Question:
Let $C \subset \mathbb{R}^n$ be a non-empty convex set and $x_0$ in $C$ closed ($x_0 \in \overline{C}$) but $x_0 \notin int(C)$. Then there exists $a \in \mathbb{R}^n \setminus \left \{ 0 \right \}$ s.t. $\forall x \in C$ it is true that $a\cdot x_0 \geq a\cdot x$
For my answer I had the idea to use the hyperplane separation theorem, but we can't use it directly as $\mathbb{R}^n \setminus C $ is not a convex set. So first I need to build a disjoint to $C$ nonempty set contening $x_0$. And here is my problem as I don't succeed to build clearly one of this type of set like a simple segment for exemple (or to show that those type sets exist).
Here my best attempt for the moment.
1/Lets suppose by absurd that it doesn't exists a segment $$S=\left \{ \lambda x_0 +(1-\lambda)y_0 | \lambda \in \left [ 0;1 \right ] \right \}$$ with $y_0 \in \mathbb{R}^n \setminus \left \{ C \right \}$ that has none of his elements/points in $C$. This will mean that at least for one $\lambda'$ we have that $\lambda' x_0 +(1-\lambda')y_0 \in C$.
On the other hand $\exists x'' \in int(C)$ and $\alpha \in \mathbb{R}$ s.t. $x_0=x'' \alpha$.
And as $x'', x' \in C$ the segment that connect them must to be in $C$ too. But in such a case this segment will contain $x_0$! That means $x_0 \in C$.2/By hyperplane separation theorem we have that $\forall s \in S$ and $\forall c \in C, \exists \beta \in \mathbb{R},a \in \mathbb{R}^n \setminus \left \{ 0 \right \} $ s.t. it is verify: $<a|s> \geq \beta \geq <a|c>$. And if it is true $\forall s \in S$ it is in particular true for $s=x_0$
Q.E.D.
Is is it correct?
In all the case if any one as a better solution to this question I will be happy to read it.
Thank you.
The two sets that you separate are $int(C)$ and $\{x_0\}$. Both are convex and disjoint, so you find a supporting hyperplane that has $a^Tx_0\ge m$ and $a^Tx<0$ for all $x\in C$.
To your attempt, in 1/ the existence of $x''$ is not guaranteed, as the ray connecting $x_0$ to the origin can be tangent to the convex set, touching it in just the point $x_0$. 2/ is true but as said, you do not need the segment outside $C$, the point set $\{x_0\}$ in itself is already convex.