Proving convexity

Question

I ask you please some help with this problem:

Let A $\subseteq$ R$^n$ be a convex set and $C(A)$ = {$\lambda$x, $\lambda \in \mathbb{R}$, $\lambda \geq 0$, x $\in$ A}. Prove that $C(A)$ is convex.

This is how I'm trying do do it:

Let $\delta \in $R, $0\leq \delta \leq 1$ and $x \in C(A)$, $y \in C(A)$.

I must show that $z_\delta = \delta x + (1-\delta)y \in C(A)$. Using the fact that $x \in C(A)$, $y \in C(A)$ then $z_\delta = \delta \lambda_1 x_1 + (1-\delta) \lambda_2 x_2$ where $\lambda_1, \lambda_2 \in \mathbb{R}$, $\lambda_1, \lambda_2 \geq 0$ and $x_1, x_2 \in A.$

I've been having some trouble trying to relate the $\lambda_1 x_1$ to $\lambda_2 x_2$. I know that $A$ is convex, so every convex combination of points of $A$ belong to $A$, so I could write

$x_1$ = $\alpha_1$a + $\beta_1$b where $\alpha_1, \beta_1 \geq 0$ and $\alpha_1 + \beta_1 = 1$

$x_2$ = $\alpha_2$a + $\beta_2$b where $\alpha_2, \beta_2 \geq 0$ and $\alpha_2 + \beta_2 = 1$

a,b $\in$ A

but I still can't show that $z_\delta$ is like some $\lambda w$, $\lambda \geq 0$, $w \in A$. Any hints? Am I in the right direction to prove this?

Thank you in advance!

Answer 1

Hint: Let $\alpha:=\delta\lambda_1$ and $\beta:=(1-\delta)\lambda_2$. The only problem with them is that $\alpha+\beta\ne 1$ generally.
So, define $\alpha'$ and $\beta'$ to get over this problem and notice that $C(A)$ is closed under multiplication by positive scalar.

Let $\ \alpha':=\displaystyle\frac{\alpha}{\alpha+\beta}\ $ and $\ \beta':=\displaystyle\frac\beta{\alpha+\beta}$.

Answer 2

Hint: It might be easier to prove that $C(A)$ satisfies a stronger condition: for any $\alpha,\beta>0$ and $x,y\in C(A)$ we have $\alpha x+\beta y\in C(A)$.