Accretive operator ($\mathrm{Re}⟨Tx,x⟩\ge 0$)

An operator on a complex Hilbert space is accretive when the displayed inequality holds throughout its domain. Its negative is a dissipative operator. For a bounded everywhere-defined operator, expansion gives $\Vert (I+T)x{\Vert }^2\ge \Vert x{\Vert }^2$. Its adjoint operator satisfies the same accretivity condition. Consequently $I+T$ has closed range and trivial kernel, and its range is dense because its adjoint has trivial kernel. Thus $I+T$ is onto and $\Vert (I+T{)}^{-1}\Vert \le 1$. Boundedness and the full domain are material here; for unbounded operators accretivity alone need not imply surjectivity.

For a bounded accretive operator, $I+T$ is invertible and the transform satisfies $\Vert S\Vert \le 1$. The identity $\Vert (I+T)x{\Vert }^2-\Vert (I-T)x{\Vert }^2=4\mathrm{Re}⟨Tx,x⟩$ proves contraction after the substitution $z=(I+T)x$. If the real part vanishes for every $x$, this identity proves norm preservation. Both $T$ and $-T$ are then accretive, making both $I+T$ and $I-T$ invertible; hence $S$ is onto and is a unitary operator.