Symmetric operator

A linear mapping $A$ of a set $D_A$ in a Hilbert space $H$ (in general, complex) into $H$ such that $\langle Ax,y\rangle =\langle x,yA\rangle$ for all $x,y\in D_A$. If $D_A$ is an everywhere-dense linear manifold in $H$ (and this is assumed in what follows), then $A$ is a linear operator. If $D_A=H$, then $A$ is bounded and hence continuous on $H$. A symmetric operator $A$ induces a bilinear Hermitian form $B(x,y)=\langle Ax,y\rangle$ on $D_A$, that is, $B(x,y)=\overline{B(x,y)}$. The corresponding quadratic form $\langle Ax,x\rangle$ is real. Conversely, if the form $\langle Ax,x\rangle$ on $D_A$ is real, then $A$ is symmetric. The sum $A+B$ of two symmetric operators $A$ and $B$ with a common domain of definition $D_A=D_B$ is again a symmetric operator, while if $\lambda$ is a real number, then $\lambda A$ is also symmetric. Every symmetric operator $A$ has a uniquely defined closure $\overline{A}$ and an adjoint $A^* \supset\overline{A}$. In general, $A^*$ is not symmetric and $A^*\neq\overline{A}$. If $A^*=A$, then $A$ is called a self-adjoint operator. This holds, for example, in the case of symmetric operators defined on the whole of $H$. If $A$ is symmetric and bounded on $D_A$, then $A$ can be extended as a bounded symmetric operator to the whole of $H$.
1. Let $\|a_{ij}\|$, $i,j=1,2,\ldots$, be an infinite matrix such that $a_{ij}=\overline{a}_{ji}$, and \begin{equation}\sum_{i,j=1}^{\infty}|a_{ij}|^2<\infty .\end{equation} Then the system of equations \begin{equation}\eta_i=\sum_{j=1}^{\infty}a_{ij}\xi_j,\quad i=1,2,\ldots,\end{equation} defining $y=\{\eta_i\}$ for an $x=\{\xi_i\}\in l_2$, defines a bounded symmetric operator, which turns out to be self-adjoint on the complex space $l_2$.
2. In the complex space $L_2(0,1)$, let $A=id/dt$ be defined on the set $D_A$ of absolutely-continuous functions $x$ on $[0,1]$ having a square-summable derivative and satisfying the condition $x(0)=x(1)=0$. Then $A$ is symmetric but not self-adjoint.