# $g(x) \ne f(x)$, then any transformation of $T(g(x)) \ne T(f(x))$?

Suppose that we have two functions $f(x)$ and $g(x)$ both are increasing (not constant) on an interval $[a,b]$. We know that $f(x) \ne g(x)$ on the interval, except for measure zero subsets of $[a,b]$. I am wondering if there is any transformation $T$ such that $T(f(x)) = T(g(x))$ almost everywhere on $[a,b]$.

I mean any meaningful transformation, not like $T(\cdot) = c$ for some constant $c$.

• There should be fairly tame pairs of continuous functions where $T$ will be forced to be trivial. I am picturing the graphs of $f$ and $g$ between two points where they agree and then trying to glue/weld values together only as necessary to construct such a $T$. These identifications will look something like orbits, ricocheting back and forth from graph to graph as they sink to the lower common value of $f$ and $g$. A perturbation should allow these trajectories to become too 'tangled' to allow a non-constant function. But you should ask someone who works in one-dimensional real dynamics. – John Samples Jul 31 '18 at 9:05
• The question specifies no requirements on $T$, so of course there always exists such a transformation. – 6005 Aug 10 '18 at 4:04

You can easily define transformations $T$ such that $T(f(x)) = T(g(x))$ holds for some $f(x) \ne g(x)$. As you point out, a trivial example of $T$ is the $0$ transformation, i.e. $T(f(x)) = 0$ for all functions $f(x)$, then for any two functions $f$ and $g$, $T(f(x)) = T(g(x))$. For a less trivial example, let $f(x) = x$, $g(x) = x + 1$, and let $T$ be differentiation. Then $f(x) \ne g(x)$ almost everywhere, but $T(f(x)) = T(g(x)) = 1$ for all $x$.
If you mean for $T$ to be a linear transformation, then what you are interested in is the nullspace of $T$. In particular, if $f(x) - g(x)$ is in the nullspace of $T$, then we will have $T(f(x) - g(x)) = 0$, which implies $T(f(x)) = T(g(x))$. It is fairly common for linear transformations to have a non-trivial nullspace, for example differentiation, like I noted above.
If you are not just looking at linear transformations, I don't know if there's a specific terminology for cases when $T(f(x)) = T(g(x))$, but you're essentially asking about the existence of non-injective transformations (since injectivity is equivalent to $f(x) \ne g(x) \implies T(f(x)) \ne T(g(x))$), but there is no reason to expect an arbitrary transformation to be injective.