Find all functions $ f: \mathbb{R} \to \mathbb{R} $ such that $ f(x) \geq f(y)\sin(x-y)$

Question

Find all functions $ f: \mathbb{R} \to \mathbb{R} $ such that $$ f(x) \geq f(y)\sin(x-y),\quad \forall x,y \in \mathbb{R} $$

My attempts

1. Let $x=y$ then $f(x)\ge0$
2. Let $y=0$ then $f(x)\ge f(0)\sin x$

Answer 1

If you define $a_n = f(\frac{\pi n}{2}+x_0)$ then

$a_{n+1} \geq a_{n} sin(\frac{\pi}{2}) = a_{n} $

$a_{n-3} \geq a_{n} sin(\frac{-3\pi}{2}) = a_{n} $

So $a_{n-3} \geq a_n \geq a_{n-1} \geq a_{n-2} \geq a_{n-3}$

Thus, $a_n = a_{n-1} = a_{n-2} = a_{n-3}$

So $f(x+\frac{\pi}{2}) = f(x)$ for all x, i.e. f is periodic.

$f(x+\frac{\pi}{2}+\Delta x) \geq f(x) sin(\frac{\pi}{2}+\Delta x)$

$f(x+\frac{\pi}{2}+\Delta x) \geq f(x) cos(\Delta x)$

$cos(\Delta x) = 1-\frac{(\Delta x)^2}{2}+\frac{(\Delta x)^4}{24} ...$

Snnce $\Delta x$ is an infinitesimal, we can take the higher powers of $\Delta x$ to be 0 and thus $cos(\Delta x)$ to be 1.

$f(x+\frac{\pi}{2}+\Delta x) \geq f(x)$

$f(x+\Delta x) \geq f(x)$

So f is nondecreasing.

Since f is a nondecreasing period function, it must be constant.