6
$\begingroup$

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!

$\endgroup$
0
0
+50
$\begingroup$

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.

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.