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Let $C\subseteq \mathbb R^n$ a convex set. Assume there is a sequence $\{c_k\}_{k\in\mathbb N}$ with $c_k\in C$, $|c_k|\to\infty$ such that $v:=\lim \frac 1{|c_k|}c_k$ exists. Does this imply that there exists $a\in C$ with $a+[0,\infty)\cdot v\subseteq C$? Is the set of such $a$ dense in $C$?

This looks intuitively clear, e.g., in the case $n=2$. I was thinking of projecting the $c_k$ on a hyperplane $\perp v$ through some inner point of $C$ and use these to find a suitable $a$. But these projections don't seem to be well-behaved ...

Hagen von Eitzen
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4 Answers4

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Yes, it is true for $a$ in a dense subset of $C$.

First, note that for any fixed $b$, $\lim_{k \to \infty} c_k/|c_k| = \lim_{k \to \infty} (c_k - b)/|c_k - b|$. Thus the set of $v$'s for $C$ is the same as the set of $v$'s for $C - b$. We can choose $b \in C$ and replace $C$ by $C-b$; equivalently, we can assume that $0 \in C$.

If $c_k \in C$ with $|c_k| > N>1$, then $u = c_k/|c_k|$ has $$ u = \left(1 - \dfrac{1}{|c_k|}\right) 0 + \dfrac{1}{|c_k|} c_k \in C$$ and $$N u = \left(1 - \dfrac{N}{|c_k|}\right) 0 + \dfrac{N}{|c_k|} c_k \in C$$

Thus for any $N > 0$ and $\varepsilon > 0$ there is $u \in C$ with $|u - v| < \varepsilon$ and $Nu \in C$.

Now suppose $a$ is an interior point of $C$, with $r > 0$ such that the ball of radius $r$ centred at $a$ is in $C$. Given $M > 0$ with $M-1 > 2 |a|/r$, take $0 < \varepsilon < r (M-1)/(2 M^2)$, and $u$ as above for this $\varepsilon$ and $N = M^2$. We have

$$ a + M v = a + M (v-u) + M u = \left(1 - \dfrac{1}{M}\right) b + \dfrac{1}{M} (N u)$$

where $ Nu \in C$ and $$ \eqalign{\left|b - a\right| &= \left|\dfrac{M}{M-1} (a + M (v-u)) - a\right| \cr &\le \left|\dfrac{a}{M-1}\right| + \left|\dfrac{M^2}{M-1} (v-u)\right| < r\cr}$$ so that $b \in C$, and therefore $a + M v \in C$.

If $C$ has empty interior, we apply the result to a minimal affine subspace that contains $C$, which then has nonempty interior relative to that affine subspace.

Robert Israel
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  • How exactly do you get the first claim (regarding $u$ with $|u-v|<\varepsilon$ and $Nu \in C$)? If you assume (w.l.o.g.?) that $0 \in C$, then you can probably take $u = \frac{c_k}{|c_k|}$ with $|c_k| > \max{1, N}$. Then (because of $0 \in C$), we get $C u \in C$ and $u \in C$ and $|u - v| < \varepsilon$ (by enlarging $k$ if necessary). If $0 \notin C$, I don't immediately see how to get that claim. – PhoemueX Aug 24 '15 at 20:32
  • Good point. I edited to address this. – Robert Israel Aug 24 '15 at 21:42
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You may look at Rockafellar, Convex Analysis where he defines the recession cone for a non-empty convex set $C$ as $$ 0^+C=\{y\colon\ x+\lambda y\in C,\ \forall\lambda\ge 0,\ \forall x\in C\}. $$ Theorem 8.2 in the book (not in the preview, unfortunately) says that for a closed convex $C$ the recession cone $0^+C$ consists of exactly your type of limit vectors $v$ making, as @user251257 has mentioned, $C+[0,\infty)v \subseteq C$.

Theorem 8.3 and Corollary 8.3.1 say that for any non-empty convex set $C$ we have $$ 0^+(\text{cl}\,C)=0^+(\text{ri}\,C), $$ which means that the points $a\in C$, that you are interested in, are at least the whole relative interior of $C$, that is for sure dense.

A.Γ.
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You probably need that $C$ is closed.

Assume $0\in C$ without loss of generality. Otherwise, if $a\in C$, then $|c_k-a|\to \infty$ and $$ \frac{c_k-a}{|c_k-a|} = \frac{c_k}{|c_k|}\frac{|c_k|}{|c_k-a|} - \frac{a}{|c_k-a|} \to v.$$

Let $r\in[0,\infty)$. As $|c_k|\to\infty$, for $k$ sufficiently large, we have $r < |c_k|$ and $\frac r{|c_k|} c_k\in C$ by convexity. If $C$ is closed than, we have $\frac r{|c_k|} c_k \to r v \in C$.

Note: If $C$ is closed, we have in fact $C+[0,\infty)v \subseteq C$.

user251257
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So here's what I myself thought of after a good night's sleep:

For $1\le m<k$ let $$\begin{align}U_{m,k}&=\operatorname{span}\{\,c_i-c_m\mid m<i\le k\,\}\\U_m&=\bigcup_{k\ge m}U_{m,k}\\U&=\bigcap_mU_m\end{align}$$ Since $U_{m,m}\subseteq U_{m,m+1}\subseteq U_{m,m+2}\subseteq\ldots$ is eventually stationary, we have $U_m=U_{m,k_m}$ for some $k_m$. Since $U_1\supseteq U_2\supseteq\ldots$ is eventually stationary, we have $U=U_{m_0}=U_{m_0,k_0}$ for some $m_0$ and $k_0$. Let $$a=\frac{c_{m_0}+c_{m_0+1}+\ldots+c_{k_0}}{k_0-m_0+1}.$$ Then for $r$ small enough $B_r(a)\cap(a+U)$, the $r$-ball around $a$ intersected with $a+U$, is in the convex hull of $c_{m_0}\,\ldots,c_{k_0}$.

For fixed $m$ we have $$\langle v,w\rangle = \lim_{k\to\infty}\left\langle \frac{c_k}{|c_k|},w\right\rangle = \lim_{k\to\infty}\left\langle \frac{c_k-c_m}{|c_k|},w\right\rangle$$ so that $w\perp U_m$ implies $w\perp v$. We conclude $v\in U_m$ and $v\in U$.

Let $t>0$. For $k\gg 0$ we have $|c_k-a| > t$ and $\left|\frac{c_k-a}{|c_k-a|}-v\right|<\frac r{2t}$ and hence find for $u=2tv-2t\frac{c_k-a}{|c_k-a|}$ that $u\in B_r(0)\cap U$. Then $a+tv$ is a convex combination of $a+u$ and $a+\frac{t}{|c_k-a|}(c_k-a)$ and the latter is a convex combination of $a$ and $c_k$. We conclude $a+tv\in C$.

The set $A:=\{\,a'\in C\mid a'+[0,\infty)\cdot v\subseteq C\,\}$ is clearly convex. Now let $c\in C$ and $a'=xc+ya$ with $x>0,y\ge 0, x+y=1$ be a convex combination of $a\in A$ and $c$. Then $a'+tv$ is a convex combination of $c$ and $a+\frac txv$, hence $a'+tv\in C$ and $a'\in A$. This shows that $A$ is a dense convex subset of $C$. Especially, $A$ contains the relative interior of $C$.

Hagen von Eitzen
  • 374,180