Interchange limit and sum in power series

Question

Suppose $(f_n)_n$ are analytic functions with power series representation $$ f_n(x) = \sum_{k=0}^\infty a_{2k}^n (x-x_0)^{2k}, \qquad x \in [a,b]. $$ Notice that for every $f_n$, all odd powers vanish in this representation. Suppose further that the $(f_n)_n$ converge uniformly on $[a,b]$ to a function $f$. Then $f$ is analytic, i.e. there are $(b_k)_k$ such that $$ f(x) = \sum_{k=0}^\infty b_k (x-x_0)^k, \qquad x \in [a,b]. $$

Is it true that all odd powers vanish as well, i.e. is $b_{2k+1} = 0$ for all $k \in \mathbb{N}$?

Answer 1

Let $g_n(x-x_0)=f_n(x).$ Let $g(x-x_0)=f(x).$ Then $g_n(y)=f_n(y+y_0)=f_n(-y-y_0)=g_n(-y).$ Therefore $g(y)=g(-y).$

So if $g(y)=\sum_{n=0}^{\infty} a_ny^n$ then $$f(y+y_0)=g(y)=(g(y)+g(-y))/2=\sum_{m=0}^{\infty}a_{2m}y^{2m}.$$ $$\text {So }\quad f(y)=\sum_{m=0}^{\infty}a_{2m}(y-y_0)^{2m}.$$

If $a< b$ then the sequence of co-efficients is unique: If $\sum_n a_n(y-x_0)^n=\sum_na'_n(y-y_0)^n$ for all $y\in [a,b]$ then $a_n=a'_n$ for every $n$.

Answer 2

Yes, it is true. Let $k$ be an odd natural number. Then$$b_k=\frac1{2\pi i}\int_\gamma\frac{f(u)}{(u-z_0)^{k+1}}\,du=\lim_{n\in\mathbb N}\frac1{2\pi i}\int_\gamma\frac{f_n(u)}{(u-z_0)^{k+1}}\,du\text,$$ where $\gamma(t)=z_0+re^{it}$ ($t\in[0,2\pi]$), for some small $r$. Since $k$ is odd, $k+1$ is even and so the Laurent series of $f_n$ near $z_0$ has no term of the type $(z-z_0)^{-1}$ (or, to be more precise, its coefficient is $0$). Therefore,$$\frac1{2\pi i}\int_\gamma\frac{f_n(u)}{(u-z_0)^{k+1}}\,du=0.$$