Understanding relation between vector valued function and function objective in an multi objective optimization problem

Question

I try to understand the relation between "vector-valued function" and "function objective" as used in optimization problem.

I understand that objective function in a multi-objective problem can be defined $min(f_1(x), f_2(x)),...f_k(x))$ with $x$ is a candidate solution taken into $\mathbb{X}$ problem space.

For example in the multi-objective test function of Schaeffer, $f1(x)$ and $f2(x)$ are two objective function the optimization algorithm search to $min(f1(x),f2(x))$

$Minimize = \begin{cases} f1(x) = x^2 \\ f2(x) = (x-2)^2 \end{cases} $

With $\mathbb{X}$ defining the problem space, i suppose each of this function have this mapping $f: \mathbb{X} \mapsto \mathbb{R}$ , so my space is $\mathbb{X} \mapsto \mathbb{R^2}$ isn't it ?

Litterature suggest that we can also write this set of objective function as "vector valued function" like this :

$f(x) = (f1(x), f2(x))^T$

with $\mathbf{f} : \mathbb{X} \mapsto \mathbb{R^n}$

I found also another multiple notation :

a) $r(t) = x^2i+ (x-2)^2j$

b) $\overrightarrow{r}(t) = x^2i+ (x-2)^2j$

c) $\overrightarrow{f}(x) = (f1(x), f2(x))^T$

d) $\overrightarrow{f}(\overrightarrow{x}) = (f1(\overrightarrow{x}), f2(\overrightarrow{x}))^T$

e) $\mathbf{f}(x) = [f1(x), f2(x)]^T$

I found also this annotation to define vector of parameters : x* = {x1*,x2*,...,xn*}

My questions are :

• Why using "vector valued function" to redefine "objective function" ? I don't understand the interest :/
• What is really a "vector valued function" on this use case and what is the best notation ? Can you explain this concept with an example or a graphic ?
• What is $^T$ and x* in this use case ?

Answer 1

I don't know much about optimization, but I believe your questions are mainly about math notations, which I will now try to explain.

• Just looking at the name, a "vector valued function" is a function with values in a vector space. It is usually specified to point out that the function output is a multi-dimensional vector, which is the same as saying it has multiple numerical outputs (in your example, those are $x^2$ and $(x-2)^2$ for input $x$). Therefore, your function is still an objective function, but with multiple outputs, that's why it is called "vector valued". The best way to represent it would be as a parametrized curve: in your example, consider the variable $x$ as time. At each instant, the function gives a point in the plane $\mathbb R^2$, so we can represent all of them as a curve. Here is the curve for your particular example, when $-2\leq x\leq 4$:

• You are free to use the notation you find the most easy to work with, as long as it is explicit that you have one input $x$ and two outputs $f_1(x)$ and $f_2(x)$. The notations with $i$ and $j$ are rather ambiguous, unless you have defined those as basis of the vector space before. The arrow notation is nice if you want to remember that the function is vector valued, but might be cumbersome to carry everywhere.

• The $(\cdot)^T$ notation stands for transpose, which means change rows to columns and columns to rows in a matrix. This is because vectors in $\mathbb{R}^n$ are usually represented as a column vector, while we usually write in rows. Specifically, $$(f_1(x), f_2(x))^T =\left(\begin{array}{c}f_1(x)\\ f_2(x)\end{array}\right).$$ The $*$ notation is more ambiguous, and could mean many different things, according to context. You can find here an explanation which seems to suit your case (although I cannot be sure of it), as the vector of solutions to the optimization problems of each coordinates. Following this rule, in the example, $$x^*=(x_1^*,x_2^*) = (0,2)$$ (because $0$ is the optimal value of $x$ for $f_1$ and $2$ is the optimal value of $x$ for $f_2$).