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\begin{document}
%\renewcommand{\thefootnote}{$*$}
\title[]{\huge A Criterion for Essential Self-Adjointness}
\author{\large B. Thaller}
%\address{Erwin Schr\"odinger International Institute for Mathematical Physics,
%Pasteurgasse 6/7, A-1090 Vienna, Austria (Permanent address:
%Inst. f. Mathematik, Universit\"at Graz, Heinrichstra\ss e 36, A-8010 Graz, Austria)}
%\email{thaller@@bkfug.kfunigraz.ac.at}
\maketitle
\bigskip\bigskip
\begin{center}
{\sc Erwin Schr\"odinger International Institute for Mathematical Physics,
Pasteurgasse 6/7, A-1090 Vienna, Austria}
\end{center}
\bigskip\bigskip
\begin{abstract}
We prove some simple facts on the essential self-adjointness of a symmetric operator $T$ in a Hilbert space $\frak H$.
The main tools are bounded commutators of $T$ with a suitable sequence of bounded self-adjoint operators.
A special case is a Theorem of J\"orgens and Chernoff stating that the essential self-adjointness of the
Dirac operator in an external field depends only on the local properties of the
potential.
\end{abstract}
\vfill
\noindent {Permanent address: Institut f\"ur Mathematik, Universit\"at Graz, Heinrichstra\ss e 36, A-8010 Graz, Austria}\\
{\it E-mail address}: thaller@@bkfug.kfunigraz.ac.at
%%%%%%%%%%%%%%%%%%%%%%%%%-------------> Erzwungener Seitenwechsel
\newpage
%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Assumptions and Main Results}\label{sec1}
%%%%%%%%%%%%%%%%%%%%%%%%%
%
\setcounter{equation}{0}
\setcounter{prop}{0}
\bigskip
\noindent Let $\frak H$ be a separable Hilbert space.
$\{A_n\}_{n=0}^\infty$ denotes a sequence of bounded and self-adjoint operators on $\frak H$.
We consider a linear operator $T$ in $\frak H$, densely defined and symmetric on
$\Dom(T)=\Dom_0 \subset \frak H$
.
\begin{nota}
If a linear operator $S$ is defined and closeable on $\Dom_0$ we denote its adjoint by $S^*$
and its closure $S^{**}$ by $S^c$.
The restriction to $\Dom_0$ of an operator $B$ defined on some larger domain
is occasionally denoted by $B|_{\Dom_0}$.
\end{nota}
\smallskip
For the operators $A_n$ and $T$ we formulate the following conditions:
\medskip
\begin{quote}
\begin{enumerate}
\item[$\bold{A_1}$:\quad]
For all $n$ there exist $k$, $m$ with
\end{enumerate}
\end{quote}
%
\vspace{-.2cm}
\begin{equation}
A_kA_n=A_n,\qquad A_mA_k=A_k.
\end{equation}
\smallskip
\begin{quote}
\begin{enumerate}
\item[$\bold{A_2}$:\quad]
For all $n$, $[T,A_n] \equiv TA_n - A_nT$ is defined on $\Dom_0$ and bounded.
\medskip
\item[$\bold{A_3}$:\quad]
For all $n$ the operator ${A_nTA_n}|_{\Dom_0}$ is essentially self-adjoint.
\medskip
\item[$\bold{A_4}$:\quad]
For all $\psi\in \Dom(T^*)$, there is a subsequence $\{A_{n_k}\}_{k=0}^\infty$ converging weakly to ${\bold 1}$ such that
\end{enumerate}
\end{quote}
\vspace{-.1cm}
\begin{equation}
\lim_{k\to \infty}(T^*A_{n_k}\psi,\psi-A_{n_k}\psi) = 0.
\end{equation}
\smallskip
\noindent One of the main results which can be proved under the conditions above (or under similar conditions discussed
below) is the following:
%%%%%%%%%%%%%%%%%%%%%-THEOREM
%
\begin{thm}\label{th1}
Assume $\bold{A_1}$--\,$\bold{A_4}$. Then $T$ is essentially self-adjoint on $\Dom_0$.
\end{thm}
%
%%%%%%%%%%%%%%%%%%%%%
\smallskip
\noindent
As an illustration consider a Hilbert space which is an infinite orthogonal sum of closed subspaces.
Define
%
\begin{equation}
\frak{H} = \bigoplus_{k=0}^\infty \frak{H}^{(k)}\, ,\qquad
\frak{H}_n = \bigoplus_{k=0}^n \frak{H}^{(k)} \, ,
\label{B1}
\end{equation}
%%
and let $A_n$ be the projection onto $\frak{H}_n$.
Hence $A_n$ converges (strongly) to $\bold 1$ and Assumption $\bold{A_1}$ is trivially satisfied.
Let $T$ be symmetric on some dense subset $\Dom_0$.
Then Theorem \ref{th1} gives conditions which assure that the essential self-adjointness of the
restrictions $A_nTA_n$ of $T$
is equivalent to the essential self-adjointness of $T$ (the equivalence follows with the help of Theorem~\ref{th3}
and Remark~\ref{r3} below).
