Problem: Show that $S:= \lbrace v \in \mathbb{R}^m \mid v=\displaystyle \sum_{j=1}^n a_j v_j, \text{ with } a_1, \dots , a_m \in [0,1], \ \sum_{j=1}^m a_j=1 \rbrace$ the Simplex of $\mathbb{R}^n$ with vertices $v_1, \dots , v_m$ is convex.
So I have to verify the following definition
Definition (convex): $M \subset V$ where $V$ is a real or complex Vectorspace is called convex if $\forall (a,b) \in M^2$ and $\forall \lambda \in \mathbb{R}$ such that $0 \leq \lambda \leq 1$ the following always holds $\lambda a + (1-\lambda)b \in M$
My approach: Geometrically this property is very intuitive, for instance the 3-Simplex or Tetrahedron is of course convex, I will never be able to construct a line that somehow leaves the defined region. I do however fail when it comes to show this property in an analytic approach.
Let $(v,v^*) \in S^2$ and $\lambda \in [0,1]$, I then need to verify that: $$\lambda v+ (1-\lambda)v^* \in S \iff \lambda \sum_{j=1}^n a_j v_j + (1-\lambda) \sum_{j=1}^m a_j^* v_j^* \in S $$ What I can see is that for sure $\lambda \cdot a_j \in [0,1]$ and similarly $(1-\lambda) a_j^* \in [0,1]$ . But that is already the only thing I can grasp from this problem. Do I need to work with the norm and try to find an upper bound for the scalars? I would appreciate some hints on how to continue/get started on this problem
There are many ways to approach this. Here is a brute force approach:
Let $u = \sum_k a_k v_k, v = \sum_k b_k v_k$, and $\lambda \in [0,1]$. We have $\sum_k a_k = \sum_k b_k = 1$ and $a_k,b_k \ge 0$.
Let $w = \lambda u + (1-\lambda) v = \sum_k (\lambda a_k + (1-\lambda) b_k) v_k$.
We have $\sum_k (\lambda a_k + (1-\lambda) b_k) = \lambda \sum_k a_k + (1-\lambda) \sum_k b_k = 1$, and $(\lambda a_k + (1-\lambda) b_k) \ge 0$ (where the latter follows because $\lambda, 1-\lambda, a_k,b_k$ are non-negative).
Hence $w \in S$.