Finite-lattice polynomial Hamiltonian data

Since $W+1\ge 2$ and every $a_j>0$, $a$ is a Hermitian positive-definite sesquilinear form on $D_0$: in particular $a(u,u)={\sum }_j{a}_j\mathrm{\parallel }{\mathrm{\partial }}_{j}u{\mathrm{\parallel }}_2^2+\mathrm{\parallel }\sqrt{W+1}u{\mathrm{\parallel }}_2^2\ge 2\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$, and $a(u,u)=0$ forces $u=0$ in $H_N$. Thus its Cauchy-sequence completion with the stated equivalence relation is a well-defined inner-product space.

To see that the completion is complete, let $(X_n)$ be a Cauchy sequence of its classes. By [A1], choose $u_n\in D_0$ with $\mathrm{\parallel }[u_n]-X_n{\mathrm{\parallel }}_Q<2^{-n}$ for each $n$. The triangle inequality makes $(u_n)$ $Q$-Cauchy; its class $X$ is defined by the Cauchy-sequence construction, and $\mathrm{\parallel }{X}_n-X{\mathrm{\parallel }}_Q\le 2^{-n}+\mathrm{\parallel }[u_n]-X{\mathrm{\parallel }}_Q\to 0$. Hence $Q$ is complete and therefore Hilbert.

For a $Q$-Cauchy representative sequence $(u_n)$, the inequalities $\mathrm{\parallel }{u}_n-u_m{\mathrm{\parallel }}_2\le \mathrm{\parallel }{u}_n-u_m{\mathrm{\parallel }}_Q$, $\sqrt{a_j}\mathrm{\parallel }{\mathrm{\partial }}_j{u}_n-{\mathrm{\partial }}_j{u}_m{\mathrm{\parallel }}_2\le \mathrm{\parallel }{u}_n-u_m{\mathrm{\parallel }}_Q$, and $\mathrm{\parallel }\sqrt{W+1}(u_n-u_m){\mathrm{\parallel }}_2\le \mathrm{\parallel }{u}_n-u_m{\mathrm{\parallel }}_Q$ show that $u_n$, each ${\mathrm{\partial }}_j{u}_n$, and $\sqrt{W+1}{u}_n$ converge in $L^2$ by [F2]; write their limits as $u$, $g_j$, and $h$. The first limit defines a continuous map $Q\to H_N$, since $\mathrm{\parallel }u{\mathrm{\parallel }}_2\le \mathrm{\parallel }[u_n]{\mathrm{\parallel }}_Q$, and equivalent Cauchy sequences have the same $L^2$ limit.

For every $\phi \in D_0$, integration by parts gives $\int u_n\overline{{\mathrm{\partial }}_j\phi }=-\int ({\mathrm{\partial }}_j{u}_n)\bar{\phi }$. Passing to the limit in both integrals by $L^2$ convergence and Cauchy–Schwarz yields $\int u\overline{{\mathrm{\partial }}_j\phi }=-\int g_j\bar{\phi }$, so $g_j$ is the stated weak derivative.

On each compact ball $B_R$, the continuous polynomial $W+1$ is bounded. Therefore $\sqrt{W+1}{u}_n\to \sqrt{W+1}u$ in $L^2(B_R)$, while the same sequence converges there to $h$; uniqueness of limits gives $h=\sqrt{W+1}u$ almost everywhere on every integer-radius ball and hence almost everywhere on ${\mathbb{R}}^N$. Taking limits of the component norms in $a(u_n,u_n)$ now gives the displayed formula for $\mathrm{\parallel }u{\mathrm{\parallel }}_Q^2$.

If $u=0$ in $L^2$, then step 3.1 says $⟨g_j,\phi ⟩=0$ for every $\phi \in D_0$; this pairing is continuous by [F3] and $D_0$ is dense by [F4], so it vanishes on all of $H_N$, and taking $\phi =g_j$ gives $g_j=0$. Step 3.2 then gives $h=0$; the norm formula yields $\mathrm{\parallel }[u_n]{\mathrm{\parallel }}_Q=0$, so the embedding is injective. Finally its image contains $D_0$, which is dense in $H_N$ by [F4]; thus the continuous injection has dense image and all claims follow.

