This is a question about a proof of a theorem in many books on $p$-adic numbers and I don't seem to understand one of the directions. The theorem is
Let $f(X) = \sum\limits_{n=0}^\infty a_nX^n \in \mathbb{Q}_p[[X]]$ be a $p$-adic power series. Let $c \in D_f$, the domain of convergence for $f$, so $f(c)$ converges. Define $g_m$ by
$g_m = \sum\limits_{n \geq m} \left(\begin{array}{c} n\\m\end{array} \right)a_nc^{n-m}$
and set $g(X) = \sum\limits_{m=0}^\infty g_mX^m.$ Then $D_f = D_g$ (the domain of convergence of $g$) and further, for all $b \in D_f$, we have $f(b+c) = g(b).$
I have no problem with the proof that $D_f \subseteq D_g$, it's the converse I don't understand. Every resource I've gone to has said the argument is symmetric and follows from reversing the roles of $f$ and $g$. I don't see why this is true. The whole argument to show that $D_f \subseteq D_g$ is given by controlling the term
$\left| \left(\begin{array}{c} n\\m\end{array} \right)a_nc^{n-m}b^m\right|_p$
where $b \in D_f$. But when I try to do this for $b \in D_g$, I can't get the same bound. In fact, what I get is that the above term is bounded by $|a_n|_p\rho^n$ where $\rho$ satisfies $|c|_p \leq \rho$ and $|b|_p \leq \rho$ and I see no reason why this term should go to zero, and further I don't see how this plays off of any symmetry in the proof. Any help to get my understanding what I'm misinterpreting would be greatly appreciated.
That must surely be the least intuitive way possible of stating that simple result. To understand what’s happening, you should do some hand computation on series. Let me show you:
If you expand out $f(X+c)$, just go as far as, maybe the $X^4$-terms, you’ll see that $g(X)=f(X+c)$. Now, the domain of convergence of $f$ is a group $S$, and your hypothesis is that $c$ is in $S$. Since $S-c=S$, the numbers that make $g$ converge are exactly the numbers that make $f$ converge. That’s all there is to it. Of course you do have to justify my statement “you’ll see that...”.
EDIT — Expansion:
If you believe your own proof that $D_f\subseteq D_g$, and that for every $z\in D_f$, $f(z+c)=g(z)$, then it still seems to me that you’re done. Let $f(X+c)=\sum_{n\ge0}b_nX^n=g(X)$, and do the same thing in reverse, $h(X)=g(X-c)=\sum_{n\ge0}\beta_nX^n$, so by your result, $D_g\subseteq D_h$. Now from the evaluation part, $h(X)-f(X)$ is a power series that vanishes at every $z\in D_f$, and thus is the zero power series, so that $h=f$, $D_h=D_f$, and thus $D_g=D_f$.