Timeline for A "uniform continuity" type condition on a Hammerstein integral equation
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Sep 9, 2021 at 21:41 | comment | added | Martin Väth | Slightly different, but easier to make explicit: For every $t\in[0,1]$ put $K(t,0)=0$, and if $s$ belongs to $I_n=(1/(n+1),1/n]$ put $K(t,s)=\max\{n^2(n+1)(1-\lvert nt-1\rvert),0\}$. Then on the one hand $\int_{1/(n+1)}^{1/n}K(1/n,s)ds=\int_{I_n}n^2(n+1)ds=n\to\infty$. On the other hand, for fixed $t$ the condition $K(t,s)>0$ holds only if $s\in I_n$ with $\lvert nt-1\rvert<1$. This holds only for finitely many $n$ so that $K(t,\cdot)$ is a simple function. | |
| Sep 8, 2021 at 18:39 | comment | added | Motaka | Vath: Can you explicit your "counterexample" more, I want to check why the two hypotheses fail. Thanks! | |
| Sep 8, 2021 at 18:34 | vote | accept | Motaka | ||
| Aug 21, 2021 at 15:46 | comment | added | Martin Väth | As I mentioned, a sufficient condition is the equi-integrability of the family $\{K(t,\cdot):t\in I\}$, that is $\sup_{t\in I}\lVert\chi_{E_n}(\cdot)K(t,\cdot)\rVert_{L_1(I)}\to0$ whenever $E_1\supseteq E_2\supseteq\ldots$ are measurable with $\bigcap_nE_n=\emptyset$. An integrable majorant $h$ (that is, $\lvert K(t,s)\rvert\le h(s)$ for almost all $s$), is sufficient for this (Lebesgue's dominated convergence theorem) but not necessary. | |
| Aug 21, 2021 at 13:35 | comment | added | Motaka | Great, thank you. I wasn't sure that those conditions are sufficient. May you suggest any improvements? (I have this idea of letting $K(t,s)\leq h(s), \;\forall t\in $I$$ with $h$ is a $L^1(I)$ function) | |
| Aug 21, 2021 at 10:55 | history | answered | Martin Väth | CC BY-SA 4.0 |