Given a finite family $A:=(A\_j)\_{j\in J}$ of operators $A\_j \in\mathcal{L}(H)$, 



> 
> It is true that for all $L\in \mathcal{L}(H)$, we have
>  $$\big\|\sum\_{j\in J}A\_j^\*LA\_j\big\|\le \big\|\sum\_{j\in J}A\_j^\*A\_j\big\|\ \|L\|?$$
> 
> 
> 


If $L\in \mathcal{L}(H)^+$, one has
$$\big\|\sum\_{j\in J}A\_j^\*LA\_j\big\|\le \big\|\sum\_{j\in J}A\_j^\*A\_j\big\|\ \|L\|.$$
 Indeed, for any $x\in H$, 
$$ \big(\sum\_{j\in J}A\_j^\*LA\_jx, \ x\big) = \sum\_{j\in J}(A\_j^\*LA\_jx,x) = \sum\_{j\in J}(LA\_jx,A\_jx)\le \sum\_{j\in J}\|L\|(A\_jx,A\_jx)$$
$$= \|L\| \sum\_{j\in J}(A\_j^\*A\_jx,x)\le\|L\|\ \big\|\sum\_{j\in J}A\_j^\*A\_j\big\|\ \|x\|^2.$$



$\big\|\sum_{j\in J}A_j^*LA_j\big\|\le \big\|\sum_{j\in J}A_j^*A_j\big\|\ \|L\|?,\;\forall \,L\in \mathcal{L}(H)$
Actually yes: it follows from the analogous inequality for any $x$ and $y$, and Cauchy-Schwarz. One has 


$$ \big(\sum\_{j\in J}A\_j^\*LA\_jx, \ y\big) = \sum\_{j\in J}(A\_j^\*LA\_jx,y) = \sum\_{j\in J}(LA\_jx,A\_jy)\le \sum\_{j\in J}\|L\|\ \|A\_jx\|\|A\_jy\|$$
$$ \le\|L\|\sqrt{ \sum\_{j\in J}\|A\_jx\|^2 \sum\_{j\in J}\|A\_jy\|^2 }
=\|L\|\sqrt{ \sum\_{j\in J}(A\_j^\*A\_jx,x) \sum\_{j\in J}(A\_j^\*A\_jy,y) }
$$


$$\le\|L\|\ \big\|\sum\_{j\in J}A\_j^\*A\_j\big\|\ \|x\|\|y\|.$$


