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In the proof of Extended Hyperplane Separation Theorem in Peter. D. Lax's book, there is a statement without proof that since $H$ and $M$ are disjoint convex subsets of linear space $X$, and at least one of them has an interior point, so does $K = H - M$. I consider that if both of $H$ and $M$ contain one interior point, denoted by $x_0$ and $y_0$ separately, then we can prove $x_0-y_0$ is interior point of $K$. But I can't figure out how to verify the statement if only one of $H$ and $K$ contains an interior point. Is there any hint on proving this statement?

zeno tsang
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1 Answers1

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Let's say $M$ contains an open set $U$ with interior point $u$. Fix any $h_0\in H$. Then $$ h_0 - u \in \{h_0\} - U \subseteq H-M. $$ As $\{h_0\} - U$ is open, $h_0-u$ is an interior point of $H-M$.

EDIT ------ after clarifying the author's definition of an interior point:

$u$ is an interior point of $M$ is for every $y\in X$ exist an $\varepsilon>0$ (depending on $y$) such that $u+ty\in M$ for every $|t|<\varepsilon$.

Ok, now if $u$ is an interior point of $M$, obviously $-u$ is an interior point of $-M$.

Now let's verify, that $h_0 -u$ is an interior point of $K = H-M$. Let $y\in X$. Then there is some $\varepsilon>0$ with $-u + ty \subset -M$ for every $|t|<\varepsilon$. Hence $h_0 -u + ty \in H-M = K$.

Vobo
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  • I only learned functional analysis on metric spaces so I have no idea what does open set mean in Topological space. In Lax's book, it says $x$ is an interior point when $\forall y \in K$, there exists $\epsilon > 0$ such that for any $|t| < \epsilon$, $x+ty \in K$. Is it possible to derive the conclusion starting from this definition? – zeno tsang Jan 20 '14 at 10:19
  • I just googled the book you mentioned, on page 123 there is a theorem basically listing all properties of open sets you need for the above. In addition, you use a very rare definition for interior points: "Standard" would be: $x$ is an interior point of $K$ if for some $\varepsilon>0$ you have $B_\varepsilon \subseteq K$, where $B_\varepsilon={x\in X| d(x,0)<\varepsilon}$. Using this definition, it would be easier for me to start from here. – Vobo Jan 20 '14 at 12:26
  • The definition you use is the open set in metric space. However, in proving the hyperplane separation theorem, the author has not introduced metric space yet. I feel strange with such definition for interior point at the first glance too. I am looking for the proof without assumption of metric or normed space. – zeno tsang Jan 20 '14 at 12:40
  • BTW, such "strange" definition is useful in proving the following property of Minkowski functional: $p_K(x) < 1$ iff $x$ is an interior point of $K$, where $K$ is convex and absorbing. I think that's why the author define interior point in this way. – zeno tsang Jan 20 '14 at 12:53
  • Hmm, difficult. Using the strange definition, every point of the set $K=(0,1)\times{0}\subset \mathbb{R}^2$ would be an interior point of $K$, but it is not in $\mathbb{R}^2$. – Vobo Jan 20 '14 at 13:01
  • Ahh, and I rechecked your definition, it is not every $y\in K$ but every $y\in X$ !!! – Vobo Jan 20 '14 at 13:06
  • Ahh!!I see.. I made a terrible mistake... You are right, and it's quite straightforward using the correct definition. Anyway, thanks a lot for your help! – zeno tsang Jan 20 '14 at 14:38