TITLE: Show that $\lim_{x \rightarrow \infty} f(x)$ exists by the given condition.
QUESTION [5 upvotes]: Suppose $f$ is a twice differentiable function such that $f''(x) \ge 0$ for all $x \in \mathbb R$. Let $0 \le f(x) \le 1$ for $x \ge 0$. Then show that $\lim_{x \rightarrow \infty} f(x)$ exists.
I have tried it but I fail. Please give me some hint to proceed in the right way.
Thank you in advance.

REPLY [2 votes]: A brief answer.
If $f:\mathbb R\to\mathbb R$ is convex, then exactly one of the following three holds:
a. $f$ is increasing,
b. $f$ is decreasing,
c. There exists an $x_0\in\mathbb R$, such that $f$ is decreasing in $(-\infty,x_0]$ and increasing in $[x_0,\infty)$.
In all three cases, $f$ is eventually monotonic (i.e., it is monotonic in some interval $[a,\infty)$, for some $a\ge 0$). Since $f$ is also bounded, in the same interval, then the limit $\lim_{x\to\infty}f(x)$ exists, and it lies in the interval $[0,1]$, since $0\le f(x)\le 1$, for $x\ge 0$.