Characterization of the interior of a convex set My question is the following: let $K$ be a convex set in $\mathbb{R}^n$ and $x$ an element of the interior of $K$. Can I affirm that there exist $z_1,...,z_n \in K$ linearly independent and $\lambda_1,....,\lambda_n > 0$ such that:
\begin{equation}
x = \displaystyle \sum_{p=1}^n \lambda_p z_p \quad ???
\end{equation}
If this statement is true, is there someone who can give me a proof ? Thank you very much and have a nice day !
 A: If a point $x$ is in the interior of $K$, then there is an open ball $B$ centered at $x$ such that $x\in B\subseteq K$. It is sufficient to prove this for a ball, where it is easy.
A: It is true iff $x \neq 0$.
If $x=0$, and $z_i$ are linearly independent, then $0 = \sum \lambda_i z_i$ implies $\lambda_i = 0$.
However, if $x \neq 0$ then it is true.
To see this, choose $b_2,..,b_n$ such that $B=\{x,b_2,...,b_n\}$ forms an orthonormal basis (that you can form a basis of $\mathbb{R}^n$ using $x$ is the key point here). Let $V = \{-(b_2+...+b_n), b_2,...,b_n \}$. Choose $\epsilon>0$ such that $\{x\}+\epsilon V \subset K$. Let $k_1 = x-\epsilon (b_2+...+b_n)$, $k_i = x+\epsilon b_i$, for $i>1$. Choosing $\lambda_i = \frac{1}{n}$, it is easy to see that
$$x = \sum_{i=1}^n \lambda_i k_i, \;\; k_i>0, \; \; k_i \in K$$
It remains to be shown that the $k_i$ are linearly independent. Suppose $\sum_{i=1}^n \alpha_i k_i = 0$. Since $B$ is orthonormal, we can easily compute the coefficients of $x, b_i$ as $\sum_{i=1}^n \alpha_i$, $\epsilon (\alpha_i-\alpha_1)$ ($i>1$). If follows from this that $\alpha_i=0$ ($\forall i$), hence the $k_i$ are linearly independent (and there are $n$ of them).
