Let $\mathscr{H}$ be a Hilbert space  and let $A\colon\mathscr{D}(A)\subset\mathscr{H}\to\mathscr{H}$ be a densely defined linear operator  . Suppose that for some $y\in\mathscr{H}$, there exists $z\in\mathscr{H}$ such that $(Ax,y)=(x,z)$ for all $x\in\mathscr{D}(A)$. Then such $z$ is unique, for if $z^{\prime}$ is another element of $\mathscr{H}$ satisfying that condition, we have $(x,z-z^{\prime})=0$ for all $x\in\mathscr{D}(A)$, which implies $z-z^{\prime}=0$ since $\mathscr{D}(A)$ is dense (http://planetmath.org/Dense). Hence we may define a new operator $A^{*}:\mathscr{D}(A^{*})\subset\mathscr{H}\to\mathscr{H}$ by
 $\displaystyle\mathscr{D}(A^{*})=$ $\displaystyle\{y\in\mathscr{H}:\text{there is}z\in\mathscr{H}\text{such that}(% Ax,y)=(x,z)\},$ $\displaystyle A^{*}(y)=$ $\displaystyle z.$
It is easy to see that $A^{*}$ is linear, and it is called the of $A$.
Remark. The requirement for $A$ to be densely defined is essential, for otherwise we cannot guarantee $A^{*}$ to be well defined.