# Prove the equation

Let $$x^0=x^*(p^0,w)$$ then $$v(p,px^0)$$ is minimized at $$p=p^0$$

What theorem are we supposed to use in order to solve this, because I am a bit lost. Thanks in advance for all the suggestions/help.

• Is this homework? Further, you should give some more details about functions $x^*(.)$ and $v(.)$. – Dayne Nov 28 '20 at 14:36
• It's not homework, but I'm trying to solve some exercises to practice myself. I wish I could give you more info but that's all that's written there. That's why I'm finding difficulties understanding the question :/ – Maybeline Lee Nov 28 '20 at 14:37

1. $$x^0=x^*(p^0,w)$$ means that $$p^0x^0\leq w$$ and that $$u(x)\leq u(x^0)$$ for all $$x$$ such that $$p^0x\leq w$$. In words $$x_0$$ is affordable and at least as good as every other affordable bundle (in many cases, better and not just at least as good).
2. $$v(p,m)=\sup\{u(x)\mid px\leq m\}$$ by definition (often one can replace $$\sup$$ by $$\max$$).
Prove first, that $$v(p,px^0)\geq u(x^0)$$ for all $$p$$. Then prove that $$u(x^0)=v(p^0,p^0x^0)$$. Combining the two gives you the result.
So $$v(p,px^0)=\sup\{u(x)\mid px\leq px^0\}.$$ Since $$x^0\in\{x\mid px\leq px^0\}$$, $$u(x_0)\in\{u(x)\mid px\leq px^0\}$$, Therefore, $$v(p,px^0)\geq u(x_0)$$.
Also, note that $$\{x\mid p^0x\leq p^0x^0\}\subseteq \{x\mid p^0x\leq w\}$$ since $$p^0x^0\leq w$$. We also now that $$u(x^0)\geq u(x)$$ for all $$x\in \{x\mid p^0x\leq w\}$$ and, therefore, also $$u(x^0)\geq u(x)$$ for all $$x\in \{x\mid p^0x\leq p^0x^0\},$$ which implies $$u(x_0)\geq\{u(x)\mid p^0x\leq p^0x^0\}=v(p^0,p^0x^0)$$. Also, since $$x^0\in\{x\mid p^0x\leq p^0x^0\}$$, we have $$u(x_0)\leq\sup\{u(x)\mid p^0x\leq p^0x^0\}=v(p^0,p^0x^0)$$. Together $$u(x^0)=v(p^0,p^0x^0)$$. It follows that $$v(p,px^0)\geq u(x_0)=v(p^0,p^0x^0)$$ for all $$p$$, which means that $$v(p,px^0)$$ is minimized at $$p$$.