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Let $X$ be a normed space. Given a nonempty convex set $C \subset X$, the relative interior of $C$, denoted by $\text{ri} C$, is the set of the interior points of $C$, considered as a subset of $\text{aff} C$ (this is the smallest affine set which contains $C$). In other words, $$ \text{ri} C = \left\{x \in X: \, \exists \varepsilon > 0 \text{ s.t. } B(x; \varepsilon ) \cap \text{aff} C \subseteq C\right\}. $$

Question n.1. How can I show that $\text{ri} C \ne \emptyset$? I have no idea on how to start, sincerely. The book gives some hint: for example, up to translations (which are allowed, since $\text{ri}(x+C)=x+\text{ri}C$ as one can easily verify), we can suppose $0 \in \text{aff} C$, i.e. the affine of $C$ is a subspace. Now we can consider a basis of this subspace, $(e_1, \ldots ,e_k)$: now how can I conclude?

Question n.2. Secondly, my book says that the concept of relative interior is important in finite dimension, since in infinite dimensions it can happen that a convex nonempty set has empty relative interior. Could you please provide an example of this? My book suggest considering a dense hyperplane with no interior points, but this is clearly a mistake (at least, it is a misleading counterexample, I cannot understand it).


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What book is this? –  mtiano May 12 '13 at 13:30

1 Answer 1

up vote 1 down vote accepted
  1. Suppose the dimension of aff $C$ is $k-1$, and choose $k$ affinely independent points of $C$. Denote them by $v_1,\ldots,v_k$. Since $C$ is convex, it contains the convex hull conv$(v_1,\ldots,v_k)$, that is, all points of the form $a_1v_1+\cdots a_kv_k$ for $a_1,\ldots,a_k\geq 1$ and $a_1+\cdots+a_k=1$. In particular it contains the point $x=(v_1+\cdots+v_k)/2$. Now, aff $C$ consists of all points of the form $a_1v_1+\cdots+a_kv_k$ for arbitrary $a_1,\ldots,a_k$. We can find a sufficiently small $\epsilon>0$ such that all points in aff $C$ at a distance at most $\epsilon$ from $x$ are contained in conv$(v_1,\ldots,v_k)$. Since conv$(v_1,\ldots,v_k)$ itself is contained in $C$, we are done. We have shown that $x\in$ Ri $C$.

  2. I think the following works. Take $l^2(\mathbb R)$ to be your infinite dimensional normed space. Write $e_k=(0,0,\ldots,0,1,0,0,\ldots)$ for all $k$, where the $1$ is in the $k$th place, and let $e_0=(0,0,\ldots)$. Let $C$ be the convex hull of all the vectors $e_0,e_1,\ldots$. Then $C$ consists of all vectors of the form $(x_1,\ldots,x_m,0,\ldots)$ where $x_i\geq 0$ and $x_1+\cdots+x_m=1$, for all $m\geq 0$, whereas aff $C$ consists of all vectors of the form $(x_1,\ldots,x_m,0,\ldots)$ for arbitrary $x_i$. No matter which such vector we pick, it will not be in the relative interior of $C$, because any ball centered at $x$ will contain the point $(x_1,\ldots,x_m,-\delta,0,\ldots)$ for some small $\delta>0$, which is in aff $C$, but it is not in $C$.

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First of all, I want to thank you for your kind reply. Your answer to question 1 is perfectly clear and I have understood your proof. Thank you. About your counterexample (i.e. question 2) I still have some doubts: indeed, who is aff $C$? It seems to be the subspace of definitely zero sequences, but I'm a bit confused: is this correct? I do not think so, but I don't manage to get a concrete "image" of this affine set. Thanks. –  Romeo May 13 '13 at 12:23
@Romeo: aff $C$ contains the zero vector $(0,0,\ldots)$, so it will just be the linear space which is spanned by $e_1,e_2,\ldots$. The span of $e_1,e_2,\ldots$ is defined as the set of finite linear combinations of $e_1,e_2,\ldots$: thus each such vector has an infinite tail of zeros, and if you think about it for a minute, you can show that aff $C$ becomes all vectors $(x_1,\ldots,x_m,0,\ldots)$ for arbitrary finite sequences $x_1,\ldots,x_m$ with arbitrarily large $m\geq 0$. –  Samuel May 13 '13 at 13:37
The point is: why aff $C$ in this case contains the zero vector? Aff $C$ is defined as the smallest affine set that contains $C$, it is not necessarily a (linear) subspace. Thanks for your help. –  Romeo May 13 '13 at 13:44
@Romeo: Oh, you're right. That's a miss on my part. Let's just add the zero vector then! I'll edit the answer. –  Samuel May 13 '13 at 14:47
Unfortunately, I made a mistake yesterday reading your previous comment: I apologize. Anyway, if a vector $(x_1, \ldots , x_m, 0,\ldots) \in C$ is s.t. $\sum_i x_i=1$ then your argument is correct and for every (small) $\delta>0$ the vector $(x_1, \ldots , x_m, 0,\delta, 0, \ldots)$ does not belong to $C$. What if $\sum_i x_i <1$ (it does happen, since $e_0 \in C$). What do you think? Thanks again. –  Romeo May 14 '13 at 12:55

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