If it is known that point I is an interior point of convex set C, would I be right in claiming that any point of the form K = pI + (1-p)X, where X is any other point in the convex set, p>0 and p<=1, is also an interior point?
Alternatively: a boundary point cannot be expressed as a convex combination of points where there is a non-trivial contribution from interior points.
Intuitively this seems to be the case, but I'm unable to prove it.
Best Answer
Since $I \in {\rm Int\,} C$, it has an open neighborhood $O$ lies completely inside $C$.
For any $K = pI + (1-p)X$ with $p \in (0,1]$, consider the map
$$\phi_p : y \mapsto X + \frac1p (y - X)$$
This map is continuous and send $K$ to $I$. The preimage $\phi^{-1}_p(O)$ will be open and contains $K$. For any $y \in \phi^{-1}_p(O)$, we have $$x = \phi_p(y) \in O \subset C \quad\implies\quad y = px + (1-p) X \in C$$ because $C$ is convex. This implies $\phi^{-1}_p(O) \subset C$. This means $K$ has an open neighborhood $\phi^{-1}_p(O)$ which lies completely inside $C$. i.e. $K \in {\rm Int\,}C$.
This argument works for any real topological vector space.