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If $S $ is a convex subset of a vector space $V$ then can we say that the null vector $\theta \in S$ ?

Actually I was reading proof a lemma for a theorem "The Pyramidal Construction for nonconvex case" , where it is directly written that " $0 \in intco E$, where E $\subseteq \mathbb R^n$ and $intco E$ stands for the interior of convex hull of E."

Please someone help..

Thank you.

Mini_me
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  • It's a peculiar notation you have for the zero vector, but the zero vector need not be in a particular convex set. – Angina Seng Aug 27 '17 at 14:16
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    No, it's not true in general that every convex set contains 0. There's obviously some missing context to your question, but unless you tell us where the quote came from we can't really help. – Brian Borchers Aug 27 '17 at 14:16
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    For some more counterexamples, all singletons are convex. – SvanN Aug 27 '17 at 14:28
  • @BrianBorchers the question has been taken from LEMMA 2.11 in the book IMPLICIT PARTIAL DIFFERENTIAL EQUATION by B.DACOROGNA and PAOLO MARCELLINI. – Mini_me Aug 27 '17 at 14:29
  • @BrianBorchers Here is the link of google book.Please see page no.-48. https://books.google.co.in/books?id=8R_vBwAAQBAJ&pg=PR4&dq=implicit+partial+differential+equation,+bernard+dacorogna,+paolo+marcellini&hl=en&sa=X&redir_esc=y#v=onepage&q=implicit%20partial%20differential%20equation%2C%20bernard%20dacorogna%2C%20paolo%20marcellini&f=false – Mini_me Aug 27 '17 at 14:37

2 Answers2

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Counterexample, $(0,\infty)$ is convex in $\mathbb{R}$ but does not contain zero. Are there any other restrictions on $E$ which would force the zero vector $\theta$ to be in it's interior of its convex hull?

CyclotomicField
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  • Here is the book https://books.google.co.in/books?id=8R_vBwAAQBAJ&pg=PR4&dq=implicit+partial+differential+equation,+bernard+dacorogna,+paolo+marcellini&hl=en&sa=X&redir_esc=y#v=onepage&q=implicit%20partial%20differential%20equation%2C%20bernard%20dacorogna%2C%20paolo%20marcellini&f=false in page 48 (lemma 2.11) at step 1 – Mini_me Aug 27 '17 at 14:47
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    @mathiu_lady They assume the open set has been translated to the origin in the second sentence of the proof which is why the zero vector is in the interior of the convex hull in this special case. – CyclotomicField Aug 27 '17 at 16:38
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No, not necessarily the null vector $\theta \in S$, when $S\subset V$, $S$ being a convex set.

Consider $S=\{(1,1)\}\subset \Bbb R^2$. S is clearly a convex subset, but the null vector $\theta = (0,0) \not \in S$.

S must be a subspace for that property to hold. For instance, defining $S=\{(x,0): x\in \Bbb R\} \subset \Bbb R^2$, that are the points in the $x$ axis of $\Bbb R^2$, it will hold that $S$ is a convex set, a subspace, and $\theta=(0,0)\in S$ in this case.

bluemaster
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  • Here is the book https://books.google.co.in/books?id=8R_vBwAAQBAJ&pg=PR4&dq=implicit+partial+differential+equation,+bernard+dacorogna,+paolo+marcellini&hl=en&sa=X&redir_esc=y#v=onepage&q=implicit%20partial%20differential%20equation%2C%20bernard%20dacorogna%2C%20paolo%20marcellini&f=false in page 48 (lemma 2.11) at step 1. Please check it and say why this is happening? – Mini_me Aug 27 '17 at 14:40
  • The page you mentioned is not available when I access the link posted. Perhaps there are other constraints on $E$ you might have not noticed in the proof, that leads to the result, as mentioned in the answer from @CyclotomicField. – bluemaster Aug 27 '17 at 14:59