Let $C$ be a cube of side $\delta >0$ and $u\in C_c^{\mathrm{\infty }}({\mathbb{R}}^N;\mathbb{C})$, with $u_C:=\mathrm{\mid }C{\mathrm{\mid }}^{-1}{\int }_{C}u$. Expanding the square and using Fubini on the finite cube gives ${\int }_C\mathrm{\mid }u-u_C{\mathrm{\mid }}^2=\frac{1}{2\mathrm{\mid }C\mathrm{\mid }}{\int }_C{\int }_C\mathrm{\mid }u(x)-u(y){\mathrm{\mid }}^{2}dxdy\mathrm{.}$ Telescope $u(x)-u(y)$ through the $N$ coordinate changes and use $\mathrm{\mid }{\sum }_{j=1}^N{z}_j{\mathrm{\mid }}^2\le N{\sum }_j\mathrm{\mid }{z}_j{\mathrm{\mid }}^2$. For the $j$th change, the fundamental theorem of calculus and Cauchy–Schwarz give $\mathrm{\mid }{D}_j(x,y){\mathrm{\mid }}^2\le \delta {\int }_{\min (x_j,y_j)}^{\max (x_j,y_j)}\mathrm{\mid }{\mathrm{\partial }}_{j}u{\mathrm{\mid }}^2$. For fixed values of the other coordinates and fixed $s\in [0,\delta ]$, the measure of pairs $(x_j,y_j)$ whose segment contains $s$ is $2s(\delta -s)\le {\delta }^2\mathrm{/}2$; integrating the displayed bound over the cube variables and dividing by $2\mathrm{\mid }C\mathrm{\mid }$ therefore yields the safe estimate $\mathrm{\parallel }u-u_C{\mathrm{\parallel }}_{L^2(C)}^2\le N{\delta }^2{\sum }_{j=1}^N\mathrm{\parallel }{\mathrm{\partial }}_{j}u{\mathrm{\parallel }}_{L^2(C)}^2\mathrm{.}$ It applies to complex $u$ because the squared-modulus Cauchy–Schwarz inequality is used.

For $R>0$ and $u\in Q$, the coercive weight gives ${\int }_{\mathrm{\mid }q\mathrm{\mid }>R}\mathrm{\mid }u(q){\mathrm{\mid }}^{2}dq\le \frac{1}{c{R}^2}{\int }_{\mathrm{\mid }q\mathrm{\mid }>R}c\mathrm{\mid }q{\mathrm{\mid }}^2\mathrm{\mid }u(q){\mathrm{\mid }}^{2}dq\le \frac{1}{c{R}^2}\mathrm{\parallel }u{\mathrm{\parallel }}_Q^2\mathrm{.}$ Thus the $L^2$ tails of the $Q$ unit ball are uniformly small for large $R$.

The estimate of step 1.1 passes to every $u\in Q$: approximate $u$ in the $Q$ norm by elements of $C_c^{\mathrm{\infty }}$ (the defining completion), restrict to $C$, and use [F1] to pass the $L^2$ and derivative limits.

Fix $\epsilon >0$. Choose $R$ so the squared tail in step 1.2 is below ${\epsilon }^2\mathrm{/}16$. For the $Q$ unit ball, [F1] gives ${\sum }_j\mathrm{\parallel }{\mathrm{\partial }}_{j}u{\mathrm{\parallel }}_2^2\le {\sum }_j{a}_j^{-1}$. Choose an integer $M$ large enough that $\delta :=2R\mathrm{/}M$ satisfies $N{\delta }^2{\sum }_j{a}_j^{-1}<{\epsilon }^2\mathrm{/}16$, and partition $[-R,R{]}^N$ into the resulting finitely many cubes of side $\delta$. Summing the squared estimates of step 2.1 over these disjoint cubes bounds the total mean-approximation error by $N{\delta }^2{\sum }_j\mathrm{\parallel }{\mathrm{\partial }}_{j}u{\mathrm{\parallel }}_2^2<{\epsilon }^2\mathrm{/}16$. The squared error outside $[-R,R{]}^N$ is the tail from step 1.2; thus replacing $u$ on each cube by its mean and setting it to zero outside the large cube has total squared $L^2$ error below ${\epsilon }^2\mathrm{/}8$, hence norm error below $\epsilon \mathrm{/}2$.

