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Let $f_{n}$ be some sequence of functions.

If $f_{n}$ uniformly converges to $f$ then is it true that $f^{\prime}_{n}\rightarrow f^{\prime}$?

Is there an example that proves/disproves this?

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  • $\begingroup$ No I think you need another condition on $f'_n$. $\endgroup$ Jan 16, 2016 at 12:35
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    $\begingroup$ Welcome to our site! $\endgroup$ Jan 16, 2016 at 12:41
  • $\begingroup$ @kjetilbhalvorsen, Thank you. $\endgroup$
    – gbd
    Jan 16, 2016 at 12:44

4 Answers 4

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Counterexample: $f_n = \frac1n\sin(n^2 x)$.

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  • $\begingroup$ What does $f_{n}$ converge to? $\endgroup$
    – gbd
    Jan 16, 2016 at 12:40
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    $\begingroup$ @gbd, $f_n\to 0$ because $|f_n|\le 1/n$. $\endgroup$ Jan 16, 2016 at 12:41
  • $\begingroup$ Does it converge to zero? $\endgroup$
    – gbd
    Jan 16, 2016 at 12:41
  • $\begingroup$ What is the other condition need that @GregoryGrant mentions? $\endgroup$
    – gbd
    Jan 16, 2016 at 12:43
  • $\begingroup$ @gbd, see the other answer. $\endgroup$ Jan 16, 2016 at 12:44
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This is false in quite a strong sense. The Weierstrass function is a uniform limit of differentiable functions, but is nowhere differentiable. By the Stone—Weierstrass Theorem, any continuous function on a closed bounded interval is a uniform limit of polynomials.

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The additional condition required is that a function $g$ exists to which the $f'_n$ uniformly converge (in which case of course $g=f'$). In particular, this PDF proves on p. 9 that this condition suffices, but on p. 8 it shows that $f_n=\frac{x}{1+nx^2}$ is a uniformly convergent sequence whose derivatives form a convergent but not uniformly convergent counterexample.

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As mentioned above this not true in general. But one might still aks what a sufficient condition for "$f_n' \to f'$ uniformly" might be. If you already know that $f_n$ converges uniformly to $f$ and $f_n'$ converges uniformly to some function say $g$, then by the fundamental theorem of calculus one indeed obtains: $g=f'$ hence $\space$$f_n' \to f'$ uniformly.

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