<p>I hope I understand correctly your question.</p>
<p>I think that, on the basis of the conditions you wrote,  you can't be sure that a maximum of <span class="math-container">$Y(a_1,a_2)$</span> exists.</p>
<p>How can we establish that a maximum of a function exists?</p>
<p>The most well-known theorem is the Weierstrass' Theorem, that ensures that a continuous function on a compact set <span class="math-container">$X$</span> to <span class="math-container">$\mathbb{R}$</span> has a   global maximum and a minimum on <span class="math-container">$X$</span>.
In your case, <span class="math-container">$Y(a_1,a_2)$</span> is defined on <span class="math-container">$\mathbb{R^2_{\geq 0}}$</span>, which is not compact, because it is unlimited.</p>
<p>So, Weierstrass' Theorem doesn't apply.</p>
<p>As the function <span class="math-container">$Y(a_1,a_2)$</span> is continously differentiable, we can resort to conditions on first order derivatives, and second order conditions on  derivatives in the case the function is twice diffentiable.</p>
<p>For a point to be of an <strong>interior</strong> point of (relative) maximum a necessary condition is that the partial derivatives are 0 in that point.  But to establish  if, in that point,  we have a maximum or not, we have to look at the Hessian, that is to the second order derivatives.</p>
<p>But you have only conditions on the first order derivatives, and no conditions on the second order derivatives, neither if they exist or not. So, nothing can be said.</p>
<p>Besides interior points of maximum, we could have points of (global) maximum on the border of the domain of the function <span class="math-container">$Y$</span>, <span class="math-container">$\mathbb{R^2_{\geq0}}$</span>, that is the points for which <span class="math-container">$a_1=0$</span> or <span class="math-container">$a_2=0$</span>.</p>
<p>But again, we cannot say anything about the existence of such points on the border.</p>
<p>Of course, there can be other ways to prove the existence of a maximum. What I mean is that proving that a maximum always exists under the given conditions is not obviuos, and proving its existence, in the case it exists, is not trivial.</p>

