extremal functions in the problem $|\Phi|\to\max_{{\mathcal{S}}_k}$ do not belong to ${\mathcal{S}}_q$ whenever ${q<k}$. Therefore, to complete the prove, it would be sufficient to show that the Becker q.c.-extension of the function $f_+$ from Theorem \[TH\_a\_3\] whose Beltrami coefficient is given by  is not extremal, i.e. that $f_+$ admits a $q$-q.c. extension to ${\mathbb{C}}$ with some $q\in(0,k)$.

Suppose on the contrary that the above mentioned Becker extension of $f_+$ is extremal. Then it would satisfy the Hamilton–Krushkal condition [@Hamilton Theorem 1], see also [@Krushkal-extremal; @HarringtonOrtel], which can be formulated as $~\sup_\varphi\big|\Lambda(\varphi)\big|=1$, where $$\Lambda(\varphi):={\frac{1}{k}\iint\limits_\Delta\varphi(z)\,\mu(z)\,{\mathrm{d}}x{\mathrm{d}}y},\quad \Delta:=\{z:1<|z|<+\infty\},$$ $\mu$ is given by , and the supremum is taken over all holomorphic differentials $\varphi(z){\mathrm{d}}z^2$ in $\Delta$ with $\|\varphi\|:=\iint_\Delta|\varphi(z)|{\mathrm{d}}x{\mathrm{d}}y\,\le\,1$. Note that $\varphi(z){\mathrm{d}}z^2$ does not have to be holomorphic at $\infty$, because the q.c.-extensions of $f_+$ that we consider are required to fix $\infty$.

The results of [@HarringtonOrtel §3] can be extended without any trouble from ${\mathbb{D}}$ to $\Delta$. In particular, by [@HarringtonOrtel Proposition 3.2], either $\big|\Lambda(\varphi_*)\big|=1$ for some $\varphi_*$ with $\|\varphi_*\|=1$ or $\big|\Lambda(\varphi_n)\big|\to1$ as $n\to+\infty$ for some sequence $(\varphi_n)$ with $\|\varphi_n\|\le1$ converging locally uniformly in $\Delta$ to zero. On the one hand, the former possibility does not hold in our case, because $\mu(z)$ is not of the 