Functional equations towers of $f$

Question

If $f(f(1))=f(1)$ does it imply that $f(1)=1$?

The problem I'm solving is $(x+y)f(y(f(x))=x^2(f(x)+f(y))$ where $f$ maps from $R^{>0}$ to $R^{>0}$.

Answer 1

In general, $f(f(1))=f(1)$ does not imply $f(1)=1$.

For your problem, you can put $x=y$ and obtain $f(xf(x))=x(f(x)$. This means that any element of the form $x(f(x)$ is a fixed point of $f$.

Let $a$ be any fixed point of $f$. Substitute $x=a,y=1$ and you will find that $f(1)=\frac{1+a-a^2}{a}$. But this means that all fixed points of $f$ satisfy this equation.

Let $b$ be a different fixed point (if any exists). Then $$\frac{1+a-a^2}{a}=\frac{1+b-b^2}{b}$$ and solving this gives $b=-\frac{1}{a}$. However we only have positive reals.

This means that for any $x$, the fixed point $xf(x)$ is $a$ i.e. $f(x)=\frac{a}{x}$.

Over to you!