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Let $S$ be a subset of $C^1([0, 1], \mathbb{R})$. It is a well-known fact that given a function $f\in C^1([0, 1], \mathbb{R})$ and a sequence $\{f_n\}\subset C^1([0,1], \mathbb{R})$ such that $f_n\to f$ uniformly, it does not necessarily hold that $f_n'\to f$ uniformly.

What condition on $S$ (or what condition on $f$) should we impose to guarantee the existence of $f_n$ such that both $f_n\to f$ and $f_n'\to f'$ in the uniform sense? Are there any non-trivial results to this question?

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    $\begingroup$ Firstly, I think there is a typo: you presumably mean that $f_n' \to f'$ $\endgroup$
    – Yemon Choi
    Commented Dec 14, 2020 at 17:46
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    $\begingroup$ Secondly: if $f_n'\to f'$ uniformly then it follows from the Fundamental theorem of calculus and basic estimates that $f_n \to f$ uniformly $\endgroup$
    – Yemon Choi
    Commented Dec 14, 2020 at 17:46
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    $\begingroup$ The holomorpgic case works well. If a sequence of holomorphic maps is locally uniformly convergence then its derivative converges too. So the real case of this situation occures in power serises. Every convergence real power series has convergence (higher order) derivative. $\endgroup$
    – Ali Taghavi
    Commented Dec 14, 2020 at 17:50
  • $\begingroup$ [deleted a comment which was based on a mis-reading] $\endgroup$
    – Yemon Choi
    Commented Dec 14, 2020 at 19:33

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As you are asking for non-trivial results in this direction: There is a nice theorem of Nachbin generalizing Weierstraß's theorem about dense subalgebras of $C(K)$ to dense subalgebras of $C^\infty(M)$ for smooth manifolds (it has also versions for $C^n(M)$), some references are given in the wikipedia entry https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem#Nachbin's_theorem

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