Rigorous proof for a ray intersecting the boundary in a convex compact set in exactly one point. As above mentioned, Iam searching for a rigourous proof of this statement:
Let $C\subseteq\mathbb{R}^2$ with $0\in int(C)$ be a compact convex set, then for every $v\in\mathbb{R}^2$ there is a unique $\lambda_v>0$ with $\lambda_v v \in\partial C$. Every proof, I have seen so far ignores something or draws only a picture. Intuitively for me there are $2$ cases, which must be contradicted. The first one is, that two isolated points on the ray must intersect $\partial C$, the second case is, that the ray is laying on the boundary, cutting $\partial C$ uncountable many times. Perhaps Iam missing something, but all proofs make assumptions I don't understand or say, the proof is trivial.
 A: Suppose that, for some $v\in\Bbb R^2\setminus\{(0,0)\}$, there are two distinct numbers $a,b\in(0,\infty)$ such that $av,bv\in\partial C$. I will assume that $a<b$. Let $w\in\Bbb R^2\setminus\{(0,0)\}$ such that $w\notin\Bbb Rv$. Since $0\in\mathring C$, there are numbers $\alpha$ and $\beta$, with $\alpha<0<\beta$, such that $(\forall t\in[\alpha,\beta]):tw\in C$. Since $C$ is convex, the whole triangle $T$ whose vertices are $\alpha w$, $\beta w$ and $bv$ (I mean by this the $2$-dimensional triangle, not just its sides) is a subset of $C$. But $\mathring T\subset\mathring C$. In particular, $av\in\mathring C$. But it was assumed that $av\in\partial C$.
A: This is true in more general spaces too (not just finite-dimensional). We can prove it with a couple of results that are helpful more generally. As a general notational note, we denote by $B[x; r]$ the closed ball centred at $x$, with radius $r$.

Proposition 1. Suppose $X$ is a normed linear space, $y, z \in X$, $r > 0$, and $\lambda \in [0, 1]$. Then
$$B[\lambda y + (1 - \lambda) z; \lambda r] = \{\lambda x + (1 - \lambda) z : x \in B[y; r]\}.$$

That is, if you take the weighted average between a fixed point $z$, and an arbitrary point in a ball $B[y; r]$, then these averages lie in the ball you'd expect.
Proof. Suppose first that $x \in B[y; r]$, which is to say, $\|x - y\| \le r$. We wish to show that
$$\lambda x + (1 - \lambda)z \in B[\lambda y + (1 - \lambda) z; \lambda r]. \tag{1}$$
Consider,
$$\|\lambda x + (1 - \lambda)z - (\lambda y + (1 - \lambda) z)\| = \|\lambda x - \lambda y\| = \lambda \|x - y\| \le \lambda r.$$
Thus, as predicted, $(1)$ holds true.
Conversely, suppose
$$p \in B[\lambda y + (1 - \lambda) z; \lambda r] \iff \|\lambda y + (1 - \lambda) z - p\| \le \lambda r. \tag{2}$$
We need to find some $x \in B[y; r]$ such that $\lambda x + (1 - \lambda)z = p$. Note that if $\lambda = 0$, then $(2)$ implies that $p = (1 - \lambda)z$, in which case the choice of $x$ is immaterial (choose any $x \in B[y; r]$ you like). If $\lambda \neq 0$, the only possible candidate for $x$ is
$$x = \frac{1}{\lambda} p - \frac{1 - \lambda}{\lambda} z.$$
It's now just a matter of showing that this $x$ lies in $B[y; r]$ as we'd expect. We have,
\begin{align*}
&\left\|\frac{1}{\lambda} p - \frac{1 - \lambda}{\lambda} z - y\right\| \\
=\,&\frac{1}{\lambda}\|p - (1 - \lambda)z - \lambda y\| \\
=\,&\frac{1}{\lambda}\|\lambda y + (1 - \lambda)z - p\| \\
\le\,&\frac{1}{\lambda}\lambda r = r.
\end{align*}
QED.
This allows us to prove the next result:

Corollary 2. Suppose $X$ is a normed linear space, $C \subseteq X$ is convex, $v, w \in C$ with $w \in \operatorname{int} C$, and $\lambda \in (0, 1]$. Then $\lambda w + (1 - \lambda)v \in \operatorname{int} C$.

