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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.

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  • $\begingroup$ Is this homework? Further, you should give some more details about functions $x^*(.)$ and $v(.)$. $\endgroup$ – Dayne Nov 28 '20 at 14:36
  • $\begingroup$ 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 :/ $\endgroup$ – Maybeline Lee Nov 28 '20 at 14:37
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You don't need to use any fancy theorem, the trick is to disentangle the definitions. Everything follows directly from the definitions.

  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.

Details:

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 $p$.

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