CC BY-SA 4.0
2022-09-16T07:15:53.577Z
While this highlight why some usual ways do not work, it does not proof that a maximum does not exist. E.g., choose the $\mathbb{R}^2$ square with corners $(0,0)$ and $(A+B,A+B)$ as your compact set. Per Weierstrass a maximum exists over this set; you only have to prove that the function $Y$ never reaches a higher value in points outside of this set.
Giskard
CC BY-SA 4.0
2022-09-16T07:34:01.703Z
@ Giskard Eli j has changed her/his question, previosly, as you can see in the comments above, She/he seemed to ask if one can be sure that a maximum exists, and a said no, for the reasons I wrote. Now his/ question is different. Of corse a maximum can exist, Weirtrass gives only sufficient condition, and a maximum exists if you restrict the function to a compact set.
BakerStreet
CC BY-SA 4.0
2022-09-16T08:08:27.797Z
@ Giskard Eli J in the comments asked me explicitly to answer the question if we can be sure that a maximum exists, on the basis of the given conditions.
BakerStreet
CC BY-SA 4.0
2022-09-16T08:43:35.960Z
1. According to the timestamps, the as of this writing the final edit of the question was at 14:42, about an hour before you first posted your answer. 2. Yes, my comment is saying that you are not proving they cannot be sure of this given the conditions. By showing you cannot prove something this and that way, you are not disproving it.
Giskard
CC BY-SA 4.0
2022-09-16T09:47:51.643Z
@ I referred to the comments, where Eli J was asking me to say if we can be sure that a maximum exists. This was the reason of my answer. Of course I haven't disproved that a maximum can exist, and it was not my purpose. I know the difference, I'm not a young student.
BakerStreet
CC BY-SA 4.0
2022-09-16T09:54:58.487Z
Okay, so to be clear: you are confident saying that we cannot be sure a maximum exists, i.e. it is impossible to prove that a maximum exist?
Giskard
CC BY-SA 4.0
2022-09-16T10:03:17.023Z
@ No, it is not impossible, why?  Of course a maximum can exist, but must be proved, and we  cannot refer here to the usual theorems, as Weierstrass. Of course we can give an exmaple of fa unction for which the maximum exists, but it is not obvious to prove that a maximum always exists  under the given conditions.
BakerStreet
CC BY-SA 4.0
2022-09-16T10:07:06.050Z
Okay. I guess I was expecting a more generalist answer, hence my misunderstanding. Thank you for the clarification.
Giskard
CC BY-SA 4.0
2022-09-16T10:13:57.823Z
You are welcome.
BakerStreet
CC BY-SA 4.0
2022-09-15T15:55:50.220Z
2022-09-17T13:48:13.717Z
2022-09-17T13:48:13.717Z
BakerStreet
<p>I have the following <strong>net utility</strong> function which is made up of one positive utility (with a bliss point) and two negative utility (i.e., disutility) functions;</p>
<p><span class="math-container">$$Y(a_1,a_2)=y(a_1)+v(a_1,a_2)+w(a_2).$$</span></p>
<p>The functions <span class="math-container">$y(a_1)$</span>, <span class="math-container">$v(a_1, a_2)$</span> may possess a trade off. For example, consuming at the bliss point in <span class="math-container">$y$</span> and minimising the disutility in <span class="math-container">$w$</span> may mean suffering a significantly high disutility in <span class="math-container">$v$</span>.</p>
<p><strong>The functions:</strong></p>
<p><span class="math-container">$$y(a_1): \mathbb {R_{\ge0}}\to \mathbb{R}$$</span></p>
<p><span class="math-container">$$v(a_1, a_2): \mathbb {R_{\ge0}^2} \to \mathbb {R_{\le0}}$$</span></p>
<p><span class="math-container">$$w (a_2): \mathbb {R_{\ge0}} \to \mathbb {R_{\le0}}$$</span></p>
<p>I would like to know if following the derivatives (see below) of the functions <span class="math-container">$y(a_1)$</span>, <span class="math-container">$v(a_1, a_2)$</span> and <span class="math-container">$w(a_2)$</span> permit the function <span class="math-container">$Y$</span> to converge to a maximum. In other words, a rational economic agent who is making the best decision at each point (by studying the marginal conditions) will eventually arrive at a maximum of <span class="math-container">$Y$</span>.</p>
<p><strong>The derivatives:</strong>
<span class="math-container">$$
y'(a_1)=
\begin{cases}
 &gt;0&amp;\text{if}\, a_1\in[0,A)\\
 0&amp;\text{if}\, a_1=A\\
      &lt;0&amp;\text{if}\ a_1 \in (A,\infty)
\end{cases}
$$</span></p>
<p><span class="math-container">$$y''(a_1)&lt;0$$</span></p>
<p><span class="math-container">$$
\frac {\partial v(a_1,a_2)}{a_i}=
\begin{cases}
 \geq0&amp;\text{if}\, a_i &lt;a_j\\\
 0&amp;\text{if}\, a_i=a_j\\
      \leq0&amp;\text{if}\ a_i&gt;a_j, \text{where} \, i\neq j
\end{cases} 
$$</span></p>
<p><span class="math-container">$$
w'(a_2)=
\begin{cases}
 \geq0&amp;\text{if}\, a_2&lt;B\\
 0&amp;\text{if}\, a_2=B\\
      \leq 0&amp;\text{if}\ a_2 &gt;B
\end{cases}
$$</span></p>
<p><strong>I would then like to say</strong>:
<span class="math-container">$$a_1^* \in \arg \max_{a_1} Y(a_1,a_2)=y(a_1)+v(a_1,a_2)+w(a_2)$$</span>
<span class="math-container">$$a_2^* \in \arg \max_{a_2} Y(a_1,a_2)=y(a_1)+v(a_1,a_2)+w(a_2),$$</span></p>
<p>where <span class="math-container">$a_1^*$</span> and <span class="math-container">$a_2^*$</span> can be found by following the marginal conditions (i.e., the derivatives) above.</p>
<p><strong>Key example:</strong></p>
<p>Suppose <span class="math-container">$a_1 = A$</span> and <span class="math-container">$a_2 = B$</span> such that <span class="math-container">$A&gt;&gt;B$</span>. In this case, the function <span class="math-container">$v (A,B)$</span> may be significantly negative, even though <span class="math-container">$y(a_1)$</span> and <span class="math-container">$w(a_2)$</span> are maximised. It may then be advantageous to bring <span class="math-container">$a_1$</span> closer to <span class="math-container">$a_2$</span> to increase (i.e., bring closer to <span class="math-container">$0$</span>) the disutility from the value <span class="math-container">$v(A,B)$</span>. The marginal incentive of such convergence is given by comparing the derivatives of the three functions.</p>

