# Can following these marginal conditions have the net utility function converge to a maximum? [Edited]

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.

• Not sure I understand your exact question. Are you asking if the conditions guarantee that a maximum exists over $\mathbb{R}^2_{\geq 0}$? Sep 15, 2022 at 11:52
• @ 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. Sep 15, 2022 at 12:32
• 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? Sep 15, 2022 at 12:36
• @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... Sep 15, 2022 at 12:53
• @Giskard I have edited the title and question significantly to more clearly reflect what I (think) I was asking. Sep 15, 2022 at 14:36

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.

• 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. Sep 16, 2022 at 7:15
• @ 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. Sep 16, 2022 at 7:34
• @ 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. Sep 16, 2022 at 8:08
• 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. Sep 16, 2022 at 8:43
• @ 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. Sep 16, 2022 at 10:03