Let $x_i \in \mathbb{R}^n $ where $i=1,…,l$ and $1\leq l \leq n$. Also, let
$$
W = \{y = \sum_1^l \alpha_i x_i \mid \alpha_i > 0 \sum_1^l \alpha_i =1 \}
$$ be the set of all strict convex combination of $x_i$'s.
We know that the following
$$
U = \{y = \sum_1^l \alpha_i x_i \mid \alpha_i \geq 0 \sum_1^l \alpha_i =1 \}
$$ is the convex hull of the set $\{x_i\}_{i=1}^{l}$.
What is the closure of $W$, i.e. $cl(W)$? Can we just say if we relax strictly greater than zero of $\alpha_i$ in $W$ we can get its closure?
Please give a counter example if it is not true or prove it if it is true?
Best Answer
$\sum \alpha_i x_i$ is the limit of $\sum \alpha_i^{k} x_i$ as $k \to \infty$ where $\alpha_i^{k}= (\alpha_i+\frac 1 k)/( \sum_j( \alpha_j+\frac 1 k))$ so $U$ is the closure of $W$.