Let $f_n(x) = \frac{x}{1+(nx)^2}$. Notice that $f_n$ are differentiable functions.
a) Show that $\{f_n\}$ converges uniformly to $0$
$\lim\sup|\frac{x}{1+(nx)^2}| =\lim |\frac1{2n}| = 0$
b) Show that $|f'_n(x)| \le 1$ for all $x$ and all $n$.
$|f'_n(x)| = |\frac1{1+(nx)^2}||1-\frac{2(nx)^2}{1+(nx)^2}| = |\frac1{1+(nx)^2}||\frac{1-(nx)^2}{1+(nx)^2}|< \frac{(nx)^2}{1+(nx)^2} \le 1$.
c) Show that $\{f'_n\}$ converges ponitwise to a function discontinuous at the origin.
Given $x$, if $n$ goes to infinity, doesn't $f'_n(x)$ converges to $0$?
d) Let $\{a_n\}$ be an enumeration of the rational numbers. Define $$g_n(x) = \sum_{k=1}^n 2^{-k}f_n (x-a_k).$$ Show that $\{g_n\}$ converges uniformly to $0$.
Shouldn't $f_n$ be replaced by $f_k$? In that case, $\lim \sup |2^{-1}||f_k(x-a_k)|^{1/k} = 0$ since $f_k$ uniformly converge to $0$. So $g_n$ uniformly converges. But, how do we know that it converges to $0$?
e) Show that $\{g'_n\}$ converges pointwise to a function $\phi$ that is discontinuous at every rational number and continuous at every irrational number. In particular, $\lim_{n\to\infty} g'_n(x) \not= 0$ for every rational number $x$.
I have no idea.
I have some difficulty in solving the last three questions. I appreciate if you give some help.
Evidently, $f_n(x)\to 0$ pointwise. Note that the $f_n$ are differentiable and defined on all of $\mathbb R$. Therefore any maximum must be attained at a point $x_n$ such that $f_n'(x_n)=0$. So we compute $$ f_n'(x) = \frac{1-n^2 x^2}{\left(1+n^2 x^2\right)^2}, $$ and note that $$f_n'(x)=0\iff 1-n^2x^2=0\iff x = \frac1n.$$ It follows that $$\limsup_{n\to\infty} f_n(x) \leqslant \limsup_{n\to\infty}f_n\left(\frac1n\right) = \limsup_{n\to\infty}\frac1{n(1+1^2)} = \limsup_{n\to\infty} \frac1n = 0, $$ and hence $f_n$ converges uniformly to zero.
It is clear that $(1+n^2x^2)\geqslant1$ and $1-n^2x^2\leqslant 1$ and therefore $$ f_n'(x) = \frac{1-n^2x^2}{(1+n^2x^2)^2}\leqslant 1. $$ For $x\ne 0$ we have $$\lim_{n\to\infty} f_n'(x) = \lim_{n\to\infty} \frac{1-n^2x^2}{(1+n^2x^2)^2} = 0,$$ but for $x=0$, $f_n'(0) = 1$, so $\lim_{n\to\infty}f_n'(0)=1$. It follows that the limit of $f_n'$ is discontinuous at $x=0$.
For d) and e) I am not sure.