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Prove that if $\{f_n : \mathbb R \rightarrow \mathbb R\}$ is a sequence of continuously differentiable functions such that the sequence of derivatives $\{f'_n : \mathbb R \rightarrow \mathbb R\}$ is uniformly convergent and the sequence $\{f_n (0)\}$ is also convergent, then $\{f_n : \mathbb R \rightarrow \mathbb R\}$ is pointwise convergent. Is the assumption that the sequence $\{f_n (0)\}$ converges necessary?

My attempt: Suppose $f'_n \rightarrow g$ uniformly and $f_n (0) \rightarrow a$. Then since $f'_n$ are integrable, $\int_0^x f'_n \rightarrow \int_0^x g$, i.e. $\lim_{n \rightarrow \infty} \int_0^x f'_n = \lim_{n \rightarrow \infty} (f_n(x)-f_n(0))=\int_0^x g$. Is this the right approach, cause I'm stuck here, if so, how to proceed?

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    $\begingroup$ $f_n=n$.${}{}{}$ $\endgroup$ Commented Apr 24, 2019 at 3:09
  • $\begingroup$ @DavidMitra, can you clarify your statement pls? $\endgroup$ Commented Apr 24, 2019 at 3:18
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    $\begingroup$ The condition is needed. A counterexample is given by the functions $f_n$ defined above ($f_n(x)=n$ for all $x$). $\endgroup$ Commented Apr 24, 2019 at 3:20

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$\newcommand{\dif}{\,\text{d}}$ By the fundamental theorem of calculus we have $f_n(x) = f_n(0) + \int_0^x f'_n(t) \dif t$. You know $f_n(0)$ converges to something and you know $\int_0^x f'_n(t) \dif t$ converges to something. Putting this together shows $f_n(x)$ converges to something.

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