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.

• What metric are you using? Jun 7 '19 at 1:50
• Not mentioned in the question. This question is from IMSc PhD entrance paper(NBHM 2019). I have tried based on $d_{\infty}.$ Jun 7 '19 at 1:53
• a. $f_n(x) = \sqrt{x^2+{1 \over n}}$ is $C^1$ and $f_n(x) \to |x|$. Jun 7 '19 at 1:55
• 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. Jun 7 '19 at 1:56
• @copper.hat You probably want $|x-1/2|$ as the limit function.
– zhw.
Jun 7 '19 at 2:59

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).

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

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