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Oct 13, 2019 at 18:09 vote accept Adam
Oct 11, 2019 at 10:08 vote accept Adam
Oct 11, 2019 at 11:58
Oct 10, 2019 at 15:33 comment added Yuval Peres Excellent. Welcome to math overflow. Here is how one accepts an answer: meta.stackexchange.com/questions/5234/…
Oct 10, 2019 at 12:19 comment added Adam Thank you. I think there is no problem.
Oct 10, 2019 at 11:58 comment added Yuval Peres Since $f$ is continuous, $ f_h$ tends to $ f$ everywhere by fundamental theorem of calculus. That is all you need. $f_h$ is differentiable everywhere. If there is still a confusing step please indicate exactly where.
Oct 10, 2019 at 11:48 comment added Adam I'm a bit confused, because when $h\rightarrow o$ , $f_{h}^{'}$ is differentiable a.e.
Oct 10, 2019 at 11:42 comment added Yuval Peres The proof shows $g$ is weakly decreasing everywhere. There is no a.e. in the conclusion. As $h$ tends to zero, $g_h(t)$ tends to $g(t)$ for all $t$.
Oct 10, 2019 at 10:53 comment added Adam :Thank you for your anwser, but your proof shows that $g$ is weakly decreasing a.e. .Because when $h\rightarrow 0$ then $g(t)$ decrease a.e.
Oct 9, 2019 at 19:58 history answered Yuval Peres CC BY-SA 4.0