# Is the second weak derivative a self-adjoint 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, $$$$\label{eq:phi n prime prime}\tag{1} \int_{\R}\psi''\phi_n = \int_\R \psi\phi_n''$$$$ for all $$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 $$D(T)=D(T^*)$$ ?

• 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. Aug 14, 2021 at 15:33
• 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. Aug 14, 2021 at 15:36
• @ChristianRemling: "'well known' (to those who know them well)" Nominated for the Mathematical Aphorism of the Week Award! :-) Aug 14, 2021 at 15:59
• 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. Aug 14, 2021 at 20:26
• 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. Aug 15, 2021 at 9:53

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 $$d=1$$.