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.

  • $\begingroup$ Is this homework? Further, you should give some more details about functions $x^*(.)$ and $v(.)$. $\endgroup$
    – Dayne
    Nov 28, 2020 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 :/ $\endgroup$ Nov 28, 2020 at 14:37

1 Answer 1


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.


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 $p$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.