How does $\arg$ relate to the $\operatorname{argmax}$ operator?

Question

I understand the definition of $\max(f(x))$: what $\operatorname{argmax}(f(x))$ means? And I also know that $\arg(z)$ is the angle between the the line passing through the points $z_o\equiv(0,0)$, $z=x+iy$ and the real axis, but what turns $\arg(z)$ and $\max(f(x))$ into $\operatorname{argmax}(f(x))$, how is this done?

Answer 1

Keep in mind that there are two different meanings of the word "argument" being used here. In "argmax" the word "argument" refers to the input to the function. In "arg$(Z)$" for a complex number, this is the angle in its polar representation. These two meanings have almost nothing to do with each other. The use of the same word for both is a historical accident.

Now to illustrate what argmax means, consider $f(x)=1-x^2$. Of course $\text{max}_xf(x)=1$ where we are maximizing the function over the real numbers. This is because the maximum value that $f$ takes is 1. On the other hand $\text{argmax}_xf(x)=0$ because the maximum value is reached for argument 0. That is to say, $f(0)=1$ and $\text{max}_xf(x)=1$, so $\text{argmax}_xf(x)=0$.

Or if you want to be more completist, we could instead say $\text{argmax}_xf(x) =\{0\}$ because in other examples, there may be many different arguments for which the function takes its maximum value. For instance,

$$ \text{argmax}_x (1-(x+1)^2(x-1)^2) = \{-1,1\} $$