Another example is provided by the Dirac operator in the Hilbert space $L^2(\Bbb{R}^3)^4$.
The theorem contains as a special case a famous result of Chernoff \cite{1,2} and J\"orgens \cite{3} stating that
the essential self-adjointness of the Dirac operator on $\cal{C}_0^\infty(\Bbb{R}^3)^4$
depends only on the local properties of the potential.
In this case the $A_n$ are multiplication operators with suitable $\cal{C}_0^\infty$-functions.
See Section~4 for a discussion and a simplified proof of this result.
We conclude this section with some remarks on the assumptions.
%%%%%%%%%%%%%%%%
\begin{rem}
Assumption $\bold{A_1}$ is satisfied, e.g., if the $A_n$ form an increasing sequence: $A_mA_n=A_n$ for all $m>n$.
One might think of a sequence of projections, but we do not require $A_n^2=A_n$.
From $A_n^*=A_n$ for all $n$ we easily conclude
that the operators $A_n$, $A_m$, and $A_k$ in $\bold{A_1}$ commute.
\end{rem}
%%%%%%%%%%%%%%%%
\begin{rem}
The commutator $[T,A_n]$ is assumed to be defined on its natural domain. Since $A_n$ is defined on all of $\frak H$,
we have $\Dom([T,A_n]) = \Dom(TA_n) \cap \Dom(T)$.
Assumption $\bold{A_2}$ requires that $\Dom([T,A_n])=\Dom_0=\Dom(T)$ which means
%
\begin{equation}
A_n\Dom_0 \subset \Dom_0.
\label{R2}
\end{equation}
Boundedness of $[T,A_n]$ on a dense domain implies the existence of
a unique bounded extension $[T,A_n]^c$ to all of $\frak H$.
\end{rem}
%%%%%%%%%%%%%%%%%%%%%
\begin{rem}
Theorem \ref{th1} can be formulated as a perturbation theoretic result.
Let $T=H_0+V$, where $H_0$ is essentially self-adjoint and $V$ is symmetric on $\Dom_0$.
If we replace $\bold{A_3}$ by the assumption that $H_0+A_nVA_n$ be essentially self-adjoint on $\Dom_0$, then
we may conclude the essential self-adjointness of $T$. See Section~5 for details.
\end{rem}
%%%%%%%%%%%%%%%%
\begin{rem}
The expression $T^*A_{n_k}\psi$ occurring in
$\bold{A_4}$ is well defined for all $\psi \in \Dom(T^*)$, provided $\bold{A_2}$ holds
(see Lemma~\ref{Lemma2} below). If $\bold{A_4}$ holds, then
\begin{equation}
\|A_{n_k}\| \le K,
\label{R2'}
\end{equation}
where the constant $K$ is independent of $k$ (any weakly convergent sequence of operators is bounded).
If the commutators $\,[T,A_{n_k}]^c$
are also bounded uniformly in $k$, then we can replace $\bold{A_4}$ by a more convenient assumption:
\end{rem}
%%%%%%%%%%%%%%%%%%%%%-LEMMA
%
\begin{lem}\label{lem1}
Assume $\bold{A_2}$. Then $\bold{A_4}$ is implied by
\end{lem}
\begin{quote}
\begin{enumerate}
\item[$\bold{A'_4}$:\quad]
For all $\psi\in \frak H$ there is a subsequence $\{A_{n_k}\}_{k=0}^\infty$ converging strongly to ${\bold 1}$ such that
$$
\|\,[T, A_{n_k}]^c \,\psi\| \le C(\psi),
$$
where $C(\psi)>0$ is independent of $k$.
\end{enumerate}
\end{quote}
%
%%%%%%%%%%%%%%%%%%%%%
\smallskip
\noindent By the uniform boundedness principle, Assumption $\bold{A'_4}$ implies
$\|\,[T,A_{n_k}]^c\| \le K'$, uniformly in $k$.
%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Lemmas and Proofs}\label{sec2}
%%%%%%%%%%%%%%%%%%%%%%%%%
%
\setcounter{equation}{0}
\setcounter{prop}{0}
\medskip
\noindent We start by proving some elementary technical lemmas.
As always we assume that $A_n$ is self-adjoint
and bounded and that $T$ is symmetric on $\Dom_0$.
\smallskip
%%%%%%%%%%%%%%%%%%%%%-LEMMA
%
\begin{lem}\label{Lemma2}
Assume $\bold{A_2}$. Then $A_n \Dom(T^*) \subset \Dom(T^*)$ and
%
\begin{equation}
[T, A_n]^c \psi = T^*A_n\psi - A_nT^*\psi\qquad\text{for all $\psi\in\Dom(T^*)$.}
\label{E2.4}
\end{equation}
%%
\end{lem}
\smallskip
%%%%%%%%%%%%%%%%%%%%%-PROOF
%
\begin{pf}
Let $\psi\in \Dom(T^*)$. For all $\phi\in\Dom_0$
%
\begin{equation}
(A_n\psi,T\phi) = (\psi,(TA_n-[T,A_n])\phi) = ((A_nT^* - [T,A_n]^*)\psi,\phi).
