Let $f: \mathbb{R} \longrightarrow \mathbb{R}$ be a bounded differentiable function. Then for any $\varepsilon > 0$, there exists $x \in \mathbb{R}$, s.t. $|f^{\prime}(x)| < \varepsilon$.
Is this statement above true? I think it's true, because if it doesn't hold, there exists $\varepsilon >0$, s.t. $|f^{\prime}(x)|$ is always larger than $\varepsilon$, then $f$ increase very fast or decrease very fast, which contradicts with the boundedness. But I don't know how to prove it.
If the statement is not true, then there exists $\varepsilon >0$, s.t. $|f^{\prime}(x)|\geq \varepsilon$, $\forall x \in \mathbb{R}$. Since $f$ is bounded, then there exists $M>0$, s.t. $|f(x)| \leq M$, $\forall x \in \mathbb{R}$. Consider $a < b \in \mathbb{R}$ with $b-a > \frac{2M}{\varepsilon}$, then by mean value thm., there exists $u \in (a,b)$, s.t. $|f^{\prime}(u)|=|\frac{f(b)-f(a)}{b-a}| \leq \frac{2M}{|b-a|} < \varepsilon$, then we get a contradiction.
$f\ $ is bounded. Suppose $\ y_1\ $ is a lower bound and $\ y_2\ $ is an upper bound of $\ f.\ $
Let $\ \varepsilon > 0.\ $ Since $\ f(0) \in [y_1,y_2]\ $ and $\ f\left( \frac{2(y_2 - y_1)}{\varepsilon} \right) \in [y_1,y_2],\ $ it follows, by the mean value theorem, that $\ \exists\ x \in \left(0,\frac{2(y_2 - y_1)}{\varepsilon} \right)\ $ such that
$$ f'(x) = \frac{ f\left(\frac{2(y_2 - y_1)}{\varepsilon} \right) - f(0) }{\frac{2(y_2 - y_1)}{\varepsilon} - 0},$$
$$\implies \left\lvert f'(x) \right\rvert = \left\lvert \frac{ f\left(\frac{2(y_2 - y_1)}{\varepsilon} \right) - f(0) }{\frac{2(y_2 - y_1)}{\varepsilon} - 0} \right\rvert \leq \frac{ y_2 - y_1}{ \frac{2(y_2 - y_1)}{\varepsilon}} = \frac{\varepsilon}{2} < \varepsilon. $$
