Expenditure function. Prove that this set is bounded 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.
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 $\epsilon&gt;0$, 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.