ユニタリ作用素
‖ λ U ( x ) − U ( λ x ) ‖ 2 = ⟨ λ U ( x ) − U ( λ x ) , λ U ( x ) − U ( λ x ) ⟩ = ‖ λ U ( x ) ‖ 2 + ‖ U ( λ x ) ‖ 2 − ⟨ U ( λ x ) , λ U ( x ) ⟩ − ⟨ λ U ( x ) , U ( λ x ) ⟩ = | λ | 2 ‖ U ( x ) ‖ 2 + ‖ U ( λ x ) ‖ 2 − λ ̄ ⟨ U ( λ x ) , U ( x ) ⟩ − λ ⟨ U ( x ) , U ( λ x ) ⟩ = | λ | 2 ‖ x ‖ 2 + ‖ λ x ‖ 2 − λ ̄ ⟨ λ x , x ⟩ − λ ⟨ x , λ x ⟩ = 0 {\displaystyle {\begin{aligned}\|\lambda U(x)-U(\lambda x)\|^{2}&=\langle \lambda U(x)-U(\lambda x),\lambda U(x)-U(\lambda x)\rangle \\&=\|\lambda U(x)\|^{2}+\|U(\lambda x)\|^{2}-\langle U(\lambda x),\lambda U(x)\rangle -\langle \lambda U(x),U(\lambda x)\rangle \\&=|\lambda |^{2}\|U(x)\|^{2}+\|U(\lambda x)\|^{2}-{\overline {\lambda }}\langle U(\lambda x),U(x)\rangle -\lambda \langle U(x),U(\lambda x)\rangle \\&=|\lambda |^{2}\|x\|^{2}+\|\lambda x\|^{2}-{\overline {\lambda }}\langle \lambda x,x\rangle -\lambda \langle x,\lambda x\rangle \\&=0\end{aligned}}}
という計算が成り立つから、斉次性が従う。
‖ U ( x + y ) − ( U x + U y ) ‖ = 0 {\displaystyle \|U(x+y)-(Ux+Uy)\|=0}
も同様に示せるから、加法性も成り立つ。