Problem A. 895. (December 2024)
A. 895. Let's call a function \(\displaystyle f\colon\mathbb{R}\to\mathbb{R}\) weakly periodic if it is continuous and satisfies \(\displaystyle {f(x+1)}=f(f(x))+1\) for all \(\displaystyle x\in \mathbb{R}\). a) Does there exist a weakly periodic function such that \(\displaystyle f(x)>x\) for all \(\displaystyle x\in\mathbb{R}\)? b) Does there exist a weakly periodic function such that \(\displaystyle f(x)<x\) for all \(\displaystyle x\in\mathbb{R}\)?
Proposed by: András Imolay, Budapest
(7 pont)
Deadline expired on January 10, 2025.
a) We claim that no such function exists. Let \(\displaystyle f^n\) denote the \(\displaystyle n^{\text{th}}\) iterate of function \(\displaystyle f\) (i.e. \(\displaystyle f^n(x)=f(f(...f(x)))\), where \(\displaystyle f\) is repeated \(\displaystyle n\) times). We will prove the following identity by mathematical induction:
\(\displaystyle f(x+k)=f^{2^k}(x)+k. \)
Indeed, this is the original condition for \(\displaystyle k=1\), and it the equality holds for \(\displaystyle k\), then
\(\displaystyle f(x+k+1)=f(f(x+k))+1=f(f^{2^k}(x)+k)+1=\left(f^{2^k}(f^{2^k}(x))+k\right)+1=f^{2^{k+1}}(x)+(k+1). \)
Next note that if we can find \(\displaystyle a\) such that \(\displaystyle f(a)=a+1\), then \(\displaystyle f(a+1)=f(f(a))+1=f(a+1)+1\), which is a contradiction. Since \(\displaystyle f\) is continuous, this means either \(\displaystyle f(x)>x+1\) for all \(\displaystyle x\) or \(\displaystyle x<f(x)<x+1\) for all \(\displaystyle x\).
In the first case \(\displaystyle f(x)=f(f(x-1))+1>f(x-1)+2\), therefore \(\displaystyle f(x-1)-(x-1)<(f(x)-2)-(x-1)=(f(x)-x)-1\). By iterating this we get \(\displaystyle f(x-k)-(x-k)<(f(x)-x)-k\), and thus \(\displaystyle f(-k)-(-k)<f(0)-k\). However, if \(\displaystyle k>f(0)\), we get a contradiction, since \(\displaystyle f(x)>x\) must hold for all \(\displaystyle x\).
In the second case, let's consider increasing sequence \(\displaystyle a_n=f^n(0)\). If this sequence is bounded from above, then it has a limit \(\displaystyle a\). However, since \(\displaystyle f\) is continuous, \(\displaystyle a_{n+1}=f(a_n)\) implies \(\displaystyle b=f(b)\), which is a contradiction. On the other hand, subsequence \(\displaystyle f^{2^k}(0)\) is bounded from above, since it's equal to \(\displaystyle f(k)-k<1\) (by the identity we've derived in the beginning of the solution), and this is a contradiction (since we are dealing with an increasing sequence). The proof is complete.
b) Such a function exists: \(\displaystyle f(x)=-\log_2(2^{-x}+1)\) satisfies the condition. Indeed,
\(\displaystyle f(f(x))=-\log_2(2^{\log_2(2^{-x}+1)}+1)=-\log_2(2^{-x}+2)=-\log_2(2\cdot(2^{-x-1}+1))=-1-\log_2(2^{-x-1}+1)=f(x+1)-1.\)
Statistics:
11 students sent a solution. 7 points: Varga Boldizsár, Xiaoyi Mo. 4 points: 3 students. 3 points: 1 student. 0 point: 5 students.
Problems in Mathematics of KöMaL, December 2024