Hilbert direct-sum construction

[Absolutely convergent pairing] Let $\psi ,\phi \in F(H)$. Coordinatewise addition preserves the defining square-summability because $\mathrm{\parallel }{\psi }_n+{\phi }_n{\mathrm{\parallel }}^2\le 2\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2+2\mathrm{\parallel }{\phi }_n{\mathrm{\parallel }}^2$, and scalar multiplication does as well, so $F(H)$ is a complex vector space. By [F2] with $a_n:=\mathrm{\parallel }{\psi }_n\mathrm{\parallel }$, $b_n:=\mathrm{\parallel }{\phi }_n\mathrm{\parallel }$, ${\sum }_n\mathrm{\mid }⟨{\psi }_n,{\phi }_n⟩\mathrm{\mid }\le {\sum }_n\mathrm{\parallel }{\psi }_n\mathrm{\parallel }\mathrm{\parallel }{\phi }_n\mathrm{\parallel }\le \left({\sum }_n\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}\right({\sum }_n\mathrm{\parallel }{\phi }_n{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}=\mathrm{\parallel }\psi \mathrm{\parallel }\mathrm{\parallel }\phi \mathrm{\parallel }<\mathrm{\infty }$, so the series defining $⟨\psi ,\phi ⟩$ converges absolutely (Series and absolute convergence in a normed space); termwise addition and scalar multiplication give sesquilinearity and linearity in the first argument, inherited from the pairings on the $H^{\otimes n}$. If $\mathrm{\parallel }\psi {\mathrm{\parallel }}^2={\sum }_n\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2=0$ then ${\psi }_n=0$ for every $n$, so the displayed form is positive definite and is an inner product.

[Completeness] Let $({\psi }^{(k)}{)}_k$ be a Cauchy sequence in $F(H)$ and let $\epsilon >0$. For each $n$, $\mathrm{\parallel }{\psi }_n^{(k)}-{\psi }_n^{(l)}\mathrm{\parallel }\le \mathrm{\parallel }{\psi }^{(k)}-{\psi }^{(l)}\mathrm{\parallel }$, so $({\psi }_n^{(k)}{)}_k$ is Cauchy in the complete space $H^{\otimes n}$ ([F1]) and converges to some ${\psi }_n\in H^{\otimes n}$; since $\mathrm{ran}{P}_{+}^{(n)}$ is closed, ${\psi }_n\in \mathrm{ran}{P}_{+}^{(n)}$ ([F1]). Choose $K$ with $\mathrm{\parallel }{\psi }^{(k)}-{\psi }^{(m)}\mathrm{\parallel }\le \epsilon$ for $k,m\ge K$. Fix $k\ge K$. For every finite $N$, ${\sum }_{n\le N}\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2\le 2{\sum }_{n\le N}\mathrm{\parallel }{\psi }_n^{(k)}{\mathrm{\parallel }}^2+2{\sum }_{n\le N}\mathrm{\parallel }{\psi }_n-{\psi }_n^{(k)}{\mathrm{\parallel }}^2\le 2\mathrm{\parallel }{\psi }^{(k)}{\mathrm{\parallel }}^2+2{\epsilon }^2$, where the last term uses the limit as $m\to \mathrm{\infty }$ in ${\sum }_{n\le N}\mathrm{\parallel }{\psi }_n^{(m)}-{\psi }_n^{(k)}{\mathrm{\parallel }}^2\le {\epsilon }^2$. Taking the supremum over finite $N$ shows $\psi =({\psi }_n{)}_n\in F(H)$. Also, for each $k\ge K$ and finite $N$, ${\sum }_{n\le N}\mathrm{\parallel }{\psi }_n^{(k)}-{\psi }_n{\mathrm{\parallel }}^2\le {\epsilon }^2$ by letting $m\to \mathrm{\infty }$ in ${\sum }_{n\le N}\mathrm{\parallel }{\psi }_n^{(k)}-{\psi }_n^{(m)}{\mathrm{\parallel }}^2\le {\epsilon }^2$; taking the supremum gives $\mathrm{\parallel }{\psi }^{(k)}-\psi \mathrm{\parallel }\le \epsilon$. Thus ${\psi }^{(k)}\to \psi$ and $F(H)$ is complete, proving (b).

