Let $A$ be some linear operator in a Hilbert space $\mathcal{H}$ and let $A^{\dagger}$ be the corresponding adjoint operator. When I look in the literature, I find that the proof of the statement $\Vert A\Vert=\Vert A^{\dagger}\Vert$ is always done by showing first that $\Vert A\Vert\leq\Vert A^{\dagger}\Vert$ and then $\Vert A\Vert\geq\Vert A^{\dagger}\Vert$.
I am just wondering why noone proves the statement directly. Is there something wrong with this proof?:
I am using that $\Vert A\Vert:=\sup_{x\in \mathcal{H},\Vert x\Vert\leq 1}\Vert Ax\Vert$ and the formula $\Vert x\Vert=\sup_{y\in\mathcal{H},\Vert y\Vert\leq 1}\vert\langle x,y\rangle\vert$.
Then we have:
$$\Vert A^{\dagger}\Vert=\sup_{x\in \mathcal{H},\Vert x\Vert\leq 1}\underbrace{\Vert A^{\dagger}x\Vert}_{\sup_{y\in\mathcal{H},\Vert y\Vert\leq 1}\vert\langle A^{\dagger}x,y\rangle\vert}=\sup_{x\in \mathcal{H},\Vert x\Vert\leq 1}\sup_{y\in \mathcal{H},\Vert y\Vert\leq 1}\vert\underbrace{\langle A^{\dagger}x,y\rangle}_{=\langle x,Ay\rangle}\vert=\sup_{y\in \mathcal{H},\Vert y\Vert\leq 1}\bigg (\underbrace{\sup_{x\in \mathcal{H},\Vert x\Vert\leq 1}\vert\langle x,Ay\rangle\vert}_{=\Vert Ay\Vert}\bigg)=\sup_{y\in \mathcal{H},\Vert y\Vert\leq 1}\Vert Ay\Vert=\Vert A\Vert$$