Derivative of bounded function Let $f$ be continous and differentiable on $[0,1]$, and say $|f(x)| \leq M$ on $[0,1]$. Does it exists some function $g$ such that $f' \leq g(M)$?? Of course, $g$ shall be continous at least. I was wandering this, having no clue how to find such a function. Thanks for any help! 
 A: The answer is no, $g$ need not exist.
Consider $f(x) = \begin{cases} x^{1.5}\sin\left(\frac{1}{x}\right) & x\neq 0 \\ 0 & x = 0\end{cases}.$  This is bounded by $1$ on $[0,1]$ since both $x^{1.5}$ and $\sin\left(\frac{1}{x}\right)$ are.
Now, this is differentible everywhere.  The only contentious points is $x=0$, where we compute $$f'(0) = \lim_{h\rightarrow 0} \frac{h^{1.5} \sin\left(\frac{1}{h}\right)}{h} = \lim_{h\rightarrow 0} h^{0.5}\sin\left(\frac{1}{h}\right) = 0$$ by the squeeze theorem.
Thus, $f'(x) = \begin{cases} 1.5 x^{0.5} \sin\left(\frac{1}{x}\right) - x^{-0.5}\cos\left(\frac{1}{x}\right) & x\neq 0\\ 0 & x = 0\end{cases}$
The term $1.5x^{0.5} \sin\left(\frac{1}{x}\right)$ is bounded on $[0,1]$.  But the term $x^{-0.5} \cos\left(\frac{1}{x}\right)$ is unbounded as $x\rightarrow 0$ from the right.  Hence, $f'(x)$ is unbounded as $x\rightarrow 0$.
Thus, there is no $g$ which depends only on $M$ which is bigger than $f'(x)$.
A: The answer depends on the interpretation of the question. If you can choose $g$ for any $f$, then the answer is true. Given that $f$ is differentiable, $f'(x)$ is bounded for each $x \in [0,1]$. Let $g$ be simply the maximum of $|f'(x)|$.
But if you want a bound that only depends on $M$ and works for any bounded function $f$, then the answer is no. Counterexample: $f(x) = -\sqrt{M^2-x^2}$ for $M > 1$. It is continuous and differentiable and $|f(x)| \leq M$ in $[0,1]$. But $f'(x) = \frac{x}{\sqrt{M^2-x^2}}$, which is bounded for any $M > 1$, but as we take $M \to 1$ we get that $f'(1) \to \infty$.
A: I assume that you define f'(0) as the limit of f'(c) as c-->0 from the right, similarly for f'(1).
There does not exist such a function $g$.  Consider $f=x^n$.  As $n$ increases, $\sup f`$ does too (therefore $g$ does), but $f$ does not.  Thus, for any selection of $M$, you can find an $n$ such that $f'$ surpasses $g(M)$.
A: Try 
$$
f(x) = 
\left\{
\begin{array}{ll}
x^{1.2} \sin\left({1 \over x}\right) & \mbox{for $0 < x \leq 1$}\\
0 & \mbox{for $x = 0$}\\
\end{array}
\right.
$$
