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Feb 8, 2023 at 23:06 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
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May 14, 2022 at 21:22 history edited LSpice CC BY-SA 4.0
Proofreading while this is on the front page; deleted "Thank you"
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Aug 17, 2021 at 9:32 history edited Pietro Majer CC BY-SA 4.0
small typo fixed
Aug 17, 2021 at 9:12 answer added Pietro Majer timeline score: 0
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Mar 11, 2021 at 14:21 answer added Iosif Pinelis timeline score: 1
Mar 11, 2021 at 8:54 history edited gmvh CC BY-SA 4.0
Removed inappropriate "descent" tag (it's used in a very different sense on all other questions so tagged)
Mar 11, 2021 at 8:02 comment added ARedder Yes, you are right, this would be sufficient for my problem. Thanks. I edited the question.
Mar 11, 2021 at 8:01 history edited ARedder CC BY-SA 4.0
added 160 characters in body
Mar 11, 2021 at 0:44 comment added Iosif Pinelis It is unclear what you want to show. Is it the following: $\forall x\in\partial X\ \exists\epsilon>0\ \forall y\in X^c\ \|y-x\|<\epsilon\implies\nabla P(y)^T(x-y)<0$? Here $X:=\mathcal X$. Placing the quantifiers $\forall$ and $\exists$ correctly and unambiguously is usually very important.
Mar 10, 2021 at 13:11 comment added Hannes Alright so a degree of explicitness is needed. I suppose you would need to assume some more on the function $P$ then. For instance, if you require the norm of $\nabla^2 P$ to be bounded on a neighborhood of $\mathcal{X}$, then you get a uniform estimate on the $o(\|x-y\|)$ term and this should lead to a more explicit result.
Mar 10, 2021 at 12:51 comment added ARedder The problem is, the point where I have the descent direction is $z=y+\lambda(x-y)$. However, for my problem, I need that $z$ is a specified point or at least arbitrarily close to some other point $z'$ (we can use continuity), which we can make arbitrarily close to the boundary.
Mar 10, 2021 at 12:35 comment added Hannes Why not the mean value theorem? For every $y \in \mathcal{X}^c$, there is $\lambda \in (0,1)$ such that $P(x) = 0 = P(y) + \nabla P(y + \lambda(x-y))^\top(x-y)$. Hence $x-y$ is a descent direction in the intermediate point $y + \lambda(x-y)$.
Mar 10, 2021 at 12:09 review First posts
Mar 10, 2021 at 12:14
Mar 10, 2021 at 12:05 comment added ARedder I reposted this from math.stackexchange.com/posts/4055496/edit. I welcome suggestions to improve the title. Thanks.
Mar 10, 2021 at 12:03 history asked ARedder CC BY-SA 4.0