\label{E2.5}
\end{equation}
%%
Hence, by the definition of the adjoint operator, $A_n\psi\in \Dom(T^*)$ and
%
\begin{equation}
T^*A_n\psi=(A_nT^* - [T,A_n]^*)\psi.
\label{E2.6}
\end{equation}
%%
Since $\rmi[T,A_n]$ is obviously symmetric on $\Dom_0$, its bounded extension is self-adjoint, i.e.,
%
\begin{equation}
[T,A_n]^* = - [T,A_n]^c.
\label{E2.7}
\end{equation}
%%
Combining this with Eq.~(\ref{E2.6}) completes the proof of the Lemma \ref{Lemma2}.
\end{pf}
\medskip
%%%%%%%%%%%%%%%%%%%%%-LEMMA
%
\begin{lem}\label{Lemma3}
Assume $\bold{A_1}$--\,$\bold{A_3}$. Then $A_n \Dom(T^*) \subset \Dom(T^c)$ and
%
\begin{equation}
T^cA_n\psi = (A_mTA_m)^cA_n\psi + [T,A_m]^cA_n\psi\qquad\text{for all $\psi\in\Dom(T^*)$,}
\label{E2.8}
\end{equation}
%%
where $m$ is chosen according to Assumption $\bold{A_1}$.
\end{lem}
\smallskip
%%%%%%%%%%%%%%%%%%%%%-PROOF
%
\begin{pf}
Let $\psi\in \Dom(T^*)$.
By Lemma \ref{Lemma2}, $A_n\psi\in\Dom(T^*)$.
Choose $m$, $k$ according to $\bold{A_1}$ and note that $A_mA_n = A_mA_kA_n = A_kA_n = A_n$.
Hence for $\phi\in\Dom_0$,
%
\begin{equation}
(A_n\psi,A_mTA_m\phi)=(A_mT^*A_n\psi,\phi).
\label{E2.9}
\end{equation}
%%
Since by Assumption $\bold{A_3}$
%
\begin{equation}
(A_mTA_m|_{\Dom_0})^* = (A_mTA_m|_{\Dom_0})^c,
\label{E2.11}
\end{equation}
%%
Eq.~(\ref{E2.9}) implies
%
\begin{equation}
A_n\psi \in \Dom\bigl( (A_mTA_m|_{\Dom_0})^c \bigr).
\label{E2.12}
\end{equation}
%%
Hence there is a sequence $\{\chi_j\}_{j=0}^\infty$ with
$\chi_j \in \Dom_0$ for all $j$, $\lim\chi_j=A_n\psi$, such that
$\{A_mTA_m\chi_j\}_{j=0}^\infty$ is convergent.
But then the sequence $\{\xi_j\}_{j=0}^\infty$ with $\xi_j=A_k\chi_j$ has the same
properties:
$\xi_j \in \Dom_0$ and (by continuity of $A_n$) $\lim\xi_j$ exists. Furthermore
%
\begin{equation}
A_mTA_m\xi_j = A_kA_mTA_m\chi_j + A_m[T,A_k]A_m\chi_j,
\label{E2.13}
\end{equation}
%%
which converges, as $j\to\infty$. Therefore
%
\begin{equation}
T\xi_j = TA_m\xi_j = A_mTA_m\xi_j + [T,A_m]\xi_j
\label{E2.14}
\end{equation}
%%
again converges in $\frak H$, as $j\to\infty$. Hence $A_n\psi=\lim\xi_j\in\Dom(T^c)$
and
%
\begin{equation}
T^cA_n\psi = \lim_{j\to\infty} T \xi_j,
\label{E2.15}
\end{equation}
%%
which is just what we wanted to prove.
\end{pf}
\medskip
%%%%%%%%%%%%%%%%%%%%%-PROOF
%
\noindent {\it{Proof of Theorem \ref{th1}}}.
Let $\psi\in\Dom(T^*)$. By Lemma \ref{Lemma3}, $A_n\psi\in\Dom(T^c)$, and
%
\begin{equation}
(\psi, T^*\psi) = (T^cA_{n_k}\psi, A_{n_k}\psi)
+ (T^cA_{n_k}\psi,\psi-A_{n_k}\psi) +
(\psi-A_{n_k}\psi,T^*\psi).
\label{E2.16}
\end{equation}
%%
Using $\bold{A_4}$ we find
%
\begin{equation}
|(T^cA_{n_k}\psi,\psi-A_{n_k}\psi)| + |(\psi-A_{n_k}\psi,T^*\psi)| @>>{n\to\infty}> 0.
\label{E2.18}
\end{equation}
%%
Note that
%
\begin{equation}
(T^cA_{n_k}\psi, A_{n_k}\psi)=(A_{n_k}\psi, T^cA_{n_k}\psi)
\label{E2.17}
\end{equation}
%%
is real, because the closure of a symmetric operator is always symmetric.
