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May 19, 2020 at 6:25 history edited leo monsaingeon CC BY-SA 4.0
removed the [lebesgue-measure] tag
May 19, 2020 at 4:37 answer added Yuval Peres timeline score: 4
May 19, 2020 at 4:22 comment added Yuval Peres The argument you suggest seems to assume a bound on $\|f_n\|_1$ rather than $\|f_n\|_\infty$.
May 19, 2020 at 4:01 comment added Parc John We have $\sup_n\|f_n\|_1<\infty$. Then, there exists $n_0\in\mathbb{N}$, such that: for all $n\geq 1$ we have $$|f_n|\leq n_0~~ a.e.~$$Then, for all $n\geq n_0$: $$F_n(f_n)=f_n$ hence, we have the desired result.
May 19, 2020 at 1:36 review First posts
May 19, 2020 at 6:26
May 19, 2020 at 1:35 history edited Parc John CC BY-SA 4.0
added 6 characters in body
May 19, 2020 at 1:27 history asked Parc John CC BY-SA 4.0