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I am struggling to see the meaning behind the theorem's statement, which is:

Let $K\subset \mathbb R^n, K\neq \emptyset$ be a convex set and $x\not\in \text{clo}(K)$. Then there is $\gamma \in \mathbb R^n, \gamma \neq 0$ such that $$\inf\{ \langle\gamma, y \rangle: y\in K\} > \langle \gamma, x \rangle$$

I know what the actual geometric interpretation's supposed to be, but I fail to see how this theorem describes it. The intepretation is the following:

There exists a half space $H=\{x\in \mathbb R^n: \langle \gamma, x \rangle \geq c \}$ such that $K\subset H$ and the point $y$ has a positive distance from $H$

I played around with the inner product and the definition of a half space ($\{x: \langle x,y \rangle \geq c \}$) and a hyperplane ($\{x: \langle x,y \rangle = c \}$) and got to understand the geometric interpretation of inner product a bit more, but I am still unable to see how this theorem states what it states.

Thank you.

Dahn
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    When you write 'I know what the actual geometric interpretation is supposed to be' and then want to get it explained you should make this interpretation explicit and stated, so readers know what kind of explanation you are actually after... – Thomas Oct 14 '15 at 16:34
  • True, will add. – Dahn Oct 14 '15 at 16:35

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Well, the statement of the theorem says that, given $x$ you can find $\gamma$ for each $y\in K$ the value $\langle \gamma, y\rangle$ is bigger than the constant $c= \langle x, \gamma \rangle$, so $K$ is completely contained in a set (which is a half space) $\langle y,\gamma \rangle >c +\varepsilon $ for some $\varepsilon >0$. The opposite half space contains $x$, since $x$ satisfies $c= \langle x,\gamma \rangle < c +\varepsilon $. The positive distance is than at least $\varepsilon$ if $||\gamma || = 1$, otherwise you nee to scale with the length of $\gamma$, but since you only want positive this should not matter.

(You know that $\gamma$ is a normal to the plane $\{z|c =\langle z, \gamma\rangle$})?

Thomas
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