For the finite partition in step 3.1, the piecewise-constant mean function $m_u$ satisfies $\mathrm{\parallel }{m}_u{\mathrm{\parallel }}_2^2={\sum }_C\mathrm{\mid }C\mathrm{\mid }\mathrm{\mid }{u}_C{\mathrm{\mid }}^2\le {\sum }_C\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2(C)}^2\le \mathrm{\parallel }u{\mathrm{\parallel }}_2^2\le 1$ by Cauchy--Schwarz [F3]. Thus each mean coordinate has modulus at most $\mathrm{\mid }C{\mathrm{\mid }}^{-1\mathrm{/}2}$, so the vectors of means lie in a bounded box in a finite-dimensional complex coordinate space. Subdivide its real-and-imaginary coordinates into finitely many boxes of sufficiently small side length so that the associated piecewise-constant functions approximate every mean vector to $L^2$ error below $\epsilon \mathrm{/}2$. Combining with step 3.1 gives a finite $\epsilon$-net for the image of the $Q$ unit ball. Scaling shows that the image of every bounded $Q$ set is totally bounded.

Let $K$ be the $H_N$ closure of the image of the $Q$ unit ball. It is complete because it is closed in the complete space $H_N$ [F1]. Suppose an open cover of $K$ has no finite subcover. For each $n\ge 1$, step 4.1 supplies a finite $2^{-n-2}$-net for the image of the $Q$ unit ball with centers in $H_N$; the corresponding closed balls of radius $2^{-n-1}$ cover its closure $K$. Starting with $C_0=K$, choose the least ball index for which $C_n:=C_{n-1}\cap \bar{B}(p_n,2^{-n-1})$ has no finite subcover. Such an index exists, since otherwise finitely many finite subcovers would cover $C_{n-1}$. Each $C_n$ is nonempty, closed, nested, and has diameter at most $2^{-n}$. For $m>n$, any point of nonempty $C_m$ lies in both selected balls, so $\mathrm{\parallel }{p}_m-p_n{\mathrm{\parallel }}_2\le 2^{-m-1}+2^{-n-1}$; hence $(p_n)$ is Cauchy and converges to some $p\in H_N$. Since $C_n$ is nonempty and is contained in both $K$ and $\bar{B}(p_n,2^{-n-1})$, the distance from $p_n$ to $K$ is at most $2^{-n-1}$. Therefore $\mathrm{dist}(p,K)\le \mathrm{\parallel }p-p_n{\mathrm{\parallel }}_2+2^{-n-1}\to 0$; closedness of $K$ gives $p\in K$. For fixed $n$ and every $m>n$, nonempty $C_m\subseteq C_n$ has a point $y\in C_m$ with $\mathrm{\parallel }{p}_m-y{\mathrm{\parallel }}_2\le 2^{-m-1}$. Thus there are points of $C_n$ arbitrarily close to $p$, and closedness gives $p\in C_n$. An open set of the cover containing $p$ contains $K\cap B(p,r)$ for some $r>0$; for large $n$, $\mathrm{diam}(C_n)<r$, so that one open set covers $C_n$, a contradiction. Thus $K$ is compact. All recursive ball choices are least admissible indices in finite lists.

Step 5.1 proves that the closed-unit-ball image under $J$ has compact closure; applying the equivalent criterion in [F2] to the linear inclusion proves that $J:Q\hookrightarrow H_N$ is compact.

In [F2] choose $\mathrm{\hslash }=1$, $m_j=(2a_j{)}^{-1}$ and ${\omega }_j=2\sqrt{a_{j}c}$. Then ${\mathrm{\hslash }}^2\mathrm{/}(2m_j)=a_j$ and $m_j{\omega }_j^2\mathrm{/}2=c$. The eigenvalue from [F2] in coordinate $j$ is ${\omega }_j(n_j+1\mathrm{/}2)=(2n_j+1)\sqrt{a_{j}c}$. Summing over the $N$ coordinates gives the displayed differential eigenvalue ${ϵ}_{\mathbf{n}}$. In the same scaling, $b_j=m_j{\omega }_j=\sqrt{c\mathrm{/}{a}_j}$ and the normalized $n_j=0$ Gaussian becomes $(b_j\mathrm{/}\pi {)}^{1\mathrm{/}4}{e}^{-b_j{q}_j^2\mathrm{/}2}$; taking the product gives the stated ground representative.

