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Jan 16, 2019 at 16:59 vote accept Y.B.
Jan 15, 2019 at 14:43 answer added Dirk timeline score: 3
Jan 15, 2019 at 14:31 comment added Y.B. @YoavKallus That is also good try, thanks. The point in choosing small $|\gamma|$ is to ensure non negativity, right? This seems an interesting example disjoint from the class indicated by iosif Pinelis. Thanks you.
Jan 15, 2019 at 14:07 comment added Yoav Kallus The form of (3) suggests $a|x| - \langle \gamma, x\rangle$ as a possibility. Doesn't that work (for small enough $|\gamma|$)?
Jan 15, 2019 at 14:05 answer added Iosif Pinelis timeline score: 4
Jan 15, 2019 at 13:20 review Close votes
Jan 17, 2019 at 13:08
Jan 15, 2019 at 13:18 comment added Y.B. @AlexandreEremenko I see your point and thanks for the comment, yet I am looking for a completely "different" example. I have not written it in the OP (to avoid being too long and verbose) but this function $f$ plays the role of integrand of a min problem in CoV, i.e. $\min \int_\Omega f(Du) dx$ among suitable competitors $u$. I am not interested into the case $f(\cdot) = \vert \cdot \vert$ (which is well-known) and, as you can now see, multiplicative constants do not play any role.
Jan 15, 2019 at 13:02 history edited Y.B. CC BY-SA 4.0
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Jan 15, 2019 at 13:00 comment added Alexandre Eremenko Then take $k|x|$, $k>0$.
Jan 15, 2019 at 12:16 history edited Y.B. CC BY-SA 4.0
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Jan 15, 2019 at 12:15 comment added Y.B. @πr8 Sure, I forgot to mention that I am looking for another function than the norm :-) You are perfectly right, thanks!
Jan 15, 2019 at 11:58 comment added πr8 Does $f(x) = |x|$ not work?
Jan 15, 2019 at 11:41 history asked Y.B. CC BY-SA 4.0