I am trying to solve the following question.
Let $K$ be an open convex subset of a normed linear space $X$ over $R$ and $0\in K$. Consider $x_0 \notin K$. Define a linear functional $$f_0 : X_0:=span(x_0) \rightarrow R, \ f_0(tx_0) = t\|x_0\|_K, \ \forall t \in R,$$ where $\|x \|_K := \inf\{t > 0 : \frac{x}{t} \in K\}$.
Using Hahn-Banach theorem to extend $f_0$, prove that there exists a functional $f \in X^*$ ($X^*$ is the dual space of $X$), $f \neq 0$ such that $f(x) \leq f(x_0)$ for all $x \in K$.
Here's what I've done so far. I have proven that $\|x\|_K$ is a sublinear. I have also proven that $f_0(x) \leq \|x\|_K$ for all $x\in span(x_0)$. By the Hahn-Banach theorem, there exists a linear extension $f$ of $f_0$ such that $f|_{X_0}=f_0$ and $f(x)\leq \|x\|_K$ for all $x\in X$.
I'm stuck here.
Edit: I know that if $x\in K$, then $\|x\|_K =0 $ because $t(\frac{x}{t} )+(1-t)0 = x \in K$ for $0<t<1$. If $x\notin K$, then $\|x\|_K\geq 1$.
Edit: Where do we need the openness of $K$?
Since $x_0\notin K$, we must have that $\| x_0 \|_K \ge 1$. Moreover, for $x\in K$, $\| x \|_K \le 1$. Thus, $$ f(x) \le \| x \|_K \le 1 \le \| x_0\|_K = f(x_0)$$ for $x\in K$.