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Let {$f_n$ } be a sequence of functions defined by $f (x) =\frac{x}{1+n^2x^2}$ on $[0, 1]$. Prove that

  1. {$f_n$} converges uniformly on $[0,1]$ to a function $f$ that is differentiable on $[0,1]$
  2. {$f′_n$ }converges pointwise on $[0,1]$ to a function $g$ that is not equal to $f′$.

I have proved that $f_n$ converges uniformly to $f(x) = 0$.

However, I've realized that {$f′_n$ } = $\frac{1 - n^2x^2}{(1+n^2x^2)^2}$ converges uniformly to $f(x) = 0$, and hence converges pointwise to $f(x) = 0$, which is equal to $f'$. Am I doing something wrong?

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The derivatives do not converge uniformly to $0$. Instead, they converge pointwise to $$g(x) = \left\{ \array{0 &\text{ if } x\in (0,1] \\ 1 & \text{ if } x=0 }\right. $$

(Note that $f^\prime_n(0) = 1$ for every $n$).

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  • $\begingroup$ That's correct. I have overlooked the case where x=0. Thanks! $\endgroup$ – Ahmad Lamaa Dec 1 '18 at 19:50

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