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Apr 13, 2017 at 12:58 history edited CommunityBot
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Oct 9, 2014 at 21:34 comment added Christian Remling @PhoemueX: You are right, I was sloppy (I mindlessly used (1) for the difference). Your argument should replace this part.
Oct 9, 2014 at 21:02 comment added PhoemueX Thanks for the answer. One technical remark: I think that it does not follow (easily) that $f_t$ is Cauchy in $L^p$; for showing that using (1), we would need to know $f_t - f_s \geq 0$ (or $\leq 0$) for $t \geq s$, which I do not see. But somewhat differently, we can conclude using (1) that $(f_t)_t$ is bounded in $L^p$, so that (by reflexivity) there is a subsequence $f_{t_n}$ with $t_n \to 0$ converging weakly in $L^p$. Using that $\mu$ can in fact be chosen to have compact support, one can then derive that the weak limit is a density of $\mu$, which yields a contradiction.
Oct 9, 2014 at 20:59 vote accept PhoemueX
Oct 8, 2014 at 0:19 history edited Christian Remling CC BY-SA 3.0
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Oct 7, 2014 at 23:10 history answered Christian Remling CC BY-SA 3.0