I'm stuck in a sentence made by Dwork in his book "An introduction to $G$-functions": Let $b$ a positive real number, define the following set $$\mathcal{G}(b)=\left\{f(X): \begin{array}{c} f(X)\in1+X\mathbb{C}_{p}[\![X]\!]\\ f \text{ converges for ord }x>-b\\ |f(x)-1|<1 \end{array} \right\}$$ (where $1+X\mathbb{C}_{p}[\![X]\!]$ means that $f(0)=1$, and $\text{ord}$ is the additive $p$-adic valuation ), then he says: if $f(X)=1+a_{1}X+a_{2}X^{2}+\cdots\in\mathbb{C}_{p}[\![X]\!]$, then $f(X)\in\mathcal{G}(b)$ if and only if $\text{ord }a_{j}\geq jb$ for every $j\geq 1$.
I have the converse, but I'm stuck dealing with the "if" direction ($\Rightarrow$). I tried arguing by contradiction supposing that there is $j$ for which $\text{ord }a_{j}<jb$ and taking the minimal $j$ with this property I tried to use the strong triangle inequality to get a contradiction with the third property of the set $\mathcal{G}(b)$, but these involves $x$ in the radius of convergence, not on the boundary.
I will appreciate any hint, thanks.
All right let me make a stab at this. I’m going to use the notation $v(x)$ for the (additive) valuation of $x$, rather than $\text{ord}(x)$, just for convenience of typing, and look at a function $f\in\mathcal G(b)$, $f=1+\sum_1^\infty a_jX^j$.
Start with the line $y=bx$, above which we hope all Newton points $(j,v(a_j))$ lie. If not, there is a line $y=\beta x$ through the origin that contains only finitely many Newton points, with no such points below the line. To see this, look at any suspect Newton point $(j_0,j_0\alpha)$ below $y=bx$ (so with $\alpha<b$), and see that there are at most finitely many points drawn below the line $y=\alpha x$. (Otherwise, the domain of convergence would be smaller than we want.)
Now we have our (orange) line $y=\beta x$, $\beta\in\Bbb Q$, with finitely many Newton points on it, and we call the rightmost of these $P=(n,v(a_n))=(n,n\beta)$. Now draw through $P$ a (blue, dashed) line of slope $\beta'\in\Bbb Q$ with $\beta<\beta'<b$, $y=\beta'x+c$, in such a way that all other Newton points lie above the line. Again, the existence of such a line is a little delicate for points to the right, but I think the fact that when $v(\xi)=-\beta>-b$ the series $f(\xi)$ is convergent, should take care of that.
Now we’re pretty much done. When you try $\xi\in\Bbb C_p$ with $v(\xi)=-\beta'>-b$, you see that the series is convergent, and all other monomials in the series other than $\xi^na_n$ have value $c=-n\beta'+v(a_n)$, so that the value of the whole series $f(\xi)$ is therefore $c<0$, contradicting the third condition for membership in $\mathcal G(b)$.
And I guess I should say that the argument would have been immeasurably easier if I had allowed myself to use the theory of the Newton copolygon, but I’m uncertain how many of our readers would have been familiar with that.