we show that the left-hand term of is less than or equal to the left-hand term of the inequality above. Expanding the left-hand of and subtracting from the one above yields $$\begin{aligned}
 &  & \frac12(a_1-c)(\Im c)^{-1}(a_1-c)^*+\frac12(a_2-c)(\Im c)^{-1}(a_2-c)^*\\
& & \mbox{}-\frac14(a_1-c)(\Im c)^{-1}(a_1-c)^*-\frac14(a_2-c)(\Im c)^{-1}(a_2-c)^*\\
& & \mbox{}-\frac14(a_1-c)(\Im c)^{-1}(a_2-c)^*-\frac14(a_2-c)(\Im c)^{-1}(a_1-c)^*\\
& = & \frac14\left[(a_1-c)(\Im c)^{-1}(a_1-c-a_2+c)^*+(a_2-c)(\Im c)^{-1}(a_2-c-a_1+c)^*\right]\\
& = & \frac14(a_1-c-a_2+c)(\Im c)^{-1}(a_1-a_2)^*=\frac14(a_1-a_2)(\Im c)^{-1}(a_1-a_2)^*\ge0.\end{aligned}$$ Since $B^+_n(c,r)$ is midpoint convex and closed, it is convex.

To conclude, observe that all the above computations hold if $c\in H^+(M_n(\mathcal A))$ is replaced by $c\otimes 1_p\in H^+(M_{np}(\mathcal A))$. Indeed, one only needs to observe that taking imaginary part, inverse and root, as well as multiplication, respect direct sums. Since $\|a\oplus b\|=
\max\{\|a\|,\|b\|\}$, we’re done. Estimates and hold on the amplifications of $c$ to any $c\otimes 1_p$, $p\in\mathbb N$, with the same constants.

The following lemma will be useful when applying Proposition \[prop:3.1\] to the proof of the main result (compare with the method used in [@BPV1 Remark 2.5]).

\[lem:3.2\] Assume that $f$ is a noncommutative self-map of the noncommutative upper half-plane of $\mathcal A$. Let $v_1,v_2>0$ in $\mathcal A$. If $${\rm wo}\text{-}\lim_{y\downarrow0}\frac{\Im f(\alpha+iyv_j)}{y}=c_j\in
\mathcal A,\quad j\in\{1,2\},$$ exist, then the set of limit points of $\Delta f(\alpha+zv_1,
\alpha+\zeta v_2)(w)$ as 