f(a_2,a_3)(c)e_{11}\Delta f(a_2,a_3)(c)^*+2\Re\left(\Delta f(a_2,a_3)(c)e_{12}
\Delta^2 f(a_2,a_3,a_4)(c,b)^*\right)}\\
& & \mbox{}+\Delta^2 f(a_2,a_3,a_4)(c,b)e_{22}\Delta^2 f(a_2,a_3,a_4)(c,b)^* \\
&=&\Delta f(a_2,a_3)(c)e_{11}\Delta f(a_2,a_3)(c)^*\\
& & \mbox{}-\Im\left(
\Delta f(a_2,a_3)(c)e_{11}\Delta f(a_3,a_4)(b)(\Im f(a_4))^{-1}
\Delta^2 f(a_2,a_3,a_4)(c,b)^*\right)\\
& & \mbox{}+\Delta^2 f(a_2,a_3,a_4)(c,b)e_{22}\Delta^2 f(a_2,a_3,a_4)(c,b)^*.\end{aligned}$$ On the right-hand side of we have the norm $$\left\|\left(\begin{array}{cc}
(\Im a_1)^{-\frac12} & 0\\
0 & (\Im a_2)^{-\frac12}
\end{array}\right)\left(\begin{array}{cc}
0 & 0\\
c & 0
\end{array}\right)\left(\begin{array}{cc}
\Im a_3 & \frac{b}{2i}\\
\left(\frac{b}{2i}\right)^* & \Im a_4
\end{array}\right)^{-\frac12}\right\|.$$ We use the properties of C${}^*$-norms to conclude that this norm in $M_2(\mathcal A)$ in fact equals the norm $\left\|(\Im a_2)^{-\frac12}c \left(\Im a_3-\frac14b(\Im a_4)^{-1}b^*
\right) c^*(\Im a_2)^{-\frac12}\right\|$ in $\mathcal A$. Thus, inequality for elements in $M_2(\mathcal A)$ translates into an inequality of elements in $\mathcal A$ as follows: $$\begin{aligned}
\lefteqn{\Delta f(a_2,a_3)(c)e_{11}\Delta f(a_2,a_3)(c)^*}\nonumber\\
& & \mbox{}-\Im\left(
\Delta f(a_2,a_3)(c)e_{12}\Delta f(a_3,a_4)(b)(\Im f(a_4))^{-1}
\Delta^2 f(a_2,a_3,a_4)(c,b)^*\right)\nonumber\\
& & \mbox{}+\Delta^2 f(a_2,a_3,a_4)(c,b)e_{22}\Delta^2 f(a_2,a_3,a_4)(c,b)^*\nonumber\\
& \leq & \left\|(\Im a_2)^{-\frac12}c \left(\Im a_3-\frac14b(\Im a_4)^{-1}b^*
\right)^{-1} c^*(\Im a_2)^{-\frac12}\right\|\Im f(a_2).\label{secondderiv}\end{aligned}$$

However, for now their form seems to be too complicated when $j>2$, and of no significant use for the purposes of this paper. Since the above proposition is applied in this paper only for $\mathcal A=\mathcal C$, from now on we shall eliminate the subscript from the notation of the C${}^*$-norm of $\mathcal A$.

\[Hyper\] Fix $n\in\mathbb N$, $r>0$ 