Diagonal trace-class operators on ${\mathrm{\ell }}^2(\mathbb{N},\mathbb{C})$

By [A2] every family in $H$ has pointwise coordinates, and for the coordinate vectors the finite-subset description of the pairing gives $⟨u_n,u_m⟩={\sum }_k{u}_n(k)\overline{u_m(k)}=u_n(n)\overline{u_m(n)}={\delta }_{nm}$, all other terms being $0$; in particular $\mathrm{\parallel }{u}_n\mathrm{\parallel }=1$ and $u_n\mathrm{\ne }0$.

For $x\in H$ and $n\in \mathbb{N}$ the pairing with $u_n$ picks out the $n$-th coordinate, $⟨x,u_n⟩={\sum }_k{x}_k\overline{u_n(k)}=x_n$, so by Cauchy–Schwarz and step 1.1 $\mathrm{\mid }{x}_n\mathrm{\mid }=\mathrm{\mid }⟨x,u_n⟩\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }{u}_n\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }$.

The span of $(u_n)$ is dense in $H$: given $x\in H$ and a real $\epsilon >0$, the finite sum ${\sum }_n\mathrm{\mid }{x}_n{\mathrm{\mid }}^2=\mathrm{\parallel }x{\mathrm{\parallel }}^2$ is finite, so by the small-tail property [A2] there is a finite $F$ with ${\sum }_{n\mathrm{\notin }F}\mathrm{\mid }{x}_n{\mathrm{\mid }}^2<{\epsilon }^2$. The vector $s:={\sum }_{n\in F}{x}_n{u}_n$ lies in the span [A8], and the difference $x-s$ has coordinates $0$ on $F$ and $x_n$ off $F$, so the splitting identity gives $\mathrm{\parallel }x-s{\mathrm{\parallel }}^2={\sum }_{n\mathrm{\notin }F}\mathrm{\mid }{x}_n{\mathrm{\mid }}^2<{\epsilon }^2$ and therefore $\mathrm{\parallel }x-s\mathrm{\parallel }<\epsilon$ by [A3]. Hence $(u_n)$ is a complete orthonormal family in $H$.

Let $(a^{(m)}{)}_{m\in \mathbb{N}}$ be a Cauchy sequence in $H$; for every $n$ step 2.1 applied to the differences $a^{(m)}-a^{(p)}\in H$ (pointwise operations, [A2]) gives $\mathrm{\mid }{a}_n^{(m)}-a_n^{(p)}\mathrm{\mid }\le \mathrm{\parallel }{a}^{(m)}-a^{(p)}\mathrm{\parallel }$, so each coordinate sequence is Cauchy in $\mathbb{C}$ and converges to a scalar $a_n$ by [A10]. Fix a real $\epsilon >0$ and $M$ with $\mathrm{\parallel }{a}^{(m)}-a^{(p)}\mathrm{\parallel }\le \epsilon$ for all $m,p\ge M$. For a finite $F\subseteq \mathbb{N}$ the sum ${\sum }_{n\in F}\mathrm{\mid }{a}_n^{(p)}-a_n^{(M)}{\mathrm{\mid }}^2$ tends, as $p\to \mathrm{\infty }$, to ${\sum }_{n\in F}\mathrm{\mid }{a}_n-a_n^{(M)}{\mathrm{\mid }}^2$ by [A11] applied to the finitely many coordinates, while every finite subsum of a square sum is at most that square sum [A2], so each value is at most $\mathrm{\parallel }{a}^{(p)}-a^{(M)}{\mathrm{\parallel }}^2\le {\epsilon }^2$; hence ${\sum }_{n\in F}\mathrm{\mid }{a}_n-a_n^{(M)}{\mathrm{\mid }}^2\le {\epsilon }^2$. Taking the supremum over finite $F$ gives $Q(a-a^{(M)})\le {\epsilon }^2<+\mathrm{\infty }$, so $a-a^{(M)}\in H$ with $\mathrm{\parallel }a-a^{(M)}\mathrm{\parallel }\le \epsilon$ by [A3], and $a=a^{(M)}+(a-a^{(M)})\in H$; then for every $m\ge M$ the triangle inequality [A4] gives $\mathrm{\parallel }{a}^{(m)}-a\mathrm{\parallel }\le \mathrm{\parallel }{a}^{(m)}-a^{(M)}\mathrm{\parallel }+\mathrm{\parallel }{a}^{(M)}-a\mathrm{\parallel }\le 2\epsilon$. Thus $a^{(m)}\to a$ and $H$ is complete, hence a complex Hilbert space.

