Solving infinite horizon optimization problems through analysis of a one-dimensional global optimization problem

Abstract

Infinite horizon optimization (IHO) problems present a number of challenges for their solution, most notably, the inclusion of an infinite data set. This hurdle is often circumvented by approximating its solution by solving increasingly longer finite horizon truncations of the original infinite horizon problem. In this paper, we adopt a novel transformation that reduces the infinite dimensional IHO problem into an equivalent one dimensional optimization problem, i.e., minimizing a Hölder continuous objective function with known parameters over a closed and bounded interval of the real line. We exploit the characteristics of the transformed problem in one dimension and introduce an algorithm with a graphical implementation for solving the underlying infinite dimensional optimization problem.

Appendix

Proof of Theorem 3

The proof that the piecewise linear extension preserves the Hölder condition can be carried out in a straightforward manner as follows. We are to show that for any pair $x,y \in [0,1/2]$,

$$\begin{aligned} |\tilde{f}(x)-\tilde{f}(y)| \le M|x-y|^\alpha . \end{aligned}$$

(20)

Case 1: if both $x,y \in Y$, (20) follows readily, since $\tilde{f}=f$ by construction of $\tilde{f}$, and f satisfies the Hölder condition by Theorem 2,

$$\begin{aligned} |\tilde{f}(x)-\tilde{f}(y)| = |f(x)-f(y)| \le M|x-y|^\alpha , \quad \text {for all } x,y\in Y. \end{aligned}$$

(21)

Case 2: if exactly one of $x,y \notin Y$, WLOG, let $x \in Y$, but $y \notin Y$. Suppose $y > x$. It is easy to check that the function g defined by

$$\begin{aligned} g(z) = \tilde{f}(x) + M(z-x)^\alpha = f(x) + M(z-x)^\alpha \end{aligned}$$

is a concave function over $z \in [x,\infty )$. Therefore, the hypograph of g on [x, 1 / 2] is a convex set. Denote this convex set by K. Since $y \notin Y$, there exist $y_1 = \text {argmin}_{u\in Y}\{|y-u|: x \le u < y\}$, and $y_2 = \text {argmin}_{u\in Y}\{|y-u|: u > y\}$. Since $y_1, y_2 \in Y$, by (21), $(y_1,\tilde{f}(y_1)), (y_2,\tilde{f}(y_2)) \in K$. Since K is convex, $(y, \tilde{f}(y))$, which is the convex combination of $(y_1,\tilde{f}(y_1))$ and $(y_2,\tilde{f}(y_2))$, is also in K. Hence,

$$\begin{aligned} \tilde{f}(y) \le g(y) = \tilde{f}(x) + M(y-x)^\alpha . \end{aligned}$$

(22)

Similarly, it is easy to check that the function h defined by

$$\begin{aligned} h(z) = \tilde{f}(x) - M(z-x)^\alpha = f(x) - M(z-x)^\alpha \end{aligned}$$

is a convex function over $z \in [x,\infty )$. Therefore, the epigraph of h on [x, 1 / 2] is a convex set. With this fact, applying the same argument, it follows that

$$\begin{aligned} \tilde{f}(y) \ge h(y) = \tilde{f}(x) - M(y-x)^\alpha . \end{aligned}$$

(23)

By (22) and (23),

$$\begin{aligned} |\tilde{f}(y)-\tilde{f}(x)| \le M(y-x)^\alpha . \end{aligned}$$

Similarly, when $y < x$,

$$\begin{aligned} |\tilde{f}(y)-\tilde{f}(x)| \le M(x-y)^\alpha . \end{aligned}$$

Hence,

$$\begin{aligned} |\tilde{f}(x)-\tilde{f}(y)| \le M|x-y|^\alpha , \quad \text {for all } x\in Y, \text { for all } y\notin Y. \end{aligned}$$

(24)

Case 3: if both $x,y \notin Y$, suppose $y > x$. The function g defined by

$$\begin{aligned} g(z) = \tilde{f}(x) + M(z-x)^\alpha \end{aligned}$$

is a concave function over $z \in [x,\infty )$. Therefore, the hypograph of g on [x, 1 / 2] is a convex set. Denote this convex set by K. Since $y \notin Y$, there exist

$$\begin{aligned} y_1 = \left\{ \begin{array}{ll} \text {argmin}_{u\in Y}\{ |y-u|: x < u < y\} &{}\quad \text {if } \{ u: x < u < y, u\in Y\} \ne \emptyset ,\\ x &{} \quad \text {if } \{ u: x < u < y, u\in Y\} = \emptyset ,\\ \end{array} \right. \end{aligned}$$

and $y_2 = \text {argmin}_{u\in Y}\{|y-u|: u > y\}$. Since $y_2 \in Y$ and $y_1 \in Y$ or $y_1 = x$, by (24), we have $(y_1,\tilde{f}(y_1)), (y_2,\tilde{f}(y_2)) \in K$. Since K is convex, $(y, \tilde{f}(y))$, which is the convex combination of $(y_1,\tilde{f}(y_1))$ and $(y_2,\tilde{f}(y_2))$, is also in K. Hence,

$$\begin{aligned} \tilde{f}(y) \le g(y) = \tilde{f}(x) + M(y-x)^\alpha . \end{aligned}$$

(25)

Similarly, the function h defined by

$$\begin{aligned} h(z) = \tilde{f}(x) - M(z-x)^\alpha \end{aligned}$$

is a convex function over $z \in [x,\infty )$. Therefore, the epigraph of h on [x, 1 / 2] is a convex set. With this fact, applying the same argument, it follows that

$$\begin{aligned} \tilde{f}(y) \ge h(y) = \tilde{f}(x) - M(y-x)^\alpha . \end{aligned}$$

(26)

By (25) and (26),

$$\begin{aligned} |\tilde{f}(y)-\tilde{f}(x)| \le M(y-x)^\alpha . \end{aligned}$$

Similarly, when $y < x$,

$$\begin{aligned} |\tilde{f}(y)-\tilde{f}(x)| \le M(x-y)^\alpha . \end{aligned}$$

Hence,

$$\begin{aligned} |\tilde{f}(x)-\tilde{f}(y)| \le M|x-y|^\alpha , \quad \text {for all } x,y\notin Y. \end{aligned}$$

(27)

By (21), (24), and (27), the theorem is proved. $\square$