I would like to prove the following theorem
Let $C$ be a non empty closed convex subset of $\mathbb{R}^n$ that does not contains the null vector and $X_0\in\mathbb{R}^n\setminus C$. Then there exists $V\in\mathbb{S}^n$ such that $ \forall X\in C : \langle V, X-X_0\rangle > 0 $
My attempt to prove this :
Consider $V = \frac{X-X_0}{\lVert X-X_0\rVert}$ then we have
$$ \langle V, X-X_0\rangle = \langle \frac{X-X_0}{\lVert X-X_0\rVert}, X-X_0\rangle = \frac{1}{\lVert X-X_0\rVert}\langle X-X_0, X-X_0\rangle > 0 $$
since $X-X_0\neq 0$ and the norme of $V$ is $1$ by construction.
Is this proof correct ?
Thank you a lot, don't hesitate to share your thoughts/proofs also I would be happy to learn more.
After a good remark from copper.hat, I realize that I need a fundamental result in order to prove this, the one above is false. The idea is first to show that if $C$ is closed convex subset that does not contain the null vector, then there exists $Y\in\mathbb{R}^n$ such that $\langle Y, X\rangle \geq \langle Y, Y\rangle > 0$ for all $X\in C$.
Attempt : We show that there exists a point in $C$ that minimizes the distance with the null vector. Consider the set $\{d(X, 0) = \lVert X\rVert : X\in C\}$, this set is not empty because $C$ is not empty and it is bounded below by $0$ so it satisfies the least upper bound property. We denote $Y$ the point that realize the least upper bound. Then we consider $X$ in $C$ and $s\in]0,1[$ : we know that that $sX + (1-s)Y$ is in $C$ and
$$ 0 < \lVert Y\rVert \leq \lVert sX + (1-s)Y\rVert\implies \lVert Y\rVert^{2} \leq \lVert sX + (1-s)Y\rVert^{2} $$
and since $\mathbb{R}^n$ is an inner product space we just have to choose the norm induced by an inner product (does not matter which one it is since they are equivalent) in order to get
$$ 0 < \langle Y, Y\rangle \leq \langle sX + (1-s)Y, sX + (1-s)Y\rangle = s^2\lVert X\rVert^{2} + 2s(1-s)\langle X, Y\rangle + (1-s)^2\lVert Y\rVert^{2} $$
$$ 0 \leq s^2\lVert X\rVert^{2} + 2s(1-s)\langle X, Y\rangle + \lVert Y\rVert^{2} -2s\lVert Y\rVert^{2} + s^2\lVert Y\rVert^{2} - \lVert Y\rVert^{2} = s[s\lVert X\rVert^{2} + 2(1-s)\langle X, Y\rangle + \lVert Y\rVert^{2} -2\lVert Y\rVert^{2} + s\lVert Y\rVert^{2} - \lVert Y\rVert^{2}] $$
which implies since $s\in]0,1[$
$$ s\lVert X\rVert^{2} + 2(1-s)\langle X, Y\rangle + \lVert Y\rVert^{2} -2\lVert Y\rVert^{2} + s\lVert Y\rVert^{2} - \lVert Y\rVert^{2}\geq 0 $$
then by taking the limit when $s$ goes to $0$ we have
$$ 2\langle X, Y\rangle \geq 2\langle Y, Y\rangle > 0 $$
Now we will use this result for the inital result I wanted to prove. First we consider the set $A = \{X- X_0 : X\in C\}$ clearly this set does not contains the null vector and it is convex since for all two points $X_1$ and $X_2$ in $C$ and $s\in [0,1]$ we have
$$ s(X_1-X_0) + (1-s)(X_2 - X_0) = sX_1 + (1-s)X_2 - X_0 \in A $$
since $C$ is convex. We need to show that $A$ is closed. Consider a sequence $(Y_n)_n = (X_n - X_0)\in A^{\mathbb{N}}$ which converges to $Y$. This means that
$$ \forall\epsilon>0, \exists N\in\mathbb{N} : n\geq N\implies \lVert Y_n - Y\rVert < \epsilon $$
which reduces to prove that $C$ is closed an hypothesis we made. Thus we can use the theorem proved previously that ensures the existence of $Y$ such that for all $X\in A$ we have $\langle Y, X\rangle \geq \langle Y, Y\rangle > 0$. We conclude by taking $V = \frac{Y}{\lVert Y\rVert}$.
Suppose $C$ is a non empty convex set and $X_0 \notin C$. (There is no need to assume that $0 \notin C$.)
Let $f(X) = \|X-X_0\|^2$ with the Euclidean norm. If we pick any $R>0$ such that $\overline{B}(X_0,R)$ intersects $C$, then we see that $\inf_{X \in C} f(X) = \inf_{X \in C \cap\overline{B}(X_0,R) } f(X)$ and since $ C \cap\overline{B}(X_0,R)$ is compact we see that $f$ has a minimiser $Y \in C$. By assumption we have $f(Y) >0$.
Hence $\|X-X_0\|^2 \ge \|X-Y\|^2$ for all $X \in C$.
Pick $X \in C$ and let $X(s) = Y+s(X-Y)$ for some $s \in [0,1]$. Then expanding & simplifying the previous inequality we get $ 2s \langle Y-X_0, X-Y \rangle + s^2 \|X-Y\|^2\ge 0$. Dividing by $s>0$ and taking limits gives $\langle Y-X_0, X-Y \rangle \ge 0$ for all $X \in C$.
Now let $V= Y-X_0$ and we see that $\langle V, X-X_0 \rangle = \langle V, X-Y+ Y-X_0 \rangle = \langle V, X-Y \rangle + \|V\|^2 \ge \|V\|^2>0$.