$f(x) \ge f(x + \sin x)$, nonconstant functions, infinite number of solutions to $f'(x) = 0$.

Question

Let $\mathcal{F}$ be the set of all the differentiable functions $f: \mathbb{R} \to \mathbb{R}$, which have the property $f(x) \ge f(x + \sin x)$, for all $x \in \mathbb{R}$.

1. Prove that $\mathcal{F}$ contains also nonconstant functions.
2. Prove that if $f \in \mathcal{F}$ then the set of the solutions of the equation $f'(x) = 0$ is infinite.

Answer 1

Prove that $\mathcal{F}$ contains also nonconstant functions.

We notice that the function$$u: \mathbb{R} \to \mathbb{R},\text{ }u(x) = x + \sin x$$is increasing,$$2k\pi \le x \le x + \sin x \le \pi + 2k\pi,\text{ }x \in [2k\pi, \pi + 2k\pi],$$and$$2\pi + 2k \pi \ge x \ge x + \sin x \ge \pi + 2k \pi,\text{ } x \in [\pi + 2k\pi, 2\pi + 2k\pi].$$This shows that any nonconstant differentiable function which is decreasing on the intervals $[2k \pi, \pi + 2k\pi]$ and increasing on the intervals $[\pi + 2k\pi, 2\pi + 2k\pi]$ would do (for instance, cosine).

Prove that if $f \in \mathcal{F}$ then the set of the solutions of the equation $f'(x) = 0$ is infinite.

Let$$g: \mathbb{R} \to \mathbb{R}, \text{ }g(x) = f(x) - f(x + \sin x).$$Obviously $g(k\pi) = 0$ and $g \ge 0$, so $k \pi$ are minimum points for $g$. It follows that $g'(k\pi) = 0$, whence$$f'(k\pi) - f'(k\pi)(1 + \cos k\pi) = 0,$$therefore $f'(k\pi) = 0$ for every integer $k$.