Hilbert-adjoint identities

First let $U:X\to Y$ be any bounded linear operator between Hilbert spaces over the same scalar field. The map $U^{\ast }:Y\to X$ exists by the adjoint construction. For $y,z\in Y$, scalars $a,b$ and $x\in X$, its identity gives $⟨x,U^{\ast }(ay+bz)⟩=⟨Ux,ay+bz⟩=\bar{a}⟨x,U^{\ast }y⟩+\bar{b}⟨x,U^{\ast }z⟩=⟨x,a{U}^{\ast }y+b{U}^{\ast }z⟩$. Positive definiteness applied to the difference proves that $U^{\ast }$ is linear. Also $\mathrm{\parallel }{U}^{\ast }y{\mathrm{\parallel }}^2=⟨U(U^{\ast }y),y⟩\le \mathrm{\mid }⟨U(U^{\ast }y),y⟩\mathrm{\mid }\le \mathrm{\parallel }U\mathrm{\parallel }\mathrm{\parallel }{U}^{\ast }y\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$, so division when $U^{\ast }y\mathrm{\ne }0$, and the trivial inequality otherwise, give $\mathrm{\parallel }{U}^{\ast }y\mathrm{\parallel }\le \mathrm{\parallel }U\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$. Thus $U^{\ast }$ is bounded and its now-defined operator norm satisfies $\mathrm{\parallel }{U}^{\ast }\mathrm{\parallel }\le \mathrm{\parallel }U\mathrm{\parallel }$. This applies to each bounded operator used below, including an adjoint once its boundedness has been established.

For uniqueness of the adjoint and conjugate-linearity in the operator, let $U,V\in \mathcal{B}(K,H)$ both satisfy the defining adjoint identity for the same operator in $\mathcal{B}(H,K)$. For $y\in K$, one has $⟨x,(U-V)y{⟩}_H=0$ for every $x\in H$; taking $x=(U-V)y$ gives $(U-V)y=0$. For scalars $a,b$, the identities $⟨(aR+bS)x,y{⟩}_K=a⟨Rx,y{⟩}_K+b⟨Sx,y{⟩}_K=⟨x,\bar{a}{R}^{\ast }y+\bar{b}{S}^{\ast }y{⟩}_H$ hold for all $x\in H$ and $y\in K$, so $(aR+bS{)}^{\ast }=\bar{a}{R}^{\ast }+\bar{b}{S}^{\ast }$.

Composition and involution: for $x\in H$ and $y\in L$, $⟨x,S^{\ast }{T}^{\ast }y{⟩}_H=⟨Sx,T^{\ast }y{⟩}_K=⟨TSx,y{⟩}_L$, so $(TS{)}^{\ast }=S^{\ast }{T}^{\ast }$ by uniqueness. Likewise, for $x\in K$ and $y\in L$, the defining identity for $T^{\ast }$ gives $⟨T^{\ast }y,x{⟩}_K=⟨y,T^{\ast \ast }x{⟩}_L$; conjugate symmetry and the defining identity for $T$ give $⟨T^{\ast \ast }x,y{⟩}_L=⟨x,T^{\ast }y{⟩}_K=⟨Tx,y{⟩}_L$, so $T^{\ast \ast }=T$ by uniqueness.

Norms: Cauchy–Schwarz gives $\mathrm{\parallel }{T}^{\ast }y{\mathrm{\parallel }}^2=⟨T^{\ast }y,T^{\ast }y{⟩}_K=⟨T(T^{\ast }y),y{⟩}_L\le \mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }{T}^{\ast }y\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$, hence $\mathrm{\parallel }{T}^{\ast }y\mathrm{\parallel }\le \mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$ (trivially when $T^{\ast }y=0$) and $\mathrm{\parallel }{T}^{\ast }\mathrm{\parallel }\le \mathrm{\parallel }T\mathrm{\parallel }$; applying this to $T^{\ast }$ and using $T^{\ast \ast }=T$ gives $\mathrm{\parallel }{T}^{\ast }\mathrm{\parallel }=\mathrm{\parallel }T\mathrm{\parallel }$. Moreover $\mathrm{\parallel }{T}^{\ast }T\mathrm{\parallel }\le \mathrm{\parallel }{T}^{\ast }\mathrm{\parallel }\mathrm{\parallel }T\mathrm{\parallel }=\mathrm{\parallel }T{\mathrm{\parallel }}^2$, while for $\mathrm{\parallel }x\mathrm{\parallel }\le 1$ one has $\mathrm{\parallel }Tx{\mathrm{\parallel }}^2=⟨x,T^{\ast }Tx{⟩}_K\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }{T}^{\ast }Tx\mathrm{\parallel }\le \mathrm{\parallel }{T}^{\ast }T\mathrm{\parallel }$, so $\mathrm{\parallel }T{\mathrm{\parallel }}^2\le \mathrm{\parallel }{T}^{\ast }T\mathrm{\parallel }$ and hence $\mathrm{\parallel }{T}^{\ast }T\mathrm{\parallel }=\mathrm{\parallel }T{\mathrm{\parallel }}^2$.