Each product Hermite function is a polynomial times a Gaussian, so it and all first derivatives have finite squared integrals after multiplication by any fixed polynomial. Choose a smooth radial cutoff ${\chi }_R(q)=\chi (q\mathrm{/}R)$ with $\chi =1$ on $\mathrm{\mid }q\mathrm{\mid }\le 1$, $\chi =0$ on $\mathrm{\mid }q\mathrm{\mid }\ge 2$, and $\mathrm{\mid }\mathrm{\nabla }{\chi }_R\mathrm{\mid }\le C\mathrm{/}R$. Such a $\chi$ can be formed from $\eta (s)=e^{-1\mathrm{/}s}$ for $s>0$ and $\eta (s)=0$ for $s\le 0$, by taking $\chi (q)=\eta (4-\mathrm{\mid }q{\mathrm{\mid }}^2)\mathrm{/}(\eta (4-\mathrm{\mid }q{\mathrm{\mid }}^2)+\eta (\mathrm{\mid }q{\mathrm{\mid }}^2-1))$. Then $0\le {\chi }_R\le 1$ and ${\chi }_R{\psi }_{\mathbf{n}}\in C_c^{\mathrm{\infty }}$. The weighted-function error is supported on $\mathrm{\mid }q\mathrm{\mid }>R$ and bounded in squared integral by the integrable tail $(c\mathrm{\mid }q{\mathrm{\mid }}^2+2)\mathrm{\mid }{\psi }_{\mathbf{n}}{\mathrm{\mid }}^2$, so its $L^2$ norm tends to zero. For each derivative, ${\mathrm{\partial }}_j((1-{\chi }_R){\psi }_{\mathbf{n}})=(1-{\chi }_R){\mathrm{\partial }}_j{\psi }_{\mathbf{n}}-({\mathrm{\partial }}_j{\chi }_R){\psi }_{\mathbf{n}};$ the first term tends to zero by its Gaussian tail and the second has norm at most $C\mathrm{\parallel }{\psi }_{\mathbf{n}}{\mathrm{\parallel }}_2\mathrm{/}R$. Thus ${\chi }_R{\psi }_{\mathbf{n}}$ converges in the form norm, proving ${\psi }_{\mathbf{n}}\in Q_{\mathrm{o}\mathrm{s}\mathrm{c}}$.

For $v\in C_c^{\mathrm{\infty }}$, integration by parts and the eigenvalue identity in step 1.1 give $a_{\mathrm{o}\mathrm{s}\mathrm{c}}({\psi }_{\mathbf{n}},v)=({ϵ}_{\mathbf{n}}+2)⟨{\psi }_{\mathbf{n}},v⟩\mathrm{.}$ The left side is continuous in the form norm of $v$ by Cauchy--Schwarz [F4] in $Q_{\mathrm{o}\mathrm{s}\mathrm{c}}$, and the right side is continuous since $\mathrm{\parallel }v{\mathrm{\parallel }}_2\le \mathrm{\parallel }v{\mathrm{\parallel }}_{Q_{\mathrm{o}\mathrm{s}\mathrm{c}}}$. Density of $C_c^{\mathrm{\infty }}$ in its defining completion extends the identity to every $v\in Q_{\mathrm{o}\mathrm{s}\mathrm{c}}$. Taking $v={\psi }_{\mathbf{m}}$ proves that ${\xi }_{\mathbf{n}}:=({ϵ}_{\mathbf{n}}+2{)}^{-1\mathrm{/}2}{\psi }_{\mathbf{n}}$ is orthonormal in $Q_{\mathrm{o}\mathrm{s}\mathrm{c}}$. If $u\in Q_{\mathrm{o}\mathrm{s}\mathrm{c}}$ is orthogonal in this form inner product to every ${\xi }_{\mathbf{n}}$, conjugating the displayed identity gives $⟨u,{\psi }_{\mathbf{n}}⟩=0$ for every $\mathbf{n}$. Completeness in $L^2$ from [F2] makes the image of $u$ zero, and the injective inclusion in [F1] makes $u=0$ in $Q_{\mathrm{o}\mathrm{s}\mathrm{c}}$. Thus $({\xi }_{\mathbf{n}})$ is complete there.

