Is there a differentiable function f which the differential function f' is bounded but has no maximum on a closed interval. Is there a differentiable function $f$ in which the differential function $f'$ is bounded but has no maximum on one closed interval? Thanks 
 A: Answer assuming looking for "at least one closed interval":
Define the sequence $a_{n}=\frac{1}{3^{n}}$. Then $s_{k}=\sum\limits_{i=k}^{\infty}a_{i}=\frac{3}{2\times 3^{k}}$. Hence $\lim\limits_{k\rightarrow\infty}\frac{s_{k}}{(\frac{1}{2^{k+1}})}=\lim\limits_{k\rightarrow\infty}3(\frac{2}{k})^{k}=0$.
Define a collection of function $g_{k}:[\frac{1}{2^{k}},\frac{1}{2^{k+1}}]\rightarrow\mathbb{R}$ to be a nonnegative continuous function that vanish at 2 endpoints and have the maximum value being $1-(\frac{1}{4})^{k}$ and also $\int\limits_{\frac{1}{2^{k}}}^{\frac{1}{2^{k+1}}}g(t)dt=a_{k}$. (it is easy to show that those $g_{k}$ exist, as we give it plenty of room to get the integral it needed, and obtaining a maximum value is just a matter of putting a small spike).
Define $g:[-1,1]\rightarrow\mathbb{R}$ to be $g(0)=0$ and $g(x)=g_{k}(|x|)$ if $|x|\in[\frac{1}{2^{k}},\frac{1}{2^{k+1}}]$.
Define $f:[-1,1]\rightarrow\mathbb{R}$ to be $f(x)=\int\limits_{0}^{x}g(t)dt$.
Now calculate $f^{\prime}(0)=\lim\limits_{h\rightarrow 0}\frac{\int\limits_{0}^{h}g(t)dt}{h}$. We have for $h\in[\frac{1}{2^{k}},\frac{1}{2^{k+1}}]$ then $0\leq\frac{\int\limits_{0}^{h}g(t)dt}{h}<\frac{s_{k}}{(\frac{1}{2^{k+1}})}$. Since $\lim\limits_{k\rightarrow\infty}\frac{s_{k}}{(\frac{1}{2^{k+1}})}=0$ by squeeze theorem we have $f^{\prime}(0)=\lim\limits_{h\rightarrow 0}\frac{\int\limits_{0}^{h}g(t)dt}{h}=0$.
As for $f^{\prime}(x)$ when $x\not=0$, notice that $g$ is continuous on $[-1,0)$ and $(0,1]$ so $f^{\prime}(x)=g(x)$.
Hence $f^{\prime}=g$.
Now notice that $g$ do not have any maximum on $[-1,1]$, since its values approach $1$ arbitrarily close, but it never take that values.
Hence $f$ is what we are looking for.
A: I found this example in "An introduction to Analysis"(in Japanese) by Ichiro Tajima.
$$f(x)=\begin{cases} x^2 \sin (\frac{1}{x} + 3 x), \  x\in [-\frac{1}{3}, 0) \cup (0, \frac{1}{3}]\\ 0, \ x = 0\end{cases}.$$
$$f'(x)=\begin{cases} 2 x \sin (\frac{1}{x} + 3 x) + (3 x^2 - 1) \cos (\frac{1}{x} + 3 x), \  x\in [-\frac{1}{3}, 0) \cup (0, \frac{1}{3}]\\ 0, \ x = 0\end{cases}.$$
If $x \neq 0$, then $$f'(x) = \sqrt {1 - 2 x^2 + 9 x^4} \sin (\frac{1}{x} + 3 x + \alpha(x)).$$
And $g(x) := 1 - 2 x^2 + 9 x^4$ attains the maximum value at $x = 0$.
So, $$\{f'(x) \mid x \in [-\frac{1}{3}, \frac{1}{3}]\} = (-1, 1).$$

