Timeline for Null sets in PDE
Current License: CC BY-SA 3.0
3 events
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| Oct 2, 2013 at 11:00 | comment | added | Piero D'Ancona | I think you are right with your questions, in the sense that my remark does not clarify these points. It only suggests that one can drastically restrict the test functions used and focus on the singularities of $u$ itself. E.g., if $V$ is separable one might use as test functions step functions taking values in a dense countable set | |
| Sep 29, 2013 at 16:16 | comment | added | aere | Do you mean the displayed equation in the OP by "the identity" that $v$ satisfies? If so then the set of test functions is the whole of $L^2(0,T;V)$ by definition, right? Presumably the sufficiency of testing with $C(0,T;V)$ comes from some density result. Also couldn't it be the case that even though testing with $v \in C(0,T;V)$ is well-defined in that $v(t)$ is well-defined for all $t$, the equality may still not hold on a null set in $[0,T]?$ Sorry for so many questions! | |
| Sep 29, 2013 at 8:06 | history | answered | Piero D'Ancona | CC BY-SA 3.0 |