# A challenging problem on continuity-MADHAVA-2020.

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.

• First thoughts: We know that $f(\frac{x}{x+1}) > f(x)$. If we assume that $f$ is monotone and decreasing, where does that lead us? Commented Jan 13, 2020 at 13:14
• @MattiP. Will you please elaborate? Commented Jan 13, 2020 at 13:16
• Tell us that MADHAVA 2020 is an exam that took place this past Sunday, so that we can tell it is now finished. madhavacompetition.com Commented Jan 14, 2020 at 13:10
• @GEdgar what do you mean?Madhava is over past sunday. Commented Jan 14, 2020 at 13:19
• I mean: you should say that in your question, so that we will know it is over. Commented Jan 14, 2020 at 13:38

Consider the function $$f(x):={2\over x}+\sin{2\pi\over x}\qquad(0 Then $$f\left({x\over x+1}\right)={2(x+1)\over x}+\sin{2\pi(x+1)\over x}=f(x)+2\qquad(0 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 $$x:={1\over u}$$ $$\>(0. This means that instead of $$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$$.

• How to show that the limit at 0+ is infinity? Commented Jan 15, 2020 at 1:49
• Brilliant new edit! +1 Commented Jan 16, 2020 at 12:34

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 $$(0,+\infty)$$.

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 $$+\infty$$.

• I have some problem with the last line $k\to \infty$ then LHS is lim_{z\to 0+}f(z) while RHS is infinity. Commented Jan 15, 2020 at 14:41
• What problem do you have with it? Commented Jan 15, 2020 at 15:02
• you must first show that $lim_{z\to 0+}f(z)$ exists then only$lim_{k\to \infty}f(g^k(x)=lim_{x\to 0+}f(z)$. Commented Jan 16, 2020 at 1:55
• please use the definition of limit to show that the limit is infinity. Commented Jan 16, 2020 at 1:56
• You can make it rigorous as follows. Let $x_0$ be such that $f(x_0) =\min_{x\in\left[\frac12,1\right]}f(x)$. Notice that the minimum exists and is attained because $f$ is continuous and $\left[\frac12,1\right]$ is compact. Then $\min_{x\in\left[\frac13,\frac 12\right]}f(x) = f(x_0) + 2$. Similarly, $\min_{x\in\left[\frac14,\frac 13\right]}f(x) = f(x_0) + 4$. You can proceed inductively. Commented Jan 16, 2020 at 2:48