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I have the equation $$h(x,k(y,z))=f(y,g(y,x)+t(y,z)), \tag{*}\label{equation}$$ where

  • $h,k,f,g,t$ are continuous real valued functions.
  • $h,f$ are strictly monotone in their second argument.
  • all functions are "sensitive" to each of the arguments - in some natural sense (if $x,y,z,$ are real numbers, just suppose that all functions are strictly monotone in each of their arguments).
  • $x,y,z$ are from some "well behaved" topological space (connected, ...) - you can assume that they are from some real connected interval.

I want to show that $g$ is additive, in the sense that: $g(y,x)=g_{y}(y)+g_{x}(x)$ (and that $h$ and $f$ are strictly monotone transformations of an additive function).

The reason I think this is true is that in the right-hand side of \eqref{equation}, $x$ is only "tied to" $y$ not $z$, while on the left hand side it is "tied to" a function of both $y$ and $z$.

More specifically, by \eqref{equation} we have:

$$k(y,z)=h^{-1}(x,f(y,g(y,x)+t(y,z))) \tag{**}\label{equation_a}$$

(where $h^{-1}$ is the the inverse of $h$ on the second argument, that is: $h(a,h^{-1}(a,v))=v$).

So, the right-hand side of \eqref{equation_a} must be independent of $x$. Is there any option but that $g$ is additive? I do not know how to go about attacking this. What theorems/tools are available?

Thanks,

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  • $\begingroup$ I've changed the formatting of your question a bit, so that typing \eqref{equation} should give a link to the equation like this \eqref{equation}. It should also work in answers as well. $\endgroup$ – Arnaud D. Jul 20 '18 at 10:16
  • $\begingroup$ I just made a few edits to the question: (1) added the assumption that all functions are "sensitive" to each of their arguments. (2) changed that $h,f$ need not be additive, (3) made some changes the explanation. $\endgroup$ – mike Jul 22 '18 at 12:20
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The statement is false. Counter example:

  • $h(a,b)=k(a,b)=a+b$
  • $g(a,b)=t(a,b)=ab$
  • $f(a,b)=b/a+a$

Then, both sides of $(*)$ are $x+y+z$.

(this is not a counter example to what I was really trying to prove, so I need to go back and rephrase the proposition).

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