Today I have came across a problem.It is harder than I think.
Given that
- (1) $f(x)$ is differentiable on $\left[ {{\rm{0}},{\rm{ + }}\infty } \right)$,
- (2) $f'(x)$ is uniformly continuous on $\left[ {{\rm{0}},{\rm{ + }}\infty } \right)$, and
- (3) $\mathop {\lim }\limits_{x \to {\rm{ + }}\infty } f(x)$ is existent, prove that $$\mathop {\lim }\limits_{x \to {\rm{ + }}\infty } f'(x){\rm{ = 0}}.$$
If not, is there any counter example? (I've been working on this for a long time.)
FIRST METHOD :
Just write $$f(x)=f(0)+\int_0^x f'(t) dt$$
Because $f$ has a limit, you deduce that the integral of $f'$ converges. And because $f'$ is uniformly continuous, this implies that $f'$ tends to $0$ (see for example A uniformly continuous function whose integral $\int_0^\infty f(x)dx$ exists converges to zero)
SECOND METHOD :
Let $\varepsilon >0$.
By uniform continuity of $f'$ applied with $\varepsilon'=\varepsilon/2 >0$, there exists $\delta >0$ such that for all $x,y \in [0,+\infty)$ such that $|x-y| \leq \delta$, then $|f'(x)-f'(y)|\leq \varepsilon/2$.
Then, you know that $f(x)$ has a limit $l \in \mathbb{R}$ when $x$ tends to $+\infty$ : in particular, with $\varepsilon''=\delta\varepsilon/4 >0$, you deduce there exists $A \in \mathbb{R}_+$ such that for all $x\geq A$, $|f(x)-l|\leq \delta\varepsilon/4$. In particular, for all $x,y \geq A$, you have $|f(x)-f(y)|\leq \delta\varepsilon/2$.
Now fix an $x \geq A$. By the MVT, there exists $c \in [x,x+\delta]$ such that $$\frac{f(x+\delta)-f(x)}{\delta}=f'(c)$$
But $|x-c|\leq \delta$, so $|f'(x)-f'(c)|\leq \varepsilon/2$. You deduce that $$|f'(x)|\leq |f'(x)-f'(c)|+|f'(c)| \leq \frac{\varepsilon}{2} + \frac{|f(x+\delta)-f(x)|}{\delta} \leq \frac{\varepsilon}{2} + \frac{\delta\varepsilon}{2\delta} = \varepsilon$$
This proves that $\forall \varepsilon >0, \exists A \in \mathbb{R}_+, \forall x \geq A, |f'(x)|\leq \varepsilon$, i.e. that $f'(x)$ tends to $0$ in $+\infty$.