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Suppose that we can factor a formal power series $f\in\mathbb{Q}_p[[ X]]$ as follows: \begin{equation*} f(x)=g(x)h(x) \end{equation*} where $g,h\in\mathbb{Q}_p[[X]]$. In addition, we have the following growth conditions on the coefficients of $f$ and $g$: \begin{equation*} \left\{\frac{|a_n|_p}{n^\alpha}\right\}_{n=1}^\infty\hspace{1cm} \text{ and }\hspace{1cm} \left\{\frac{|b_n|_p}{n^\alpha}\right\}_{n=1}^\infty \end{equation*} are bounded sequences (as sequences in $\mathbb{R}$). *Edit: Any smaller value of $\alpha$, say $\alpha^*$, will make the sequences unbounded.

Prove or Disprove: $h$ has bounded coefficients. I am hoping this is something we can prove, but I'm not sure if it's possible in full generality. To simplify the matter, you may assume that $\alpha=1$.

To clarify \begin{align*} f(x)&=\sum_{j=0}^\infty a_nx^n\\ g(x)&=\sum_{j=0}^\infty b_nx^n\\ h(x)&=\sum_{j=0}^\infty c_nx^n \end{align*}

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    It's hard to come up with a sequence $(a_n)_n$ like that. My guess is that this isn't true and you can make an example of the form $f(x) = \sum_{n=0}^{\infty} a_n x^n$ and $g(x) = (p-x) \sum_{n=0}^{\infty} a_n x^n$ – 2017-01-29
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    Try $a_n=\frac{1}{n}$. This sequence satisfies the conditions for $\alpha=1$. – 2017-01-29
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    Hmm you're right, I guess it isn't as hard as I thought. In that case I think you have your counterexample since $b_n = \frac{p}{n} - \frac{1}{n-1}$ should also satisfy the conditions? – 2017-01-29

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