adjointAdjoint52003-07-28 10:56:132006-06-15 19:02:16Definition% this is the default PlanetMath preamble. as your knowledge
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\newcommand{\Per}{\operatorname{Per}}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 $z'$ is another element of $\mathscr{H}$ satisfying that condition, we have $(x,z-z') = 0$ for all $x\in \mathscr{D}(A)$, which implies $z-z'=0$ since $\mathscr{D}(A)$ is \PMlinkname{dense}{Dense}. Hence we may define a new operator $A^*:\mathscr{D}(A^*)\subset\mathscr{H}\to\mathscr{H}$ by
\begin{align*}
\mathscr{D}(A^*) =& \{y\in \mathscr{H} : \text{there is} z\in \mathscr{H} \text{such that} (Ax,y) = (x,z)\},\\
A^*(y) =& z.
\end{align*}
It is easy to see that $A^*$ is linear, and it is called the \textbf{adjoint} of $A$.
\textbf{Remark.} The requirement for $A$ to be densely defined is essential, for otherwise we cannot guarantee $A^*$ to be well defined.