Prove the set $M=\{x\in\mathbb{R}^2_+ \mid \alpha x_1+\gamma x_2\leq \beta\}$ is convex

Let $\alpha\gt 0$, $\gamma\gt 0$, and $\beta\gt 0$ be real numbers. Let $$M=\{x\in\mathbb{R}^2_+ \mid \alpha x_1+\gamma x_2\leq \beta\}$$ Prove $M$ is a convex set. Prove that $M$ is bounded. What does this set resemble (in economics)?

Attempt: If $(x_1,x_2),(y_1,y_2)\in M$ we get \begin{align*} \alpha x_1 + \gamma x_2&\leq \beta\\ \alpha y_1 + \gamma y_2 &\leq \beta \end{align*}

We want to prove $$\alpha(ax_1 + (1-a)y_1) + \gamma(ax_2 + (1-a)y_2)\leq \beta.$$

The question is how do I prove this inequality?

• I think the best proof is just to say: the set is a triangle! (I don't what if anything triangles "resemble in economics") – Omar Antolín-Camarena Apr 25 '12 at 3:16
• nope. not convincing enough. – Koba Apr 25 '12 at 14:06

Algebra! (pronounced like Jon Lovitz's Master Thespian character)

\begin{align*} \alpha(ax_1 + (1-a)y_1) + \gamma(ax_2+(1-a)y_2) &= \alpha ax_1 + \gamma ax_2 + \alpha(1-a)y_1 + \gamma(1-a)y_2\\ &= a(\alpha x_1+\gamma x_2) + (1-a)(\alpha y_1 + \gamma y_2)\\ &\leq a\beta + (1-a)\beta. \end{align*}

• ok you expanded it and rearranged terms.I did the same thing, but the last expression a(αx1+γx2)+(1−a)(αy1+γy2) should be leq than β. I do not understand why you are saying that a(αx1+γx2)+(1−a)(αy1+γy2)≤aβ+(1−a)β – Koba Apr 21 '12 at 23:48
• oh wait so if I expand aβ+(1−a)β it will β, right? – Koba Apr 21 '12 at 23:50
• @Dostre: $\alpha x_1+\gamma x_2\leq \beta$ by assumption; multiplying through by $a$ we get $a(\alpha x_1+\gamma x_2)\leq a\beta$. Similarly, form $\alpha y_1+\gamma y_2\leq \beta$, multiplying through by $(1-a)$ we get $(1-a)(\alpha y_1+\gamma y_2)\leq (1-a)\beta$. Add both inequalities to get the one I have; finally, $a\beta + (1-a)\beta = (a+(1-a))\beta = \beta$. – Arturo Magidin Apr 21 '12 at 23:50
• @Dostre: That's the last step, yes; but you said you didn't understand the last step I did do; I explained it in the comment just above this one. – Arturo Magidin Apr 21 '12 at 23:51
• I see now thank you very much. This problem occupied me for a long time. Thanks. – Koba Apr 21 '12 at 23:53

Same thing Arturo posted in more detail:

We know that the below two inequalities on the far left are true. So lets use them to prove the one we need to prove$[α(ax_1+(1−a)y_1)+γ(ax_2+(1−a)y_2)≤β]$:

$αx_1+γx_2≤β\;\;|*a\Rightarrow a(\alpha x_1+\gamma x_2)\leq a\beta$

$αy_1+γy_2≤β\;\;|*(1-a)\Rightarrow (1-a)(αy_1+γy_2)\leq (1-a)\beta$

Now add the inequalities on the far right side and we get:

$$a(\alpha x_1+\gamma x_2) + (1-a)(\alpha y_1 + \gamma y_2)\leq a\beta+(1-a)\beta$$

After expanding the expressions in parenthesis on the LHS and rearranging the terms we get:

$$α(ax_1+(1−a)y_1)+γ(ax_2+(1−a)y_2)≤a\beta+(1-a)\beta$$

Which almost looks exactly like the one we need to prove. The RHS after expanding:

$$a\beta+(1-a)\beta=a\beta +\beta -a\beta =\beta \Rightarrow$$

$$\Rightarrow a(\alpha x_1+\gamma x_2) + (1-a)(\alpha y_1 + \gamma y_2)\leq\beta$$

Which is what we needed to show.

2&3 questions:

This set M={$x∈ℝ^2_+∣αx_1+γx_2≤β$} looks like a budget constraint and is bounded by:

if $x_1=0;\;$ $\gamma x_2\leq \beta$;$\;\;x_2\leq \frac{\beta}{\gamma}$

if $x_2=0;\;$ $\alpha x_1\leq \beta$;$\;\;x_1\leq \frac{\beta}{\alpha}$