Let $f$ be a bounded twice differentiable real valued function on $\mathbb R$ such that $f''(x) \geq 0$ for all $x$. Then $f$ is constant Let $f$ be a bounded twice differentiable real valued function on $\mathbb  R$ such
that $f''(x) \geq 0$ for all $x$. Then how to show that $f$ is a constant function.
please give me some hint.
 A: Given that $f''\geq0$, $f$ is convex. This implies that $f$ lies weakly above its tangent line at any point. Formally, for any $x_0\in\mathbb R$,
$$f(x)\geq f(x_0)+f'(x_0)(x-x_0).$$
Now, if $f'(x_0)>0$, then the right-hand side goes to infinity as $x\to\infty$, which contradicts $f$ being bounded. Similarly, if $f'(x_0)<0$, then the right-hand side diverges to $\infty$ as $x\to-\infty$, which implies that $\lim_{x\to-\infty}f(x)=\infty$, another contradiction. It follows that $f'(x_0)=0$. Since this is true for any $x_0\in\mathbb R$, $f'=0$, so that $f$ is constant.
A: Because $f'' \geq 0$, $f'$ is nondecreasing. Suppose that there exists $x_0 \in \mathbb{R}$ such that $f'(x_0) \neq 0$. If $f'(x_0)>0$, because $f'(x)\geq f'(x_0)$ for all $x \geq x_0$, we have $$f(x)=f(x_0)+ \int_{x_0}^x f'(t)dt \geq f(x_0)+f'(x_0)(x-x_0) \underset{x \to + \infty}{\longrightarrow}+ \infty.$$
If $f'(x_0)<0$, because $f'(x) \leq f'(x_0)$ for all $x \leq x_0$, we have $$f(x)=f(x_0)-\int_{x}^{x_0} f'(t)dt \geq f(x_0)+f'(x_0)(x-x_0) \underset{x \to - \infty}{\longrightarrow} + \infty.$$
Therefore, if $f$ is bounded, necessarily $f'=0$, that is $f$ is constant.
