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.


1 Answer 1


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.

  • $\begingroup$ But how do you determine x*? $\endgroup$ Jun 23, 2020 at 17:51
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    $\begingroup$ 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. $\endgroup$ Jun 23, 2020 at 18:01

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