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Consider the following theorem:

Theorem 6.4.3.:Let $f_n$ be differentiable on an interval $[a,b]$ and assume $\sum f_n' (x)$ converges uniformly to a limit $g(x)$ on $[a,b]$. If there exists at least one $x_0 \in [a,b]$ where $\sum f_n (x_0)$ converges then the series $\sum f_n (x)$ converges uniformly to a differentiable function $f$ with $f' = g$ on $[a,b]$.

My question is does the following theorem also hold:

If $\sum_n f_n(x)$ converges uniformly on $[a,b]$ to some limit function $f$ and $f$ is differentiable then $f_n$ are differentiable and the sum $\sum_n f_n'$ converges to $f'$ on [a,b]?

blue
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    No. Let $f_1(x)=|x|$, $f_2(x)=-|x|/2$, $f_3(x)=-|x|/4$, $\cdots$. – David Mitra May 24 '14 at 12:29
  • @DavidMitra Thank you for your comment! I asked this question because I got this answer to one of my previous questions. The author seems to argue using the converse so I thought it might hold. Does it mean the answer is wrong? – blue May 24 '14 at 12:32

1 Answers1

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No.

Take for example: $$f_n(x)=\left\{ \begin{array}{lcl} \frac1{2^n}&\text{ if }&x\in[0,1]\\ \frac1{n(n+1)}&\text{ if }&x\in(1,2] \end{array}\right.$$

Then $\sum f_n\to 1$ uniformly, but $f_n$ are not even continuous (except for $n=1$).

ajotatxe
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  • Thank you for your comment! I asked this question because I got this answer to one of my previous questions. The author seems to argue using the converse so I thought it might hold. Does it mean the answer is wrong? – blue May 24 '14 at 12:32