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

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

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 $n\in\N$. But note that $\lVert\phi_n\rVert_{L^2}\le\lVert\phi_n-\psi\rVert_{L^2}+\lVert\psi\rVert_{L^2}$ and $\lVert\phi_n''\rVert_{L^2}\le\lVert\phi_n''-\psi''\rVert_{L^2}+\lVert\psi''\rVert_{L^2}$ for all $n\in\N$. Now, from the Cauchy-Schwartz inequality, we get \begin{equation*} \left\lvert\int_\R \psi'' \phi_n - \int_\R \psi'' \bar{\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 $D(T)=D(T^*)$ ?

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

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. (Note that my application of the dominated convergence Theorem is fishy.) I think I can prove 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 $W^{2,2}$ and pointwise to $\tilde\psi$. (Such a sequence exists. For instance, by Lemma 23 of https://terrytao.wordpress.com/2009/04/30/245c-notes-4-sobolev-spaces/, $C_{\text c}^\infty(\R)$ is dense in $W^{2,2}(\R)$ in the $W^{2,2}$-sense. But by Evans, chapter 5.6.3, Theorem 6, this implies that $C_{\text c}^\infty(\R)$ is also dense in the "uniform convergence on every compact interval"-sense, which implies pointwise convergence.)

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 $n\in\N$. But note that $\lVert\phi_n\rVert_{L^2}\le\lVert\phi_n-\psi\rVert_{L^2}+\lVert\psi\rVert_{L^2}$ and $\lVert\phi_n''\rVert_{L^2}\le\lVert\phi_n''-\psi''\rVert_{L^2}+\lVert\psi''\rVert_{L^2}$ for all $n\in\N$. Therefore, by Lebesgue's dominated convergence Theorem (we have $\lvert\psi''\tilde\psi\rvert\le\frac12(\lvert\psi''\rvert^2+\lvert\tilde\psi\rvert^2)$ as well as $\lvert\psi\tilde\psi''\rvert\le\frac12(\lvert\psi\rvert^2+\lvert\tilde\psi''\rvert^2)$), we see by taking $n\to\infty$ on both sides of \eqref{eq:phi n prime prime} that \begin{equation*} \int_{\R} \psi''\tilde\psi=\int_\R\psi\tilde\psi'' \end{equation*} holds.

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

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 $n\in\N$. But note that $\lVert\phi_n\rVert_{L^2}\le\lVert\phi_n-\psi\rVert_{L^2}+\lVert\psi\rVert_{L^2}$ and $\lVert\phi_n''\rVert_{L^2}\le\lVert\phi_n''-\psi''\rVert_{L^2}+\lVert\psi''\rVert_{L^2}$ for all $n\in\N$. Now, from the Cauchy-Schwartz inequality, we get \begin{equation*} \left\lvert\int_\R \psi'' \phi_n - \int_\R \psi'' \bar{\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 $D(T)=D(T^*)$ ?