We conclude that
%
\begin{equation}
(\psi, T^*\psi) = \lim_{k\to\infty}(A_{n_k}\psi, T^cA_{n_k}\psi)
\label{E2.19}
\end{equation}
%%
is real as a limit of real numbers. Hence (by the polarization identity)
$T^*$ is symmetric which proves that $T$ is essentially self-adjoint.\qed
%
%%%%%%%%%%%%%%%%%% END PROOF
\medskip
%%%%%%%%%%%%%%%%%%%%%-PROOF
%
\noindent {\it{Proof of Lemma \ref{lem1}}}.
Assuming $\bold{A_2}$ we find with the help of Lemma \ref{Lemma2}
that for all $\psi\in\Dom(T^*)$
%
\begin{equation}
\|T^*A_{n_k}\psi\| \le \|A_{n_k}\|\,\|T^*\psi\| + \|\,[T,A_{n_k}]^c\,\psi\|.
\label{E2.20}
\end{equation}
%%
Using Eq~(\ref{R2'}) and
$\bold{A'_4}$ we find
%
\begin{equation}
\|T^*A_{n_k}\psi\| \le K \|T^*\psi\| + C(\psi) \equiv C_1(\psi).
\label{E2.21}
\end{equation}
%%
Hence we obtain
%
\begin{equation}
|(T^cA_{n_k}\psi,\psi-A_{n_k}\psi)| \le C_1(\psi)\|\psi-A_{n_k}\psi\|,
\label{E2.21a}
\end{equation}
%%
which tends to zero because the sequence $A_{n_k}$ is assumed to converge strongly to $\bold 1$.
This proves Lemma \ref{lem1}.\qed
%
%%%%%%%%%%%%%%%%%% END PROOF
%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Some Remarks and Further Results}\label{sec3}
%%%%%%%%%%%%%%%%%%%%%%%%%
%
\setcounter{equation}{0}
\setcounter{prop}{0}
\medskip
\noindent %
\begin{rem}
A slight modification of the proof of Theorem \ref{th1} allows to replace the condition
$\bold{A_4}$ resp. $\bold{A_4'}$ with, e.g.,
\end{rem}
\begin{quote}
\begin{enumerate}
\item[$\bold{B_4}$:\quad]
For some subsequence $\{A_{n_k}\}_{k=0}^\infty$ and for all $\psi\in \frak H$,
%
$$\sum_{k=1}^\infty A_{n_k}={\bold 1},$$
where the series converges in the strong sense, and
%
$$\|\sum_{k=1}^j [T, A_{n_k}]^c\psi\|\le C(\psi),$$
where $C(\psi)>0$ is independent of $j$.
\end{enumerate}
\end{quote}
\smallskip
\noindent Repeat the calculation Eqs.~(\ref{E2.16})--(\ref{E2.21a}) with
$A_n$ replaced by
$\tilde{A}_j \equiv \sum_{k=1}^j A_{n_k}$
to arrive at the conclusion of Theorem \ref{th1}.
\medskip
\begin{rem}
An immediate consequence of the essential self-adjointness of $T$ on $\Dom_0$
and the boundedness of the commutator is the
following.
\end{rem}
\medskip
%%%%%%%%%%%%%%%%%%%%%-LEMMA
%
\begin{cor}\label{Lemma4}
Assume $\bold{A_2}$ and let $T$ be essentially self-adjoint on $\Dom_0$.
Denote
%
\begin{equation}
\psi(t) \equiv \exp(- {\rom{\rmi}} t T^c)\,\psi, \quad\text{for all $\psi\in\frak H$.}
\label{E2.22}
\end{equation}
%%
Then for all $\psi$, $\phi\in \frak H$ the function $t\to(\psi(t),A_n\phi(t))$ is continuously differentiable with
\vspace{-.5cm}
%
\begin{equation}
\frac{d}{dt} (\psi(t),A_n\phi(t)) = (\psi(t),{\rom{\rmi}} [T,A_n]^c\phi(t)).
\label{E2.23}
\end{equation}
%%
\end{cor}
\smallskip
%%%%%%%%%%%%%%%%%%%%%-PROOF
%
\begin{pf} For convenience we give a proof of this simple fact:
For $\psi$, $\phi\in \frak H$ choose $\{\psi_j\}$, $\{\phi_j\}$ in $\Dom_0$ with
$\psi=\lim\psi_j$, $\phi=\lim\phi_j$. By the strong continuity of the unitary group,
$f_j(t) \equiv (\psi_j(t),A_n\phi_j(t))$
is continuous, and by $\bold{A_2}$ even continuously differentiable with
$f_j'(t) \equiv (\psi_j(t),\rmi [T,A_n]\phi_j(t))$.
Define $g(t)\equiv (\psi(t), \rmi [T,A_n]^c\phi(t))$. Then
%
\begin{equation}
|f_j'(t)-g(t)| \le
\|[T,A_n]^c\|( \|\phi_j\|\|\psi_j-\psi\|+\|\psi\|\|\phi_j-\phi\|) < \varepsilon,
\label{E2.25}
\end{equation}
%%
where $\varepsilon$ can be chosen independently of $t$ and is arbitrarily small
for $j$ large. Hence we can exchange the differentiation and the limit to conclude
that $f(t)=\lim f_j(t)$ is differentiable with $f'(t)=\lim f_j'(t)$.
