c\xi,\xi\rangle+\left\|(\Im c)^{-\frac12}\Re(a-c)\xi\right\|_2^2+\left\|(\Im 
c)^{-\frac12}\Im a\xi\right\|_2^2\\
& & \mbox{}-2\Im\left\langle(\Im c)^{-\frac12}\Re(a-c)\xi,(\Im c)^{-\frac12}\Im 
a\xi\right\rangle\\
&\leq &(2+r^2)\langle\Im a\xi,\xi\rangle.\end{aligned}$$ Since this is true for all vectors $\xi\in\mathcal H^n$, we obtain $\Im c\leq(2+r^2)\Im a$, implying .

That $B^+_n(c,r)$ is closed in norm follows even easier: if $a_m\in B^+_n(c,r)$ and $\lim_{m\to\infty}
\|a_m-a\|=0$, then $\lim_{m\to\infty}\|a_m^*-a^*\|=0$, and thus $\lim_{m\to\infty}
\|\Im a_m-\Im a\|=0$. This also implies that $\Im a\ge\frac{1}{2+r^2}\Im c>0$, so that, by analytic functional calculus, $\lim_{m\to\infty}\left\|(\Im a_m)^{-\frac12}-(\Im a)^{-\frac12}\right\|=0$. A few succesive applications of the product-norm inequalities in C${}^*$-algebras provides $$\begin{aligned}
\left\|(\Im a)^{-\frac12}(a-c)(\Im c)^{-\frac12}\right\|&\leq&\left\|(\Im a)^{-\frac12}\right\|\|a_m-a\|
+\left\|(\Im a_m)^{-\frac12}-(\Im a)^{-\frac12}\right\|\\
& & \mbox{}\times\left\|(a_m-c)(\Im c)^{-\frac12}\right\|+\left\|(\Im a_m)^{-\frac12}(a_m-c)
(\Im c)^{-\frac12}\right\|.\end{aligned}$$ First and second right-hand terms converge to zero as $m\to\infty$, and the last is no more than $r$. Thus, $\left\|(\Im a)^{-\frac12}(a-c)(\Im c)^{-\frac12}\right\|\leq r$, which implies $a\in B^+_n(c,r)$.

Midpoint convexity of $B^+_n(c,r)$ follows easily from a direct computation: let $a_1,a_2\in B_n^+(c,r)$. We show that $(a_1+a_2)/2$ is in $B_n^+(c,r)$. As in , this is equivalent to showing that $$\left(\Im\frac{a_1+a_2}{2}\right)^{-\frac12}\left(\frac{a_1+a_2}{2}-c\right)(\Im c)^{-1}
\left(\frac{a_1+a_2}{2}-c\right)^*\left(\Im\frac{a_1+a_2}{2}\right)^{-\frac12}\leq r^2\cdot1,$$ which is in its own turn equivalent to $$\label{conv}
\left(\frac{a_1-c}{2}+\frac{a_2-c}{2}\right)(\Im c)^{-1}\left(\frac{a_1-c}{2}+\frac{a_2-c}{2}\right)^*\leq
\frac{r^2}{2}\Im(a_1+a_2).$$ However, adding the inequalities $(a_1-c)(\Im c)^{-1}(a_1-c)^*\leq r^2\Im a_1$ and $(a_2-c)(\Im c)^{-1}(a_2-c)^*\leq r^2\Im a_2$ (assumed to be true by hypothesis) and dividing by 2, we obtain $$\frac12((a_1-c)(\Im c)^{-1}(a_1-c)^*+\frac12(a_2-c)(\Im c)^{-1}(a_2-c)^*\leq\frac{r^2}{2}\Im(a_1+a_2).$$ Thus, our statement is proved if 