Let $c$ be bounded with $C:={\sup }_n\mathrm{\mid }{c}_n\mathrm{\mid }$, and let $x\in H$. The coefficients $b_n:=c_n⟨x,u_n⟩$ satisfy ${\sum }_n\mathrm{\mid }{b}_n{\mathrm{\mid }}^2\le C^2{\sum }_n\mathrm{\mid }⟨x,u_n⟩{\mathrm{\mid }}^2\le C^2\mathrm{\parallel }x{\mathrm{\parallel }}^2<+\mathrm{\infty }$ by Bessel [A5], so by the summation theorem [A6] applied to the orthonormal family $(u_n)$ in the Hilbert space $H$ the net ${\sum }_{n\in F}{b}_n{u}_n$ converges to a vector $D_{c}x\in H$ with $\mathrm{\parallel }{D}_{c}x{\mathrm{\parallel }}^2={\sum }_n\mathrm{\mid }{b}_n{\mathrm{\mid }}^2\le C^2\mathrm{\parallel }x{\mathrm{\parallel }}^2$. Thus $D_c$ is a well-defined map $H\to H$ satisfying $\mathrm{\parallel }{D}_{c}x\mathrm{\parallel }\le C\mathrm{\parallel }x\mathrm{\parallel }$ for every $x$.

Additivity and homogeneity: for $x,y\in H$ and finite $F$, additivity of the pairing in its first argument [A2] gives ${\sum }_{n\in F}{c}_n⟨x+y,u_n⟩u_n={\sum }_{n\in F}{c}_n⟨x,u_n⟩u_n+{\sum }_{n\in F}{c}_n⟨y,u_n⟩u_n$; the two nets on the right converge to $D_{c}x$ and $D_{c}y$, so by [A13] the left net converges to $D_{c}x+D_{c}y$, and uniqueness of limits [A13] together with the defining series of step 4.1 gives $D_c(x+y)=D_{c}x+D_{c}y$. The same computation with $\lambda x$ in place of $x+y$ gives $D_c(\lambda x)=\lambda D_{c}x$, so $D_c$ is linear and therefore bounded with $\mathrm{\parallel }{D}_c\mathrm{\parallel }\le C$ by step 4.1. For a coordinate vector the coefficient family of $u_m$ is supported at $m$ with value $c_m$ (unique coefficients, [A8]), so its finite-subset net is eventually constant at $c_m{u}_m$ and $D_c{u}_m=c_m{u}_m$; in particular $D_{(1)}$ is the identity of $H$ by the Fourier expansion [A7], the constant sequence $1$ being bounded.

Operator identities: fix $x\in H$ and a bounded $c^{\mathrm{\prime }}$. Applying the coefficient clause of [A6] to the vector $D_{c^{\mathrm{\prime }}}x$ gives $⟨D_{c^{\mathrm{\prime }}}x,u_n⟩=c_n^{\mathrm{\prime }}⟨x,u_n⟩$ for every $n$, so by step 4.1 the coefficient family of $D_{c^{\mathrm{\prime }}}x$ for the operator $D_c$ is $c_n{c}_n^{\mathrm{\prime }}⟨x,u_n⟩$ and $D_c{D}_{c^{\mathrm{\prime }}}x={\sum }_n{c}_n{c}_n^{\mathrm{\prime }}⟨x,u_n⟩u_n=D_{c{c}^{\mathrm{\prime }}}x$. Likewise the finite sums for $D_{c+c^{\mathrm{\prime }}}$ and for $D_c+D_{c^{\mathrm{\prime }}}$ have equal terms, since $(c_n+c_n^{\mathrm{\prime }})⟨x,u_n⟩=c_n⟨x,u_n⟩+c_n^{\mathrm{\prime }}⟨x,u_n⟩$, so $D_c+D_{c^{\mathrm{\prime }}}=D_{c+c^{\mathrm{\prime }}}$; and $D_{\lambda c}x=\lambda D_{c}x$ by homogeneity of the coefficients. As $x$, $c$ and $c^{\mathrm{\prime }}$ were arbitrary, $D_c{D}_{c^{\mathrm{\prime }}}=D_{c{c}^{\mathrm{\prime }}}$, $D_c+D_{c^{\mathrm{\prime }}}=D_{c+c^{\mathrm{\prime }}}$ and $\lambda D_c=D_{\lambda c}$.

