Timeline for $\|1_{\{f>M\}}f\|_{L^\infty_t L^p}\rightarrow 0$ as $M\rightarrow \infty$ for $f\in L^\infty([0,T],L^p)$?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 17, 2021 at 5:24 | comment | added | Lev Bahn | Thank you! That is really helpful | |
| Feb 17, 2021 at 5:06 | comment | added | Willie Wong | The idea is very similar to the basic analysis statement that a continuous function on a compact is uniformly continuous (and hence bounded), but a continuous function on a open can be unbounded. For example, if your underlying space is $(0,T)$ and not $[0,T]$, then the $f(t,x)$ I defined in the comments is actually in $C^0((0,T); L^p)$, but cannot be extended to $C^0([0,T];L^p)$. But that's as far as the intuitions go. | |
| Feb 17, 2021 at 4:54 | vote | accept | Lev Bahn | ||
| Feb 17, 2021 at 4:54 | comment | added | Lev Bahn | You are right! Thank you so much! Do you know by any chance the general reason why that limit hold for continuous function but not $L^\infty$ function? | |
| Feb 17, 2021 at 4:46 | comment | added | Willie Wong | The form of $\chi_s$ is chosen so that $\|\chi_s\|_{L^p(\mathbb{R}^N)} = \|\chi\|_{L^p(\mathbb{R}^N)}$. You probably did your change of variables wrong. | |
| Feb 17, 2021 at 4:44 | comment | added | Lev Bahn | Thank you for the answer! I have question on the last equality. I have computed that $$\|f\|_{L^\infty([0,T],L^p)}=ess\sup_{t\in [0,T]}\left| t^{-\frac{N}{p}}\|\chi\|_p \right|$$ The last value seems depends on $p$ and $N$. I am not sure if I computed rightly. Could you check it out for me? | |
| Feb 17, 2021 at 4:18 | comment | added | Willie Wong | Actually, come to think of it, the Dyadic decomposition is a bit of a red herring. You can just define $f(t,x) = \chi_{t^{-1}}(x)$ for $t \neq 0$ and $f(0,x) = 0$ and get the same effect. | |
| Feb 17, 2021 at 4:16 | history | answered | Willie Wong | CC BY-SA 4.0 |