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I want to know if there's a way of saying: "The price that sets the output of a function equal to a constant" in a way that resembles $\arg \max$.

What I'm trying to express is close to saying: "x is part of the set of arguments that maximize a function", except I'm looking for an $x$ that solves an equality, not a maximization problem.

I came up with the following: $$ \hat{p} \in \{p|k=f(p)\} $$ But I'm looking for something that resembles $\hat{p} \in \arg \text{ } eq$. Does such a thing exist?

sam wolfe
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Arturo Sbr
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    The notation $p\in f^{-1}(k)$ is quite standard. Or you can define for yourself something along the lines of $\operatorname{lvlset}(f;k)$ (level set of $f$ for the value $k$) –  Oct 02 '19 at 13:39
  • For the record, $f^{-1}(k)$ is a set and is defined to be ${p \mid f(p) = k}$. This is defined regardless of whether $f$ is invertible. If $f$ has multiple points that take the value $k$, then $f^{-1}(k)$ has more than one element, while if there is a unique point that takes the value $k$ then $f^{-1}(p)$ is a singleton set. – kccu Oct 02 '19 at 13:56
  • Awesome. I was concerned that $f(p)$ is not invertible, but now I see this answers my question. – Arturo Sbr Oct 02 '19 at 14:04

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