<p>Let us prove it by contradiction. Suppose that $x^{*}$ is optimal for the first problem but not for the second. Since $x^{*}$ is not optimal for the expenditure minimization problem, there exists a bundle $x$ such that 
\begin{equation*}
u(x) \geq u(x^{*}) \text{ and } p \cdot x &lt; p \cdot x^{*}
\end{equation*}</p>

<p>Since the function $y \rightarrow p.y$ is continuous, and $p \cdot x &lt; p \cdot x^{*}$, there exists $\epsilon&gt;0$ such that $\|y-x\|&lt;\epsilon \Rightarrow p \cdot y &lt; p \cdot x^{*}$. In addition, since preferences are locally nonsatiated, there exists $y$ such that $\|y-x\|&lt;\epsilon$ and $u(y)&gt;u(x)$. Take such a $y$. We have \begin{equation*}
u(y) &gt; u(x) \geq u(x^{*}) \text{ and } p \cdot y &lt; p \cdot x^{*} \leq w
\end{equation*}</p>

<p>Or, in other words,
\begin{equation*}
u(y)&gt;u(x^{*}) \text{ and } p.y &lt; w
\end{equation*}
which contradicts the assumption that $x^{*}$ is a solution of the utility maximization problem.</p>

CC BY-SA 3.0
2016-03-05T14:01:04.970Z
I think you may want to use "\cdot" instead of "."
(The reason I do not upvote your correct answer is not this. It is because I think this was a homework question with zero effort shown.)
Giskard
CC BY-SA 3.0
2016-03-05T14:29:13.790Z
@denesp You are right, I shouldn't have answered. Your use of incentives makes sense^^.
Oliv
CC BY-SA 3.0
2016-03-05T13:12:04.360Z
2016-03-05T14:28:16.350Z
2016-03-05T14:28:16.350Z
Oliv
<p><em>For solving this question, which assumptions/axioms should be used and should Lagrange multipliers be equal or different?</em> </p>

<p>Think of an agent with utility function u() and wealth w. Assume that u() is continuus and represents a locally non-satiated preference relation. The utility maximization problem of this agent can be stated as:</p>

<p>max  u(x)
 s.t. p.x ≤ w</p>

<p>On the other hand, when the agent’s constraint is to attain a utility level h, her expenditure minimization problem can be stated as:</p>

<p>min  p.x 
s.t. u(x) ≥ h</p>

<p><strong>Show that if x* is optimal in the utility maximization problem when w > 0, then x* is optimal in the expenditure minimization problem when the constraint is to attain a utility level u(x∗).</strong></p>

CC BY-SA 3.0
2016-03-08T18:41:14.497Z
Have you tried the question yourself? What have you done and where did you get stuck?
Herr K.
<duality>
Duality Proof - Consumer Theory
Ktp
# Duality Proof - Consumer Theory

(Asked by: username_0 on 2016-03-05)

*For solving this question, which assumptions/axioms should be used and should Lagrange multipliers be equal or different?*

Think of an agent with utility function u() and wealth w. Assume that u() is continuus and represents a locally non-satiated preference relation. The utility maximization problem of this agent can be stated as:

max u(x)
s.t. p.x ≤ w

On the other hand, when the agent’s constraint is to attain a utility level h, her expenditure minimization problem can be stated as:

min p.x
s.t. u(x) ≥ h

**Show that if x\* is optimal in the utility maximization problem when w > 0, then x\* is optimal in the expenditure minimization problem when the constraint is to attain a utility level u(x∗).**

**Tags:** duality

> **Comments:**
> **username_1**: Have you tried the question yourself? What have you done and where did you get stuck?  

---

## 1 answers

### Answer by username_2 on 2016-03-05. Score: 1 | Accepted answer

Let us prove it by contradiction. Suppose that $x^{\*}$ is optimal for the first problem but not for the second. Since $x^{\*}$ is not optimal for the expenditure minimization problem, there exists a bundle $x$ such that
\begin{equation\*}
u(x) \geq u(x^{\*}) \text{ and } p \cdot x < p \cdot x^{\*}
\end{equation\*}

Since the function $y \rightarrow p.y$ is continuous, and $p \cdot x < p \cdot x^{\*}$, there exists $\epsilon>0$ such that $\|y-x\|<\epsilon \Rightarrow p \cdot y < p \cdot x^{\*}$. In addition, since preferences are locally nonsatiated, there exists $y$ such that $\|y-x\|<\epsilon$ and $u(y)>u(x)$. Take such a $y$. We have \begin{equation\*}
u(y) > u(x) \geq u(x^{\*}) \text{ and } p \cdot y < p \cdot x^{\*} \leq w
\end{equation\*}

Or, in other words,
\begin{equation\*}
u(y)>u(x^{\*}) \text{ and } p.y < w
\end{equation\*}
which contradicts the assumption that $x^{\*}$ is a solution of the utility maximization problem.

> **Comments:**
> **username_3**: I think you may want to use "\cdot" instead of "." (The reason I do not upvote your correct answer is not this. It is because I think this was a homework question with zero effort shown.)  
> **username_2**: @denesp You are right, I shouldn't have answered. Your use of incentives makes sense^^.  

---

