How to prove convex combination of two vectors also convex set?

Question

I am not able to prove below theorem, any ideas to prove it?

Answer 1

Since $v$ and $v^\prime$ are convex, let

$$v=\sum \lambda_i v_i$$

and

$$v^\prime = \sum \lambda_i ^\prime v_i$$

where $\sum \lambda_i = \sum \lambda _i ^\prime = 1$, and $\lambda, \lambda^\prime \in [0,1]$. Now we consider the convex combination $\mu v +(1-\mu)v^\prime$, for $\mu \in [0,1]$. Then we have

$$ \begin{align} \mu v +(1-\mu)v^\prime &= \mu\sum \lambda_i v_i+(1-\mu)\sum \lambda_i ^\prime v_i\\ &=\sum (\mu\lambda_i+(1 -\mu) \lambda_i ^\prime)v_i \end{align} $$

Notice that $\mu \lambda_i \ge 0$, and $(1-\mu)\lambda_i ^\prime\ge 0$. Now,

$$\sum (\mu\lambda_i+(1 -\mu)\lambda_i ^\prime) = \mu\sum\lambda _i + (1-\mu)\sum\lambda _i ^\prime = \mu(1) + (1-\mu)(1) = 1$$

Hence all the coefficients are positive and sum to one. Hence each $\mu\lambda _i + (1-\mu)\lambda _i ^\prime\in [0,1]$. So $\mu v+ (1-\mu)v^\prime$ is also a convex combination of $v_1, v_2, \dots, v_k$.