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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.

David C. Ullrich
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Hendrra
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2 Answers2

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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 $\lambda \alpha_1+(1-\lambda)\alpha_2 \in [a,b]$.

Kavi Rama Murthy
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I would like to show now that $\lambda \omega_1 + (1 - \lambda) \omega_2 \in \Omega$.

Hint: Since you know that $\omega_1 \in \alpha_1 C$ and $\omega_2 \in \alpha_2 C$, you should be able to find an $\alpha'$ such that $(\lambda \omega_1 + (1 - \lambda) \omega_2)\in\alpha'C$ by that same constant scaling concept.