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I am trying to solve the following problem. Let $C$ be a convex subset of a topological vector space $X$. Let x be in the interior $C^\circ$ and let y be in the closure $\bar{C}$. I am asked to prove that in the line segment \begin{equation} z(\lambda)= \lambda x + (1-\lambda) y \end{equation} points with $\lambda \in (0,1]$ all belong to the interior $C^\circ$.

If another point $x^{'}$ is in the interior, we have the open set $\lambda N_x + (1-\lambda) N_{x^{'}}$ to cover the points in between $\lambda x + (1-\lambda) x^{'}$ for $\lambda\in [0,1]$ ($N_x$ and $N_{x^{'}}$ are open neighborhoods of $x$ and $x^{'}$ in $C$). Therefore the preimage $z^{-1}(C^\circ)$ is an open and convex (connected) subset of $[0,1]$, which contains $1$ by definition. I want to rule out the possibility that for some $t>0$, $z([0,t))$ is contained in the closure $\bar{C}$ but not in $C$, presumably using the fact that $y\in \bar{C}$ and convexity of $C$. I don't know how to proceed though.

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  • $\begingroup$ Maybe we should consider the linear homeomorphism $L_\epsilon (\zeta) = (1-\epsilon)\zeta + \epsilon y$, which maps $x$ to a point $x^{'}=(1-\epsilon)x + \epsilon y$ closer to $y$. For $\lambda\in(t,1]$, $\lambda x^{'} + (1-\lambda)y = \lambda(1-\epsilon)x+(1-\lambda(1-\epsilon)y) \in L_\epsilon(C)^\circ \subset C^\circ$. $\endgroup$ Dec 2, 2023 at 8:18

3 Answers 3

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Here is my trial. I will try to explain my line of thought first. Preimages are a red herring here.


Let $O$ be an open neighborhood of zero such that $x+O \subset C$. We want to prove that $z+ \lambda O \subset C$.

Since $y \in \bar C$, for every open neighborhood $U$ of zero we can find $x' \in ( y+ U) \cap C$ . The problem is to find $U$ that works. The idea is to ensure that $\lambda (x+O)+ (1-\lambda)x'$ contains $z$.

We still have to find $U$. Since $x' \in ( y+ U) \cap C$, there is $u\in U$ with $x'=y+u$. Then $$ \lambda (x+O)+ (1-\lambda)x' = z + \lambda O +(1-\lambda)u. $$ So we need that $0\in \lambda O +(1-\lambda)u$, or equivalently $ u \in -\frac{\lambda}{1-\lambda}O$. The latter set is our $U$.


Lets start the proof.

Let $O$ be an open neighborhood of zero such that $x+O \subset C$. Since $y\in \bar C$, there is $x'\in ( y-\frac{\lambda}{1-\lambda}O) \cap C$. Let $u:=x'-y \in -\frac{\lambda}{1-\lambda}O$.

Then $$ \lambda (x+O)+ (1-\lambda)x' = z + \lambda O +(1-\lambda)u. $$ Since $(1-\lambda)u \in -\lambda O$, it follows that $0 \in \lambda O +(1-\lambda)u$, so that $z\in \lambda (x+O)+ (1-\lambda)x'$. Since $x'\in C$ and $x+O \subset C$, the open set $\lambda (x+O)+ (1-\lambda)x'$ is contained in $C$. And $z$ is an interior point of $C$.

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It seems there is a simple proof. Let $L_t, t\in [0,1)$ be a linear homeomorphism \begin{equation} L_t(\xi) = (1-t)\xi + t y \end{equation} Clear $L_0$ is the identity map and it maps $x$ closer to $y$ in the line segment as $t$ increases.

By convexity of $C$, $L_t(C)\subset C$. Let $O_x$ be an open neighborhood of $x$ that is contained in $C$, $L_t(O_x)$ is an open set contained in $C$, so is the union \begin{equation} \cup_{t\in[0,1)} L_t(O_x) \end{equation} We have $\lambda x + (1-\lambda)y$ is contained in the open set above for $\lambda \in (0,1]$ (set $\xi=x,t=1-\lambda$), so they are all interior points of $C$.

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    $\begingroup$ You need to explain why $L_t(C) \subset C$ if $y$ is in $\bar C\setminus C$. $\endgroup$
    – daw
    Dec 3, 2023 at 16:37
  • $\begingroup$ @daw Thank you for pointing it out. Indeed, not really a proof. $\endgroup$ Dec 4, 2023 at 13:42
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Suppose that there is a $z$ in the relative interior of $[x,y]$ that is in the boundary of $C$. The boundary of $\overline{C}$ is the union of all the proper faces of $\overline{C}$. In particular, $z\in F$ for some proper face $F$ of $\overline{C}$. As the relative interior of $[x,y]$ meets $F$ and $F$ is a face of $\overline{C}$, we have that $[x,y]\subseteq F$. Since $F$ is disjoint from the interior of $\overline{C}$, which contains the interior of $C$, we conclude that $x$ is not in the interior of $C$. The resulting contradiction implies that all the points of $[x,y)$ are in the interior of $C$.

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