Does the pointwise convergence of a sequence of Taylor series imply the convergence of the coefficients?

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For each fixed $n\in\mathbb N$, let $(a_k^{(n)})_{k\in\mathbb N\cup\{0\}}$ be a real sequence and let $(a_k)_{k\in\mathbb N\cup\{0\}}$ be another real sequence.


Suppose the following: there exists some $r>0$ such that the Taylor series: \begin{align*}f_n(x)\equiv&\,\sum_{k=0}^{\infty}a_k^{(n)}x^k\quad\text{for each $n\in\mathbb N$,}\\f(x)\equiv&\,\sum_{k=0}^{\infty}a_k x^k\end{align*} all converge whenever $x\in[-r,r]$. (The formal power “$0^0$” is interpreted to be $1$.)


Suppose furthermore that $$\lim_{n\to\infty}f_n(x)=f(x)$$ pointwise for each $x\in[-r,r]$.


Question: Does it necessarily follow that the coefficients $$\lim_{n\to\infty}a_k^{(n)}=a_k$$ converge for each $k\in\mathbb N\cup\{0\}$? Any hints would be greatly appreciated.

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Set $f_n(x) = \frac{\sin (nx+n)}{n}$. Now $f_n \to 0$ pointwise. The derivatives are $$ f'_n(x) = \cos (nx+n)\,, $$ and thus the pointwise limits $\lim_{n \to \infty}f_n'(x)$ do not exist.

Assume that your claim is true. Now $f_n'(0) = a_1^{(n)}$ should converge to something...