# What are the necessary conditions on $f$ if $f(x)=f(\sin(\pi x)+x)\iff x\in\Bbb{Z}$?

I am aware 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 $$x^x$$. I have been told that 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?

Here is a condition that $$f$$ must satisfy if it happens to be periodic: The period must be larger than $$1$$. Otherwise, the function $$f$$ satisfies $$f(x+T)=f(x)$$ for all $$x$$ where $$T\in (0,1]$$. But then there exists $$x_0\in\Bbb{R}-\Bbb{Z}$$ with $$\sin(\pi x_0)=T$$ for which $$f(x_0+\sin(\pi x_0))=f(x_0)$$.

Another necessary condition for a continuous $$f$$ would be: $$f$$ should not have a unique global maximum or global minimum at any $$0 over $$[0,t]$$ where $$t=\arg\max_{0\leq x\leq 1}(x + \sin(\pi x)]$$. Also, let $$g(x) = f(x+\sin(\pi x))$$.

Suppose it does have a global maximum over $$[0,t]$$ at some $$0. Let $$b$$ be the smallest positive real such that $$b+\sin(\pi b)=a$$. Certainly, $$0. Now, $$g(b)=f(a)>f(b)$$, since $$f(a)$$ is the maximum. Similarly, $$g(a). This implies that $$f$$ and $$g$$ must intersect somewhere within $$[b,a]$$, a contradiction. A similar argument hold is there existed a global minimum.

This condition can be applied to $$[2m,2m+t]$$, for any integer $$m$$. Hope this helps.

• Thanks for the answer! I tried to test this theory on Desmos, and, unless I am mistaken, $f(x)=\sin(x)$ does not have a max/min at any $a$ over $[0,t]$. But I am certain that this $f$ does not satisfy the statement. I calculated $t=-\frac{\cos^{-1}\left(-\frac{1}{\pi}\right)}{\pi}$, which means $[0,t]$ should actually be written $[t,0]$. Is this a case of me not understanding your answer or of an error in the theory? Commented Apr 30, 2020 at 16:44
• Sorry about the oversight, and thanks for catching that. It is $t=\arg\max_{0\leq x\leq 1}(x + \sin(\pi x)]$ instead of $t=\arg\min_{0\leq x\leq 1}(x + \sin(\pi x)]$. Hope it helps.
– DSM
Commented May 1, 2020 at 4:24

WARNING: INCORRECT see comments for why
If f is continuous on some connected sets it must be monotonic on them. Say it is not. Then there is some point $$x^*$$ such that in some of its neighborhood V such that $$x^*=\sup_V f$$. Then it can be shown easily that there are $$x_1,x_2$$ in some neighborhood of $$x^*$$ such that $$f(x_1)=f(x_2)$$ and $$d(x_1,x_2)\lt 1$$ causing a contradiction.

note: here monotone is defined as that there is no $$x^*$$ such that it is the superior of all $$f(x)$$ in some of its neighborhood, which can be shown to be equivalent to the concept of monotone in real valued functions of one real variable.

• Why the difference between $x_1$ and $x_2$ should be the same as $\sin \pi x_1$? Commented May 3, 2020 at 3:56
• @KhashF well I think that I have managed to mess up... I just read your answer and thought, emm, there mustn't be two x with their distance less than 1 and f(x) the same. Would this be a fixable issue (or non fixable which probably means that I need to delete this answer)? Commented May 3, 2020 at 4:03