Let $W^{2,2}(\mathbb R)\newcommand{\R}{\mathbb R}\newcommand{\C}{\mathbb C}\newcommand{\N}{\mathbb N}$ denote the Sobolev space as defined in chapter 5 of [Evans' PDE book](https://bookstore.ams.org/gsm-19-r) and consider the linear operator


\begin{equation\*}\begin{split}T: D(T)&\to L^2(\mathbb R), \\ \phi&\mapsto \phi''.\end{split}\end{equation\*}


Here, $D(T):=W^{2,2}(\mathbb R)$ is a dense subset of $L^2(\mathbb R)$ and $T$ will be considered as a densily defined operator on $L^2(\mathbb R)$.




---


**My question.** Is $T$ self-adjoint? By self-adjoint I mean that $\langle T\psi, \tilde\psi\rangle = \langle \psi, T\tilde\psi\rangle$ for all $\psi,\tilde\psi\in D(T)$, and that $D(T)=D(T^\*)$, where $D(T^\*)$ is defined as the set of all $\tilde\psi\in L^2(\mathbb R)$ such that the linear operator $$T^\*\_{\tilde\psi}: D(T)\to\mathbb R, \psi\mapsto\langle{T\psi, \tilde\psi}\rangle$$ is bounded.




---


**My attempt.** A proof of the first property: Let $\psi, \tilde\psi\in W^{2,2}(\R)$. We thus need to show that
\begin{equation\*}
\int\_{\R} \psi''\tilde\psi=\int\_\R \psi\tilde\psi''.
\end{equation\*}


Let $(\phi\_n)\_{n\in\N}$ be a sequence of functions in $C\_{\text c}^\infty(\R)$ converging in $W^{2,2}$ to $\tilde\psi$. (Such a sequence exists, see for instance Lemma 23 of <https://terrytao.wordpress.com/2009/04/30/245c-notes-4-sobolev-spaces/.>)
Now, by Definition of the weak derivative,
\begin{equation}\label{eq:phi n prime prime}\tag{1}
\int\_{\R}\psi''\phi\_n = \int\_\R \psi\phi\_n''
\end{equation}
for all $n\in\N$. Now, from the Cauchy-Schwarz inequality, we get
\begin{equation\*}
 \left\lvert\int\_\R \psi'' \phi\_n - \int\_\R \psi'' \tilde\psi\right\rvert \le \int\_\R \lvert\psi''\rvert\lvert\phi\_n-\tilde\psi\rvert\le \lVert\psi''\rVert\_{L^2} \lVert\phi\_n-\tilde\psi\rVert\_{L^2}.
\end{equation\*}
Analogously,
\begin{equation\*}
 \left\lvert\int\_\R \psi \phi\_n''-\int\_\R \psi \tilde\psi''\right\rvert\le \int\_\R \lvert\psi\rvert \lvert\phi\_n''-\tilde\psi''\rvert\le \lVert\psi\rVert\_{L^2} \lVert\phi\_n''-\tilde\psi''\rVert\_{L^2}.
\end{equation\*}
But $\phi\_n\to \tilde\psi$ in $W^{2,2}$ implies that $\lVert\phi\_n-\tilde\psi\rVert\_{L^2}\to 0$ and $ \lVert\phi\_n''-\tilde\psi''\rVert\_{L^2}\to 0$ as $n\to\infty$. Therefore, we do indeed have
\begin{equation\*}
 \int\_{\R} \psi''\tilde\psi=\int\_\R\psi\tilde\psi''.
\end{equation\*}




---


However, how can one prove that $D(T)=D(T^\*)$ ?



Is the second weak derivative a self-adjoint operator?
In Theorem 4.7b) of [these lecture notes](https://www.math.kit.edu/iana3/%7Eschnaubelt/media/st-skript15.pdf), the following general result is proved:



> 
> Let $X$ be a Hilbert space and let $A$ be densely defined, closed and symmetric. Then $A$ is self-adjoint if and only if the spectrum of $A$ is contained in the real axis.
> 
> 
> 


In Example 4.8d) of the same lecture notes, this is applied to prove the self-adjointness of the Laplacian $A = \Delta$ on $L^2(\mathbb{R}^d)$ with domain $D(A) = W^{2, 2}(\mathbb{R}^d)$, which specializes to the second derivative for $d=1$.


