I'm currently working through Convex Polytopes by Arne Brøndsted. Exercise 3.1 asks the following
Let $P=\text{conv}\ \{x_1,\ldots,x_n\}$ be a polytope in $\mathbb R^d$. Show that a point $x$ is in $\text{ri}\ P$ (the relative interior of $P$) if and only if $x$ is a convex combination of $x_1,\ldots,x_n$ with strictly positive coefficients, i.e. there are $\lambda_1,\ldots,\lambda_n$ such that $$x={\sum_{i=1}^n}{^c}\lambda_i x_i$$ and $\lambda_i>0$ for $i=1,\ldots,n$. (The $c$ indicates a convex combination).
I was able to prove that any point resulting from a convex combination with positive coefficients was in the relative interior: informally, you can "tweak" the coefficients in enough independent directions to get an open set surrounding it contained in the polytope.
I however am completely stumped in proving the converse. Sure, any point in the interior can be written down as a convex combination of this set, but nothing guarantees nonzero coefficients. If I had an affine independent set of $d+1$ points with nonzero coordinates adding up to this point, I could use affine dependence to "insert" the other points into this sum, but I don't know how to guarantee this either.
Best Answer
Let's assume wlog. $P$ has full rank by restricting ourselves to the affine hull if necessary. Let $x$ be a point in the interior of $P$.
Claim: For every $i$, there exists a convex combination $x = \sum_j \lambda^i_j x_j$ such that $\lambda^i_i \neq 0$.
Then $x = \sum_j \frac 1n(\sum_i\lambda^i_j)x_j$ is a positive convex combination of $x$, as desired.
Proof of claim: Since $x$ is in the interior of $P$, there exists some $\varepsilon > 0$ such that $y := x + \varepsilon(x - x_i)$ is in $P$. Hence there exists a convex combination $y = \sum_j\mu_jx_j$. Then $$x = \frac{1}{1+\varepsilon}(y + \varepsilon x_i) = \frac{\varepsilon}{1+\varepsilon}x_i + \sum_j\frac{1}{1+\varepsilon}\mu_j x_j,$$ so
$$\lambda_j^i := \frac{1}{1+\varepsilon}\begin{cases}\varepsilon+\mu_j&\text{ if }i=j\\\mu_j&\text{ else }\end{cases}$$ satisfies $x = \sum_j \lambda^i_j x_j$ and $\lambda^i_i \neq 0$.