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Dec 2, 2018 at 21:33 answer added Iosif Pinelis timeline score: 1
Dec 2, 2018 at 19:37 comment added user64494 Thank you. Don't you miss condition (1.4) from the manuscript in your question?
Dec 2, 2018 at 19:09 comment added user2714795 No h is finite. I am pasting a link to the place from where the basic problem has arisen. statistics.stanford.edu/sites/default/files/EFS%20NSF%20159.pdf. On pg 9 of this the authors are trying to prove that $ \frac{1}{h^2}\int_0^h (f -f_h)^2 - \frac{1}{12} \int_{0}^{h}f^{'}^{2}$ can be small. The $1/12$ fraction only comes if we are able to get some kind of inequality as above. Since the author has not explicity mentioned hence I was trying to derive it. The $1/12 $ fraction appears to be coming from the min value that the expression can have.
Dec 2, 2018 at 18:58 comment added user64494 The question is unclearly formulated. In particular, could $h$ be $\infty$?
Dec 2, 2018 at 17:33 answer added Alexandre Eremenko timeline score: 0
Dec 2, 2018 at 15:06 history edited Joe Silverman CC BY-SA 4.0
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Dec 2, 2018 at 14:45 history edited user2714795 CC BY-SA 4.0
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Dec 2, 2018 at 14:27 history edited Pietro Majer CC BY-SA 4.0
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Dec 2, 2018 at 14:22 comment added fedja You certainly are in trouble if $f$ is concentrated near $0$, say $f=1$ on $[0,\delta]$ and $0$ on $(\delta,h]$ with $\delta\ll h$.
Dec 2, 2018 at 14:07 comment added user2714795 right, you are correct here
Dec 2, 2018 at 14:06 history edited user2714795 CC BY-SA 4.0
edited body
Dec 2, 2018 at 14:05 comment added Fedor Petrov I guess you mean $0\le u,v\le h$
Dec 2, 2018 at 14:05 review First posts
Dec 2, 2018 at 14:32
Dec 2, 2018 at 14:00 history asked user2714795 CC BY-SA 4.0