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Suppose we are given that a sequence of functions $f_n(z)$ convergences pointwise to $f(z)$ on the interval $[0,1]$. Suppose further that all of these functions are given by power series centered at 0 with radius of convergence $R > 1$.

To fix notation, say $f_n(z) = a_{n, 0} + a_{n, 1}z + a_{n, 2} z^2 + \dots$

and $f(z) = a_0 + a_1 z + a_2 z^2 + \dots$.

We are given $f_n(z) \to f(z)$ as $n \to \infty$, for each fixed $z \in [0,1]$.

Is it true that $a_{n, k} \to a_k$ as $n \to \infty$, for each $k$? Why? We can't use complex analysis it seems, since we are only on $[0,1]$.

If not, is it true under the additional assumption that all the $a_{n, k}$ and $a_n$ are uniformly bounded by some $M$?

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No, this is false. Define $f_n(z) = \sin(nz)/n, n= 1,2,\dots$ Then $f_n(z) \to 0$ pointwise on $\mathbb R$ (in fact, uniformly on $\mathbb R.$) But $f_n'(0) = 1$ for all $n,$ while $f'(0) = 0.$

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  • $\begingroup$ So under what conditions is it true? $\endgroup$ – keej May 9 '18 at 18:16
  • $\begingroup$ Could it still be true if all of the coefficients are uniformly bounded? $\endgroup$ – keej May 9 '18 at 18:17
  • $\begingroup$ I think if all coefficients are uniformly bounded, then the answer is yes, using a normal families argument (complex analysis). $\endgroup$ – zhw. May 9 '18 at 18:34
  • $\begingroup$ @MarkViola Your two varibables for Moore-Osgood appear to be $n$ and $K,$ as in $k=0,1,\dots, K.$ So uniform convergence for $z\in [0,1]$ seems irrelevant. Furthermore, not even pointwise convergence with respect to $n$ has been established; in fact, this is the very question at hand. $\endgroup$ – zhw. May 10 '18 at 15:58
  • $\begingroup$ @zhw. Yes, you're correct. We needed that $\sum_{k=0}^\infty C_k(n)z^k$ converges uniformly in $n\in \mathbb{N}$ for each $z$, and we needed piecewise convergence of the $C_k(n)$. We have been given neither of those. I'll delete my answer. $\endgroup$ – Mark Viola May 10 '18 at 16:19

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