If $f(g(x)) = g(x)$, what is $f(x)$?

Question

If $f(g(x)) = g(x)$ then can we conclude $f(x)=x$? (In fact, $g(x) = x + 2f(x)$.)

Or which properties should $g(x)$ satisfy? (like one-to-one, etc.)

Answer 1

If $f(g(x)) = g(x)$, then for any $y$ that occurs as $y = g(x)$ for some $x$, we have (by putting that particular $x$ in the equation) the fact that $f(y) = y$.

In other words, $f$ is the identity function on the range of $g$.

However, at values that don't occur in the range of $g$ (i.e. for values $y$ which don't occur as $y = g(x)$ for any $x$), the function $f$ could take any arbitrary value, and we would not be able to tell the difference by just considering $f(g(x)) = g(x)$, because this fact is simply agnostic and irrelevant to such values.

In the example in marty cohen's answer, with $g(x) = x^2$, the range of $g$ is the nonnegative numbers (assuming that we're working over the reals or a subset thereof). So on the set of nonnegative numbers, we can say that $f(x) = x$. But $f(x)$ could take any value at all for negative numbers. His example of $f(x) = |x|$ is one where $f(x) = x$ for nonnegative numbers, but $f(x) = -x$ for negative numbers. We could also take functions like $\displaystyle f(x) = x\frac{1 + x}{1 + |x|} + (x - |x|)^3$ or indeed any arbitrary function $f$ satisfying $f(x) = x$ for nonnegative numbers.

Answer 2

It depends on the range of $g$.

For example, if $g(x) = x^2$, and $f(x) = |x|$, then $f(g(x)) = g(x)$ for all real $x$.

Answer 3

This is the question I am answering:

If some functions $f$ and $g$ are such that $f(g(x)) = g(x)$ and $g(x) = x + 2f(x)$ for every $x$, can we conclude that $f(x)=x$ for every $x$? More generally, which properties should such a function $g$ satisfy, such as, being one-to-one, etc.?

If $f\circ g=g$ and $g(x)=x+2f(x)$ for every $x$ then $2f\circ g=g\circ g-g$ hence $g\circ g=3g$. Conversely, if $g\circ g=3g$, define $f$ by $f(x)=\frac12(g(x)-x)$, then $f\circ g=g$ and $g(x)=x+2f(x)$ for every $x$. This shows one is reduced to study the solutions of the functional equation $g\circ g=3g$.

To answer succinctly the main question in the post, the obvious solution $g:x\mapsto3x$ is not the only one hence no, one cannot conclude that $f:x\mapsto x$.

For a counterexample, let $X\subset\mathbb R$ and $h:X\to \mathbb R\setminus X$ and consider the function $g$ defined by $g(x)=3x$ for every $x$ not in $X$ and $g(x)=h(x)$ for every $x$ in $X$. Then $g\circ g=3g$ but $g$ need not be the function $x\mapsto3x$, nor a regular function, nor one-to-one, nor onto.

Note that $X$ can be as large as one wants, as long as $X\ne\mathbb R$, and that $h$ can be any function whose image is included in $\mathbb R\setminus X$. This leaves plenty of room for "pathological" examples.

To be fully specific (while being not too pathological), consider the function $f$ defined by $f(x)=x$ for every $x$ not in $\mathbb Q$ and $f(x)=\frac12(\pi-x)$ for every $x$ in $\mathbb Q$.