<p>You don't need to use any fancy theorem, the trick is to disentangle the definitions. Everything follows directly from the definitions.</p>
<ol>
<li><span class="math-container">$x^0=x^*(p^0,w)$</span> means that <span class="math-container">$p^0x^0\leq w$</span> and that <span class="math-container">$u(x)\leq u(x^0)$</span> for all <span class="math-container">$x$</span> such that <span class="math-container">$p^0x\leq w$</span>. In words <span class="math-container">$x_0$</span> is affordable and at least as good as every other affordable bundle (in many cases, better and not just at least as good).</li>
<li><span class="math-container">$v(p,m)=\sup\{u(x)\mid px\leq m\}$</span> by definition (often one can replace <span class="math-container">$\sup$</span> by <span class="math-container">$\max$</span>).</li>
</ol>
<p>Prove first, that <span class="math-container">$v(p,px^0)\geq u(x^0)$</span> for all <span class="math-container">$p$</span>. Then prove that <span class="math-container">$u(x^0)=v(p^0,p^0x^0)$</span>. Combining the two gives you the result.</p>
<p>Details:</p>
<p>So <span class="math-container">$$v(p,px^0)=\sup\{u(x)\mid px\leq px^0\}.$$</span>
Since <span class="math-container">$x^0\in\{x\mid px\leq px^0\}$</span>, <span class="math-container">$u(x_0)\in\{u(x)\mid px\leq px^0\}$</span>,
Therefore, <span class="math-container">$v(p,px^0)\geq u(x_0)$</span>.</p>
<p>Also, note that <span class="math-container">$\{x\mid p^0x\leq p^0x^0\}\subseteq \{x\mid p^0x\leq w\}$</span> since <span class="math-container">$p^0x^0\leq w$</span>. We also now that <span class="math-container">$u(x^0)\geq u(x)$</span> for all <span class="math-container">$x\in \{x\mid p^0x\leq w\}$</span> and, therefore, also <span class="math-container">$u(x^0)\geq u(x)$</span> for all <span class="math-container">$x\in \{x\mid p^0x\leq p^0x^0\},$</span> which implies <span class="math-container">$u(x_0)\geq\{u(x)\mid p^0x\leq p^0x^0\}=v(p^0,p^0x^0)$</span>. Also, since <span class="math-container">$x^0\in\{x\mid p^0x\leq p^0x^0\}$</span>, we have <span class="math-container">$u(x_0)\leq\sup\{u(x)\mid p^0x\leq p^0x^0\}=v(p^0,p^0x^0)$</span>.  Together <span class="math-container">$u(x^0)=v(p^0,p^0x^0)$</span>. It follows that
<span class="math-container">$$v(p,px^0)\geq u(x_0)=v(p^0,p^0x^0)$$</span> for all <span class="math-container">$p$</span>, which means that <span class="math-container">$v(p,px^0)$</span> is minimized at <span class="math-container">$p$</span>.</p>

CC BY-SA 4.0
2020-11-28T15:03:51.897Z
2020-11-28T15:03:51.897Z
Michael Greinecker
<p>Let <span class="math-container">$$ x^0=x^*(p^0,w)$$</span> then <span class="math-container">$$v(p,px^0)$$</span> is minimized at <span class="math-container">$$p=p^0$$</span></p>
<p>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.</p>

CC BY-SA 4.0
2020-11-28T14:36:12.967Z
Is this homework? Further, you should give some more details about functions $x^*(.)$ and $v(.)$.
Dayne
CC BY-SA 4.0
2020-11-28T14:37:45.387Z
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
<microeconomics><demand><walrasian>
Prove the equation
Maybeline Lee
# Prove the equation

(Asked by: username_0 on 2020-11-28)

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.

**Tags:** microeconomics, demand, walrasian

> **Comments:**
> **username_1**: Is this homework? Further, you should give some more details about functions $x^*(.)$ and $v(.)$.  
> **username_0**: 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 :/  

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## 1 answers

### Answer by username_2 on 2020-11-28. Score: 6 | Accepted answer

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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