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Let $K \subset \mathbb{R}^d$ be compact and convex. Consider a point $x \in K$ such that $\lVert x \rVert$ is maximized and let $H$ be the hyperplane with equation $\langle y, x \rangle = \lVert x \rVert$. Is it true that $H \cap K = \{ x \}$ i.e. does $H$ intersect $K$ only at $x$? I'm unsure that $\forall a \in K$, $\langle a, x \rangle < \langle x, x \rangle$.

What I'm trying to show is that every compact, convex set has a $0$-dimensional face.

blm
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1 Answers1

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This is false. For instance if $0\in K$ then there are lots of examples where $H\cap K$ contains way more points than only $x$.

For your more general question you might like to have a look at Krein-Milman's theorem.

Popyaitte
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  • In the book I'm reading, this exercise comes before the Krein-Milman theorem so I suspect it's not necessary. Do you have any insight as to how I might prove this without Krein-Milman? – blm Oct 15 '19 at 23:08
  • I just had a look, I think you are in the right direction by looking at a point with maximal norm. By using the reverse triangular inequality you get that if such a point $z$ is on a segment of $K$, then the extremal points of the segment (say $x$ and $y$) have the same maximal norm as $z$. Then if $x$ and $y$ are distinct gives you that $||z|| < ||x||=||y||$ which is absurd by hypothesis. So $z$ is an extremal point. – Popyaitte Oct 15 '19 at 23:38