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Let $K\subset \mathbb R^n$ be a convex set which is also compact with $dim K=n$ and $H \subset \mathbb R^n$ a hyperplane.

Proof that for any Hyperplane H there exist exactly two supporting hyperplanes of K parallel to H

Can anyone give me a hint how to proof the statement above ?

I know that any given hyperplane can be written as $H=\{x\in R^n : c^tx=\delta \}$ and I know the hyperplane seperation theorem.

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  • $\begingroup$ Suppose there were $3$ such planes. Each plane has at least one point of $K$. How could the middle one (the one with the middle $\delta$) be a supporting plane? $\endgroup$ – quasi Jun 24 '17 at 21:42
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    $\begingroup$ To get exactly two, you need to know that $K$ is bounded, else the claim is false. $\endgroup$ – quasi Jun 24 '17 at 21:46
  • $\begingroup$ @quasi: Thanks for the reply. I just got one question. We defined the dimension of any set to be the dimension of it`s affine Hull. Is $dimK=n$ necessary for the statement ? I forgot to mention that $K$ is compact... $\endgroup$ – XPenguen Jun 24 '17 at 21:47
  • $\begingroup$ Yes, otherwise suppose $K$ is a subset of $H$. $\endgroup$ – quasi Jun 24 '17 at 21:48
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Outline:

Assume $K$ is a compact, convex subset of $\mathbb{R}^n$ with $\dim(K)=n$.

Define $f \colon \mathbb{R}^n \to \mathbb{R}$ by $f(x) = c^tx$.

Let $J=f(K)$.

Clearly $f$ is continuous, hence, since $K$ is compact, so is $J$.

Since $K$ is convex and $f$ is linear, $J$ is convex.

Thus, $J$ is a compact, convex subset of $\mathbb{R}$.

Since $\dim(K)=n$, argue that $J$ contains more than one point.

Thus, $J = [a,b]$, for some $a,b \in \mathbb{R}$, with $a < b$.

Argue that $f^{-1}(\delta)$ is a supporting hyperplane if and only if $\,\delta=a\;\,\text{or}\;\,\delta=b$.

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