CC BY-SA 4.0
2022-09-15T11:52:26.160Z
Not sure I understand your exact question. Are you asking if the conditions guarantee that a maximum exists over $\mathbb{R}^2_{\geq 0}$?
Giskard
CC BY-SA 4.0
2022-09-15T12:32:04.703Z
@ Giskard Yes, and that the function $Y$ is continuously differentiable over $\mathbb {R_{\ge 0}^2}$.; that is, $\frac {\partial Y}{\partial a_i}$ is defined across the whole domain as per the functions $y$, $v$, and $w$. So for example, if $a_1=0$, then $y'(0) >0$, $\frac {\partial v}{\partial a_1} \ge 0$ if $a_1<a_2$, and so on... as per the conditions. Essentially, I want to make sure that I didn't miss anything in any function that would violate continuous differentiability and the obtainment of a maximum.
Eli J
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2022-09-15T12:36:43.877Z
Hi! I still don't understand :| These things that you list in your comment are the assumptions/conditions again, right? So, you are asking if, given these assumptions, the function has a maximum? Or are you also asking something else that I do not get?
Giskard
CC BY-SA 4.0
2022-09-15T12:53:56.447Z
@Giskard. Sorry, the fault is mine for being unclear. I have been working on this problem for a while and am getting a little off. I think I just want to make sure the function I have created behaves in the way I want it to. I want the conditions for the derivatives to make sense across the whole of $Y$, one key consequence of which will allow me to obtain a maximum of $Y$. In "baby language", I want to be able to say that any change of $a_1$ and $a_2$ will be well defined in the function $Y$ and that there is some $a_1$ and $a_2$ that maximises $Y$. I hope this helps...
Eli J
CC BY-SA 4.0
2022-09-15T14:36:40.550Z
@Giskard I have edited the title and  question significantly to more clearly reflect what I (think) I was asking.
Eli J
CC BY-SA 4.0
2022-09-15T15:06:59.080Z
@ Eli J I 'm not sure to understand the question in your title. What's the meaning of 'converge' to a maximum'? Convergence is the language of limits. Maybe you want to know if you can be sure that a maximum of $Y(a_1,a_2)$ exists, isn't it?
BakerStreet
CC BY-SA 4.0
2022-09-15T15:19:37.673Z
@BakerStreet You may be right, I’m not sure to be honest. If you can kindly provide such a solution then I may be able to say.
Eli J
CC BY-SA 4.0
2022-09-16T12:37:05.390Z
Are you asking about the existence of a global maximum, a local maximum, or whether a gradient ascent process would always converge to such a maximum?
VARulle
<mathematical-economics><optimization>
Can following these marginal conditions have the net utility function converge to a maximum? [Edited]
Eli J
# Can following these marginal conditions have the net utility function converge to a maximum? [Edited]

(Asked by: username_0 on 2022-09-15)

I have the following **net utility** function which is made up of one positive utility (with a bliss point) and two negative utility (i.e., disutility) functions;

$$Y(a\_1,a\_2)=y(a\_1)+v(a\_1,a\_2)+w(a\_2).$$

The functions $y(a\_1)$, $v(a\_1, a\_2)$ may possess a trade off. For example, consuming at the bliss point in $y$ and minimising the disutility in $w$ may mean suffering a significantly high disutility in $v$.

