Timeline for Counterexample of non-negative sequence weakly converging in $\mathscr{M}^1$ but not $L^1$
Current License: CC BY-SA 3.0
8 events
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| May 27, 2013 at 6:27 | comment | added | Ayman Moussa | Thanks Bill, I eventually understood my mistake and erased the corresponding part of my post. | |
| May 24, 2013 at 16:20 | comment | added | Bill Johnson | For $x$ in $D$ let $N(x)$ be the set of $n$ s.t. $f_n(x) < f(x) + 1$. Since $D$ is infinite, $\cap_{x \in D} N(x)$ can be empty. | |
| May 24, 2013 at 12:48 | comment | added | Ayman Moussa |
Sorry, my last comment is about your previous comment (about the intervals) ! About the inequality in Step 3, I am also missing something : for each $x\in D$ there is infinitely many integer $n$ such as the inequality holds, so, for each $x$, I may extract a subsequence so that the inequality holds for each $n$, and since $D$ is countable, I may do this extraction diagonally. At the end of the day, the inequality holds for each $n$, more precisely it holds $f_{\sigma(n)}(x) < f(x)+1$, where $\sigma$ is the mentionned diagonal extraction, no ?
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| May 24, 2013 at 12:43 | comment | added | Ayman Moussa |
But if I do this, I feel that I am losing the approximation of the Dirac mass in each $x\in F_n$, at least, it is not clear for me !
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| May 24, 2013 at 12:24 | comment | added | Bill Johnson | Well, one mistake in your proof is that the final inequality in Step 3 only holds for $n$ sufficiently large, where "sufficiently large" depends on $x$. | |
| May 24, 2013 at 12:11 | comment | added | Bill Johnson | The intervals do intersect, but the functions are a small perturbation of a disjoint sequence of functions (multiply $f_n$ by the characteristic function of the complement of the union of the supports of $f_m$ for $m>n$). | |
| May 24, 2013 at 11:48 | comment | added | Ayman Moussa |
Okay, I do understand the idea but I don't see how you manage to insure that the intervalls $I_{k,n}$ do not overlap. Also, this seems to be in contradiction with the proof that I have written above (which is supposed to work in the case where the limit is continuous - here the limit is $\mathbf{1}_{[0,1]}$), but there is probably a mistake in my reasoning !
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| May 24, 2013 at 4:18 | history | answered | Bill Johnson | CC BY-SA 3.0 |