[Vacuum, density and embeddings] The vacuum has $\mathrm{\parallel }\mathrm{\Omega }\mathrm{\parallel }=\mathrm{\mid }1\mathrm{\mid }=1$. For $\psi \in F(H)$ the truncations ${\psi }^{(N)}:=({\psi }_0,\dots ,{\psi }_N,0,0,\dots )\in F_0$ satisfy $\mathrm{\parallel }\psi -{\psi }^{(N)}{\mathrm{\parallel }}^2={\sum }_{n>N}\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2\to 0$, so $F_0$ is dense by [F4]; $F_0$ is a linear subspace because each $\mathrm{ran}{P}_{+}^{(n)}$ is one ([F1]). Each ${\iota }_n$ is linear and $\mathrm{\parallel }{\iota }_n(u){\mathrm{\parallel }}^2=\mathrm{\parallel }u{\mathrm{\parallel }}^2$, and its range is the set of vectors supported in coordinate $n$, a closed subspace identified isometrically with $\mathrm{ran}{P}_{+}^{(n)}$.

[The direct sum is closed] Let $(T_n)$ be as in (f) and let $({\psi }^{(k)}{)}_k\subseteq D(T)$ with ${\psi }^{(k)}\to \psi$ and $T{\psi }^{(k)}\to \eta$ in $F(H)$. For each $n$, ${\psi }_n^{(k)}\to {\psi }_n$ and $T_n{\psi }_n^{(k)}=(T{\psi }^{(k)}{)}_n\to {\eta }_n$; since $T_n$ is self-adjoint, hence closed ([F3]), ${\psi }_n\in D(T_n)$ and $T_n{\psi }_n={\eta }_n$. For every finite $N$, continuity of the norm and the finite sum give ${\sum }_{n\le N}\mathrm{\parallel }{T}_n{\psi }_n{\mathrm{\parallel }}^2={\lim }_k{\sum }_{n\le N}\mathrm{\parallel }{T}_n{\psi }_n^{(k)}{\mathrm{\parallel }}^2\le {\lim }_k\mathrm{\parallel }T{\psi }^{(k)}{\mathrm{\parallel }}^2=\mathrm{\parallel }\eta {\mathrm{\parallel }}^2$. Taking the supremum over finite partial sums shows ${\sum }_n\mathrm{\parallel }{T}_n{\psi }_n{\mathrm{\parallel }}^2\le \mathrm{\parallel }\eta {\mathrm{\parallel }}^2<\mathrm{\infty }$, so $\psi \in D(T)$ and $T\psi =\eta$; thus $T$ is closed. It is symmetric because for $\psi ,\phi \in D(T)$ the middle series below converges absolutely by [F2] applied to the square-summable families $(\mathrm{\parallel }{T}_n{\psi }_n\mathrm{\parallel }{)}_n$, $(\mathrm{\parallel }{\phi }_n\mathrm{\parallel }{)}_n$ and $(\mathrm{\parallel }{T}_n{\phi }_n\mathrm{\parallel }{)}_n$: $⟨T\psi ,\phi ⟩={\sum }_n⟨T_n{\psi }_n,{\phi }_n⟩={\sum }_n⟨{\psi }_n,T_n{\phi }_n⟩=⟨\psi ,T\phi ⟩$.

