Extreme Value Theorem in Economics

Question

Would appreciate some thoughts on proving the existence of a solution (a maximum) in an applied context.

Suppose the objective function is one of maximising utility:

$$\max\Sigma\beta U(C_t)$$

Subject to some constraints on resources, say:

$$\\C_t+K_{t+1} = F_F(K_F,E_F,S_t)$$

Where the constraint is the economy's resource constraint, inclusive of current consumption $C_t$ and investment possibilities $K_{t+1}$, set equal to the production function $F_F$, which includes arguments on capital $K_t$, energy, $E_t$ and the emissions stock $S_t$ (i.e. emissions affect production costs).

The extreme value theorem states 'that a continuous function throughout a non-empty compact set will have a max (min. respectively)'.

In this respect, my questions is: what are the requirements for the feasible set of solutions to be on a compact set?

• non-negativity of variables?
• terminal conditions on the variables (i.e. they are all exhausted in the future)?

Would appreciated some thoughts, for sure.

Answer 1

One can prove the existence of such optimal plans using the extreme value theorem of Weierstrass, but it requires some advanced math.

Here is a toy version of the model without energy and emissions. Both the instantaneous utility functions $u:\mathbb{R}_+\to\mathbb{R}$ and the production function $f:\mathbb{R}_+\to\mathbb{R}$ are assumed to be continuous and nondecreasing. Moreover, $u$ is assumed to be bounded (!). There is given initial capital stock $k_1\geq 0$. The space of feasible consumption and production plans is defined as $$F=\big\{(c_1,k_1,c_2,k_2,\ldots)\mid 0\leq k_{t+1}\leq f(k_t-c_t)\}, k_t\geq c_t\geq 0\big\}.$$ $F$ describes all feasible paths of consumption and capital. This set is a nonempty compact subset of $\mathbb{R}^\infty$ endowed with the product topology. Here is why: By Tychonoff's theorem, the set $$\prod_{t=1}^\infty [0,f^t(k_1)]^2$$ with $f^0$ the identity function $f^{t+1}=f\circ f^t$ is compact and $F$ is a subset of this compact set. By definition, all coordinate functions are continuous. Each of the relevant inequalities defines a closed set and $F$ is, therefore, a closed subset of a compact set and, therefore, compact itself. Also, the plan that never invests and consumes everything is in $F$, so $F$ is nonempty.

The utility function $U:F\to\mathbb{R}$ given by $$U(c_1,k_1,c_2,k_2,\ldots)=\sum_{t=1}^\infty \beta^t u(c_t)$$ is well-defined and continuous in the product topology. Since the utility function is bounded, $U$ will always be finite and, therefore, well-defined. To see that it is continuous, note that since $u$ is bounded, there exists for each $\epsilon>0$ some $T$ such that the utility of any two paths that only differ at times later than $T$ can differ by at most $\epsilon/2$. Since all coordinate functions are continuous, if the two consumption plans are close enough in the first $T$ coordinates, they will differ at most by $\epsilon/2$. So if the paths are close enough at finitely many coordinates, the corresponding utility will be close enough. So $U$ is continuous in the product topology.

So $U$ is a continuous function on the nonempty compact set $F$. By the extreme value theorem of Weierstrass, $U$ takes on a maximum at some point in $F$ and such a point is an optimal plan.

You can find a more general proof of the existence of optimal plans along these lines in the book "Dynamic Programming in Economics" by Le Van and Dana. The assumption that $F$ is increasing is not needed, you can replace the product of intervals that includes $F$ by bounding it with the largest values that can be produced instead of the one that comes from always reinvesting capital. That $u$ is bounded was assumed to guarantee that $U$ is finite. One can replace this by an assumption that guarantees that the instantaneous utility cannot increase to fast along a feasible path.

Answer 2

At present, I would need more information to answer this question. (Missing details might be obvious to an economist, my background is in applied math.)

The extreme value theorem (e.g. Theorem 4.16 of Rudin’s Principles of Mathematical Analysis) says that if $f$ is a continuous real function on a compact metric space, then for a compact subset $M$, then the supremum and infimum of $f$ are achieved at some point(S) within $M$.

Examples to keep in mind.

• Compact sets are not infinite. The function $f(x) = 1 - \frac{1}{x}$ does not achieve its supremum on the set $x \geq 1$.
• It is not an if and only if condition. E.g. $f(x) = x^2$, it achieves its infimum on $(-1,1)$, even though the set is not compact.

The question does not specify all the constraints for the mathematical system. All I can offer is the following points if one wants to apply the theorem.

• The time horizon has to be finite, as otherwise the feasible solution set would not be compact (if it is non-empty).
• At each time point, it is necessary to show that the feasible values at that time point is compact. If we have linear constraints (budget constraints), a non-negative condition on variables may be enough for this.
• Constraints $F_F$ have to either hold with equality, or be non-strict inequalities. A strict inequality could be disqualifying (it would be OK if it was not binding.)
• It might only be necessary to show that the feasible set lies with a bounded set, and is non-empty (e.g. find one solution).
• Continuity of the utility function should be straightforward.