Diagonal adjoint and graph-core lemma

[Well-definedness and density] $D_0$ is the linear span of the family and is a linear subspace; by [F1] the coefficients in a finite expansion are unique, so $S$ is well defined, and it is linear and maps $D_0$ into $H$. By [F1] the finite partial sums of the expansion of any $x\in H$ converge to $x$, so the span $D_0$ is dense in $H$.

[Symmetry and the series for $A$] Let $x={\sum }_{n\in F}{c}_n{e}_n$ and $y={\sum }_{m\in G}{d}_m{e}_m$ lie in $D_0$, with $c_n=⟨x,e_n⟩$ and $d_m=⟨y,e_m⟩$ by [F1]. Then $⟨Sx,y⟩={\sum }_{n\in F}{\lambda }_n{c}_n\overline{d_n}={\sum }_{n\in F}{c}_n\overline{{\lambda }_n{d}_n}=⟨x,Sy⟩$ because ${\lambda }_n$ is real: $S$ is symmetric. For $x\in D(A)$ the sequence $({\lambda }_n⟨x,e_n⟩{)}_n$ is square summable with ${\sum }_n{\lambda }_n^2\mathrm{\mid }⟨x,e_n⟩{\mathrm{\mid }}^2<\mathrm{\infty }$, so its finite partial sums are Cauchy: $\parallel {\sum }_{m\le M<N}{\lambda }_m⟨x,e_m⟩e_m{\parallel }^2={\sum }_{M<m\le N}{\lambda }_m^2\mathrm{\mid }⟨x,e_m⟩{\mathrm{\mid }}^2\to 0$ by [F1]; by completeness of $H$ the series ${\sum }_n{\lambda }_n⟨x,e_n⟩e_n$ converges, its sum $Ax$ satisfies $\mathrm{\parallel }Ax{\mathrm{\parallel }}^2={\sum }_n{\lambda }_n^2\mathrm{\mid }⟨x,e_n⟩{\mathrm{\mid }}^2$ (the finite-subset net of squared norms has the same supremum), and $A$ is linear because coefficient extraction is linear. If moreover $x\in D_0$, then $⟨x,e_n⟩=0$ for all but finitely many $n$ and $Ax=Sx$; hence $S\subseteq A$.

[$S^{\ast }=A$] Let $y\in D(S^{\ast })$ and $w:=S^{\ast }y$. For each $n$, ${\lambda }_n\overline{⟨y,e_n⟩}=⟨S{e}_n,y⟩=⟨e_n,w⟩=\overline{⟨w,e_n⟩}$, so ${\lambda }_n⟨y,e_n⟩=⟨w,e_n⟩$. By [F1] applied to $w$, ${\sum }_n{\lambda }_n^2\mathrm{\mid }⟨y,e_n⟩{\mathrm{\mid }}^2={\sum }_n\mathrm{\mid }⟨w,e_n⟩{\mathrm{\mid }}^2=\mathrm{\parallel }w{\mathrm{\parallel }}^2<\mathrm{\infty }$, so $y\in D(A)$; and ${\sum }_n{\lambda }_n⟨y,e_n⟩e_n={\sum }_n⟨w,e_n⟩e_n=w$ by [F1], so $S^{\ast }y=Ay$. Hence $S^{\ast }\subseteq A$. Conversely let $y\in D(A)$ and $x={\sum }_{n\in F}{c}_n{e}_n\in D_0$; then $⟨Sx,y⟩={\sum }_{n\in F}{\lambda }_n{c}_n\overline{⟨y,e_n⟩}=⟨x,Ay⟩$, so $x\mapsto ⟨Sx,y⟩$ is the restriction to the dense subspace $D_0$ of the bounded functional $⟨\cdot ,Ay⟩$ and is therefore bounded on $(D_0,\mathrm{\parallel }\cdot \mathrm{\parallel })$; by [F2] $y\in D(S^{\ast })$ and $S^{\ast }y=Ay$. Thus $S^{\ast }=A$.