\end{pf}
%
%%%%%%%%%%%%%%%%%%%% END PROOF
\medskip\noindent
The following theorem is a converse of Theorem~\ref{th1}.
We give conditions for the essential self-adjointness of
$A_nTA_n$ as a consequence of the self-adjointness of $T^c$.
First we state a lemma similar to Lemma \ref{Lemma2}.
\smallskip
%
%%%%%%%%%%%%%%%%%%%%%-LEMMA
%
\begin{lem}\label{Lemma1}
Assume $\bold{A_2}$. Then $A_n \Dom(T^c) \subset \Dom(T^c)$ and
%
\begin{equation}
[T, A_n]^c \psi = T^cA_n\psi - A_nT^c\psi\qquad\text{for all $\psi\in\Dom(T^c)$.}
\label{E2.1}
\end{equation}
%%
\end{lem}
\smallskip
%%%%%%%%%%%%%%%%%%%%%-PROOF
%
\begin{pf}
By the definition of closure, for any $\psi\in\Dom(T^c)$ there is a sequence
$\{\psi_k\}_{k=0}^\infty$ with $\psi_k\in\Dom_0$, $\lim \psi_k =\psi$, and $\lim T\psi_k=T^c\psi$.
Using the continuity of $A_n$ and $[T,A_n]$ we find
%
\begin{equation}
TA_n\psi_k = A_nT\psi_k + [T,A_n]\psi_k @>>{k\to\infty}>
A_nT^c\psi + [T,A_n]^c\psi
\label{E2.2}
\end{equation}
%%
Hence the sequences $\{A_n\psi_k\}$ and $\{TA_n\psi_k\}$ both converge. This implies
%
\begin{equation}
\lim_{k\to\infty}A_n\psi_k=A_n\psi \in \Dom(T^c),\qquad
T^c A_n\psi = \lim_{k\to\infty}TA_n\psi_k.
\label{E2.3}
\end{equation}
%%
Now Eq.~(\ref{E2.1}) follows immediately.
\end{pf}
%
\medskip
%%%%%%%%%%%%%%%%%%%%%-THEOREM
%
\begin{thm}\label{th3}
Assume $\bold{A_2}$. Let $T$ be essentially self-adjoint on $\Dom_0$.
In addition we assume for all $\psi$ that
$A_n^2\psi\in \Dom(T^c)$ implies $A_n\psi\in\Dom(T^c)$.
Then $A_nTA_n|_{\Dom_0}$ is essentially self-adjoint.
\end{thm}
%
%%%%%%%%%%%%%%%%%%%%%-END THEOREM
%%%%%%%%%%%%%%%%%%%%%-PROOF
%
\begin{pf}
Denote $B=A_nTA_n|_{\Dom_0}$. Let $\psi\in \Dom(B^*)$, $\phi\in\Dom_0$. Then
%
\begin{equation}
(B^*\psi,\phi) = (\psi,A_nTA_n\phi) = (A_n^2\psi,T\phi) + ([T,A_n]^*A_n\psi,\phi)
\label{E2.26}
\end{equation}
%%
shows that $A_n^2\psi\in \Dom(T^*)=\Dom(T^c)$ and
%
\begin{equation}
T^cA_n^2\psi = B^*\psi - [T,A_n]^*A_n\psi.
\label{E2.27}
\end{equation}
%%
The additional assumption implies $A_n\psi\in\Dom(T^c)$ and
with Lemma \ref{Lemma1} we obtain
%
\begin{equation}
T^cA_n^2\psi = A_nT^cA_n\psi + [T,A_n]^cA_n\psi.
\label{E2.28}
\end{equation}
%%
Using Eq.~(\ref{E2.7}) we find that
%
\begin{equation}
B^*\psi = A_nT^cA_n\psi\quad\text{for all $\psi\in \Dom(B^*)$.}
\label{E2.29}
\end{equation}
%%
Hence $B^*$ is symmetric which is equivalent to the essential self-adjointness of $B$.
\end{pf}
\begin{rem}\label{r3}
The additional assumption in Theorem~\ref{th3} is trivially satisfied if the operators $A_n$ are projections.
\end{rem}
%%%%%%%%%%%%%%%%%%%%%%%%%
\section{An Example: Dirac Operators}\label{sec4}
%%%%%%%%%%%%%%%%%%%%%%%%%
%
\setcounter{equation}{0}
\setcounter{prop}{0}
\medskip
\noindent
In the Hilbert space ${\frak H}= L^2({\Bbb R}^3)^4$ of ${\Bbb C}^4$-valued square integrable functions on ${\Bbb R}^3$
we define the free Dirac operator as the closure of the operator
%
\begin{equation}
H_0 = -\rmi \alpha\cdot\nabla + \beta \quad\text{on}\quad
\Dom_0 = {\cal C}_0^\infty({\Bbb R}^3\setminus \{0\})^4.
\label{E3.1}
\end{equation}
%%
Here $\alpha = (\alpha_1,\alpha_2,\alpha_3)$ and $\beta$ are the Hermitian $4\times 4$ Dirac matrices (see \cite{4} for details).
