Are these two versions of identity theorem of power series equivalent?

Question

I have two versions of identity theorem of power series:

1. From PMA - Rudin

2. From Analysis I - Amann

It seems to me that the one from PMA - Rudin is more general since it does not require the limit point to be $0$.

My question: How to generalize the theorem from Analysis I - Amann to not require that $(y_j)$ converges to $0$?

Thank you for your help!

Answer 1

Lemma. Let the power series $h(x) = \sum_kc_kx^k$ converge for $x\in G := (-R,R)$ and set $A := \{x\in G\,|\,\forall n\in\mathbb N_0 : h^{(n)}(x) = 0\}$. Then either $A=\emptyset$ or $A = G$.

Proof. We only have to prove that $A$ is both open and closed in $G$. The closedness of $A$ follows immediately from $A = \bigcap_n (h^{(n)})^{-1}(\{0\})$. Now, let $x_0\in A$. Then $h(x) = \sum_k\frac{h^{(k)}(x_0)}{k!}(x-x_0)^k$ vanishes completely for $x$ in a neighborhood $U$ of $x_0$. In particular $h^{(n)}(x) = 0$ for $x\in U$ and thus $U\subset A$. $\quad\square$

Now, let $h(x) = \sum_kc_kx^k$ be as above and assume the existence of a sequence $(z_j)\subset G$ such that $z_j\to x_0$, where $x_0\in G$, and $h(z_j) = 0$ for all $j$. Write $h(x) = \sum_kd_k(x-x_0)^k$ for $x$ in a neighborhood $U$ of $x_0$ and $z_j = x_0+e_j$. Then $e_j\to 0$ and $\sum_kd_ke_j^k = 0$ for all $j$. By Amann's result, $d_k = 0$ for all $k$. In particular $h(x) = 0$ for $x\in U$ and thus the set $A$ from the lemma is not empty, which implies $A = G$. Once more in particular, $0\in A$, which implies that $c_k = \tfrac{h^{(k)}(0)}{k!} = 0$ for all $k$.

Let me remark that Amann's result does not help that much here because one can easily show by hand that the existence of $(z_j)$ leads to $x_0\in A$.