Let $C \subset \mathbb{R}^n$ be a convex set. Moreover let's fix two real numbers $a, b$ such that $0 \le a \le b$.
I am to prove that
$$ \Omega = \bigcup_{\alpha \in [a, b]} \alpha C$$is convex,
where $\alpha C = \{ x \in \mathbb{R}^n: x = \alpha y, y \in C \}$.
I tried to complete the proof using the definition of convex sets.
Let $\omega_1, \omega_2 \in \Omega$. That means that there exists $\alpha_1, \alpha_2 \in [a, b]$ such that $\omega_1 = \alpha_1 C$ and $\omega_2 = \alpha_2 C$. Let's choose $\lambda \in [0, 1]$.
I would like to show now that $\lambda \omega_1 + (1 – \lambda) \omega_2 \in \Omega$. Thus
$$\lambda \omega_1 + (1 – \lambda) \omega_2 = \lambda \alpha_1 y_1 + (1 – \lambda) \alpha_2 y_2 \tag{1}$$
I know that $(1)$ should be equal to $\beta z$, where $\beta \in [a, b]$ and $z \in C$ but I have no idea how I can achieve that goal. I would appreciate any hints or tips.
Best Answer
Hint: $\omega_1=\alpha_1 y_1$ and $\omega_w=\alpha_2 y_2$ for some $y_1,y_2 \in C$ (and not as you have stated).
Now $\lambda \omega_1+(1-\lambda) \omega_2= [\lambda \alpha_1 +(1-\lambda) \alpha_2] [ty_1+(1-t)y_2]$ where
$$ t=\frac {\lambda \alpha_1} {\lambda \alpha_1 +(1-\lambda) \alpha_2}$$.
Now use the fact that $ty_1+(1-t)y_2 \in C$ and the fact that $\lambda \alpha_1+(1-\lambda)\alpha_2 \in [a,b]$.