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I was just deadly curious about a solution of a certain functional equation, or better, a question.

Given $f(x)$ and $g(x)$ as two functions domained in $A$ and $B$, and given the equation for f(x)

$$f(x)=g(f(x))$$

I found out, with some weird reasoning that

$f(x)=g^{-1}(g^{-1}(g^{-1}...(x))..)$

Now I want to ask, with which conditions is this conclusion correct? Because withour any doubt some conditions are to be met and satisfied, and certainly I have not satisfied them or mentioned them because....I do not know them!!! Any insight?

us er
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  • What happens when $g$ is the identity? Are you missing an $f$ in the second equation? It is not clear how should we interpret the second equation. – Carl Christian Oct 09 '19 at 20:12
  • $g(x)$ is a generic function, so it can be supposedly any function. – us er Oct 09 '19 at 20:13
  • When g is the identity the solution satisfies the question. – us er Oct 09 '19 at 20:15
  • The second equation has as RHS a composed function, the inverse of g is self-composed infinitely. – us er Oct 09 '19 at 20:16
  • If $g$ is the identity, the second equation appears to read as $f(x)=x$. – Carl Christian Oct 09 '19 at 20:21
  • The dots mean that there are infinite parentheses – us er Oct 09 '19 at 20:22
  • $g$ has to be the identity function on the range of $f.$ When that is the case, $f$ could be anything. – Adrian Keister Oct 09 '19 at 20:31
  • If in this case g is the indentity, then also f(x) is the IDENTITY. I'M looking for a explicit expression of x, not to make the main equation as an IDENTITY. That's different. – us er Oct 09 '19 at 20:56
  • *of f(x) I was meaning – us er Oct 09 '19 at 20:58
  • The thing is to find f. G is supposed to be known. In fact I wrote FOR f(x) – us er Oct 09 '19 at 20:59
  • If $f(x) = g(f(x))$ for all $x \in a$, then $y = g(y)$ for all $y \in f(A)$. You are left with the tautology that $f(x) = f(x)$ for all $x \in A$. – Carl Christian Oct 09 '19 at 21:01
  • I mean your solution may satisfy the above equation as an IDENTITY, and in the same case itself f(x) happens to be an identity function. – us er Oct 09 '19 at 21:02
  • I'm not looking for tautologies, I'm looking to find f(x), which is unknown, in terms of g(x), which is known. In order to understand better, I can ask to find f(x)=e^(f(x)). Find f(x) – us er Oct 09 '19 at 21:04
  • You are looking for fixed points of $g$ and functions $f$ whose range consists of fixed points of $g$. A point of etiquette: using caps lock is the equivalent of shouting. It is not going work to your advantage here. – Carl Christian Oct 09 '19 at 21:19
  • Caps lock was not meant for shouting by my person. And by the way "It's not going to work to your advantage here" seems more like an arrogant reprimanding here, even thought I didn't mean to be arrogant in anyway. I honestly don't like this kind of challenging warnings, especially when I don't mean to commit a certain act. This may seem arrogant from your part. – us er Oct 09 '19 at 21:43

1 Answers1

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The statement $$ f(x) = g(f(x))$$ is simply the demand that each $y = f(x)$ is a fixed point of $g$, i.e., $y=g(y)$. If $S$ is the set of fixed points of $g$, then the set of solutions consists of all functions $f$ for which $f(x) \in S$ for all $x$. With no restrictions placed on $g$ there is no reason to assume that $g$ has any fixed points. In this case, there are no solutions $f$.

The existence of fixed points is a separate field of study with several theorems ensuring the existence of at least one fixed point. The fixed point theorems of Banach, Brouwer and Schauder are three distinct examples.
Carl Christian
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