# Help with weierstrass’ theorem

Question: Use the Weierstrass Theorem to show that a solution exists to the expenditure minimization problem of subsection 2.3.2, as long as the utility function II is continuous on $$\mathbb{R}$$ and the price vector p satisfies p » O. What if one of these conditions fails?

$$X(\bar u)$$ = $$\{x\in$$ $$\mathbb{R}^n_+$$| $$u(x)$$ $$\geq$$ $$\bar u$$

The objective is to solve:

Minimize $$p\cdot x$$ Subject to $$x\in X(\bar u)$$.

I do not understand how to actually apply the weierstrass theorem here. Can anyone solve this exercise?

• This question was crossposted. – Giskard Jun 14 '20 at 6:34

The function $$p\mathbin{\cdot} x$$ is continuous. The problem is that $$X(\bar{u})$$ is not compact. So, an additional trick is needed. Since we are working with a subset of $$\mathbb{R}^n$$ compact means the same as closed and bounded (Heine-Borel theorem). The continuity of $$u$$ is used to guarantee that $$X(\bar{u})$$ is closed. But it could be unbounded.
This is where $$p\gg 0$$ becomes useful. Take an arbitrary point $$x_0 \in X(\bar{u})$$ and consider the set $$Y$$ defined by $$Y = \{\,y\in X(\bar{u}) \mid p\cdot y\leq p\cdot x_0\, \}.$$ Since $$x_0\in Y$$, we know that $$Y$$ is nonempty. Using the fact that $$p\gg 0$$, you can show that $$Y$$ is bounded (I'll let you figure out the details). Since $$Y$$ is the intersection of two closed sets, it is also closed. Hence, it is compact. Therefore, by Weierstrass theorem, $$p\cdot x$$ has a minimum in $$Y$$.
I will also let you figure out the last step on your own, which is to show that the minimum of $$p\cdot x$$ in $$Y$$ is also a minimum in $$X(\bar{u})$$.