Invertibility criterion: suppose ${\inf }_n\mathrm{\mid }{c}_n\mathrm{\mid }>0$. Then no $c_n$ vanishes, the reciprocal sequence $c^{-1}=(c_n^{-1})$ is bounded with ${\sup }_n\mathrm{\mid }{c}_n^{-1}\mathrm{\mid }=({\inf }_n\mathrm{\mid }{c}_n\mathrm{\mid }{)}^{-1}$, and step 6.1 gives $D_{c^{-1}}{D}_c=D_{c^{-1}c}=D_{(1)}=I$ and $D_c{D}_{c^{-1}}=I$ by step 5.1, so $D_c$ is boundedly invertible with inverse $D_{c^{-1}}$. Conversely let $S$ be a bounded two-sided inverse of $D_c$. If $\mathrm{\parallel }S\mathrm{\parallel }=0$ then $S=0$ and $I=S{D}_c=0$, contradicting $u_0=I{u}_0\mathrm{\ne }0$ from step 1.1; hence $\mathrm{\parallel }S\mathrm{\parallel }>0$, and for every $n$ the operator bound [A9] gives $1=\mathrm{\parallel }{u}_n\mathrm{\parallel }=\mathrm{\parallel }S{D}_c{u}_n\mathrm{\parallel }\le \mathrm{\parallel }S\mathrm{\parallel }\mathrm{\parallel }{c}_n{u}_n\mathrm{\parallel }=\mathrm{\parallel }S\mathrm{\parallel }\mathrm{\mid }{c}_n\mathrm{\mid }$, so $\mathrm{\mid }{c}_n\mathrm{\mid }\ge \mathrm{\parallel }S{\mathrm{\parallel }}^{-1}>0$ for all $n$ and ${\inf }_n\mathrm{\mid }{c}_n\mathrm{\mid }>0$.

Now assume ${\sum }_n\mathrm{\mid }{c}_n\mathrm{\mid }<+\mathrm{\infty }$. Put $c_n^{(m)}:=c_n$ for $n<m$ and $c_n^{(m)}:=0$ for $n\ge m$, and $R_m:=D_{c^{(m)}}$, so that $R_{m}x={\sum }_{n<m}{c}_n⟨x,u_n⟩u_n$ lies in the span of $u_0,\dots ,u_{m-1}$ and $R_m$ has finite rank and is compact [A8, A14]. By step 6.1 applied to the bounded sequences $c^{(m)}$ and $c-c^{(m)}$ one has $D_c-R_m=D_{c-c^{(m)}}$, so step 4.1 bounds $\mathrm{\parallel }{D}_c-R_m\mathrm{\parallel }\le {\sup }_{n\ge m}\mathrm{\mid }{c}_n\mathrm{\mid }\le {\sum }_{n\ge m}\mathrm{\mid }{c}_n\mathrm{\mid }$, and this tail tends to $0$ by the small-tail property [A2] of the summable family $(\mathrm{\mid }{c}_n\mathrm{\mid })$; therefore $D_c$ is a norm limit of compact operators and is compact. Indexing the same finite-rank sums by the positive integers, $R_m={\sum }_{j=1}^m⟨\cdot ,u_{j-1}⟩c_{j-1}{u}_{j-1}$ with ${\sum }_{j\ge 1}\mathrm{\parallel }{u}_{j-1}\mathrm{\parallel }\mathrm{\parallel }{c}_{j-1}{u}_{j-1}\mathrm{\parallel }={\sum }_n\mathrm{\mid }{c}_n\mathrm{\mid }<+\mathrm{\infty }$, so the nuclear-series characterization [A15] makes $D_c$ trace class with $\mathrm{\parallel }{D}_c{\mathrm{\parallel }}_1\le {\sum }_n\mathrm{\mid }{c}_n\mathrm{\mid }$.

