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

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 $$f(x)\ge f(0)\sin x$$
• What about $x=0$? – Mr Pie Apr 12 '18 at 14:47
• @user477343: $f(y)\sin y\ge -f(0)$ – Roman83 Apr 12 '18 at 16:55
• So $-f(y)\sin y \leqslant f(0)$ and thus by multiplying both sides by $\sin x$, we get that $$f(x)\geqslant -f(0)\sin y\sin x.$$ Ugh didn't help :/ – Mr Pie Apr 12 '18 at 16:59

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.

• Why if $f(x+\frac{\pi}{2}+\Delta x) \geq f(x) \cos(\Delta x)$ then $f(x+\frac{\pi}{2}+\Delta x) \geq f(x)$ ? – Roman83 Apr 14 '18 at 8:03
• You've shown that $f(\pi n/2)$ is independent of $n$, but I don't see how this implies that $f$ is periodic. – B. Mehta Apr 16 '18 at 16:12
• @B.Mehta It's not just $f(\frac{\pi n}{2})$ is independent of n, but $f(\frac{\pi n}{2}+x_0)$ is independent of n for all x_0. – Acccumulation Apr 16 '18 at 16:52
• @Roman83 Powers of infinitesimals greater than 1 can be taken to be zero. If you take f(x+π2+h) and write it as f(x+π2+Δx)≥f(x)sin(π2+nΔx) where Δx = h/n, then the amount f can differ by changing by Δx goes like Δx^2, so taking n of these results in n(Δx)^2 difference. Since Δx = h/n, the max difference is (h^2)/n. As we take increasing n for fixed h, the amount it can differ goes to zero. – Acccumulation Apr 16 '18 at 16:59
• @Acccumulation Thanks, your edit clears this up. – B. Mehta Apr 16 '18 at 17:00