Inner product at the boundary

Question

Let $K\subset \mathbb{R}^n$ be a convex set with nonempty interior, i.e. $K^{\circ}\neq \emptyset$. Suppose there exist a nonzero vector $v$ such that

$$\langle x, v\rangle\geq 0$$

for all $x\in K^{\circ}$. I want to show that in this case the property holds for all $x\in K$.

It is supposed to be a continuity argument. Am having trouble because $x\in K\setminus K^{\circ}$ doesn't imply that there exist a sequence in $K^{\circ}$ converging to $x$.

Any help is greatly appreciated.

Answer 1

But $x \in K \setminus K^\circ$ does imply that there exists a sequence in $K^\circ$ converging to $x$.

Choose any open ball $B \subset K^\circ$ centered at $p$.

Let $\overline{ab}$ be the diameter of $B$ perpendicular to the segment $\overline{xp}$.

The interior of the entire triangle $\triangle xab$ is therefore contained in $K^\circ$, and in particular the half-open segment $[p,x)$ is contained in $K^\circ$.