Give an example of a sequence of smooth equicontinuous functions whose derivatives are unbounded As the title suggests. I want the function to be from $[a, b]$ to $\mathbb R$. By unbounded derivative, I mean if $f_n$ is a sequence of functions and $M_n$ is the sup norm of $f_n'$ on $[a, b]$, then $\{M_n\}$ is an unbounded sequence of reals. I really have no idea how to do these kind of questions (the ones where you are asked to give an example). Can anyone help me out?
 A: Consider the function $f$ defined on $[0,1]$ by 
$$
f(x)=x\sin\left(\frac{1}{x}\right)
$$
for $x\neq 0$ and 
$$
f(0)=0.
$$
Then for each $n$, take a smooth function $\phi_n$ such that
$$
\phi_n(x)=0\quad\forall x\in[0,\frac{1}{ n}]\quad \phi_n(x)=1\quad\forall x\in[\frac{2}{n},1]
$$
and 
$$
0\leq \phi_n(x)\leq 1\quad \forall x\in [\frac{1}{n},\frac{2}{n}].
$$
This can be achieved with appropriate modifications of $e^{-1/x^2}$.
Now consider the sequence
$$
f_n=f\phi_n.
$$
I believe this works.
1) Equicontinuity: 
First fix $x$ in $(0,1]$ and a small $r>0$. For $n$ large enough, namely $2/n<x-r$, we have $f_n(y)=f(y)$ for all $y\in(x-r,x+r)$. 
Take $\epsilon>0$. By continuity of $f$ at $x$ there is $0<\delta<r$ such that 
$$
|f_n(y)-f_n(x)|=|f(y)-f(x)|<\epsilon \qquad \forall n\geq N=\lfloor\frac{2}{x-r}\rfloor+1.
$$
Now $f_1,f_2,\ldots,f_{N-1}$ are all continuous at $x$ and in finite number, so there exist a common $\delta'>0$ such that $|f_n(y)-f_n(x)|\epsilon$ for these $n$ and all $y\in(x-\delta',x+\delta')$.
It only remains to take the minimum of $\delta$ and $\delta'$ to get what you want and equicontinuity at $x$ follows.
Now take $x=0$. We have $|f_n(y)|\leq y\sin (1/y)\leq y$ for all $y\in[0,1]$ and all $n$. So equicontinuity at $0$ follows.
By compactness of $[0,1]$, we even have uniform equicontinuity.
2) Uniform unboundedness of the derivatives:
Compute $f'(x)$ and check it is unbounded as $x$ approaches $0$.
The claim follows easily.
3) Smoothness is obvious.
