Q:

Extensions of maximally complete fields

Recall that an immediate extension of a (rank 1, non-archimedean) valued field $(K,v)$ is another valued field $(F,w)$ extending $K$, such that $w|_K = v$, and such that $F$ and $K$ have the same residue fields and value groups. We say $K$ is maximally complete if its only immediate extension is itself. 

Suppose $K$ is maximally complete, and let $H$ be a completion of
  $K(T)$ with respect to a valuation extending that on $K$, where $T$ is
  an indeterminate. Suppose we know either of two things:

$|H| = |K|$ and $\widetilde{H} = \widetilde{K}(T)$, or
$\widetilde{H} = \widetilde{K}$ and $|H^\times|$ is generated in $\mathbb{R}$ by $|K^\times|$ and one extra element $\rho \in
\mathbb{R}$. 

Can we conclude from either of these assumptions that $H$ is also
  maximally complete?

I'll appreciate (partial) answers referencing assumptions 1 or 2, or any counterexample.

A:

Those conditions do not imply that $H$ be maximally complete as can the following counter-examples illustrate.
2. Take $F:=\mathbb{R}((x^{\mathbb{R}}))$ with its natural valuation, and consider its subfield $K :=\mathbb{R}((x^{\mathbb{Q}}))$. Then $x^{\sqrt{2}}$ is transcendantal over $K$ and the completion $H$ of $K(x^{\sqrt{2}})$ in $F$ has residue field $\mathbb{R}$ and value group $\Gamma$ generated by $\mathbb{Q}$ and $\sqrt{2}$, but $H$ is not maximally complete since it does not contain $\sum \limits_{n \in \mathbb{N}} x^{\sqrt{2}(1-\frac{1}{n+1})}$ and thus admits $\mathbb{R}((x^{\Gamma}))$ as a proper immediate extension.
1. This time consider the subfield $K' = \mathbb{Q}((x^{\mathbb{R}}))$ of $F$, and its extension $K'(\pi)$ in $F$, and take $H'$ as the completion of $K'(\pi)$. Then likewise $H'$ does not contain $\sum \limits_{n \in \mathbb{N}} \pi^nx^{1-\frac{1}{n+1}}$ so it admits $\mathbb{Q}(\pi)((x^{\mathbb{R}}))$ as a proper immediate extension. 