It is well known that $H_0$ is essentially self-adjoint on $\Dom_0$, its closure $H_0^c$ is
self-adjoint on the first Sobolev space
%
\begin{equation}
\Dom(H_0^c) = H^1({\Bbb R}^3)^4 \equiv
\left\{\psi\in L^2({\Bbb R}^3)^4\mid \alpha\cdot\nabla \psi\in L^2({\Bbb R}^3)^4\right\}.
\label{E3.2}
\end{equation}
%%
Here $\nabla$ denotes the distributional derivative.
We denote by $g$ the bounded operator of multiplication by a function $g\in {\cal C}_0^\infty({\Bbb R}^3)$,
%
\begin{equation}
\psi\to g\psi,\quad(g\psi)(x) = g(x)\psi(x),\quad \text{all $\psi\in{\frak{H}}$, all $x\in{\Bbb R}^3$.}
\label{E.3.3}
\end{equation}
%%
It is easy to see that
%
\begin{equation}
[H_0,g]^c = -\rmi\alpha\cdot(\nabla g)
\label{E3.4}
\end{equation}
%%
is a bounded operator of multiplication with a Hermitian matrix-valued ${\cal C}_0^\infty$-function.
\smallskip
%
%%%%%%%%%%%%%%%%%%%%%-LEMMA
%
\begin{lem}\label{Lemma3.1}
The operator $gH_0g$ is essentially self-adjoint on $\Dom_0$.
%%
\end{lem}
%%%%%%%%%%%%%%%%%%%%%-PROOF
%
\begin{pf}
Since $\beta$ is a bounded operator in ${\frak{H}}$,
it is sufficient to consider the operator
%
\begin{equation}
T_0 \equiv -\rmi \alpha\cdot\nabla
\label{E3.2a}
\end{equation}
%%
when investigating the self-adjointness-properties. Again, $T_0$ is essentially self-adjoint on $\Dom(T_0) = \Dom_0$,
self-adjoint on $\Dom(T_0^c) = \Dom (H_0^c)$, and $[T_0,g]^c=[H_0,g]^c$.
Our proof uses the self-adjointness of $T_0^c$ on $H^1({\Bbb R}^3)^4$ and the
boundedness of $[T_0,g]^c$. Denote $B=gT_0g|_{\Dom_0}$.
As in the calculation leading to Eq.~(\ref{E2.27}) we conclude for all $\psi\in\Dom(B^*)$
that $g^2\psi\in \Dom(T_0^*)=\Dom(T_0^c)$
and
%
\begin{equation}
B^*\psi = -\rmi\alpha\cdot\nabla g^2\psi + \rmi g\alpha\cdot(\nabla g) \psi.
\label{E3.5}
\end{equation}
%%
By the Leibniz rule which holds for the product of a distribution with a smooth function we obtain
%
\begin{equation}
-\rmi\alpha\cdot\nabla g^2\psi = -2\rmi g\alpha\cdot(\nabla g)\psi-\rmi g^2 \alpha\cdot\nabla \psi.
\label{E3.6}
\end{equation}
%%
This shows that for all $\psi\in \Dom(B^*)$ the distribution
$\alpha\cdot\nabla\psi\in H^{-1}({\Bbb R}^3)^4$ satisfies
$g^2\alpha\cdot\nabla\psi\in L^2({\Bbb R}^3)^4$.
Therefore we can perform the following calculation for arbitrary $\psi,\phi\in\Dom(B^*)$
%
\begin{align}
(\phi, B^*\psi) &= (\phi, -\rmi\alpha\cdot\nabla g^2\psi + \rmi g\alpha\cdot(\nabla g) \psi)\\
&= (-\rmi g^2\alpha\cdot\nabla \phi, \psi) + (-\rmi g\alpha\cdot(\nabla g)\phi, \psi)\\
&= (-\rmi \alpha\cdot\nabla g^2\phi + \rmi g\alpha\cdot(\nabla g) \phi,\psi) = (B^*\phi, \psi)
\label{E3.7}
\end{align}
%%
Hence $B^*$ is symmetric, i.e., $B$ is essentially self-adjoint.
\end{pf}
\medskip\noindent
Now, let consider a function $f\in {\cal C}_0^\infty({\Bbb R})$ with the properties $f(0)=1$, and $f(r) = 0$ if $r\ge 1$. Define
the sequence of multiplication operators $A_n$, $n=1,2,\ldots$ by
%
\begin{equation}
\bigl(A_n\psi\bigr)(x) = \begin{cases}
\psi(x) & \text{if $|x|\le n$,}\\
f(|x|-n)\psi(x) & \text{if $|x| \ge n$.}
\end{cases}
\label{E3.8}
\end{equation}
%%
The sequence $\{A_n\}$ satisfies $\bold{A_1}$ and $\bold{A_4'}$,
the commutators $[H_0,A_n]^c$ are bounded uniformly in $n$ by sup$|f'(r)|$.
By Lemma~\ref{Lemma3.1} all the operators $A_nH_0A_n$ are essentially self-adjoint on $\Dom_0$.
