# About continuous functions on $p$-adic fields

Consider $$K/ \mathbb{Q}_p$$ a finite extension of the field of $$p$$-adic numbers. If for every such an extension $$f_K: K \to K$$ is continuous can we extend these functions to $$\mathbb{C}_p$$? My idea was that since $$x \in \mathbb{C}_p$$ then $$x= lim_{n \to \infty}x_n$$ where $$x_n$$ is an element of a finite extension on $$\mathbb{Q_p}$$ then $$\mathbb{C}_p \subset \prod_{K/Q_p finite}K$$ the map given by the sequence $$F=(f_K)_K$$ is continuous since all the components are continuous. And so $$F_{|C_p}$$ is also continuous.

This method could work?

• Do you mean that for each finite $K\vert \Bbb Q_p$ contained in $\Bbb C_p$ you have a continuous $f_K$, such that if $K_1 \subseteq K_2$ then the restriction of $f_{K_2}$ to $K_1$ is $f_{K_1}$? Then I would think yes, there is a continuous $F$ on $\Bbb C_p$ whose restriction to $K$ is $f_K$, for all $K$. However, I do not think your "inclusion in product" formula is valid, and I don't understand what "map given by the sequence" should mean. Rather, you can indeed define $F$ by $F(x) = \lim F(x_n)$ for $x_n \to x$, just have to show this is independent from the choice of the sequence $x_n$. – Torsten Schoeneberg Feb 27 at 4:48
• @TorstenSchoeneberg How do you extend $f_L(x) = f (\frac{1}{[\mathbb{Q}_p(x):\mathbb{Q}_p]} Tr_{\mathbb{Q}_p(x)/\mathbb{Q}_p}(x)), x \in L$ to $\overline{\mathbb{Q}_p}$ and $\mathbb{C}_p$ ? – reuns Feb 27 at 17:39
• @reuns: Good point. I'll think about it and take back my "I would think yes" from the comment for the time being. The main point I wanted to make is that the question seems not well-posed, and that the attempt in there does not make sense to me. – Torsten Schoeneberg Feb 27 at 20:20
• @reuns: Well if those maps are compatible in the sense of my comment, they already well-define a map on $\overline{\Bbb Q_p} = \bigcup_{\Bbb Q_p \subseteq K \subset \overline {\Bbb Q_p}, K\vert \Bbb Q_p \text{ finite}} K$. What I realise now as indeed unclear, in the general situation and your example, is whether that map is continuous. – Torsten Schoeneberg Feb 27 at 22:41

Your question seems not well-posed to me. User reuns seems to interpret it as:

Let $$K$$ be one fixed finite extension of $$\Bbb Q_p$$. If $$f: K \rightarrow K$$ is continuous, is there a continuous map $$\hat{f}: \Bbb C_p \rightarrow K$$ such that $$\hat{f}_{\vert K} = f$$?

and in his answer simultaneously restricts to the special case $$\Bbb Q_p$$ for the domain of $$f$$, but generalises to an arbitrary topological space $$X$$ for the codomain of $$f$$ and $$\hat{f}$$. Both the restriction and the generalisation are harmless though, and indeed this boils down to the question whether there is a continuous map $$\phi: \Bbb C_p\rightarrow K$$ (or in the special case $$\rightarrow \Bbb Q_p$$) with $$\phi_{\vert K} = id_K$$, which I think he indeed constructs. So the answer to that interpretation of the question is yes.

I however interpret the question differently, namely as:

For each finite extensions $$K\vert \Bbb Q_p$$ (contained in $$\Bbb C_p$$), let a continuous map $$f_K : K \rightarrow K$$ be given. Then is there a continuous map $$F: \Bbb C_p \rightarrow \Bbb C_p$$ such that $$F_{\vert K} = f_K$$ for all $$K$$ as above?

The answer to this is no, for two reasons.

• Quite obviously the given maps $$f_K$$ need to be compatible in the sense that for every $$K_1, K_2$$, we need $$f_{K_1 \vert (K_1 \cap K_2)} = f_{K_2 \vert (K_1 \cap K_2)}$$ (or something similar).
• Now if the condition in 1. is satisfied, then indeed the maps do define a map $$f$$ on $$\overline{\Bbb Q_p} = \bigcup_{K\vert \Bbb Q_p \text{ finite}} K$$, and maybe that is what you have in mind when you write that product in the OP. I am not sure if that map necessarily is continuous on $$\overline{\Bbb Q_p}$$; however, even if it is continuous, I think that such a map does not necessarily extend to $$\Bbb C_p$$. A counterexample is:
• Let $$c \in \Bbb C_p \setminus \overline{\Bbb Q_p}$$ such that there exists a sequence $$(x_n)_n$$ with $$x_n \to c$$ and $$v_p(x_n-c) \in \Bbb Z$$ for all but finitely many $$n$$. Then for all $$K$$ as above, set $$f_K(x) = \phi (\dfrac{1}{x-c})$$, where $$\phi: \Bbb C_p \rightarrow \Bbb Q_p$$ is the map constructed in reuns' answer. While the $$f_K$$ give a map $$f$$ on $$\overline{\Bbb Q_p}$$ as above, we have $$\lim_{n\to \infty} v_p(f(x_n)) = -\infty$$, which means $$f$$ cannot be extended to $$c$$.
• Thank you @Torsten Schoeneberg for the answer, now I understand the problem, for me it is enough to extend the map from $\mathbb{Q}_p \to \mathbb{Q}_p$ to $\mathbb{C}_p$ I'll try to prove it. – andres Mar 9 at 0:45

If $$f$$ is continuous $$\mathbb{Q}_p \to X$$ then $$f \circ \phi$$ is continuous $$\mathbb{C}_p\to X$$ and extends $$f$$.

Where $$\phi$$ sends $$x \in \mathbb{C}_p$$ to the closest point in $$\mathbb{Q}_p$$ :

let $$u(x) = \sup_{t \in \mathbb{Q}_p} v_p(x-t)$$ and $$\phi(x) = \cases{ x \text{ if } x\in \mathbb{Q}_p \\ 0 \text{ if } v_p(x)=u(x) \not \in \mathbb{Z} \\ p^{v_p(x)} \min \{n \in \mathbb{Z}_{\ge 0}, v_p(x-p^{v_p(x)} n) = u(x)\} \text{ otherwise}}$$

Since $$v_p(x-\phi(x)) = u(x)$$ then $$v_p(\phi(x)-\phi(y)) \ge v_p(x-y)$$

• Why is the composition of those projections continuous? Also, to extend to $\mathbb C_p$ you need $\phi$ to be uniformly continuous, why does that hold? – Wojowu Feb 27 at 21:38
• (ignoring just about everything you said, existence of such a projection follows from nonarchimedean Hahn-Banach theorem, since $\mathbb Q_p$ is spherically complete; I meant to post this as an answer to your question, but you've deleted it before I could post.) – Wojowu Feb 27 at 21:40
• @Wojowu Tks, your idea of using Hahn-Banach is good. What about this one ? Is it possible to make $\phi$ a $\mathbb{Q}_p$-linear map ? – reuns Feb 28 at 0:11