<p>The set need not be bounded. To see this, just take <span class="math-container">$U$</span> to be constant. Then the set will either be empty or equal to <span class="math-container">$\mathbb{R}_+^n$</span>.</p>
<p>It is also possible that no minimum exists. Let <span class="math-container">$n=2$</span>, <span class="math-container">$p=(0,1)$</span>, <span class="math-container">$v=1$</span>, and <span class="math-container">$U$</span> be given by <span class="math-container">$U(x)=U(x_1,x_2)=x_1\cdot x_2$</span>. Clearly, <span class="math-container">$px=0$</span> is only possible if <span class="math-container">$x_1=0$</span>, which would lead to a utility of <span class="math-container">$0$</span>. But for every <span class="math-container">$\epsilon&gt;0$</span>, the bundle <span class="math-container">$(1/\epsilon,\epsilon)$</span> has utility <span class="math-container">$1$</span> and a price of <span class="math-container">$\epsilon$</span>. So no expenditure minimizing bundle exists. Of course, there is also no expenditure minimzing bundle if there exists no <span class="math-container">$x$</span> at all such that <span class="math-container">$U(x)\geq v$</span>. This can happen if <span class="math-container">$U$</span> is bounded above.</p>
<p>However, an expenditure minimizing bundle exists if it holds that prices are strictly positive and there is at least one bundle <span class="math-container">$x^*$</span> such that <span class="math-container">$U(x^*)\geq v$</span>. To see this, note that minimizing expenditure on the set
<span class="math-container">$$\{x\in\mathbb{R}_+^n\mid U(x)\geq v\}$$</span>
is equal to minimizing expenditure on the set
<span class="math-container">$$\{x\in\mathbb{R}_+^n\mid U(x)\geq v, px\leq px^*\}.$$</span>
The latter set is closed and bounded.</p>

CC BY-SA 4.0
2020-06-23T17:51:49.453Z
But how do you determine x*?
mmendina
CC BY-SA 4.0
2020-06-23T18:01:46.593Z
Any bundle that is good enough will do. Whether such a bundle exists will generally depend on the utility function. If the utility function is not bounded above, such a bundle exists necessarily. Otherwise, we have to assume it exists.
Michael Greinecker
CC BY-SA 4.0
2020-06-23T17:18:22.047Z
2020-06-23T17:18:22.047Z
Michael Greinecker
<p>I need to prove that the following set is bounded in order to derive the expenditure function:</p>
<p><span class="math-container">$e(p,v)=min_x px$</span>
ST
<span class="math-container">$\{x \in R^n_+$</span> such that <span class="math-container">$U(x)\geq v\}$</span>.</p>
<p>Knowing that <span class="math-container">$U(x):R^n \longrightarrow R$</span> is a continous function.</p>
<p>I already proved that the set is closed and I think that if the set is closed for being <span class="math-container">$U(x)$</span> continous, the set must be bounded. However I'm not sure.</p>
<p>Please help.
Thanks in advance.</p>

<mathematical-economics><utility><consumer-theory><preferences>
Expenditure function. Prove that this set is bounded
mmendina
# Expenditure function. Prove that this set is bounded

(Asked by: username_0 on 2020-06-23)

I need to prove that the following set is bounded in order to derive the expenditure function:

$e(p,v)=min\_x px$
ST
$\{x \in R^n\_+$ such that $U(x)\geq v\}$.

Knowing that $U(x):R^n \longrightarrow R$ is a continous function.

I already proved that the set is closed and I think that if the set is closed for being $U(x)$ continous, the set must be bounded. However I'm not sure.

Please help.
Thanks in advance.

**Tags:** mathematical-economics, utility, consumer-theory, preferences

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## 1 answers

### Answer by username_1 on 2020-06-23. Score: 4 | Accepted answer

The set need not be bounded. To see this, just take $U$ to be constant. Then the set will either be empty or equal to $\mathbb{R}\_+^n$.

It is also possible that no minimum exists. Let $n=2$, $p=(0,1)$, $v=1$, and $U$ be given by $U(x)=U(x\_1,x\_2)=x\_1\cdot x\_2$. Clearly, $px=0$ is only possible if $x\_1=0$, which would lead to a utility of $0$. But for every $\epsilon>0$, the bundle $(1/\epsilon,\epsilon)$ has utility $1$ and a price of $\epsilon$. So no expenditure minimizing bundle exists. Of course, there is also no expenditure minimzing bundle if there exists no $x$ at all such that $U(x)\geq v$. This can happen if $U$ is bounded above.

However, an expenditure minimizing bundle exists if it holds that prices are strictly positive and there is at least one bundle $x^\*$ such that $U(x^\*)\geq v$. To see this, note that minimizing expenditure on the set
$$\{x\in\mathbb{R}\_+^n\mid U(x)\geq v\}$$
is equal to minimizing expenditure on the set
$$\{x\in\mathbb{R}\_+^n\mid U(x)\geq v, px\leq px^\*\}.$$
The latter set is closed and bounded.

> **Comments:**
> **username_0**: But how do you determine x*?  
> **username_1**: Any bundle that is good enough will do. Whether such a bundle exists will generally depend on the utility function. If the utility function is not bounded above, such a bundle exists necessarily. Otherwise, we have to assume it exists.  

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