Diagonal trace-class Fredholm determinant

The sequence $d$ is bounded, since $\mathrm{\mid }{d}_0\mathrm{\mid }=0$ and $\mathrm{\mid }{d}_n\mathrm{\mid }=2^{-n}\le 1\mathrm{/}2$ for $n\ge 1$. By [A2] the operator $T=D_d$ is bounded with $T{u}_n=d_n{u}_n$ for every $n$, so $T{u}_0=0$ and $T{u}_n=2^{-n}{u}_n$ for $n\ge 1$. By [A5], ${\sum }_n\mathrm{\mid }{d}_n\mathrm{\mid }={\sum }_{n\ge 1}{2}^{-n}=1<+\mathrm{\infty }$ and ${\sum }_n{d}_n=1$; hence [A3] makes $T$ trace class with $\mathrm{\parallel }T{\mathrm{\parallel }}_1={\sum }_n\mathrm{\mid }{d}_n\mathrm{\mid }=1$ and $\mathrm{tr}(T)={\sum }_n{d}_n=1$.

Apply [A3] with $c=d$. The nonzero eigenvalues of $T$, repeated according to algebraic multiplicity, are the nonzero scalars of the list $(d_n)$, the value $\lambda \mathrm{\ne }0$ occurring $\mathrm{\#}\{n:d_n=\lambda \}$ times. The nonzero scalars are the values $2^{-n}$ with $n\ge 1$, and for $\lambda =2^{-r}$ the index set is $\{n\ge 1:2^{-n}=2^{-r}\}=\{r\}$ by [A7]; hence the eigenvalue list with algebraic multiplicities is $(2^{-1},2^{-2},2^{-3},\dots )$, each value occurring once. In particular $T\mathrm{\ne }0$, because $T{u}_1=2^{-1}{u}_1$ and $u_1\mathrm{\ne }0$ by [A2].

The sequence $1+zd$ is bounded, so [A2] gives $D_{1+zd}$ with $D_{1+zd}{u}_n=(1+z{d}_n)u_n$, and the operator calculus of [A2] gives $I+zT=D_{(1)}+z{D}_d=D_{1+zd}$. If $z=-2^r$ for some $r\ge 1$, then $1+z{d}_r=0$, so $D_{1+zd}{u}_r=0$ with $u_r\mathrm{\ne }0$ and $I+zT$ is not injective, hence not boundedly invertible. Conversely let $z\mathrm{\ne }-2^n$ for every $n\ge 1$. For $z=0$ all coefficients $1+z{d}_n$ equal $1$. For $z\mathrm{\ne }0$, [A6] gives $N\ge 1$ with $\mathrm{\mid }z\mathrm{\mid }{2}^{-n}\le \mathrm{\mid }z\mathrm{\mid }{2}^{-N}<1\mathrm{/}2$ for every $n\ge N$ by [A7], so [A11] gives $\mathrm{\mid }1+z{d}_n\mathrm{\mid }\ge 1-\mathrm{\mid }z\mathrm{\mid }{2}^{-n}>1\mathrm{/}2$ there, while the finitely many remaining coefficients $1+z{d}_0=1,1+z{d}_1,\dots ,1+z{d}_{N-1}$ are all nonzero and therefore have a positive minimum by [A10]. Hence ${\inf }_n\mathrm{\mid }1+z{d}_n\mathrm{\mid }>0$, and the invertibility criterion of [A2] makes $D_{1+zd}=I+zT$ boundedly invertible, with inverse $D_{(1+zd{)}^{-1}}$.

By [A8] the arbitrary-Hilbert determinant $D_T(z)=D_H(I+zT)$ is independent of the support and equals the locally uniform product over the nonzero eigenvalues of $T$ repeated according to algebraic multiplicity; substituting the list of step 1.2 gives $D_T(z)={\prod }_{n\ge 1}(1+z{2}^{-n})$, locally uniformly on $\mathbb{C}$, and $D_T(0)=1$ by [A8].

By step 1.3 the operator $I+zT$ is boundedly invertible exactly when $z\mathrm{\notin }\{-2^n:n\ge 1\}$. Since $H$ is a separable complex Hilbert space by [A2] and [A4], and since $H$ itself is a closed support with $T(H)\subseteq H$ and $T{\mathrm{\mid }}_{H^{\perp }}=0$, the support-independence clause of [A8] identifies the arbitrary-Hilbert value $D_H(I+zT)$ with the locally constructed separable determinant of $T$ on $H$; applying [A9] therefore gives $D_T(z)=0$ exactly for $z=-2^r$, $r\ge 1$. For such an $r$ the eigenvalue $2^{-r}$ has algebraic multiplicity $m_{\mathrm{a}\mathrm{l}\mathrm{g}}(2^{-r};T)=1$ by step 1.2, so [A9] makes each zero simple; at $z=0$, not a zero, step 2.1 gives $D_T(0)=1$. The basis, the coefficient sequence and the parameter are explicit and no interval or endpoint occurs. AC is used exactly through the hypotheses of the suppliers [A2], [A3], [A8] and [A9], which are stated under AC, and the example makes no further choice. Both directions of the zero characterization are proved in steps 1.3 and 3.1. [A1, A2, A4, A8, A9, step 1.2, step 2.1, step 1.3] \qed