[Density of the direct-sum domain] Let $\psi \in F(H)$ and $\epsilon >0$. Choose $N$ so that ${\sum }_{n>N}\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2<{\epsilon }^2\mathrm{/}4$. For each of the finitely many $n\le N$, the self-adjoint operator $T_n$ is densely defined, so choose ${\phi }_n\in D(T_n)$ with $\mathrm{\parallel }{\phi }_n-{\psi }_n\mathrm{\parallel }<\epsilon \mathrm{/}(2\sqrt{N+1})$; put ${\phi }_n=0$ for $n>N$. The vector $\phi$ belongs to $D(T)$ because it has finite support and every retained coordinate belongs to its operator domain. Moreover, $\mathrm{\parallel }\phi -\psi {\mathrm{\parallel }}^2={\sum }_{n\le N}\mathrm{\parallel }{\phi }_n-{\psi }_n{\mathrm{\parallel }}^2+{\sum }_{n>N}\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2<{\epsilon }^2\mathrm{/}2<{\epsilon }^2$. Thus $D(T)$ is dense in $F(H)$.

[The adjoint is the same diagonal sum] Let $\phi \in D(T^{\ast })$ with $C:=\mathrm{\parallel }{T}^{\ast }\phi \mathrm{\parallel }$, so $\mathrm{\mid }⟨T\psi ,\phi ⟩\mathrm{\mid }\le C\mathrm{\parallel }\psi \mathrm{\parallel }$ for all $\psi \in D(T)$. Testing with $\psi$ supported in a single coordinate $n$ shows that the functional ${\psi }_n\mapsto ⟨T_n{\psi }_n,{\phi }_n⟩$ on the dense domain $D(T_n)$ is bounded by $C\mathrm{\parallel }{\psi }_n\mathrm{\parallel }$; hence ${\phi }_n\in D(T_n^{\ast })=D(T_n)$ and $T_n{\phi }_n=(T^{\ast }\phi {)}_n$. Since $T^{\ast }\phi \in F(H)$, the coordinate identity gives ${\sum }_n\mathrm{\parallel }{T}_n{\phi }_n{\mathrm{\parallel }}^2=\mathrm{\parallel }{T}^{\ast }\phi {\mathrm{\parallel }}^2<\mathrm{\infty }$. Therefore $D(T^{\ast })\subseteq D(T)$ and $T^{\ast }\phi =T\phi$; with $T\subseteq T^{\ast }$ from step 1.4, $T=T^{\ast }$ is self-adjoint, and $D(T)$ is dense by step 1.5. If every $T_n$ is bounded by $M:={\sup }_n\mathrm{\parallel }{T}_n\mathrm{\parallel }<\mathrm{\infty }$, then $\mathrm{\parallel }T\psi {\mathrm{\parallel }}^2={\sum }_n\mathrm{\parallel }{T}_n{\psi }_n{\mathrm{\parallel }}^2\le M^2\mathrm{\parallel }\psi {\mathrm{\parallel }}^2$ for every $\psi \in F(H)$ and testing on unit vectors in individual nonzero sectors where $\mathrm{\parallel }{T}_n\mathrm{\parallel }$ approaches $M$ gives the reverse inequality.

[Separability] Assume $(e_i{)}_{i\in I}$ is a complete orthonormal family of $H$ with $I$ at most countable. For $n\ge 1$ the elementary tensors $e_{i_1}\otimes \cdots \otimes e_{i_n}$, $(i_1,\dots ,i_n)\in I^n$, have dense linear span in $H^{\otimes n}$ by the product-bases statement of Hilbert tensor construction: inner product, completion and product bases; since $P_{+}^{(n)}$ is bounded, the image $P_{+}^{(n)}$ of a dense set is dense in $\mathrm{ran}{P}_{+}^{(n)}$ (given $u=P_{+}^{(n)}u$ in the range and approximants $v_j\to u$, continuity gives $P_{+}^{(n)}{v}_j\to u$). Hence the at most countable set $D:=\{{\iota }_n(P_{+}^{(n)}(e_{i_1}\otimes \cdots \otimes e_{i_n})):n\ge 1,i_1,\dots ,i_n\in I\}\cup \{\mathrm{\Omega }\}$ has dense span in $F(H)$: any $\psi \in F(H)$ is the limit of its truncations, and each ${\psi }_n$ is approximated in $\mathrm{ran}{P}_{+}^{(n)}$ by finite combinations from $D$. A normed space with a countable set whose finite linear combinations are dense is separable.