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As we all know, a classic optimization problem can be represented in the following way:

Given: a function $f: A \rightarrow \mathbb{R}$ from some set $A$ to the real numbers
Sought: an element $x_0 ∈ A$ such that $f(x_0) \le f(x)$ for all $x \in A$ ("minimization") or such that $f(x_0) \ge f(x)$ for all $x \in A$ ("maximization").

However, I am now facing an optimization problem, in which the objective function returns a function instead of a real number. The function has two variables, one is the input variable $x$, the other is an unknown variable $z$ within a known interval. The problem can be described as follows:

Given: a function $f: A \rightarrow \mathcal{C}(x, z)$ from some set $A$ to a function space $\mathcal{C}(x, z)$ where $z \in [a, b]$
Sought: a maximum subset $B \subseteq A$, where for each element $x_0 ∈ B$, there exists at least one interval of $z \in [c, d]$ $(a \le c \le d \le b)$, such that $f(x_0, z) \le f(x, z)$ for all $x \in A$ ("minimization") or such that $f(x_0, z) \ge f(x, z)$ for all $x \in A$ ("maximization").

To give a concrete example, say I want to minimize the objective function $f(x, z) = xz^3 - \frac{1}{x} z^2 + z$ where $z \in [-1, 1]$. In the end, I want to get the lowest curve of $f(x, z)$ (scan $z$ from $-1$ to $1$ and connect all lowest parts with different $x$). In my case, I have a finite set of $x$s. The expected result is an optimal subset of $x$s and the corresponding intervals at which $f(x,z)$ is the lowest.

Can anyone suggest how should I approach this kind of problem? I cannot find any literature related to this.

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    $\begingroup$ I think you're looking for optimal control theory, the simplest form of which is Pontryagin's maximum principle. $\endgroup$
    – user7868
    Commented Nov 9, 2021 at 23:18
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    $\begingroup$ I don't understand the notation $f : A \to f(x, z)$. Does it mean $f : A \times [a, b] \to \mathbb R$, where we are using $x$ as the name for an element of $A$ and $z$ as the name for an element of $[a, b]$? Also, since you allow $c = d$, it seems that your condition could be expressed more concisely as: for each $x_0 \in B$, there exists at least one $z = c = d \in [a, b]$ such that $f(x_0, z) \le f(x, z)$ for all $x \in A$, or $f(x_0, z) \ge f(x, z)$ for all $x \in A$. (And presumably you mean to pick a fixed sense of inequality, right, not allow it to differ depending on $x_0$?) $\endgroup$
    – LSpice
    Commented Nov 10, 2021 at 0:30
  • $\begingroup$ @LSpice Yes I agree the notation is confusing, but you understand it correctly. I have changed the notation from $f(x,z)$ to a function space $\mathcal{C}(x, z)$. As for $f(x_0,z) \le f(x,z)$, I mean for a given fixed interval $[c,d]$ of $z$, the curve of $f(x_0,z)$ is the lowest, i.e., $x_0$ minimizes $\int_c^d f(x, z) dz$ $\endgroup$
    – Shaun Han
    Commented Nov 10, 2021 at 1:01

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