A challenging problem on continuity-MADHAVA-2020.

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 $\lim_{x\to0+} f(x)=\infty$.

I am led to believe that the answer is no.

Let $g:(0,+\infty) \longrightarrow (0,1)$ be given by $g(x) = x/(x+1)$. The statement of the problem gives us that $f(g(x)) = f(x) + 2$, and letting $x\to+\infty$ we get $\lim_{x\to+\infty}f(x) = f(1) - 2$.

Observe that $g$ maps $[1,+\infty)$ to $\left[\frac12, 1\right)$ and that it maps each interval $\left[\frac1{n+1},\frac1n\right]$ to the interval $\left[\frac1{n+2},\frac1{n+1}\right]$. Notice that $g$ is a monotonically increasing bijection, so in each case these mappings are themselves monotonically increasing bijections.

Define $f$ continuously on $[1,+\infty)$ however you wish (not necessarily monotonically), except that we require that $\lim_{x\to+\infty}f(x)$ exists and equals $f(1) - 2$.
With the relationship $f(g(x)) = f(x) +2$, one mapping of $g$ propagates the values of $f$ to $\left[\frac12, 1\right)$ and the limit condition ensures continuity at $x=1$.
Another mapping of $g$ propagates the values of $f$ to $\left[\frac13, \frac12\right]$, and in this manner we define $f$ inductively on all of $(0,+\infty)$.

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To see that $\lim_{x\to 0^+} f(x) = +\infty$, pick any $x>0$. We have

\begin{align} f(x) &= f\Big(g(x)\Big) - 2 \\&= f\Big(g^2(x)\Big) - 4 \\&= f\Big(g^3(x)\Big) - 6 \\&=\,\,\dots \\&= f\Big(g^k(x)\Big) - 2k \end{align}

for all $k\geqslant 0$. Equivalently,

$$f\Big(g^k(x)\Big) = f(x) + 2k$$

and letting $k\to\infty$ the LHS is $\lim_{z\to 0^+}f(z)$ while the RHS is $+\infty$.