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Let $X \subset \mathbb R^d$ be a simplex and $x \notin X$

Prove that there exists a unique face $F \subseteq X $ with minimal dimension $\dim(F)=\dim(aff(F))$ such that $x \in K_F$ where $K_F:=\{x+\lambda(y-x):\lambda \ge 0 \;x\in F,y\in X \}$ denotes the tangent cone of $F$

Note: $\mathrm{aff}(F)$ is supposed to be the affine hull.

Would appreciate any help.

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  • $\begingroup$ What are your approaches? Also where does the "tangent cone" from the title appear in the question? $\endgroup$ – M. Winter Jun 8 '18 at 21:48
  • $\begingroup$ @M.Winter The tangent cone in the title is $K_F$. At least that's my definition of a tangent cone given a face $F$ of $X$ $\endgroup$ – XPenguen Jun 8 '18 at 21:53
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Here is one way to prove this by contradiction.

Let $F_1\neq F_2$ be two faces of $X$ that satisfy your constraints. Because $x\in K_{F_i}$, there are $p_i\in F_i$ and $q_i\in X$ such that $x$ belongs to the ray from $p_i$ through $q_i$. The properties of $p_i$ and $q_i$ are enough to derive a contradiction.

Hint 1:

$X$ is a simplex and is thus convex, it must fully contain the segment between $q_1$ and $q_2$.

Hint 2:

Because $x\notin X$, thus $x$ cannot be in the segment between $q_1$ and $q_2$. You can look at rays from $x$ and through points of that segment.

Hint 3:

The existence of $p_1$ and $p_2$ guarantee that for each rays mentioned in hint 2, the last point of $X$ that is met by the ray is not in the segment $[q_1q_2]$, but somewhere further. Look at those points.

The three hints above should be more than enough to get you started. If you're still struggling, here are some more.

Hint 4:

The points from hint 3 form a continuous path, let's call it $\gamma$. This path $\gamma$ entirely lies in the boundary of $X$. And each boundary point of $X$ belongs to some face of $X$ of dimension (strictly) less than $d$.

Hint 5:

If you look at the points in $\gamma$, you obtain a sequence of faces of $X$ that starts from $F_1$ and ends with $F_2$. Because you go from one face to another face of the same dimension, at some point $\gamma$ must cross the boundary of $F_1$ and $F_2$. What property do those boundary points (of $F_1$/$F_2$) have?

The last hint is one step away from the conclusion.

Hint 6:

Let $p$ be one of the points mentioned in hint 5. Point $p$ must belong to another face, $F_3$, whose dimension is (strictly) less than those of $F_1/F_2$. And it also belongs to $\gamma$.

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  • $\begingroup$ Thanks for the reply. I just have one question. While this shows the uniqueness but what about existence ? I mean why does $x \in K_F$ for some face $F$ of $X$ ? $\endgroup$ – XPenguen Jun 9 '18 at 16:46
  • $\begingroup$ @XPenguen $X$ is the (unique) $d$-dimensional face of $X$, and $F_X=\mathbb R^d$. If you exclude $X$ as a face, the result is false. $\endgroup$ – N.Bach Jun 9 '18 at 16:48
  • $\begingroup$ Now I feel stupid :P Thanks for taking your time. I appreciate it $\endgroup$ – XPenguen Jun 9 '18 at 16:51

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