This is a follow-up from my previous question.

I now know that the statement: $$f(x)=f(\sin(\pi x)+x)\iff x\in\Bbb{Z}$$ is not true for all $f$. For example, $f$ can be $x$ to any constant power or any constant to the $x$th power but it cannot be the gamma function $\Gamma(x)$ or $\sin(x)$ or $x^x$. According to the answer I received, it is important to note whether or not $f$ is injective. However, $f(x)=x^2$ is not injective, yet it satisfies the statement. If being injective is only a sufficient condition as opposed to a necessary condition, what exactly do we know about the class of functions that makes this statement true?

Thanks in advance!


1 Answer 1


We assume that $f$ is a function from $\Bbb R$ to$\Bbb R$. Let $\mathcal P_f$ be a partition of $\Bbb R$ into the equivalence classess of a relation $\sim_f$, where $x\sim_f x’$ iff $f(x)=f(x’)$ for any $x,x’\in\Bbb R$. The function makes the statement true iff for each $P\in P_f$ and each $x\in P\setminus\Bbb Z$ we have $x+\sin(\pi x)\not\in P$. This is clearly satisfied when $f$ is injective, because in this case each $P\in\mathcal P_f$ is a one-point set.

Now consider an other binary relation $\sim_s$ on $\Bbb R$ defined as follows. For any $x,x’\in\Bbb R$ we have $x\sim_s x’$ iff there exists a finite sequence $x=x_1, x_2,\dots, x_n=x’$ such that for each $1\le i\le n-1$ we have $x_{i+1}=x_i+\sin (\pi x_i)$ or $x_{i}=x_{i+1}+\sin (\pi x_{i+1})$. Then $\Bbb R$ splits into equivalence classes of the relation $\sim_s$, providing a partition $\mathcal P_s$ of $\Bbb R$. Then the function $f$ satisfies the condition iff for each $P\in\mathcal P_f$ and each $P’\in\mathcal P_s$ we have $P\cap P’$ is empty or an one-point set.

So we have to study the partition $\mathcal P_s$. The graph of the function $x+\sin(\pi x)$ shown below suggests that for each integer $n$, the set $\{2n\}$ is an one-point member of $\mathcal P_s$, but other members of $\mathcal P_s$ can have more complicated structure.

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