Find all real function $f(x)$ satisfy this functional equation $|f(x) - f(y) | \leq |\sin(x-y) -x+y| $ for all $x, y \in \mathbb{R}$?
My ideas: I showed that the constant functions are a class of such functions. However, I have not known all solutions. Would you please give some comments (or hints) for this problems?
Thank you so much!
On
Assume $|f(x) - f(y) | \leq |\sin (x-y) -x+y|$ for all real $x,y.$ For $y$ held constant, put $x=y+h$ where $h \neq 0.$ Then the inequality, after division by $|h|,$ becomes $|\frac{f(y+h)-f(y)}{h}| \leq | \frac{\sin h}{h}-1|.$ Now let $h \to 0$ through nonzero values from either side, then the right side goes to $0$ from which follows $f'(y)=0$ for any real $y.$ Thus only constants can satisfy your inequality.
Hint 1: Notice that \begin{align} \sin z = z-\frac{z^3}{3!}+\frac{z^5}{5!}+\ldots. \end{align}
Hint 2: For what functions do we have \begin{align} |f(x)-f(y)|\le C|x-y|^3? \end{align}