Given a closed, convex, full-dimensional cone $K$, how do I show that $x\in int(K) \iff y^Tx>0 \quad \forall y\in K^*- \{0\} $ ?
I've thought about applying the Hahn-Banach separation theorem
If $C\subseteq H$ is a closed, convex set and if $b\notin S$ then $\exists y\in H $ and $\beta\in\mathbb{R}$ such that $y^Tx \geq \beta$ and $y^Tb <\beta$.
But I'm not sure if that is usefull.
Anyone that can help?
Since $K$ is closed convex cone, the half-space $H=\{x \mid y^Tx \leq \beta\}(y \neq 0)$ will contain $K$, where $$\beta = \delta^*(y \mid K) = \sup_{x \in K}{y^Tx}.$$ Due to $K \subset \{x \mid y^Tx \leq \beta\}$ and $K$ is full-dimensional, we have \begin{equation} \mathrm{int}~K \subset \mathrm{int}~\{x \mid y^T x \leq \beta\} = \{x \mid y^Tx < \beta\}. \end{equation}
Furthermore, due to $K$ is cone, then \begin{equation} \beta = \left\{ \begin{array}{cl} +\infty & {\text{if}~y \in \{y \mid y^Tx >0, \forall x \in K\},} \\ 0 & {\text{if}~y \in \{y \mid y^Tx \leq 0, \forall x \in K\} = K^{\circ}.} \end{array} \right. \end{equation} Finally, for any $x \in \mathrm{int}~K$, we have $y^Tx < 0, \forall y \in K^{\circ} \setminus \{0\}$. Thus $x \in \mathrm{int}~{K} \rightarrow y^Tx > 0, \forall y \in K^* \setminus \{0\}$.
On the contrary, if we have $y^Tx >0, \forall y \in K^* \setminus \{0\}$. According the proceeding remarks, this means that $$\forall y \in K^* \setminus \{0\}, y^Tx < \beta, \quad \beta = \sup_{x \in K}y^Tx.$$ We can prove by contradiction, suppose $x \notin \mathrm{int}~K$, then we can find $y' \neq 0$, and $\beta'$, such that $y'^Tx > \beta'$, and $y'^Tz \leq \beta'$ for any $z \in \mathrm{int}~K$. It is obvious that this cannot be held for $\beta'$ since it could not be held for $\beta = \sup_{x \in K} y'^T x$.
(1) Here we prove that if $K \subset H= \{x \mid y^T x \leq \beta\}$ and $K$ is full-dimensional, we have $\text{int}~K \subset \text{int}~H$. Due to $K$ is full-dimensional, thus $\text{int}~K$ is non-emptyset. Suppose there exists $z \in \text{int}~K$ and $y^T z = \beta$, according the definition of interior, there exists $\varepsilon > 0$ such that $z + \varepsilon B \subset \text{int}~K$, where $B$ is the Euclidean unit ball $B = \{x \mid \|x\| \leq 1\}$. Obviously, $z + \varepsilon y \in z + \varepsilon B$ and it is not in half-space $H$. Thus we have $z \in \{x \mid y^Tx < \beta\}$.
If we don't have the full-dimension of $K$, we could not guarantee that $\text{int}~K$ is non-empty.
(2) There are two kinds of definition of cone, one is that the set is closed to positive scalar multiplication, i.e., $\forall x \in K$, we have $\lambda x \in K, \forall \lambda > 0$. In this sense, the cone may not contain the origin. The other kind is the definition that you are using, i.e., $\forall x \in K$, we have $\lambda x \in K, \forall \lambda \ge 0$. It means that the set is closed to non-negative scalar multiplication.
(3) If for any $x \in \text{int}~K$, we have $y^Tx < 0, \forall y \in K^{\circ} \setminus \{0\}$. Next, we use $-y$ instead of the original $y$. It means that $\forall (-y) \in K^{\circ} \setminus 0, (-y)^Tx < 0$. Furthermore, we can conclude that $\forall y \in K^* \setminus 0, y^Tx>0$.
I am very apologize that my proof is not clear enough, but hope this can help you.