Proof. Let $r > 0$ such that $B[w; r] \subseteq C$, which exists because $w \in \operatorname{int} C$. Using Proposition 1, we see that
$$B[\lambda w + (1 - \lambda)v; \lambda r] = \{\lambda x + (1 - \lambda)v : x \in B[w; r]\}.$$
But, since $B[w; r] \subseteq C$, $v \in C$ and $C$ is convex, every point in the latter set belongs to $C$, so we have,
$$B[\lambda w + (1 - \lambda)v; \lambda r] \subseteq C,$$
where $\lambda r > 0$. Thus, $\lambda w + (1 - \lambda)v \in \operatorname{int} C$.
QED.
Finally, we can prove what you want proven. We don't need two dimensions, but we do want a closed, bounded convex set $C$, containing $0$ in its interior. Suppose $v \neq 0$, and let
$$\lambda_v = \sup\{\lambda \in [0, \infty) : \lambda v \in C\}.$$
Since $C$ is bounded, this supremum must be finite. If we have
$$\lambda_n \in \{\lambda \in [0, \infty) : \lambda v \in C\}$$
such that $\lambda_n \to \lambda_v$, then $\lambda_n v \in C$ and
$$\|\lambda_n v - \lambda_v v\| = (\lambda_v - \lambda n)v \to 0,$$
hence $\lambda_v v$ belongs to the closure of $C$, and hence $C$.
Note that, for all $\varepsilon > 0$, $(\lambda_v + \varepsilon)v \notin C$, by definition of $\lambda_v$. This implies that $\lambda_v v$ does not lie in the interior of $C$, and hence lies in the boundary of $C$.
Now, if we have $\lambda \ge 0$ such that $\lambda \neq \lambda_v$, then either $\lambda > \lambda_v$ (in which case, as before, $\lambda v \notin C$, or indeed its boundary, since $C$ is closed), or $\lambda < \lambda v$. In the latter case, we get
$$\lambda v = \frac{\lambda}{\lambda_v} (\lambda_v v) + \left(1 - \frac{\lambda}{\lambda_v}\right) 0.$$
Note that $1 - \frac{\lambda}{\lambda_v} \in (0, 1]$. By Corollary 2, $\lambda v \in \operatorname{int} C$, which means it cannot lie in the boundary of $C$. Thus, $\lambda_v v$ is the unique non-negative multiple of $v$ that lies in the boundary of $C$. That is, the ray intersects the boundary uniquely, as required.
A: First, the statement is true if $\lambda_v$ is required to be positive. It's not hard to see that if $\lambda_v$ is just real, there are in fact two possible coefficients.

There are some interesting ways one could go about proving this, but I will answer your particular questions, out of order.


*The ray might lay on the boundary.

This is impossible, as $0 \in \text{int}(C)$. After all, if the ray were on the boundary, then $0v$ would be as well, ergo $0 \in \partial C$.

*

*Two isolated points on the ray might intersect $\partial C$.

Suppose that there are two positive numbers $\lambda, \xi$ so that $\lambda v$ and $\xi v$ are in $\partial C$. WLOG, $\xi > \lambda$. By definition of the boundary, there is $\epsilon > 0$ so that $(\lambda + \epsilon)v \notin C$ (we may choose this point from the $\epsilon$-ball in particular because we know that the ray is not on the boundary, by my first answer). But since $C$ is convex, and $\lambda + \epsilon \in (\lambda, \xi)$, we have a contradictory fact that $(\lambda + \epsilon) \in C$.

Here is a different proof sketch that you might find more palatable.
Assume without loss of generality $v \in C$. This can be achieved by scaling $v$ down, then scaling the coefficient $\lambda$ we obtain appropriately.
We triangulate $C$ so that $0$ is a vertex. Then $v$ is a convex combination of some triangle $0,a,b$:
$$v = \lambda_1a + \lambda_2 b + \lambda_3 0$$
where $\lambda_1 + \lambda_2 = 1-\lambda_3$, and $\lambda_i \in [0,1]$. Then:
$$\frac{1}{1-\lambda_3}v = \frac{\lambda_1}{1-\lambda_3}  a + \frac{\lambda_2}{1-\lambda_3} b$$
Where the coefficients sum to 1. Thus $(1-\lambda_3)^{-1}$ is the $\lambda_T$ we seek for the triangulation $T$. It's also not hard to see that this is unique.
Then just create finer and finer triangulations, notice that $\lambda_T$ is continuous in $T$ and unique at each triangulation, and conclude the result.