Hence the following corollary is an immediate consequence of Theorem~\ref{th1}
%
%%%%%%%%%%%%%%%%%%%%%-Corollary
%
\begin{cor}\label{Cor4.1}
Let $A_n$ be defined as above and let $V$ be a symmetric operator on $\Dom_0$ such that $\bold{A_4}$ holds with $T=V$,
and assume that $A_n (H_0 +V) A_n$ is essentially self-adjoint on $\Dom_0$.
Then $H_0 + V$ is essentially self-adjoint on $\Dom_0$.
%%
\end{cor}
%
\medskip\noindent
The assumptions of the corollary are trivially satisfied,
if $V$ is multiplication by a locally bounded Hermitian matrix-valued function (no matter how fast it grows at infinity),
because in this case $A_nVA_n$ is a bounded perturbation of the essentially self-adjoint operator $A_nH_0A_n$.
As noted by Chernoff \cite{2} this result is in marked contrast to the situation for the second-order Schr\"odinger operator
and is related to the existence of a limiting velocity for the propagation of wavepackets according to the Dirac equation.
Because of the Kato-Rellich theorem only the relative boundedness of $A_nVA_n$
with respect to $A_nH_0A_n$ is needed and one could also consider potentials
with local singularities. Other examples include nonlocal potentials. See
\cite{3,4} for details.
A variant of the preceeding proof
is obtained by using a partition of unity $\{f_n\}$ on ${\Bbb R}^3$ with $\sup_{x,n}|\nabla f_n(x)|\le M<\infty$
to define operators $A_n$ satisfying $\bold{B_4}$.
%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Perturbation Theory}\label{sec5}
%%%%%%%%%%%%%%%%%%%%%%%%%
%
\setcounter{equation}{0}
\setcounter{prop}{0}
\medskip
\noindent
Let $H_0$ and $V$ be symmetric operators defined on a dense subset $\Dom_0$ of some Hilbert space $\frak H$.
Let $\{A_n\}_{n=1}^\infty$ be a sequence of bounded self-adjoint operators satisfying $\bold{A_1}$.
Assume that $\bold{A_2}$ holds with $T$ replaced by $H_0$ and $V$, respectively.
Instead of $\bold{A_3}$ let us now assume
\begin{quote}
\begin{enumerate}
\item[$\bold{C_3}$:\quad]
$H_0$ is essentially self-adjoint on $\Dom_0$, and for all $n$ the operator $H_0 + {A_nVA_n}$
is essentially self-adjoint on $\Dom_0$ with
\end{enumerate}
\end{quote}
%
\vspace{-.1cm}
\begin{equation}
\Dom\bigl( (H_0 + A_nVA_n)^c \bigr) = \Dom(H_0^c).
\label{E4.1}
\end{equation}
%%
The operator $T\equiv H_0 + V$ is well defined and symmetric on $\Dom_0$ and we assume that it satisfies $\bold{A_4}$.
%%%%%%%%%%%%%%%%%%%%%-THEOREM
%
\begin{thm}\label{th4}
Under the above assumptions, $T$ is essentially self-adjoint on $\Dom_0$.
\end{thm}
%
%%%%%%%%%%%%%%%%%%%%%
\smallskip
With the help of the following lemma, the proof of Theorem~\ref{th4} is an easy modification of the proof of Theorem~\ref{th1}.
%%%%%%%%%%%%%%%%%%%%%-LEMMA
%
\begin{lem}\label{lemma4.1}
Assume $\bold{A_1}$, $\bold{C_3}$, and $\bold{A_2}$ with $H_0$ and $V$.
Then $\psi\in \Dom\bigl(T^*\bigr)$ implies that $A_n\psi\in\Dom\bigl(T^c\bigr)$ and
%
\begin{equation}
T^cA_n\psi = H_0^cA_n\psi + V^cA_n\psi,
\quad\text{for all $\psi\in \Dom\bigl(T^*\bigr)$.}
\label{E4.2}
\end{equation}
%%
\end{lem}
%
%%%%%%%%%%%%%%%%%%%%%
\begin{pf}
Let $\psi\in \Dom\bigl(T^*\bigr)$. Since $T$ satisfies $\bold{A_2}$ we find with
Lemma~\ref{Lemma2} that $A_n\psi\in\Dom\bigl(T^*\bigr)$. Hence for $\phi\in \Dom_0$ we obtain using $A_mA_n=A_n$
%
\begin{equation}
(T^* A_n\psi,\phi) = (A_n\psi, \{ T_m - [V,A_m] \}\phi),
\qquad T_m\equiv H_0 + A_mVA_m.
\label{E4.3}
\end{equation}
%%
This shows that $A_n\psi\in \Dom(T_m^*) = \Dom(T_m^c) = \Dom (H_0^c)$, where we have used $\bold{C_3}$.
By definition of closure, there is a sequence $\chi_j\in \Dom_0$ with $\chi_j\to A_n\psi$ and $H_0^c\chi_j \to H_0^cA_n\psi$,
i.e., $\{\chi_j\}$ converges in the Hilbert space $\Dom(H_0)$ equipped with the graph norm
$\|\psi\|_0^2 = \|H_0^c\psi\|^2 + \|\psi\|^2$. Since $T_m^c$ is closed on $(\Dom(H_0),\|\cdot\|_0)$,
it is bounded and hence the sequence $\{T_m^c\chi_j\}$ is again convergent.
