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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 $$ \Vert f_1\cdot S_1^*+f_2\cdot S_2^*\Vert^2=\sup\{|(f_1\cdot S_1^*+f_2\cdot S_2^*)(y)|^2:\Vert y\Vert\leq 1\}\geq $$ $$ \sup\left\{\left|(f_1\cdot S_1^*+f_2\cdot S_2^*)\left(S_1\cdot\frac{y_1}{\sqrt{2}}+S_2\cdot\frac{y_2}{\sqrt{2}}\right)\right|^2:\Vert y_1\Vert\leq 1,\Vert y_2\Vert\leq 1\right\}= $$ $$ 0.5 \sup\left\{|f_1(y_1)+f_2(y_2)|^2:\Vert y_1\Vert\leq 1,\Vert y_2\Vert\leq 1\right\}= $$ $$ 0.5 \sup\left\{|f_1(y_1)|^2+|f_2(y_2)|^2+2\operatorname{Re}(f_1(y_1)\overline{f_2(y_2)}):\Vert y_1\Vert\leq 1,\Vert y_2\Vert\leq 1\right\}=??? $$

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Are you also assuming that $S_1$ and $S_2$ are isometries? If so, you should probably add that to the question. –  user16299 Feb 6 '12 at 22:33
    
Oh, my mistake, I didn't read carefully enough. My apologies. –  user16299 Feb 6 '12 at 22:47
    
Hint: Your final supremum is equal to $0.5( \|f_1\| + \|f_2\|)^2$. Actually, it's easiest to see this from the line before. This isn't what you want, but suppose instead you had taken the supremum over all $y_1,y_2$ with $\|y_1\|^2 + \|y_2\|^2\leq 1$... –  Matthew Daws Feb 7 '12 at 7:54
    
So you propose the following $$\Vert f_1\cdot S_1^*+f_2\cdot S_2^*\Vert^2=$$$$\sup\{|(f_1\cdot S_1^*+f_2\cdot S_2^*)(y)|^2:\Vert y\Vert\leq 1\}\geq$$ $$\sup\left\{\left|(f_1\cdot S_1^*+f_2\cdot S_2^*)\left(S_1\cdot y_1+S_2\cdot y_2\right)\right|^2:\Vert y_1\Vert^2+\Vert y_2\Vert^2\leq 1\right\}=$$$$\sup\left\{|f_1(y_1)+f_2(y_2)|^2:\Vert y_1\Vert^2+\Vert y_2\Vert^2\leq 1\right\}$$ But, why the last supremum equals to $\Vert f_1\Vert^2+\Vert f_2\Vert^2$? This supremum is taken over dependent variables. –  no identity Feb 7 '12 at 9:55
    
@MatthewDaws I was thinking about your hint but haven't come to any conclusions. Do you know a complete solution, or that hint was just an idea that you think may be helpful? –  no identity Feb 8 '12 at 18:48

1 Answer 1

up vote 2 down vote accepted

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!

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Ah, the old proof that the p-norm is a sup over pairing with things in the ball of the q-norm. (Flashbacks to a certain AJW course, perhaps?) –  user16299 Feb 8 '12 at 22:23
    
@Norbert, erm, yes, I guess. But the result is true for more than two summands, which would be less clear from your proof. Actually, all I was proving I guess is that the dual of the vector-valued space $\ell^2(F)$ is $\ell^2(F^*)$ for the obvious dual pairing... –  Matthew Daws Feb 9 '12 at 7:27

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