Parseval in $Q_{\mathrm{o}\mathrm{s}\mathrm{c}}$ with the complete orthonormal family $({\xi }_{\mathbf{n}})$ gives $a_{\mathrm{o}\mathrm{s}\mathrm{c}}(u,u)={\sum }_{\mathbf{n}}({ϵ}_{\mathbf{n}}+2)\mathrm{\mid }⟨u,{\psi }_{\mathbf{n}}⟩{\mathrm{\mid }}^2\mathrm{.}$ Parseval in $L^2$ gives $\mathrm{\parallel }u{\mathrm{\parallel }}_2^2={\sum }_{\mathbf{n}}\mathrm{\mid }⟨u,{\psi }_{\mathbf{n}}⟩{\mathrm{\mid }}^2$. Subtracting $2\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$ from the first identity yields the asserted expansion of the unshifted oscillator form.

For fixed $f\in H_N$, the functional $v\mapsto ⟨v,f{⟩}_{H_N}$ is linear and satisfies $\mathrm{\mid }⟨v,f⟩\mathrm{\mid }\le \mathrm{\parallel }v{\mathrm{\parallel }}_2\mathrm{\parallel }f{\mathrm{\parallel }}_2\le \mathrm{\parallel }v{\mathrm{\parallel }}_Q\mathrm{\parallel }f{\mathrm{\parallel }}_2$ by [F1] and Cauchy–Schwarz. Thus [F3] applied to the Hilbert space $(Q,a)$ gives a unique $Rf\in Q$ with the stated variational identity. For $\alpha ,\beta \in \mathbb{C}$, conjugate-linearity in the second argument gives $a(v,\alpha Rf+\beta Rg)=⟨v,\alpha f+\beta g⟩$, so uniqueness makes $R$ linear.

Taking $v=Rf$ in the identity gives $\mathrm{\parallel }Rf{\mathrm{\parallel }}_Q^2=a(Rf,Rf)=⟨Rf,f⟩$. Hence $\mathrm{\parallel }Rf{\mathrm{\parallel }}_Q^2\le \mathrm{\parallel }Rf{\mathrm{\parallel }}_2\mathrm{\parallel }f{\mathrm{\parallel }}_2\le \mathrm{\parallel }Rf{\mathrm{\parallel }}_Q\mathrm{\parallel }f{\mathrm{\parallel }}_2$, so $\mathrm{\parallel }Rf{\mathrm{\parallel }}_Q\le \mathrm{\parallel }f{\mathrm{\parallel }}_2$ (with the zero case immediate). Consequently $R:H_N\to H_N$ is bounded by [F5].

The same identity and $a(Rf,Rf)\ge 2\mathrm{\parallel }Rf{\mathrm{\parallel }}_2^2$ give $2\mathrm{\parallel }Rf{\mathrm{\parallel }}_2^2\le \mathrm{\parallel }Rf{\mathrm{\parallel }}_2\mathrm{\parallel }f{\mathrm{\parallel }}_2$; division when $Rf\mathrm{\ne }0$ yields $\mathrm{\parallel }R\mathrm{\parallel }\le \frac{1}{2}$. If $Rf=0$, the variational identity says $⟨v,f⟩=0$ for every $v\in Q$; since $Q$ is dense in $H_N$ by [F1], continuity of the inner product gives $⟨f,f⟩=0$, hence $f=0$. Thus $R$ is injective.

For $f,g\in H_N$, the variational identity and conjugate symmetry of $a$ yield $⟨Rf,g⟩=a(Rf,Rg)=⟨f,Rg⟩$, so $R$ is self-adjoint by [F7]. Also $⟨Rf,f⟩=a(Rf,Rf)\ge 0$, so it is positive.

The map $f\mapsto Rf$ is bounded from $H_N$ into $Q$ by step 2.1, and the inclusion $J:Q\to H_N$ is compact by [F2]. Therefore for every bounded $B\subseteq H_N$, $R(B)=J(R(B))$ has compact closure, which is exactly compactness by [F6].

If $z\perp \mathrm{ran}R$, then $⟨Rf,z⟩=0$ for every $f\in H_N$. Self-adjointness from step 2.3 gives $⟨f,Rz⟩=0$ for every $f$, so $Rz=0$ and step 2.2 gives $z=0$. Thus $(\mathrm{ran}R{)}^{\perp }=\{0\}$, and [F8] gives $\overline{\mathrm{ran}R}=(\mathrm{ran}R{)}^{\perp \perp }=\{0{\}}^{\perp }=H_N$. Therefore $R$ has dense range.

