Let $\{f_n\}_{n=1}^\infty$ be a sequence of functions in $C^1[0, 1]$, Which of the following statements are true? Let $\{f_n\}_{n=1}^\infty$ be a sequence of functions in $C^1[0, 1]$ such that $f_n(0) = 0$ for
all $n ∈ \mathbb N$. Which of the following statements are true?
a. If the sequence $\{f_n\}$ converges uniformly on the interval $[0, 1]$, then the limit function is in $C^1[0, 1]$.
b. If the sequence $\{f'_n\}$ is uniformly convergent over the interval $[0, 1]$, then
the sequence $\{f_n\}$ is also uniformly convergent over the same interval.
c. If the series $\sum_{n=1}^\infty f'_n$ converges uniformly over the interval $[0, 1]$ to a
function $g$, then $g$ is Riemann integrable and
$\int_0^1 g(t)dt=\sum_{n=1}^\infty f_n(1)$
I am able to prove (c) I know $g$ is integrable. Moreover,  $\int_0^1\sum_{n=1}^\infty f'_n dt=\sum_{n=1}^\infty \int_{0}^1f'_n dt(\because$  the series $\sum_{n=1}^\infty f'_n$ converges uniformly over the interval $[0, 1]$ to a
function $g$)
$\sum_{n=1}^\infty \int_{0}^1f'_n dt= \sum_{n=1}^\infty (f_n(1)-f_n(0))=\sum_{n=1}^\infty f_n(1)$
b. Suppose that $\{f'_n\}$ converges uniformily to $h$. So, For $\epsilon>0,$ there is a natural number $N=N(\epsilon):\forall n\ge N\implies |f'_n(x)-h(x)|<\epsilon, \forall x\in[0,1].$ I wanted to prove For $\exists g:\epsilon>0,$ there is a natural number $K=K(\epsilon):\forall n\ge K\implies |f_n(x)-g(x)|<\epsilon, \forall x\in[0,1].$ First I want to define $g$. How do I define $g$?
a. I know it is wrong, I am not able to find counter examples.
 A: There is a sequence $(p_n)$ of polynomials converging uniformly to $\sqrt x$. Take $f_n(x)=p_n(x)-p_n(0)$ for a).
For b) use the fact that $f_n(x)=\int_0^{x} f_n'(t)dt$ so $|f_n(x)-f_m(x)| \leq \int_0^{x} |f_n'(t)-f_m'(t)|dt$. Hence $(f_n(x))$  is (uniformly) Cauchy and define $g(x)$ as $\lim f_n(x)$.
For c) let $f(x)=\int_0^{x} g(t)dt$. Then $\sum f_n$ converges uniformly to $f$: if $s_n(x)=\sum\limits_{k=1}^{n} f_k'(x)$ then $|\sum\limits_{k=1}^{n} f_k(x)-f(x)| =|\sum\limits_{k=1}^{n} \int_0^{x} f_k'(t)dt-\int_0^{x}g(t)dt|=| \int_0^{x} \sum\limits_{k=1}^{n} f_k'(t)dt-\int_0^{x}g(t)dt|\leq \epsilon$ for all $x$ if $|\sum\limits_{k=1}^{n} f_k'(t)dt-g(t)| <\epsilon$ for all $t$ (which is true for $n$ sufficiently large).
A: f'n(x) converges uniformly on [0,1].fn(0) converges to 0.There is a theorem(page no.152) in Rudin's Principles of Mathematical Analysis stating that"If f'n(x) converges uniformly on [a,b] and fn(x') converges for some x' in [a,b] ,then fn(x) converges uniformly on [a,b]".In our case ,fn(x) converges for x=0.
For option a
Hint: use Stone - Weierstrass theorem
