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Suppose we have a function $f$ defined on a disk in the field $\mathbb Q_p$, given by an absolutely convergent power series, with $f(0)=0$ and $f'(0)=1$:

$$\mathbb Q_p\ni f(x)=x+\sum_{n\geq2}a_nx^n,\quad|x|\leq R>0$$ $$\mathbb R\ni\sum_{n\geq2}|a_n|R^n<\infty$$

Are there disks of (positive) radii $S$ and $r\leq R$, and a function $g$, also given by an absolutely convergent power series, such that $g(f(x))=x$ for all $|x|\leq r$, and $f(g(y))=y$ for all $|y|\leq S$?

$$g(y)=y+\sum_{n\geq2}b_ny^n,\quad|y|\leq S$$ $$\sum_{n\geq2}|b_n|S^n<\infty$$

(Assuming it is true) I'd actually like a uniform proof that works for $\mathbb R$, $\mathbb C$, or $\mathbb Q_p$. See Is the inverse of a real analytic function still analytic? and If $f'(z_0)\neq 0$ then $f$ has an holomorphic inverse. for proofs specific to the complex field. See Composition of real-analytic functions is real-analytic for a uniform proof of the type I'm looking for.

By expanding the formal power series $g(f(X))$, I can see that the required coefficients $b_n$ exist and are unique. But I can't see that $g(y)$ converges for some $y\neq0$.

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  • $\begingroup$ I don't know very much about analysis in the non-Archimedean setting, but over $\mathbb{R}$ or $\mathbb{C}$, one thing you can do to prove this is use the fact that the function is $C^1$ and has a differential which is a similarity of the (complex) tangent space (rotation + scaling), so the ordinary inverse function theorem works and the property of being a similarity is preserved. I wonder if there is any similar thing you can do in this setting? $\endgroup$ Commented Jun 30, 2022 at 2:07
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    $\begingroup$ Check out these lecture notes of Peter Schneider's,proposition 6.4 (relying especially on proposition 5.9). $\endgroup$ Commented Jun 30, 2022 at 2:30
  • $\begingroup$ @TorstenSchoeneberg - Thanks. But that definition of analyticity only requires convergence, while I require absolute convergence. To show the difference: $$f(x)=x+2x^2+2x^3+4x^4+4x^5+4x^6+4x^7\\+8x^8+8x^9+8x^{10}+8x^{11}+8x^{12}+8x^{13}+8x^{14}+8x^{15}+16x^{16}+\cdots$$ converges $2$-adically at $f(1)=-1/3$, but doesn't converge absolutely: $$1+\frac12+\frac12+\frac14+\frac14+\frac14+\frac14\\+\frac18+\frac18+\frac18+\frac18+\frac18+\frac18+\frac18+\frac18+\frac1{16}+\cdots$$ $$=\infty$$ $\endgroup$
    – mr_e_man
    Commented Jun 30, 2022 at 18:36
  • $\begingroup$ On second thought, convergence for $|x|\leq R$ implies absolute convergence for $|x|<R$, so the two definitions should be equivalent. But the proof of 5.9 still uses the strong triangle inequality. I think I should edit the question, to put less focus on $p$-adic fields. Thanks anyway for those notes. $\endgroup$
    – mr_e_man
    Commented Jun 30, 2022 at 18:50
  • $\begingroup$ (There may not be an $x_*$ with $|x_*|_p=R$. I should have said this instead: Convergence at $x_*$ implies absolute convergence at $x$ for $|x|<|x_*|$, so the two definitions should be equivalent.) $\endgroup$
    – mr_e_man
    Commented Jul 3, 2022 at 23:36

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