[$A$ is self-adjoint] Let $z\in D(A^{\ast })$ and $v:=A^{\ast }z$. For $y\in D(A)$ the pairing $⟨Ay,z⟩={\sum }_n{\lambda }_n⟨y,e_n⟩\overline{⟨z,e_n⟩}$ is absolutely convergent by [F4] and $\mathrm{\parallel }Ay\mathrm{\parallel }\mathrm{\parallel }z\mathrm{\parallel }$. Testing $y=e_m$ gives ${\lambda }_m⟨e_m,z⟩=⟨A{e}_m,z⟩=⟨e_m,v⟩$, that is ${\lambda }_m⟨z,e_m⟩=⟨v,e_m⟩$. By [F1], ${\sum }_m{\lambda }_m^2\mathrm{\mid }⟨z,e_m⟩{\mathrm{\mid }}^2={\sum }_m\mathrm{\mid }⟨v,e_m⟩{\mathrm{\mid }}^2=\mathrm{\parallel }v{\mathrm{\parallel }}^2<\mathrm{\infty }$, so $z\in D(A)$; and then $Az={\sum }_m{\lambda }_m⟨z,e_m⟩e_m={\sum }_m⟨v,e_m⟩e_m=v=A^{\ast }z$ by [F1]. Hence $D(A^{\ast })\subseteq D(A)$; the reverse inclusion was proved in step 1.3 as $A\subseteq S^{\ast }$ with $S^{\ast }=A$, and directly: for $z\in D(A)$ and $y\in D(A)$, $⟨Ay,z⟩={\sum }_n{\lambda }_n⟨y,e_n⟩\overline{⟨z,e_n⟩}=⟨y,Az⟩$ by absolute convergence and realness of ${\lambda }_n$, so $z\in D(A^{\ast })$ with $A^{\ast }z=Az$. Therefore $A^{\ast }=A$.

[Graph core] Let $x\in D(A)$ and $x_N:={\sum }_{n=0}^N⟨x,e_n⟩e_n\in D_0\subseteq D(A)$. By [F1], $\mathrm{\parallel }x-x_N{\mathrm{\parallel }}^2={\sum }_{n>N}\mathrm{\mid }⟨x,e_n⟩{\mathrm{\mid }}^2\to 0$, and by step 1.2, $\mathrm{\parallel }Ax-A{x}_N{\mathrm{\parallel }}^2={\sum }_{n>N}{\lambda }_n^2\mathrm{\mid }⟨x,e_n⟩{\mathrm{\mid }}^2\to 0$ because the series converges. Thus $(x_N,A{x}_N)\to (x,Ax)$ in $H\oplus H$; since $x\in D(A)$ was arbitrary, $D_0$ is dense in $(D(A),\mathrm{\parallel }\cdot {\mathrm{\parallel }}_A)$, that is, $D_0$ is a core for $A$.

[Closure, essential self-adjointness and uniqueness] Since $S$ is symmetric, [F2] makes $S$ closable with symmetric closure $\bar{S}$, and [F3] gives $\bar{S}=S^{\ast \ast }$. Since $A=S^{\ast }$ is the adjoint of a densely defined operator, $A$ is closed by [F2]; as $S\subseteq A$, the least closed extension satisfies $\bar{S}\subseteq A$. Conversely, for $x\in D(A)$ step 1.2 supplies $x_N\in D_0$ with $x_N\to x$ and $S{x}_N=A{x}_N\to Ax$, so $(x,Ax)\in \overline{\mathrm{\Gamma }(S)}=\mathrm{\Gamma }(\bar{S})$ and $A\subseteq \bar{S}$. Hence $\bar{S}=A$, and $S$ is essentially self-adjoint because $\bar{S}=A$ is self-adjoint. If $T$ is any self-adjoint extension of $S$, then $T$ is closed and contains $\bar{S}=A$; since $A$ is self-adjoint and $A\subseteq T$ with $T$ symmetric, [F2] forces $T=A$. So $A$ is the unique self-adjoint extension of $S$.

[Positivity and eigenvalues] For $x\in D(A)$ the series ${\sum }_n{\lambda }_n\mathrm{\mid }⟨x,e_n⟩{\mathrm{\mid }}^2$ is absolutely convergent by [F4] and step 1.2: ${\sum }_n\mathrm{\mid }{\lambda }_n\mathrm{\mid }\mathrm{\mid }⟨x,e_n⟩{\mathrm{\mid }}^2\le ({\sum }_n{\lambda }_n^2\mathrm{\mid }⟨x,e_n⟩{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}({\sum }_n\mathrm{\mid }⟨x,e_n⟩{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}<\mathrm{\infty }$, and $⟨Ax,x⟩={\sum }_n{\lambda }_n\mathrm{\mid }⟨x,e_n⟩{\mathrm{\mid }}^2$ by continuity of the inner product in the first argument applied to the convergent series of step 1.2. Since the ${\lambda }_n$ are real the value is real, and ${\lambda }_n\ge c$ for all $n$ gives $⟨Ax,x⟩\ge c\mathrm{\parallel }x{\mathrm{\parallel }}^2$ by [F1]. Finally $A{e}_n={\lambda }_n{e}_n$ directly from the definition of $A$ and [F1], so each ${\lambda }_n$ is an eigenvalue of $A$ with eigenvector $e_n\mathrm{\ne }0$.