Function with bounded $f''$ on $(0,\infty)$ and $f(x) \to 0$ as $ x\to \infty$ then $f'(x) \to 0$ as $ x\to \infty$

Question

Suppose $f$ is twice differentiable function on $(0,\infty)$ and $f''$ is bounded on $(0,\infty)$ and $f(x) \to 0$ as $ x\to \infty$. Then I want to prove $$\lim_{x\to\infty} f'(x) = 0.$$

My attempt:
On contrary suppose that$|f'(x)| > 0$ for $ x\to \infty$. As $f''$ is bounded on $(0,\infty)$ then $f'(x) $ is uniformly continuous on $(0,\infty)$.
As $\lim_{x\to \infty}f(x)=0$ and sequence $x_n \to \infty$, $f(x_n)\to 0$. Similary another equivalent sequnce to $x_n$ sequence $y_n \to \infty ,f(y_n)\to 0$ As $f'$ is uniformly continous on $(0,\infty)$ then $|f'(x_n)-f'(y_n)|<\epsilon$, for some small $\epsilon >0$ as $|(x_n)-(y_n)|<\delta$.
I was expecting some contradiction, but I did not get one. Where is my mistake? or any other argument needed. Any Help will be appreciated.

Answer 1

Writing $f(x) = f(0) + \int_0^x f'(t) dt$, and taking into account that the limit of $f$ exists at infinity, we have that the integral $\int_0^\infty f'(t) dt$ converges.

Observe, that the boundedness of the 2nd derivative of $f$ implies uniform continuity of $f'$ in $(0,\infty)$. Now assume that $f'$ does not converge to $0$ at infinity. Hence, there exists a sequence $t_k \to \infty$ such that $f'(t_k) > a$, where $a>0$ is fixed (without loss of generality we assume $a>0$, as otherwise we could have taken $f'(t_k) < a$, with $a<0$). In view of the uniform continuity of $f'$ there exists $\delta>0$ such that for all $k\in \mathbb{N}$ one has $$ |f'(t) - f'(t_t) | \leq \frac 12 a, \ \text{ if } |t-t_k| \leq \delta, $$ and hence $$ f'(t) - f'(t_k) \geq -\frac 12 a, \text{ for all } t \in [t_k - \delta, t_k + \delta]. $$ Since $f'(t_k) \geq \frac 12 a$ , from the last inequality we get $$ f'(t) \geq \frac 12 a, \qquad \text{ for all } t \in [t_k - \delta, t_k + \delta]. $$

It follows that the integrals $$ \int_{t_k - \delta}^{t_k + \delta} f'(t) dt \geq \frac{1}{2} a 2 \delta = a \delta>0 $$ do not converge to $0$, which contradicts to the convergence of the integral $\int_0^\infty f'(t) dt$ and proves the original claim.

Answer 2

There is a problem from the start. You seem to think that if you deny $\lim_{x\to\infty}f'(x)=0$ then you have $f'(x)>0$ if $x$ is large enough (or, I suppose, that you have $f'(x)<0$ if $x$ is large enough). This is not true. Suppose that $f(x)=\sin x$. Then you don't have $\lim_{x\to\infty}f'(x)=0$. However, it is not true that $f(x)>0$ (or that $f(x)<0$) if $x$ is large enough.