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

Question

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: $$ y'(a_1)= \begin{cases} >0&\text{if}\, a_1\in[0,A)\\ 0&\text{if}\, a_1=A\\ <0&\text{if}\ a_1 \in (A,\infty) \end{cases} $$

$$y''(a_1)<0$$

$$ \frac {\partial v(a_1,a_2)}{a_i}= \begin{cases} \geq0&\text{if}\, a_i <a_j\\\ 0&\text{if}\, a_i=a_j\\ \leq0&\text{if}\ a_i>a_j, \text{where} \, i\neq j \end{cases} $$

$$ w'(a_2)= \begin{cases} \geq0&\text{if}\, a_2<B\\ 0&\text{if}\, a_2=B\\ \leq 0&\text{if}\ a_2 >B \end{cases} $$

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.

Answer 1

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.