As in the proof of Lemma~\ref{Lemma3} we can replace $\chi_j$ by the sequence $\xi_j=A_k\chi_j$,
which has the same properties, if $A_k$ is chosen according to $\bold{A_1}$.
In particular, the sequences $\{T_m\xi_j\}$ and $\{H_0\xi_j\}$ are convergent.
But then
%
\begin{equation}
V \xi_j = V A_m \xi_j = A_mVA_m \xi_j + [V,A_m] \xi_j = (T_m-H_0) \xi_j + [V,A_m]^c \xi_j
\label{E4.4}
\end{equation}
%%
is convergent, i.e., $\lim\xi_j = A_n\psi\in\Dom(V^c)$. Finally,
%
\begin{equation}
T\xi_j = H_0\xi_j + V\xi_j \to H_0^c A_n\psi + V^cA_n\psi,
\label{E4.5}
\end{equation}
%%
which implies $A_n\psi\in \Dom(T^c)$ together with Eq.~(\ref{E4.2}).
\end{pf}
\smallskip
%%%%%%%%%%%%%%%%%%%%%-THEOREM
%
\begin{thm}\label{th5}
Let $H_0$ be essentially self-adjoint on $\Dom_0$ and $X$ be self-adjoint on $\Dom(X)$, such that
$[H_0^c,X]$ is well defined on $\Dom_0$ and bounded.
Let $V$ be a real-valued function on $\Bbb R$, which is locally bounded.
Define $V(X) = \int V(\lambda)\,dE_X(\lambda)$ and assume $\Dom_0\subset \Dom(V(X))$.
Then $H_0+V(X)$ is essentially self-adjoint on $\Dom_0$.
\end{thm}
%
%%%%%%%%%%%%%%%%%%%%%
\begin{pf}
Let $g$ be a real-valued function, which can be written as the Fourier-transform of a function $\tilde g$, such
that ${(1+|\xi|)\tilde g(\xi)}$ is integrable:
%
\begin{equation}
g(\lambda) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\rmi \lambda \xi}\,\tilde g(\xi)\,d\xi.
\label{E4.A}
\end{equation}
%%
Define $g(X)$ by the weak integral
%
\begin{equation}
g(X)\psi = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\rmi X \xi}\psi\,\tilde g(\xi)\,d\xi,\quad\text{for all $\psi\in\frak H$}.
\label{E4.B}
\end{equation}
%%
Then
%
\begin{equation}
\|[H_0^c,g(X)]^c\| \le \frac{1}{\sqrt{2\pi}}\,\|[H_0^c,X]^c\|\int_{-\infty}^\infty |\xi|\,|\tilde g(\xi)|\,d\xi.
\label{E4.C}
\end{equation}
%%
Let $f\in {\cal C}_0^\infty(\Bbb R)$ be real-valued, with $f(\lambda)=1$, if $|\lambda|\le 1/2$ and $f(\lambda) = 0$, if $|\lambda|\ge 1$.
Define $A_n:= f(X/n)=\int_{-n}^n f(\lambda/n)\,dE_X(\lambda)$, where $E_X$ is the spectral measure of $X$.
Now it is easy to see that $T=H_0$ and $A_n$ satisfy the assumptions $\bold{A_1}$--$\bold{A_4}$, and even $\bold{A_4'}$.
By our assumptions,
$V(X) = \int V(\lambda)\,dE_X(\lambda)$ is a densely defined self-adjoint operator
which commutes with all $A_n$. Moreover, $A_nV(X)A_n$ is bounded and symmetric on $\Dom_0$.
Hence $H_0 +A_nV(X)A_n$ is essentially self-adjoint on $\Dom_0$.
Essential self-adjointness of $H_0 + V(X)$ now follows immediately from Theorem~\ref{th4}.
\end{pf}
\medskip
\begin{ack}
I would like to thank T. Hoffmann-Ostenhof for his kind invitation to the Erwin Schr\"odinger Institute,
where part of this work was done.
\end{ack}
\medskip
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\bibitem{1} P. R. Chernoff,
{\em Essential self-adjointness of powers of generators of hyperbolic equations}. J. Func. Anal. {\bf 12}, 401--414 (1973).
\bibitem{2} P. R. Chernoff,
{\em Schr\"odinger and Dirac operators with singular potentials and hyperbolic equations}. Pacific J. Math. {\bf 72}, 361--382 (1977).
\bibitem{3} K. J\"orgens,
{\em Perturbations of the Dirac operator}. In ``Proceedings of the conference on the theory of ordinary and partial differential equations,
Dundee (Scotland)'',
W. N. Everitt and B. D. Sleeman (eds.),
Lecture Notes in Mathematics {\bf 280}, 87--102, Springer Verlag, Berlin (1972).
\bibitem{4} B. Thaller, {\em The Dirac Equation}, Texts and Monographs in Physics, Springer Verlag, Berlin (1992).
\end{thebibliography}
\end{document}