|K(x,y)|dy\le\|T\|$, how can we take the $\sup$ to get the inequality? You have to notice that if you want to take $\sup_x$ for $(Tf_x)(x)$, then $f_x$ should be a function independent of $x$, which is the definition of the operator of $T$. | |
| Jul 5, 2013 at 6:12 | comment | added | Pietro Majer | It is enough: as I said, take the supremum over all x. | |
| Jul 4, 2013 at 20:14 | answer | added | Bazin | timeline score: 1 | |
| Jul 4, 2013 at 19:22 | comment | added | Danqing | Thank you. But I don't think the inequality is true, since if we fix $x$, then $\int K(\cdot,y)f_x(y)dy$ is really an $L^{\infty}$ function whose norm is bounded by $\|T\|$, but it's not enough to show that the inequality is valid for some specific point. | |
| Jul 4, 2013 at 19:07 | comment | added | Pietro Majer | For $f_x(y):=\operatorname{sgn}K(x,y)$ we have $(Tf_x)(x)=\int |K(x,y)|dy\le \|T\|_{\infty,\infty}$; taking $\sup_x$ gives the inequality. | |
| Jul 4, 2013 at 16:45 | history | asked | Danqing | CC BY-SA 3.0 |