By [F1], $R$ is compact and self-adjoint. Applying [F2] and using $\ker R=\{0\}$ shows that the closed span of its nonzero eigenspaces is all of $H_N$. If $R\psi =\mu \psi$ with $\psi \mathrm{\ne }0$, then $\mu \mathrm{\parallel }\psi {\mathrm{\parallel }}^2=⟨R\psi ,\psi ⟩\ge 0$ by positivity, while injectivity rules out $\mu =0$; hence every nonzero eigenvalue is positive.

The domains defined by the squared coefficient sums for $E_n$ and ${\lambda }_n=E_n-(C_0+1)$ coincide. Indeed ${\lambda }_n^2\le 2E_n^2+2(C_0+1{)}^2,E_n^2\le 2{\lambda }_n^2+2(C_0+1{)}^2\mathrm{.}$ Multiplying by $\mathrm{\mid }{c}_n{\mathrm{\mid }}^2$ and summing preserves finiteness in both directions because ${\sum }_n\mathrm{\mid }{c}_n{\mathrm{\mid }}^2=\mathrm{\parallel }u{\mathrm{\parallel }}_2^2<\mathrm{\infty }$ by [F4]. This proves the displayed coefficient-domain description of $D(h)$.

Each eigenspace is finite-dimensional by [F2], and so has an orthonormal basis by [F3]. The set of nonzero eigenvalues is at most countable, and [A1] selects these finite bases; their union is an orthonormal family whose closed span is $H_N$ by step 1.1. The space is infinite-dimensional: the indicators of the pairwise disjoint unit cubes $[r,r+1)\times [0,1{)}^{N-1}$, $r\in \mathbb{N}$, are nonzero mutually orthogonal $L^2$ vectors. Thus the family can be indexed by $n\ge 1$. The bound $\mathrm{\parallel }R\mathrm{\parallel }\le \frac{1}{2}$ gives $0<{\mu }_n\le \frac{1}{2}$ and $E_n\ge 1$. The compact spectral theorem says only finitely many eigenvalues satisfy $\mu \ge \epsilon$ for each $\epsilon >0$; recursively taking a largest remaining eigenvalue (which exists because above half its positive supremum there are only finitely many) orders them nonincreasingly with multiplicity and gives ${\mu }_n\to 0$, hence $E_n\to \mathrm{\infty }$.

The family $({\psi }_n)$ is complete by step 1.1, so [F4] gives $u={\sum }_n⟨u,{\psi }_n⟩{\psi }_n$ in norm for every $u\in H_N$ and Parseval. In particular $D(S)$ is dense because it contains all finite linear combinations of the ${\psi }_n$. Since each $E_n$ is real, the diagonal formula for $S$ gives $⟨Su,v⟩=⟨u,Sv⟩$ for finite sums, and then for all $u,v\in D(S)$ by [F7] and convergence of the coefficient series; thus $S$ is symmetric. For $u\in D(S)$ with coefficients $c_n$, Cauchy--Schwarz [F7] makes ${\sum }_n{E}_n\mathrm{\mid }{c}_n{\mathrm{\mid }}^2$ absolutely convergent, and step 2.1 gives $E_n\ge 1$, so $⟨Su,u⟩={\sum }_n{E}_n\mathrm{\mid }{c}_n{\mathrm{\mid }}^2\ge {\sum }_n\mathrm{\mid }{c}_n{\mathrm{\mid }}^2=\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$.

To compute the adjoint domain, let $v={\sum }_n{d}_n{\psi }_n\in D(S^{\ast })$ and write $S^{\ast }v=z={\sum }_n{z}_n{\psi }_n$. Testing the adjoint identity on each ${\psi }_n$ gives $z_n=E_n{d}_n$. Because $z\in H_N$, Parseval gives ${\sum }_n{E}_n^2\mathrm{\mid }{d}_n{\mathrm{\mid }}^2<\mathrm{\infty }$, so $v\in D(S)$. Conversely, if this sum is finite, then $z:={\sum }_n{E}_n{d}_n{\psi }_n\in H_N$ and the coefficient identity $⟨Su,v⟩=⟨u,z⟩$ holds first for finite sums and then for every $u\in D(S)$ by [F7]; hence $v\in D(S^{\ast })$ and $S^{\ast }v=Sv$. Therefore $D(S^{\ast })=D(S)$ and $S^{\ast }=S$ as operators; by the unbounded self-adjointness definition in [F5], $S$ is self-adjoint.