# Why the definition for optimal value is the $\inf{f_0(x)}$ rather than $\min{f_0(x)}$?

Suppose an optimization problem

\begin{equation*} \begin{aligned} & \underset{x}{\text{minimize}} & & f_0(x) \\ & \text{subject to} & & f_i(x) \leq b_i, \; i = 1, \ldots, m. \end{aligned} \end{equation*}

Then, the optimal value is defined as

$$p^{\star} = \inf\{f_{0}(x) \: | x \in \mathcal{A} \}$$

where $A$ is the feasible set. My question is that why we use $\inf$ rather than $\min$ for representing the optimal value?

• Are you familiar with the definition of supremum/infimum? – Math1000 Oct 29 '17 at 12:23
• @Math1000: Yes. – math14 Oct 29 '17 at 12:23

However, in the context of the problem statement, if $f$ is also a continuous function and $A$ (the obtainable set) is compact, then the minimum and infimum are the same and you can attain the infimum in the attainable set.
• Consider the example $f(x)=e^{x}$ on $R$. This is a continuous function, and the inf is 0, but the minimum is never attained. Also, consider $f(x)=x^{2}$ $R^{++}=\left\{ x | x > 0 \right\}$. Again, the inf is 0, but the minimum is not attained. It's true that if $f$ is a continuous function on a closed and bounded set, then the minimum is attained. – Brian Borchers Oct 29 '17 at 15:50