Let be a Hilbert space and let be a densely defined linear operator. Suppose that for some , there exists such that $(Ax,y) = (x,z)$ for all . Then such $z$ is unique, for if $z'$ is another element of satisfying that condition, we have $(x,z-z') = 0$ for all , which implies $z-z'=0$ since is dense. Hence we may define a new operator by

	there is   such that

Remark. The requirement for $A$ to be densely defined is essential, for otherwise we cannot guarantee $A^*$ to be well defined.