**The functions:**

$$y(a\_1): \mathbb {R\_{\ge0}}\to \mathbb{R}$$

$$v(a\_1, a\_2): \mathbb {R\_{\ge0}^2} \to \mathbb {R\_{\le0}}$$

$$w (a\_2): \mathbb {R\_{\ge0}} \to \mathbb {R\_{\le0}}$$

I would like to know if following the derivatives (see below) of the functions $y(a\_1)$, $v(a\_1, a\_2)$ and $w(a\_2)$ permit the function $Y$ to converge to a maximum. In other words, a rational economic agent who is making the best decision at each point (by studying the marginal conditions) will eventually arrive at a maximum of $Y$.

**The derivatives:**
$$
y'(a\_1)=
\begin{cases}
>0&\text{if}\, a\_1\in[0,A)\\
0&\text{if}\, a\_1=A\\
<0&\text{if}\ a\_1 \in (A,\infty)
\end{cases}
$$

$$y''(a\_1)<0$$

$$
\frac {\partial v(a\_1,a\_2)}{a\_i}=
\begin{cases}
\geq0&\text{if}\, a\_i a\_j, \text{where} \, i\neq j
\end{cases}
$$

$$
w'(a\_2)=
\begin{cases}
\geq0&\text{if}\, a\_2B
\end{cases}
$$

**I would then like to say**:
$$a\_1^\* \in \arg \max\_{a\_1} Y(a\_1,a\_2)=y(a\_1)+v(a\_1,a\_2)+w(a\_2)$$
$$a\_2^\* \in \arg \max\_{a\_2} Y(a\_1,a\_2)=y(a\_1)+v(a\_1,a\_2)+w(a\_2),$$

where $a\_1^\*$ and $a\_2^\*$ can be found by following the marginal conditions (i.e., the derivatives) above.

**Key example:**

Suppose $a\_1 = A$ and $a\_2 = B$ such that $A>>B$. In this case, the function $v (A,B)$ may be significantly negative, even though $y(a\_1)$ and $w(a\_2)$ are maximised. It may then be advantageous to bring $a\_1$ closer to $a\_2$ to increase (i.e., bring closer to $0$) the disutility from the value $v(A,B)$. The marginal incentive of such convergence is given by comparing the derivatives of the three functions.

**Tags:** mathematical-economics, optimization

> **Comments:**
> **username_1**: Not sure I understand your exact question. Are you asking if the conditions guarantee that a maximum exists over $\mathbb{R}^2_{\geq 0}$?  
> **username_0**: @ username_1 Yes, and that the function $Y$ is continuously differentiable over $\mathbb {R_{\ge 0}^2}$.; that is, $\frac {\partial Y}{\partial a_i}$ is defined across the whole domain as per the functions $y$, $v$, and $w$. So for example, if $a_1=0$, then $y'(0) >0$, $\frac {\partial v}{\partial a_1} \ge 0$ if $a_1<a_2$, and so on... as per the conditions. Essentially, I want to make sure that I didn't miss anything in any function that would violate continuous differentiability and the obtainment of a maximum.  
> **username_1**: Hi! I still don't understand :| These things that you list in your comment are the assumptions/conditions again, right? So, you are asking if, given these assumptions, the function has a maximum? Or are you also asking something else that I do not get?  
> **username_0**: @username_1. Sorry, the fault is mine for being unclear. I have been working on this problem for a while and am getting a little off. I think I just want to make sure the function I have created behaves in the way I want it to. I want the conditions for the derivatives to make sense across the whole of $Y$, one key consequence of which will allow me to obtain a maximum of $Y$. In "baby language", I want to be able to say that any change of $a_1$ and $a_2$ will be well defined in the function $Y$ and that there is some $a_1$ and $a_2$ that maximises $Y$. I hope this helps...  
> **username_0**: @username_1 I have edited the title and  question significantly to more clearly reflect what I (think) I was asking.  
> **username_2**: @ username_0 I 'm not sure to understand the question in your title. What's the meaning of 'converge' to a maximum'? Convergence is the language of limits. Maybe you want to know if you can be sure that a maximum of $Y(a_1,a_2)$ exists, isn't it?  
> **username_0**: @username_2 You may be right, I’m not sure to be honest. If you can kindly provide such a solution then I may be able to say.  
> **username_3**: Are you asking about the existence of a global maximum, a local maximum, or whether a gradient ascent process would always converge to such a maximum?  

