How do I show a function on 2-adic units is continuous (using the 2-adic metric)?
I'd be happy to learn the general rule or definition. But in particular I need to show that $f(x)=\dfrac{3x+1}{2^{v_2(3x+1)}}$ is continuous at all odd numbers.
Since the function is isometric in $\lvert\cdot\rvert_2$, i.e. since $\lvert f(x)\rvert_2=\lvert x\rvert_2$ every orbit of the function is duplicated, multiplied by any power of $2$. This can therefore be formulated in various ways; as a function though the odd numbers, through the dyadic rationals, but the following seems good to work with. If we let $n$ be a positive integer and $k$ be the power of $2$ we can define:
$f(2^k(2n+1))=(3(2n+1)+1)\cdot2^k\cdot\lvert3(2n+1)+1\rvert_2$
So $f(2^k(2n+1)=(3n+2)\times2^k\times\lvert3n+2\rvert_2$
Here is what I have so far:
I think to prove continuity in the 2-adic metric I need $\lvert x_n-x\rvert_2=0\implies\lvert f(x_n)-f(x)\rvert_2=0$
I think $k$ and $n$ are independent so I think I can examine them independently. Taking $k$ to infinity brings both $x$ and $f(x)$ to zero so that seems to satisfies the continuity requirement.
Moving on to $n$; this seems to be an exercise in proving convergence within odd integers which are in a sense a subset of the 2-adic units. I know all Cauchy sequences in these converge to 2-adic units but not much more than that.
However I do have a little insight into this particular function. For example if we examine the inputs $x$ which map to any given output of $f(x)$, it can fairly easily be shown that these $x$'s take the form of a set $\left\{4^mp+\dfrac{4^m-1}{3}:m,n\in\mathbb{N}\right\}$ so, at least for any given output $f(x)$ the inputs always converge to $x=\frac{-1}{3}$ as the intervals between them become large powers of $2$.
In fact the orbit of the function $a(2^kx)=2^k(4x+1)$ on variation of $x$ and holding $k$ fixed is in a sense orthogonal to $f(x)$; which is a restatement of the above except not in closed form.
As for the map
$$f: x \mapsto \frac{3x+1}{2^{v_2(3x+1)}} = (3x+1)\cdot |3x+1|_2,$$
it is the composition of $x\mapsto 3x+1$ and $y\mapsto y |y|_2$, so we want to enquire where these are continuous, the only interesting part being actually the absolute value map $| \cdot|_2$ itself. Viewed as map $(\Bbb{Q}_2, |\cdot|_2)\rightarrow (\Bbb{Q}_2, |\cdot|_2)$, the absolute value is not continuous at $0$ (because $|2^n|_2 =2^{-n}$ does not converge $2$-adically for $n\to \infty$), but outside of $0$, it is actually locally constant and hence continuous. So the composite map $g: (\Bbb{Q}_2, |\cdot|_2)\rightarrow (\Bbb{Q}_2, |\cdot|_2)$ is continuous everywhere except at $x=-\frac{1}{3}$. (Note, however, that this point $-1/3$ w.r.t. the $2$-adic metric does lie in every neighbourhood of $\mathbb{N}$, even in every neighbourhood of the odd natural numbers, as mentioned here recently.)
With a similar argument, the function $\tilde f$ in your answer -- which, if I understand it correctly, is nothing else than $x\mapsto |x|_2^{-1}\cdot f(x\cdot |x|_2)$ -- is continuous as function $(\Bbb{Q}_2, |\cdot|_2)\rightarrow (\Bbb{Q}_2, |\cdot|_2)$, except at the points $-\frac{2^k}{3}, k \in \mathbb{Z}$.
As for the function $g$, which I would rewrite as $x\mapsto (3x+|x|_2^{-1})\cdot |3x+|x|_2^{-1}|_2$, it looks as if it is continuous except at $0$ and all $-\frac{2^k}{3}, k \in \mathbb{Z}$.