Difference between minimizing the absolute value of a function and minimizing subject to a non-negative constraint

Question

Problem 1: Minimizing the absolute value of a function:

$$\min_x |f(x)|$$

Problem 2: Minimizing the function subject to a non-negative constraint:

$$\min_x f(x)\;:\;f(x) \ge 0$$

Could someone please provide an example where the solutions ($x^* = \arg \min f(x)$) of Problem 1 and Problem 2 differ OR a proof why the solutions are the same for any $f(x)$? I understand that Problem 1 allows negative $f(x)$, while negative $f(x)$ are not considered in Problem 2. I'm wondering if these two formulations can lead to different solutions in certain cases.

Thank you in advance for your help!

Answer 1

$f(x) = x + \frac{1}{x} + 1$ is a more useful example for your question. $0$ is not in the range of $f(x)$ and $f(x)$ is not continuous. Problem 1 would yield a minimum of $1$ at $x^* = -1$. But problem 2 would yield a minimum of $3$ at $x^* = 1$.

Answer 2

Consider the constant function $f(x)\equiv-1$. Problem 1 yields a minimum of $|-1|=1$, attained for all $x$. Problem 2 is infeasible.

For a more complete understanding, consider the following three mutually exclusive cases for continuous $f$:

• $f(x) > 0$ always: the two problems are obviously equivalent.
• $f(x) = 0$ for some $x$: the minimum for both problems is $0$, attained by the roots of $f(x)$.
• $f(x) < 0$ always: Problem 1 is feasible, but Problem 2 is not.