Give an example of a sequence of functions $f_n$ which is uniformly convergent on $(-1,1)$ and for which $\{f'_n(0)\}$ is unbounded.

Question

Give an example of a sequence of functions $f_n$ which is uniformly convergent on $(-1,1)$ and for which $\{f'_n(0)\}$ is unbounded.

I thought of making the $f'_n$ part first.

If I take, $f'_n(x) = \frac{n}{1+n^2x}$ then $f'_n(0) = n$ which is unbounded.

But from here if I integrate it to make the $f_n(x)$ then it's becoming something like this :

$$f_n(x) = \frac{\ln(1+n^2x)}{n}$$, which not even defined on the $(-1,1)$.

Every time I'm fixing one part, problem keep coming on the other parts.

Can anyone please help me with this example?

Edit : I am thinking of another function I made in a similar manner.

I'm going directly to $f_n(x)$ part.

If $f_n(x) = \arctan nx$ then $f'_n(0) = n$, which is unbounded sequence.

But I'm having difficulty to figure out the uniform convergence of $f_n(x)$ here.

Answer 1

You can take$$f_n(x)=\begin{cases}\frac{(1-x)^{n^2}}n&\text{ if }x\in[0,1)\\\frac2n-\frac{(1+x)^{n^2}}n&\text{ if }x\in(-1,0).\end{cases}$$Then $(f_n)_{n\in\Bbb N}$ converges uniformly to the null function, since$$(\forall x\in(-1,1))(\forall n\in\Bbb N):0<f_n(x)<\frac2n$$and $\lim_{n\to\infty}\frac2n=0$. Besides,$$(\forall n\in\Bbb N):f_n'(0)=-n.$$To give you an idea, the graph of $f_3$ can be seen below:

Answer 2

A piecewise linear map defined by $f(-1) = -1$, $f(-1/n^2)=-1/n$, $f(1/n^2)=1/n$ and $f(1)=1$ would work.