Eigenvalues and multiplicities: let $b$ be any bounded sequence and let $\lambda \in \mathbb{C}$. By steps 5.1 and 6.1, $D_b-\lambda I=D_{b-\lambda }$ and $(D_b-\lambda I{)}^k=D_{(b-\lambda {)}^k}$ for every $k\ge 1$. For $x\in H$ the norm identity of [A6] applied to the coefficient family $((b_n-\lambda {)}^k⟨x,u_n⟩{)}_n$ gives $\mathrm{\parallel }(D_b-\lambda I{)}^{k}x{\mathrm{\parallel }}^2={\sum }_n\mathrm{\mid }(b_n-\lambda {)}^k⟨x,u_n⟩{\mathrm{\mid }}^2$, so $(D_b-\lambda I{)}^{k}x=0$ exactly when $⟨x,u_n⟩=0$ for every $n$ with $b_n\mathrm{\ne }\lambda$; by the Fourier expansion [A7] these are exactly the vectors of the closed span of $\{u_n:b_n=\lambda \}$, and conversely every vector of that closed span is killed by $D_{(b-\lambda {)}^k}$, because each such $u_n$ is and bounded operators are continuous [A9]. Now take $b=c$ summable and $\lambda \mathrm{\ne }0$. The index set $\{n:c_n=\lambda \}$ is finite: it is contained in $\{n:\mathrm{\mid }{c}_n\mathrm{\mid }\ge \mathrm{\mid }\lambda \mathrm{\mid }\}$, and if the latter were infinite the finite subsums of the nonnegative family $(\mathrm{\mid }{c}_n\mathrm{\mid })$ would be unbounded, contradicting ${\sum }_n\mathrm{\mid }{c}_n\mathrm{\mid }<+\mathrm{\infty }$ [A2]. Hence its closed span is the algebraic span of finitely many orthonormal vectors, of dimension $\mathrm{\#}\{n:c_n=\lambda \}$ by [A8] and [A19], and since this kernel is the same for every $k\ge 1$ it is the stabilized kernel of [A17], so $m_{\mathrm{a}\mathrm{l}\mathrm{g}}(\lambda ;D_c)=\mathrm{\#}\{n:c_n=\lambda \}$. These are all the nonzero eigenvalues: if $D_{c}x=\lambda x$ with $x\mathrm{\ne }0$ and $\lambda \mathrm{\ne }0$, then some coefficient $⟨x,u_n⟩$ is nonzero by [A7], and the identity above forces $c_n=\lambda$.

Trace norm and trace: by step 7.2 the operator $D_c$ is compact and trace class with $\mathrm{\parallel }{D}_c{\mathrm{\parallel }}_1\le {\sum }_n\mathrm{\mid }{c}_n\mathrm{\mid }$, and step 8.1 identifies its nonzero eigenvalues with algebraic multiplicity, so their moduli are the terms $\mathrm{\mid }{c}_n\mathrm{\mid }$ over the indices with $c_n\mathrm{\ne }0$. Weyl's inequality [A18] therefore gives ${\sum }_n\mathrm{\mid }{c}_n\mathrm{\mid }={\sum }_j\mathrm{\mid }{\lambda }_j\mathrm{\mid }\le \mathrm{\parallel }{D}_c{\mathrm{\parallel }}_1$, and with step 7.2 $\mathrm{\parallel }{D}_c{\mathrm{\parallel }}_1={\sum }_n\mathrm{\mid }{c}_n\mathrm{\mid }$. Since $(u_n)$ is a supplied Hilbert basis of the Hilbert space $H$ by steps 3.1 and 2.2, the trace theorem [A16] identifies $\mathrm{tr}(D_c)={\mathrm{tr}}_{(u_n)}(D_c)={\sum }_n⟨D_c{u}_n,u_n⟩={\sum }_n{c}_n$, the family $(c_n)$ being absolutely summable by hypothesis.

Collecting the parts: claim 1 is steps 3.1 and 2.2, claim 2 is steps 4.1–7.1, and claim 3 is steps 7.2–9.1. The zero sequence $c=0$ gives $D_0=0$ with $\mathrm{\parallel }{D}_0{\mathrm{\parallel }}_1=0={\sum }_n\mathrm{\mid }{c}_n\mathrm{\mid }$, $\mathrm{tr}(D_0)=0$ and an empty list of nonzero eigenvalues, so the conventions hold there; the finite-support and one-dimensional cases are instances of the general argument, and $H\mathrm{\ne }\{0\}$ because $u_0\mathrm{\ne }0$. AC is consumed only through the declared hypotheses of the cited suppliers: the Countable Choice clause of the summation theorem [A6], of the Fourier expansion [A7] and of the norm-limit clause of [A14], the Countable Choice hypotheses of the nuclear-series characterization [A15] and of the trace theorem [A16], and the AC hypotheses of the algebraic-multiplicity definition [A17] and of Weyl's inequality [A18]; the coordinate family $(u_n)$ is explicit and the argument selects nothing further. Both directions of the invertibility criterion are proved in step 7.1, and the equality of the trace norm is obtained from the two inequalities of steps 7.2 and 9.1. No interval, endpoint or degenerate parameter occurs in the statement.