Let $K$ be a complete field with respect to a non-trivial non-archimedean absolute value $|\cdot|$. Let $E$ be a vector space over $K$. A norm $||\cdot||$ on $E$ is a map $E \rightarrow \mathbb{R}$ satisfying the following properties.

1) $||x|| = 0$ if and only if $x = 0$.

2) $||\alpha x|| = |\alpha|||x||$ for all $\alpha \in K$ and all $x \in E$.

3) $||x + y|| \le max(||x||, ||y||)$ for all $x, y \in E$.

Clearly $||x - y||$ defines a metric on $E$. A vector space over $K$ equipped with a norm is called a normed vector space. If $E$ is complete with respect to this metric, $E$ is called a Banach space.

Let $E, F$ be normed vector spaces over $K$. Let $U$ be an open subset of $E$. Let $a \in U$. Let $f\colon U \rightarrow F$ be a map. Suppose there exists a continuous linear map $L\colon E \rightarrow F$ such that $$\frac {||f(x) - f(y) - L(x - y)||}{||x - y||} \rightarrow 0$$ when $(x, y) \rightarrow (a, a)$. Then $f$ is called strictly differentiable at $a$. It is easy to see that $L$ is uniquely determined by $f$ and $a$. We denote $L$ by $Df(a)$.

The following proposition is stated without a proof in Bourbaki, Variete differentielles et analytiques. How do we prove it?

Proposition Let $E, F$ be Banach spaces over $K$. Let $U$ be an open subset of $E$. Let $a \in U$. Let $f\colon U \rightarrow F$ be a map. Suppose $f$ is strictly differentiable at $a$ and $Df(a)$ is an isomorphism $E \rightarrow F$. Then there exist an open neighborhood $U_0$ of $a$ such that $U_0 \subset U$ and an open neighborhood $V_0$ of $f(a)$ with the following properties.

1) $f|U_0$ is a homeomorphism $U_0 \rightarrow V_0$.

2) Let $g$ be the inverse of $f|U_0$. Then $g$ is strictly differentiable at $f(a)$.

3) $Dg(f(a)) = Df(a)^{-1}$.

  • $\begingroup$ What have you tried? For example, how might a proof over the reals fail for the $p$-adics? $\endgroup$ – Andrew Oct 23 '12 at 1:22
  • $\begingroup$ Actually, have you tried this book: books.google.ca/… $\endgroup$ – Andrew Oct 23 '12 at 1:33
  • $\begingroup$ @Andrew No. Does the book treat Banach spaces over a non-archimedean field? $\endgroup$ – Makoto Kato Oct 23 '12 at 1:37
  • $\begingroup$ It treats Banach spaces over $\Bbb R,$ but I thought that maybe the proof could be tweaked to work more generally. $\endgroup$ – Andrew Oct 23 '12 at 1:56

I've a quick reference for your question. Chapter 2 of Igusa's book on local zeta functions has proofs for the implicit function theorems over an arbitrary complete field. The non-archimedean case is treated in section 2.2. Good luck!!

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.