<p>I assume, by a <strong>strongly monotonically decreasing</strong> preference <span class="math-container">$\succsim$</span>, you mean: for all <span class="math-container">$x,y\in\mathbb R^2_+$</span>,
<span class="math-container">\begin{equation}
x\succ y \text{ only if $x\le y$ and $x\ne y$}. 
\end{equation}</span>
Then it is clear that <span class="math-container">$(0,0)$</span> is the most preferred bundle in <span class="math-container">$\mathbb R^2_+$</span>, i.e. <span class="math-container">$(0,0)\succ x$</span> for all <span class="math-container">$x\in\mathbb R^2$</span> such that <span class="math-container">$x\ne (0,0)$</span>. </p>

<p>Therefore, any utility function representing <span class="math-container">$\succsim$</span> must also satisfy <span class="math-container">$u(0,0)\ge u(x)$</span> for all <span class="math-container">$x\in\mathbb R^2_+$</span>. Hence, 
<span class="math-container">\begin{equation}
(0,0)=\arg\max_{x} u(x)
\end{equation}</span></p>

CC BY-SA 4.0
2018-10-06T16:09:04.883Z
Nice one! Thanks! So you mean that $(0,0)$ would always be a solution but not necessarily the unique one?
PostDocing
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2018-10-06T16:29:44.183Z
@Alessandro if preference is *strongly* monotone then solution must be unique.
Herr K.
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2018-10-06T18:29:21.620Z
But what if $w>0$? I mean $(0,0)$ can be not included in the budget set
PostDocing
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2018-10-06T19:20:50.427Z
@Alessandro: By specifying $\mathbb R^2_+$ as the consumption set, you're already assuming implicitly that all bundles in $\mathbb R^2_+$ are feasible. Plus, why would anyone spend a positive amount on a "bad"?
Herr K.
CC BY-SA 4.0
2018-10-06T06:46:45.070Z
2018-10-06T06:46:45.070Z
Herr K.
<p>I am working with a consumption set <span class="math-container">$X = R_+^2$</span> and preferences that are complete, transitive, continuous and strongly monotonically decreasing. The economy is characterized by the presence of two economic bads. I need to show that "there always exist a solution to the consumer's utility maximization problem". </p>

<p>My point is that if I had a continuous utility function over a compact budget set, using Weierstrass theorem I have the certainty of the function reaching its extrema. The problem here, if I am right, is that the budget set is no longer bounded and thus neither compact. </p>

<p>Are there other results I can use to show the statement? </p>

CC BY-SA 4.0
2018-10-06T05:12:21.830Z
I apologize if this is naive: But if the budget set B is a subset of $R^2_+$ and the utility function U is continuous in $R^2_+$, then the function $U:B->R^2_+$ is bounded. So the sup and inf of U[.] in B exist. Which means that in the budget set the utility function has a solution. I do not understand the problem with this logic. Again, sorry if this is naive.
erik
CC BY-SA 4.0
2018-10-06T06:56:20.900Z
@erik: Note that the consumption set is the entire $\mathbb R^2_+$, not a subset of it. And since $\mathbb R^2_+$ is not bounded and thus not compact, we can't directly apply Weierstrass theorem.
Herr K.
CC BY-SA 4.0
2018-10-06T06:58:43.770Z
@ Herr K: I saw X = $R^2_+$ and assumed that this meant that all goods are measured on the positive quadrant...I glossed over the consumption set before it. Thank you for you clarification.
erik
<microeconomics><utility>
Existence of maximum utility with two bads
PostDocing
# Existence of maximum utility with two bads

(Asked by: username_0 on 2018-10-05)

I am working with a consumption set $X = R\_+^2$ and preferences that are complete, transitive, continuous and strongly monotonically decreasing. The economy is characterized by the presence of two economic bads. I need to show that "there always exist a solution to the consumer's utility maximization problem".

My point is that if I had a continuous utility function over a compact budget set, using Weierstrass theorem I have the certainty of the function reaching its extrema. The problem here, if I am right, is that the budget set is no longer bounded and thus neither compact.

Are there other results I can use to show the statement?

**Tags:** microeconomics, utility
> **Comments:**
> **username_1**: I apologize if this is naive: But if the budget set B is a subset of $R^2_+$ and the utility function U is continuous in $R^2_+$, then the function $U:B->R^2_+$ is bounded. So the sup and inf of U[.] in B exist. Which means that in the budget set the utility function has a solution. I do not understand the problem with this logic. Again, sorry if this is naive.  
> **username_2**: @erik: Note that the consumption set is the entire $\mathbb R^2_+$, not a subset of it. And since $\mathbb R^2_+$ is not bounded and thus not compact, we can't directly apply Weierstrass theorem.  
> **username_1**: @ Herr K: I saw X = $R^2_+$ and assumed that this meant that all goods are measured on the positive quadrant...I glossed over the consumption set before it. Thank you for you clarification.  

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## 1 answers

### Answer by username_2 on 2018-10-06. Score: 3 | Accepted answer

I assume, by a **strongly monotonically decreasing** preference $\succsim$, you mean: for all $x,y\in\mathbb R^2\_+$,
\begin{equation}
x\succ y \text{ only if $x\le y$ and $x\ne y$}.
\end{equation}
Then it is clear that $(0,0)$ is the most preferred bundle in $\mathbb R^2\_+$, i.e. $(0,0)\succ x$ for all $x\in\mathbb R^2$ such that $x\ne (0,0)$.

Therefore, any utility function representing $\succsim$ must also satisfy $u(0,0)\ge u(x)$ for all $x\in\mathbb R^2\_+$. Hence,
\begin{equation}
(0,0)=\arg\max\_{x} u(x)
\end{equation}

> **Comments:**
> **username_0**: Nice one! Thanks! So you mean that $(0,0)$ would always be a solution but not necessarily the unique one?  
> **username_2**: @Alessandro if preference is *strongly* monotone then solution must be unique.  
> **username_0**: But what if $w>0$? I mean $(0,0)$ can be not included in the budget set  
> **username_2**: @Alessandro: By specifying $\mathbb R^2_+$ as the consumption set, you're already assuming implicitly that all bundles in $\mathbb R^2_+$ are feasible. Plus, why would anyone spend a positive amount on a "bad"?  

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