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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?

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  • $\begingroup$ Are you familiar with the definition of supremum/infimum? $\endgroup$ – Math1000 Oct 29 '17 at 12:23
  • $\begingroup$ @Math1000: Yes. $\endgroup$ – math14 Oct 29 '17 at 12:23
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Not every set has a minimum value, but every set that’s bounded below has an infimum value. If you used min instead of inf, you would lose the ability to talk about many functions

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As mentioned, the minimum isn't always attainable in the feasible set. In this case you need to look for the infimum.

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.

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  • $\begingroup$ 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. $\endgroup$ – Brian Borchers Oct 29 '17 at 15:50
  • $\begingroup$ I forgot to add that the attainable set is compact, thanks. $\endgroup$ – abnry Oct 29 '17 at 16:15

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