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 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 $D(T)=D(T^*)$ ?

  • 5
    $\begingroup$ Yes, this is true. It can be shown directly by methods similar to those discussed in my lecture notes here, example 11.1: math.ou.edu/~cremling/teaching/lecturenotes/fa-new/ln11.pdf In this particular case though, it would be much quicker to note that $T$ is multiplication by $-k^2$ on its natural domain after taking Fourier transforms. $\endgroup$ Aug 14, 2021 at 15:33
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    $\begingroup$ Or you could observe that the deficiency indices of $T$ are zero. All these arguments are classical and "well known" (to those who know them well), so a walk to the library should really help you here. $\endgroup$ Aug 14, 2021 at 15:36
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    $\begingroup$ @ChristianRemling: "'well known' (to those who know them well)" Nominated for the Mathematical Aphorism of the Week Award! :-) $\endgroup$ Aug 14, 2021 at 15:59
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    $\begingroup$ For (1), use Cauchy-Schwarz instead to estimate $\int \psi'' \phi_n - \int \psi'' \tilde{\psi} = \int \psi'' (\phi_n - \tilde{\psi})$. Notice that $\phi_n \to \tilde{\psi}$ in $L^2$. For the other side, proceed similarly and note that $\phi_n'' \to \tilde{\psi}''$ in $L^2$ by virtue of $W^{2,2}$ convergence. DCT isn't needed here. $\endgroup$ Aug 14, 2021 at 20:26
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    $\begingroup$ The F.T. is a Hilbert space isomorphism on $L^2$ of the line. When I learned Sobolev spaces, they were defined as the inverse images of weighted $L^2$-spaces under the F.T. These are, in turn. the domains of definition of multiplication operators and it is a well-known standard fact that the latter are self-adjoint. This method , of course, works in a much more general setting than that of your question. $\endgroup$ Aug 15, 2021 at 9:53

1 Answer 1


In Theorem 4.7b) of these lecture notes, 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 $d=1$.


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