Timeline for Operators from $L^{\infty}$ to $L^{\infty}$
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 6, 2013 at 1:55 | comment | added | Danqing | Actually I finished the discussion until the case $K$ is a characteristic function of arbitrary measurable set, but I don't see that rest cases follow. | |
| Jul 5, 2013 at 22:10 | comment | added | Nik Weaver | I don't think you read my hint. | |
| Jul 5, 2013 at 19:26 | comment | added | Danqing | Somebody may argue that if we can take $ess\sup$ of $\int K(t,y)f_t(y)dy$ as a function of $t$, but we have to notice that we cannot exclude the case that for each fixed t, $\int K(t,y)f_t(y)dy=\infty$. Thank you for your discussion about the reverse, but there is a more direct argument where we don't need the duality. We need only to notice that $|\int K(x,y)f(y)dy|\le \|f\|_{L^{\infty}}\int |K(x,y)|dy$ for each point $x$. | |
| Jul 5, 2013 at 19:18 | comment | added | Danqing | Thanks for your answer. I read it carefully but I still don't know how to solve my question by your hint, since I think there is a essential difference between the discrete case and continuous case. Let me make it clear and use notations $x$ and $y$ in both cases. If we fix $t$ and define $f_t(y)=\textrm{sgn } K(t,y)$, then $\|\int K(\cdot,y)f_t(y)dy\|\le \|T\|$, in particular, in the discrete case, $\int K(t,y)f_t(y)dy\le\|T\|$, therefore we get the conclusion. But obviously the reasoning doesn't work for the continuous case since $\int K(t,y)f_t(y)dy$ may be any number. | |
| Jul 5, 2013 at 17:34 | history | answered | Nik Weaver | CC BY-SA 3.0 |