A symmetric closed operator that is not self-adjoint

An absolutely continuous representative is continuous and unique in its almost-everywhere class: a nonzero difference at a point would stay nonzero on a relative interval of positive length. It is bounded on [0,1] and thus belongs to $L^2$; its a.e. derivative is independent of the representative. The stated domains are linear and define linear operators. Complex absolutely continuous calculus is obtained componentwise from the real theory. If $F,G$ are complex-valued with absolutely continuous real and imaginary parts, then $F(x)-F(0)={\int }_0^x{F}^{\mathrm{\prime }}$ for all $x$, and ${\int }_0^1{F}^{\mathrm{\prime }}\bar{G}=F(1)\overline{G(1)}-F(0)\overline{G(0)}-{\int }_0^{1}F\overline{G^{\mathrm{\prime }}}$; if $h\in L^2(0,1)$ then $x\mapsto {\int }_0^{x}h$ is of this kind with derivative $h$ almost everywhere by the first $L^1$ fundamental theorem, using the inclusion $L^2\subseteq L^1$ in [A5]. These are the uses of the stated Countable Choice and Dependent Choice through the calculus and Hilbert-space suppliers.

Given $u\in H$, extend it by zero to $u_0$ on $\mathbb{R}$. By [A3] and Countable Choice, choose complex ${\phi }_n\in C_c^{\mathrm{\infty }}(\mathbb{R})$ with $\mathrm{\parallel }{\phi }_n-u_0{\mathrm{\parallel }}_2<1\mathrm{/}n$, using real and imaginary approximations if necessary. For $n\ge 5$ put ${\zeta }_n(x)=(1-\chi (nx))(1-\chi (n(x-1)))$ on $(0,1)$ and extend it by zero outside. It vanishes in neighborhoods of both endpoints, so this extension is smooth with compact support in $(0,1)$. Also $0\le {\zeta }_n\le 1$ and ${\zeta }_n(x)\to 1$ for every $x\in (0,1)$. Then ${\psi }_n={\zeta }_n{\phi }_n\in C_c^{\mathrm{\infty }}(0,1)\subseteq D(T)$ and $\mathrm{\parallel }{\psi }_n-u{\mathrm{\parallel }}_2\le 1\mathrm{/}n+\mathrm{\parallel }({\zeta }_n-1)u{\mathrm{\parallel }}_2\to 0$ by domination by $\mathrm{\mid }u{\mathrm{\mid }}^2$ for the squared error. Thus $D(T)$ is dense.

Symmetry: for $f,g\in D(T)$ the boundary term in 1.1 vanishes because $f(0)=f(1)=0$, so $⟨Tf,g⟩=-i{\int }_0^1{f}^{\mathrm{\prime }}\bar{g}={\int }_0^{1}f\overline{-i{g}^{\mathrm{\prime }}}=⟨f,Tg⟩$; hence $T\subseteq T^{\ast }$, and $T$ is symmetric.

Closedness: let $f_n\in D(T)$ with $f_n\to f$ and $T{f}_n\to w$ in $L^2$. Then $f_n^{\mathrm{\prime }}\to iw$ in $L^2$, so by 1.1 the functions $G_n(x):={\int }_0^x{f}_n^{\mathrm{\prime }}$ converge uniformly to $G(x):={\int }_0^{x}iw$ on $[0,1]$, since ${\sup }_x\mathrm{\mid }{G}_n(x)-G(x)\mathrm{\mid }\le \mathrm{\parallel }{f}_n^{\mathrm{\prime }}-iw{\mathrm{\parallel }}_2$ by [A5], with $G_n=f_n-f_n(0)=f_n$; hence $f_n\to G$ pointwise and in $L^2$, so $f=G$, that is, $f$ is absolutely continuous with $f^{\mathrm{\prime }}=iw$ almost everywhere, $f(0)=G(0)=0$ and $f(1)=G(1)={\lim }_n{f}_n(1)=0$. Thus $f\in D(T)$ and $Tf=-i{f}^{\mathrm{\prime }}=-i(iw)=w$, so $T$ is closed. The sequential graph criterion applies in this metric product under the assumed Countable Choice.

Adjoint computed. If $g\in AC[0,1]$ has $g^{\mathrm{\prime }}\in L^2(0,1)$, then for every $f\in D(T)$ the identity in step 1.1 read backwards gives $⟨Tf,g⟩=⟨f,-i{g}^{\mathrm{\prime }}⟩$, so $g\in D(T^{\ast })$ and $T^{\ast }g=-i{g}^{\mathrm{\prime }}$.

