proved that$f (x)$ is uniformly continuous on $[0,+\infty)$ if and only if $\lim\limits_{x\to+\infty}\frac{f(x)}{x}$ exists.

Question

Let $f(x)$ be second-order differentiable on $[0,+\infty)$,$f''(x)\geqslant\sin x(\forall x\in[0,+\infty))$.Prove that $f (x)$ is uniformly continuous on $[0,+\infty)$ if and only if $\lim\limits_{x\to+\infty}\frac{f(x)}{x}$ exists.This is our monthly exam question,I try to expand it with Taylor's theorem: $$f(x+h)=f(x)+f'(x)h+\frac{f''(\xi)}{2}h^2\geqslant f(x)+f'(x)h +\frac{\sin x}{2}h^2.$$ But then I don't know how to deal with it, I also know that if $f (+\infty)$ exists, then $f (x)$ is uniformly continuous, but it doesn't seem to help the problem.

Answer 1

Some comments:

(1) We cannot drop the continuity from the right at $x=0$. Here is a counterexample: the function $f(x):= \frac{1}{x}-\sin{x}$ is not uniformly continuous on $(0,\infty)$. We have $f^{\prime\prime}(x)=\frac{2}{x^3}+\sin{x}>\sin{x}$ for every $x\in (0,\infty)$. And obviously $\lim_{x\rightarrow \infty}\frac{f(x)}{x}=0$.

(2) It is also enough to prove that $f$ is uniformly continuous on $[1,\infty[$ since it is, by assumption, continuous on any closed interval
$[0,N]$ with $N>0$, and consequently uniformly continuous thereat.

(3) We know that, since $\lim_{x\rightarrow +\infty}\frac{f(x)}{x}$ exists,
$$|f(x)|\leq C |x|$$ for large $x$, meaning the function can grow at most linearly.

But the statement does not apply to functions of the form $f(x)=x^\alpha$ for any $\alpha\in (0,1]$ because the restriction $f^{\prime\prime}(x)\geq \sin{x}$ is not satisfied by such functions.

(4) I suggest the following simplification of the problem: We can write any smooth $f:[0,\infty)\rightarrow \mathbb{R}$ as $f(x)=g(x)-\sin{x}$ where $g:[0,\infty)\rightarrow \mathbb{R}$ is smooth. Now, the peculiar condition $f^{\prime\prime}(x)\geq \sin{x}$ translates into $g^{\prime\prime}(x)\geq 0$. We also have $\lim_{x\rightarrow +\infty}\frac{g(x)}{x}=\lim_{x\rightarrow +\infty}\frac{f(x)}{x}$ exists. Since $x\mapsto \sin{x}$ is uniformly continuous on $\mathbb{R}$, we only need to show that $g$ is uniformly continuous on $[N,\infty)$ with $N$ large.

Answer 2

proof:

We think $f^{''}(x)=sin(x)+g(x)$. Cuase $f^{''}(x)>sin(x)$, $g(x)>0$ on $[0,\infty)$.

So we can get $f(x)=-sin(x)+(f^{'}(0)+1)x+f(0)+\int_{0}^{x}\,dy \int_{0}^{y} g(t)\, dt$

It's easy to konw $f(x)$ being uniformly continuous on $[0,\infty)$ is equivalent to $f^{'}(x)$ being bounded, which is equivalent to $\int_{0}^{\infty} g(x)\,dx$ is convergent, under the conditions given in the problem.

It's easy to find that $\lim_{x \to \infty} \frac{f(x)}{x}=\lim_{x \to \infty} \frac{-sin(x)+(f^{'}(0)+1)x+f(0)+\int_{0}^{x}\,dy \int_{0}^{y} g(t)\, dt}{x}=f^{'}(0)+1+\lim_{x \to \infty}\frac{\int_{0}^{x}\,dy \int_{0}^{y} g(t)\, dt}{x}=f^{'}(0)+1+\int_{0}^{\infty} g(t)\,dt$

This shows that $\lim_{x \to \infty} \frac{f(x)}{x}$ being exist is equivalent to $\int_{0}^{\infty} g(x)\,dx$ being convergent, which means that $\lim_{x \to \infty} \frac{f(x)}{x}$ being exist is equivalent to $f(x)$ being uniformly continuous on $[0,\infty)$.