2
$\begingroup$

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?

Thanks in advance!

$\endgroup$
0

3 Answers 3

2
$\begingroup$

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 $f(x_0+\sin(\pi x_0))=f(x_0)$.

$\endgroup$
2
$\begingroup$

Another necessary condition for a continuous $f$ would be: $f$ should not have a unique global maximum or global minimum at any $0<a<1$ 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 $0<a<1$. Let $b$ be the smallest positive real such that $b+\sin(\pi b)=a$. Certainly, $0<b<a$. Now, $g(b)=f(a)>f(b)$, since $f(a)$ is the maximum. Similarly, $g(a)<f(a)$. 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.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$ Commented Apr 30, 2020 at 16:44
  • 1
    $\begingroup$ 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. $\endgroup$
    – DSM
    Commented May 1, 2020 at 4:24
0
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ Why the difference between $x_1$ and $x_2$ should be the same as $\sin \pi x_1$? $\endgroup$
    – KhashF
    Commented May 3, 2020 at 3:56
  • $\begingroup$ @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)? $\endgroup$ Commented May 3, 2020 at 4:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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