TITLE: A challenging problem on continuity-MADHAVA-2020.
QUESTION [8 upvotes]: Suppose $f$ be a continuous function on $(0,\infty)$ such that $f(\frac{x}{x+1})=f(x)+2$.Is the function monotonically decreasing?I have tried to prove it in different ways but still I could not find any way out.Also how to show $f(x) \to \infty$ as $x\to 0+$.Actually it is a question from MADHAVA-2020 which took place on 12 Jan,2020.

REPLY [6 votes]: Consider the function
$$f(x):={2\over x}+\sin{2\pi\over x}\qquad(0<x<\infty)\ .$$
Then
$$f\left({x\over x+1}\right)={2(x+1)\over x}+\sin{2\pi(x+1)\over x}=f(x)+2\qquad(0<x<\infty)\ .$$
Furthermore
$$f'(x)=-{2\over x^2}-{2\pi\over x^2}\cos{2\pi\over x}=-{2\over x^2}\left(1+\pi \cos{2\pi\over x}\right)\ .$$
Here the RHS is positive on the intervals where $\cos{2\pi\over x}<-{1\over\pi}$; hence $f$ is not decreasing in these intervals.
In order to find the general solution to the given functional equation we replace the variable $x$ by $x:={1\over u}$ $\>(0<u<\infty)$. This means that instead of $f$ we now look at the function
$$g(u):=f\left({1\over u}\right)\ .\tag{1}$$
We now have
$$g(u+1)=f\left({1\over u+1}\right)=f\left({{1\over u}\over {1\over u}+1}\right)=f\left({1\over u}\right)+2=g(u)+2\ .$$
This implies that
$$g(u)=2u+h(u)\ ,$$
where $h$ is periodic with period $1$. From $(1)$ we then get
$$f(x)=g\left({1\over x}\right)={2\over x}+h\left({1\over x}\right)\ .$$
If $f$ is continuous then $h$ has to be continuous as well, hence the periodic $h$ is bounded. This implies $\lim_{x\to0+} f(x)=\infty$.