Proof of inequality$\Vert f_1S_1^*+f_2 S_2^*\Vert^2\geq\Vert f_1\Vert^2+\Vert f_2\Vert^2$ for functionals I was trying to prove Remark 3.7 in this article, but failed. Here some necessary definitions.
Let $\mathcal{A}$ be a unital $C^*$-algebra such that there exist $S_1,S_2\in\mathcal{A}$ with property $S_1^*S_1=S_2^*S_2=1_{\mathcal{A}}$, $S_1^*S_2=S_2^*S_1=0$ (i.e. isometries with pairwise orthogonal images)
Let $Y$ be contractive (i.e. for all $a\in\mathcal{A}$, $y\in Y$ we have $\Vert a\cdot y\Vert\leq\Vert a\Vert\Vert y\Vert$) normed (i.e. normed space) unital (i.e. for all $y\in Y$ we have $1_\mathcal{A}y=y$) left module over $\mathcal{A}$.
Such a normed module is called semi-Ruan if for all $y_1, y_2\in Y$ the following holds
$$
\Vert S_1\cdot y_1+S_2\cdot y_2\Vert^2\leq\Vert y_1\Vert^2+\Vert y_2\Vert^2
$$
My question. 
Let $Y^*$ be dual module of semi-Ruan module $Y$ (i.e. dual space with the structure of right unital contractive normed module with outer multiplication defined by $(f\cdot a)(y)=f(a\cdot y)$ ), then the following inequality holds
$$
\Vert f_1\cdot S_1^*+f_2\cdot S_2^*\Vert^2\geq\Vert f_1\Vert^2+\Vert f_2\Vert^2
$$
My attempt. 
Note that $\Vert y_1\Vert\leq 1$,$\Vert y_2\Vert\leq 1$ implies $\Vert S_1\cdot y_1+S_2\cdot y_2\Vert\leq\sqrt{2}$, so
$$
\Vert f_1\cdot S_1^*+f_2\cdot S_2^*\Vert^2=\sup\{|(f_1\cdot S_1^*+f_2\cdot S_2^*)(y)|^2:\Vert y\Vert\leq 1\}\geq
$$
$$
\sup\left\{\left|(f_1\cdot S_1^*+f_2\cdot S_2^*)\left(S_1\cdot\frac{y_1}{\sqrt{2}}+S_2\cdot\frac{y_2}{\sqrt{2}}\right)\right|^2:\Vert y_1\Vert\leq 1,\Vert y_2\Vert\leq 1\right\}=
$$
$$
0.5 \sup\left\{|f_1(y_1)+f_2(y_2)|^2:\Vert y_1\Vert\leq 1,\Vert y_2\Vert\leq 1\right\}=
$$
$$
0.5 \sup\left\{|f_1(y_1)|^2+|f_2(y_2)|^2+2\operatorname{Re}(f_1(y_1)\overline{f_2(y_2)}):\Vert y_1\Vert\leq 1,\Vert y_2\Vert\leq 1\right\}=???
$$
 A: This is maybe a comment, but it became too long.  We seem to agree that
$$ \|f_1\cdot S_1^*+f_2\cdot S_2^*\|^2 \geq \sup\big\{ |f_1(y_1)+f_2(y_2)|^2 : \|y_1\|^2+\|y_2\|^2\leq 1 \big\}. $$
Now, if $a_1,a_2\geq 0$ are such that $a_1^2+a_2^2\leq 1$ then for each $\epsilon>0$, we can pick $z_k$ with $|f_k(z_k)| = f_k(z_k) > \|f_k\|-\epsilon$, and with $\|z_k\|=1$.  So setting $y_k = a_kz_k$, we see that
$$ \sup\big\{ |f_1(y_1)+f_2(y_2)|^2 : \|y_1\|^2+\|y_2\|^2\leq 1 \big\} \geq
\sup\big\{ (a_1\|f_1\|+a_2\|f_2\|)^2 : a_1^2+a_2^2\leq 1 \big\}. $$
But this supremum involves only scalars, and it's standard (I guess via Cauchy-Schwarz-- think about the proof that $(\ell^2)^*=\ell^2$) that this equals
$$ \|f_1\|^2+\|f_2\|^2. $$
Conversely, $|f_1(y_1)+f_2(y_2)|^2 \leq ( \|f_1\|\|y_1\| + \|f_2\|\|y_2\| )^2
\leq (\|f_1\|^2+\|f_2\|^2)(\|y_1\|^2+\|y_2\|^2)$ by Cauchy-Schwarz.  So actually
$$ \sup\big\{ |f_1(y_1)+f_2(y_2)|^2 : \|y_1\|^2+\|y_2\|^2\leq 1 \big\}
= \|f_1\|^2+\|f_2\|^2. $$
And we're done!
