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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.

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  • $\begingroup$ What metric are you using? $\endgroup$
    – copper.hat
    Jun 7 '19 at 1:50
  • $\begingroup$ Not mentioned in the question. This question is from IMSc PhD entrance paper(NBHM 2019). I have tried based on $d_{\infty}.$ $\endgroup$
    – Math geek
    Jun 7 '19 at 1:53
  • $\begingroup$ a. $f_n(x) = \sqrt{x^2+{1 \over n}}$ is $C^1$ and $f_n(x) \to |x|$. $\endgroup$
    – copper.hat
    Jun 7 '19 at 1:55
  • $\begingroup$ How do you know that the $f'_n$ in c. are integrable? For example, $f(x) = x^2 \sin {1 \over x^2}$ (and zero at zero) is differentiable by not integrable. Take $f_n$ to be the $n$th Taylor series. $\endgroup$
    – copper.hat
    Jun 7 '19 at 1:56
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    $\begingroup$ @copper.hat You probably want $|x-1/2|$ as the limit function. $\endgroup$
    – zhw.
    Jun 7 '19 at 2:59
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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).

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  • $\begingroup$ Sir , is't the $b$ part solution in your answer is actually the answer of $c$ part of OP's question? $\endgroup$
    – TheStudent
    Aug 6 '21 at 6:55
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    $\begingroup$ @TheStudent You are right. I have added a sketch of proof for b). $\endgroup$ Aug 6 '21 at 7:22
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    $\begingroup$ For (a) I gave a different example in a comment to the Q that you may find amusing. $\endgroup$ Aug 6 '21 at 12:58
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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

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