Let us consider a function $f:[0,1]\rightarrow R^+$. Now if we consider a functional of $f$ as $$g=\mathop{\arg\max}_{x\in[0,1]}f(x).$$

Then I define inverse of $g$ is $g^{-1}(x)=\{\forall f :g=\mathop{\arg\max}_{x\in[0,1]}f(x) \}$. I am not sure if I am right or not. Any help will be highly appreciable.

  • $\begingroup$ How are you defining the argmax function? In particular, does it return the set of all maximising x, or a particular such x? If you are choosing only one such x, how have you made it well defined? $\endgroup$ – James Jun 11 '15 at 14:20
  • $\begingroup$ I think there is a typo in your question. You write: "Then the I am inverse". What do you mean by that? $\endgroup$ – Andrea Jun 11 '15 at 14:22
  • $\begingroup$ Thanks @AndreaDiBiagio. I corrected it. $\endgroup$ – Janak Jun 11 '15 at 14:23
  • $\begingroup$ @James. Thanks for your attention. Here I am defining the $argmax$ as the set of all maximizing $x$. $\endgroup$ – Janak Jun 11 '15 at 14:25
  • $\begingroup$ NP, although I deleted it because Andrea is right. Maybe something more like $g^{-1}(y) = \{ f : y \in arg max f(x)\}$. The element of symbol to emphasize arg max is set valued. $\endgroup$ – muaddib Jun 11 '15 at 14:33

First off, let $F = \{f: [0,1] \to \mathbb{R}^+\}$. Then $g : F \to \mathcal{P}([0,1])$. That is $g$ maps from the set of all such functions $f$ to the powerset of the interval $[0,1]$. That should help you to understand the problem.

Then $g^{-1} : \mathcal{P}([0,1]) \to \mathcal{P}(F)$ given by $g^{-1}(\{x\}) = \{f \in F: g(f) = \{x\}\}$. Note that here I am using $\{x\}$ to denote an element of $\mathcal{P}([0,1])$.


Your function will return all $[0,1] \rightarrow \Bbb R^+ $ which peak at a given real number between $0$ and $1$, if you managed to define it properly.

However you didn't define it properly. If you look at your expression for $g^{-1}(x)$ you'll see that $x$ appears as a dummy variable...


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