<p>This is the version of the theorem I will use: </p>

<p><strong>Weierstrass theorem</strong>: <em>Every continuous function defined over a compact set attains a minimum.</em> </p>

<p>The function <span class="math-container">$p\mathbin{\cdot} x$</span> is continuous. The problem is that <span class="math-container">$X(\bar{u})$</span> is not compact. So, an additional trick is needed. Since we are working with a subset of <span class="math-container">$\mathbb{R}^n$</span> compact means the same as closed and bounded (Heine-Borel theorem). The continuity of <span class="math-container">$u$</span> is used to guarantee that <span class="math-container">$X(\bar{u})$</span> is closed. But it could be unbounded. </p>

<p>This is where <span class="math-container">$p\gg 0$</span> becomes useful. Take an arbitrary point <span class="math-container">$x_0 \in X(\bar{u})$</span> and consider the set <span class="math-container">$Y$</span> defined by  <span class="math-container">$$Y = \{\,y\in X(\bar{u}) \mid p\cdot y\leq p\cdot x_0\,   \}.$$</span>
Since <span class="math-container">$x_0\in Y$</span>, we know that <span class="math-container">$Y$</span> is nonempty. Using the fact that <span class="math-container">$p\gg 0$</span>, you can show that <span class="math-container">$Y$</span> is bounded (I'll let you figure out the details). Since <span class="math-container">$Y$</span> is the intersection of two closed sets, it is also closed. Hence, it is compact. Therefore, by Weierstrass theorem, <span class="math-container">$p\cdot x$</span> has a minimum in <span class="math-container">$Y$</span>. </p>

<p>I will also let you figure out the last step on your own, which is to show that the minimum of <span class="math-container">$p\cdot x$</span> in <span class="math-container">$Y$</span> is also a minimum in <span class="math-container">$X(\bar{u})$</span>.</p>

CC BY-SA 4.0
2020-06-06T14:49:45.713Z
2020-06-08T17:01:35.550Z
2020-06-08T17:01:35.550Z
brunosalcedo
<p>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 <span class="math-container">$\mathbb{R}$</span> and the price vector p satisfies p » O. What if one of these conditions fails?</p>

<p><span class="math-container">$X(\bar u)$</span> = <span class="math-container">$\{x\in$</span> <span class="math-container">$\mathbb{R}$$^n$$_+$</span>| <span class="math-container">$u(x)$</span> <span class="math-container">$\geq$</span> <span class="math-container">$\bar u$</span></p>

<p>The objective is to solve:</p>

<p>Minimize <span class="math-container">$p\cdot x$</span> Subject to <span class="math-container">$x\in X(\bar u)$</span>. </p>

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

CC BY-SA 4.0
2020-06-14T06:34:16.477Z
This question was [crossposted](https://math.stackexchange.com/questions/3708263/using-the-weierstrass-theorem-to-prove-a-solution-to-a-minimization-problem-exis).
Giskard
<mathematical-economics>
Help with weierstrass’ theorem
Tony456
# Help with weierstrass’ theorem

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

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?

**Tags:** mathematical-economics
> **Comments:**
> **username_1**: This question was [crossposted](https://math.stackexchange.com/questions/3708263/using-the-weierstrass-theorem-to-prove-a-solution-to-a-minimization-problem-exis).  

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## 1 answers

### Answer by username_2 on 2020-06-06. Score: 5

This is the version of the theorem I will use:

**Weierstrass theorem**: *Every continuous function defined over a compact set attains a minimum.*

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 $X(\bar{u})$.


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