Essentially self-adjoint operator

Remarks

The following conditions are equivalent for a symmetric operator $A$:

1. $A$ is essentially self-adjoint;
2. $A$ has exactly one self-adjoint extension;
3. $A^*$ is self-adjoint, and then $A^* = \overline{A}$;
4. the kernels $\ker(A^* - iI)$ and $\ker(A^* + iI)$ are both $\set{0}$.

Example: on $L^2(\R)$, let $A = -i \, d/dx$ on the smooth functions with compact support. Then $A^*$ is $-i \, d/dx$ on the Sobolev space $H^1(\R)$. The equation $A^* u = \pm i u$ means $u' = \mp u$, with solutions $c e^{\mp x}$, which are not in $L^2(\R)$ unless $c = 0$; so condition 4 holds. Non-example: on $L^2([0,1])$, let $B = -i \, d/dx$ on the smooth functions with compact support in $(0,1)$. Now $e^{-x}$ and $e^{x}$ lie in $L^2([0,1])$ and solve $B^* u = i u$ and $B^* u = -i u$, so condition 4 fails, and $B$ has many self-adjoint extensions, one for each boundary condition $u(1) = e^{i \theta} u(0)$. A self-adjoint operator is essentially self-adjoint; a core of a self-adjoint operator $T$ is a domain on which the restriction of $T$ is essentially self-adjoint.