# Extreme Value Theorem in Economics

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.

• I’m a bit confused. You are attempting to apply the theorem? The theorem says nothing about “non-negativity.” To apply it, you need to prove that the set of feasible solutions is compact (and the function is continuous). – Brian Romanchuk Oct 29 '20 at 18:41
• Thank you Brian. You put your finger on it better than I could've. How would you prove/state that the set of feasible solutions is compact? What would be the economic analog required for this to be upheld? – EB3112 Oct 29 '20 at 18:52
• I realise after I wrote that what you meant. You probably need to add the non-negative constraints to ensure compactness. I’ll make a stab at an answer. I’m an applied mathematician, and not an economist, and it might help if you specified the meaning of the variables, as well any missing dynamics. That’s why I had a hard time parsing the question. – Brian Romanchuk Oct 29 '20 at 19:11
• Thank you Brian, your time is very appreciated. The variables in the constraint are capital, energy, and the subsequent emissions. (i.e. it is assumed that a producer, uses the two resources, but emissions impose a cost; so they're included with a constraint). The overall objective function however, is maximising welfare/utility. – EB3112 Oct 29 '20 at 19:15
• At the minimum, it needs to be a finite horizon problem, as the solution set would not be compact. After that, you would need to show that the set of feasible solutions is compact at each finite time point, which is not going to be possible with an arbitrary nonlinear constraint function $F_F$. – Brian Romanchuk Oct 29 '20 at 19:19

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.

• Thank you Michael, very informative! Alongside that, thank you for the book recomendation! – EB3112 Oct 30 '20 at 20:16
• Why is $F$ compact? – Michael Oct 31 '20 at 0:41
• @Michael The set $\prod_{t=1}^\infty [0,f^{t-1}(k)]^2$ is compact by Tychonoff's theorem and the continuity of $f$ implies that $F$ is a closed subset. – Michael Greinecker Oct 31 '20 at 1:11
• Hi Michael, could I just confirm with you that I am understanding the point above? Namely, the utility function is well-defined and continuous; and the production function is compact (by Tychonoff's theorem). Therefore the EVT is confirmed/an optimal plan exists. – EB3112 Oct 31 '20 at 13:48
• @AndrewCoyle It is not the production function that is compact, it is the set of feasible production ad consumption paths. – Michael Greinecker Nov 1 '20 at 14:16

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.
• Thank you Brian, your answer is very informative. – EB3112 Oct 29 '20 at 20:15
• As an economics point, it doesn't matter if the maximum utility is negative. The maximum is still the maximum. Behaviorally, people should still act the same, at least up until prospect theory. – RegressForward Oct 30 '20 at 0:56
• My understanding is that the non-negative condition referred to is on variables like consumption, investment, etc. Otherwise, the feasible set would be unbounded. – Brian Romanchuk Oct 30 '20 at 1:31
• I am a bit confused by your statement "Compact sets are not infinite". Clearly there are compact, infinite sets, did you mean under the discrete topology? Or are you referring to the "length" of the set? – Alba Mar 9 at 12:24
• @Alba - I am using the definition of compact as in Rudin. On the real line, a set that is infinite is not compact. As my example f(x) = 1 - 1/x (x=>1) shows, the supremum is not achieved (since “infinity” is not normally considered a point on the real line), thus showing the IVT does not apply to such a set. – Brian Romanchuk Mar 9 at 12:40