---

## 1 answers

### Answer by username_2 on 2022-09-15. Score: 0

I hope I understand correctly your question.

I think that, on the basis of the conditions you wrote, you can't be sure that a maximum of $Y(a\_1,a\_2)$ exists.

How can we establish that a maximum of a function exists?

The most well-known theorem is the Weierstrass' Theorem, that ensures that a continuous function on a compact set $X$ to $\mathbb{R}$ has a global maximum and a minimum on $X$.
In your case, $Y(a\_1,a\_2)$ is defined on $\mathbb{R^2\_{\geq 0}}$, which is not compact, because it is unlimited.

So, Weierstrass' Theorem doesn't apply.

As the function $Y(a\_1,a\_2)$ is continously differentiable, we can resort to conditions on first order derivatives, and second order conditions on derivatives in the case the function is twice diffentiable.

For a point to be of an **interior** point of (relative) maximum a necessary condition is that the partial derivatives are 0 in that point. But to establish if, in that point, we have a maximum or not, we have to look at the Hessian, that is to the second order derivatives.

But you have only conditions on the first order derivatives, and no conditions on the second order derivatives, neither if they exist or not. So, nothing can be said.

Besides interior points of maximum, we could have points of (global) maximum on the border of the domain of the function $Y$, $\mathbb{R^2\_{\geq0}}$, that is the points for which $a\_1=0$ or $a\_2=0$.

But again, we cannot say anything about the existence of such points on the border.

Of course, there can be other ways to prove the existence of a maximum. What I mean is that proving that a maximum always exists under the given conditions is not obviuos, and proving its existence, in the case it exists, is not trivial.

> **Comments:**
> **username_1**: While this highlight why some usual ways do not work, it does not proof that a maximum does not exist. E.g., choose the $\mathbb{R}^2$ square with corners $(0,0)$ and $(A+B,A+B)$ as your compact set. Per Weierstrass a maximum exists over this set; you only have to prove that the function $Y$ never reaches a higher value in points outside of this set.  
> **username_2**: @ username_1 Eli j has changed her/his question, previosly, as you can see in the comments above, She/he seemed to ask if one can be sure that a maximum exists, and a said no, for the reasons I wrote. Now his/ question is different. Of corse a maximum can exist, Weirtrass gives only sufficient condition, and a maximum exists if you restrict the function to a compact set.  
> **username_2**: @ username_1 username_0 in the comments asked me explicitly to answer the question if we can be sure that a maximum exists, on the basis of the given conditions.  
> **username_1**: 1. According to the timestamps, the as of this writing the final edit of the question was at 14:42, about an hour before you first posted your answer. 2. Yes, my comment is saying that you are not proving they cannot be sure of this given the conditions. By showing you cannot prove something this and that way, you are not disproving it.  
> **username_2**: @ I referred to the comments, where username_0 was asking me to say if we can be sure that a maximum exists. This was the reason of my answer. Of course I haven't disproved that a maximum can exist, and it was not my purpose. I know the difference, I'm not a young student.  
> **username_1**: Okay, so to be clear: you are confident saying that we cannot be sure a maximum exists, i.e. it is impossible to prove that a maximum exist?  
> **username_2**: @ No, it is not impossible, why?  Of course a maximum can exist, but must be proved, and we  cannot refer here to the usual theorems, as Weierstrass. Of course we can give an exmaple of fa unction for which the maximum exists, but it is not obvious to prove that a maximum always exists  under the given conditions.  
> **username_1**: Okay. I guess I was expecting a more generalist answer, hence my misunderstanding. Thank you for the clarification.  
> **username_2**: You are welcome.  

---

