So I have looked through a handful of "What is arg min" questions on here before posting this, but I haven't found anything about using an inequality as an index. Specifically, I would like to know what this means:
$$t=\arg\min_{0\leq x\leq 1}(f(x))$$
What happens if the global minimum of the function lies outside of $(0,1)?$ I have also seen:
$$t=\arg\min_{x\in\Bbb{X}}(f(x))$$
where $\Bbb{X}$ is some set. How would we even evaluate this?
Thanks in advance!
That's pretty straight forward isn't it? According to Wikipedia the difference between "min f(x)" and "arg min f(x)" is that "min f(x)" returns the minimum value of f while "arg min f(x)" returns the value of x at which f(x) is min.
For example, if $f(x)= x^2+ 1$ then "min f(x)" is 1 while "arg min f(x)" is 0 because f(0)= 1 while f(x)> 1 for any other value of x.
Now, what about "$arg min_{1\le x\le 2}$"? The global minimum is not in that set but that is irrelevant! We are not asking about a global minimum, we are asking about the smallest value for x in that set. f(1)= 2 and f(x)> 2 for all other x in [1, 2] so $min_{1\le x\le 2} f(x)= 2$ while $arg min_{1\le x\le 2} f(x)= 1$.