Conversely let $g\in D(T^{\ast })$ and put $h:=T^{\ast }g$; by 1.1 the function $H(x):={\int }_0^{x}h$ is absolutely continuous with $H^{\mathrm{\prime }}=h$ and $H(0)=0$. For every $f\in D(T)$ integration by parts in the form of 1.1 gives ${\int }_0^{1}f\bar{h}=f(1)\overline{H(1)}-f(0)\overline{H(0)}-{\int }_0^1{f}^{\mathrm{\prime }}\bar{H}=-{\int }_0^1{f}^{\mathrm{\prime }}\bar{H}$, while ${\int }_0^{1}f\bar{h}=⟨f,T^{\ast }g⟩=⟨Tf,g⟩=-i{\int }_0^1{f}^{\mathrm{\prime }}\bar{g}$. Hence ${\int }_0^1{f}^{\mathrm{\prime }}(\bar{H}-i\bar{g})=0$ for every $f\in D(T)$.

The derivatives of elements of $D(T)$ are exactly $E=\{\phi \in L^2(0,1):{\int }_0^1\phi =0\}$: one inclusion follows from step 1.1, and conversely $f(x)={\int }_0^x\phi$ has derivative $\phi$, vanishes at both endpoints, and belongs to $D(T)$. Set $k=H+ig\in L^2$ and $m={\int }_0^{1}k$. Step 2.5 says $\int \phi \bar{k}=0$ for every $\phi \in E$. Taking $\phi =k-m$ gives $0=\int (k-m)\bar{k}=\int \mathrm{\mid }k-m{\mathrm{\mid }}^2$, since $\int (k-m)=0$. Thus $k=m$ as a class and $g=iH-im$. This supplies an absolutely continuous representative of $g$ with $g^{\mathrm{\prime }}=ih$ a.e., so $h=-i{g}^{\mathrm{\prime }}$. No unproved orthogonal-hyperplane assertion is needed.

By steps 2.4 and 3.1, $D(T^{\ast })=\{g\in AC[0,1]:g^{\mathrm{\prime }}\in L^2(0,1)\}$ and $T^{\ast }g=-i{g}^{\mathrm{\prime }}$. This is strictly larger than $D(T)$: the constant function $\mathbf{1}$ lies in $D(T^{\ast })$ but not in $D(T)$. Hence $T\mathrm{\ne }{T}^{\ast }$, and $T^{\ast }$ is a proper closed extension of $T$ by [A4]. Moreover $T^{\ast }$ is not symmetric: for $g_1(x)=x$, $g_2=\mathbf{1}$ one computes from step 1.1 that $⟨T^{\ast }{g}_1,g_2⟩-⟨g_1,T^{\ast }{g}_2⟩=-i(g_1(1)\overline{g_2(1)}-g_1(0)\overline{g_2(0)})=-i\mathrm{\ne }0$.

Let $Sg=-i{g}^{\mathrm{\prime }}$ on the periodic domain $D_2$. It is densely defined because it extends $T$. For $f,g\in D_2$ the boundary form in step 1.1 vanishes, so $S$ is symmetric. If $g\in D(S^{\ast })$, the adjoint identity restricted to $D(T)$ gives $g\in D(T^{\ast })$ and $S^{\ast }g=T^{\ast }g=-i{g}^{\mathrm{\prime }}$ by [A4] and step 4.1. For arbitrary $f\in D_2$, integration by parts therefore gives $0=-i(f(1)\overline{g(1)}-f(0)\overline{g(0)})$. Choose $f=\mathbf{1}\in D_2$: then $g(1)=g(0)$, so $g\in D_2$. Conversely the boundary form vanishes for every periodic $g$, giving $g\in D(S^{\ast })$ and $S^{\ast }g=Sg$. Hence $S=S^{\ast }$ with equality of domains; $S$ is self-adjoint and therefore closed by [A1]. The constant function lies in $D_2\setminus D(T)$, and $x\mapsto x$ lies in $D(T^{\ast })\setminus D_2$, proving both strict inclusions.

Every claim is witnessed: $T$ is densely defined and closed by steps 2.1 and 2.3, symmetric by step 2.2, and $T\mathrm{\ne }{T}^{\ast }$ by step 4.1; the failure of "symmetric implies self-adjoint" is therefore established, and no claim is made that a self-adjoint extension does not exist: the periodic domain of step 5.1 provides one by its explicitly computed adjoint.