= Accretive operator
{title2=$\operatorname{Re}\langle Tx,x\rangle\geq0$}
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 $\|(I+T)x\|^2\geq\|x\|^2$. 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 $\|(I+T)^{-1}\|\leq1$. Boundedness and the full domain are material here; for unbounded operators accretivity alone need not imply surjectivity.
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