Fredholm Determinants and the Lidskii Trace Formula: Examples

Matrix multiplication gives $T{e}_1=e_1$ and $T{e}_2=e_1$, so for all $a,b\in \mathbb{C}$, $T(a{e}_1+b{e}_2)=(a+b)e_1$.

For $x=a{e}_1+b{e}_2$, $b{e}_2\in M^{\perp }$ and $a{e}_1\in (M^{\perp }{)}^{\perp }$, so the orthogonal decomposition in [A3] gives $Qx=P_{M^{\perp }}x=b{e}_2$.

Every $m=a{e}_1\in M$ satisfies $Tm=a{e}_1\in M$, so $T(M)\subseteq M$ and $M$ is invariant by [A1]. Since $e_2\in M^{\perp }$ but $T{e}_2=e_1\mathrm{\notin }{M}^{\perp }$, we also have $T(M^{\perp })⊈M^{\perp }$ by [A2] and step 1.1.

For every $x=a{e}_1+b{e}_2$, step 1.2 and step 1.1 give $QTQx=QT(b{e}_2)=Q(b{e}_1)=0$, while step 1.1 gives $T{e}_2=e_1\mathrm{\ne }0$. Thus the orthogonal compression vanishes although $T$ has a nonzero block from $M^{\perp }$ into $M$; combining this with step 2.1 proves the stated counterexample.

Build a countable dense subset of $H$ by pairing restricted rational-box steps. [A2, A3, A4] Let $S$ be the countable dense family of real rational-box steps in $L^2(\mathbb{R})$ from [A3]. For $f\in H$, choose a measurable representative $f=u+iv$. By [A2], both real components belong to real $L^2([0,1])$. Extend each by zero to $\mathbb{R}$. For a Borel set $B\subseteq \mathbb{R}$, the extension's inverse image is $(u^{-1}(B)\cap [0,1])$, together with $\mathbb{R}\setminus [0,1]$ exactly when $0\in B$; these sets are Lebesgue measurable, so the extension is measurable. Its norm agrees with the norm on $[0,1]$ by the integral-over-a-set definition [A4]. Restriction from $\mathbb{R}$ to $[0,1]$ is contractive by [A4]. Thus, for any $\epsilon >0$, choose $s,t\in S$ with each component's restriction error less than $\epsilon \mathrm{/}\sqrt{2}$. The set $D:=\{{(s+it)\mid }_{[0,1]}:s,t\in S\}$ is countable by [A3], and its members are bounded step functions. The identity $\mathrm{\mid }a+ib{\mathrm{\mid }}^2=a^2+b^2$ and additivity in [A2] give $\mathrm{\parallel }f-{(s+it)\mid }_{[0,1]}{\mathrm{\parallel }}_2^2=\mathrm{\parallel }u-{s\mid }_{[0,1]}{\mathrm{\parallel }}_2^2+\mathrm{\parallel }v-{t\mid }_{[0,1]}{\mathrm{\parallel }}_2^2<{\epsilon }^2\mathrm{.}$ Hence $D$ is dense and $H$ is separable.

Compute the triangle area and identify its indicator kernel with $V$. [A4, A5, A7, A9] Put $\mathrm{\Delta }:=\{(x,t)\in [0,1{]}^2:0\le t\le x\le 1\}\mathrm{.}$ It is closed in ${\mathbb{R}}^2$, hence Borel and product-measurable by [A5]. The two restricted Lebesgue factors are finite, so their product measure $\rho$ exists by [A4, A5], and the rectangle formula gives $\rho ([0,1{]}^2)=1$. Tonelli and [A4] give $\rho (\mathrm{\Delta })={\int }_0^1\lambda ([0,x])dx={\int }_0^{1}xdx=\frac{1}{2}\mathrm{.}$ For the last equality, $x\mapsto x$ is continuous, the primitive $x^2\mathrm{/}2$ has derivative $x$ by [A9], Newton–Leibniz evaluates the Riemann integral, and [A9] identifies it with the Lebesgue integral. Thus $k_1={\mathbf{1}}_{\mathrm{\Delta }}$ is in $L^2(\bar{\rho })$ with $\mathrm{\parallel }{k}_1{\mathrm{\parallel }}_2^2=1\mathrm{/}2$. The kernel theorem [A7] identifies $V=T_{k_1}$ and gives $\mathrm{\parallel }V{\mathrm{\parallel }}_{HS}=\mathrm{\parallel }{k}_1{\mathrm{\parallel }}_2=1\mathrm{/}\sqrt{2}\mathrm{.}$

Prove $V^2$ is trace class. [A1, A7, A8] By [A7], AC supplies a Hilbert basis $E$ of $H$ and $V$ is Hilbert–Schmidt relative to it. By [A1], AC supplies Countable Choice for [A8]; hence [A8] makes $V$ compact and then $V^2=V\circ V$ compact. Apply the Hilbert–Schmidt product theorem [A8] with both factors $V$ and the same basis $E$. It follows that $V^2$ is trace class and $\mathrm{\parallel }{V}^2{\mathrm{\parallel }}_1\le \mathrm{\parallel }V{\mathrm{\parallel }}_{HS}^2=\frac{1}{2}\mathrm{.}$

Bound the kernel operators $T_{k_n}$. [A4, A5, A7, step 1.2] For each integer $n\ge 1$, define $k_n(x,t):=\frac{(x-t{)}^{n-1}}{(n-1)!}{\mathbf{1}}_{\mathrm{\Delta }}(x,t)\mathrm{.}$ Each $k_n$ is product-measurable by [A5]. Since $\mathrm{\mid }x-t\mathrm{\mid }\le 1$ on $\mathrm{\Delta }$, [A4] gives $\mathrm{\parallel }{k}_n{\mathrm{\parallel }}_2^2\le \frac{\rho (\mathrm{\Delta })}{((n-1)!{)}^2}=\frac{1}{2((n-1)!{)}^2}\le \frac{1}{((n-1)!{)}^2},$ using $\rho (\mathrm{\Delta })=1\mathrm{/}2$ from step 1.2. The kernel theorem [A7] therefore defines a bounded operator $T_{k_n}$ with $\mathrm{\parallel }{T}_{k_n}\mathrm{\parallel }\le 1\mathrm{/}(n-1)!$.

Prove the power-kernel identity and factorial norm estimate by induction. [A2, A3, A5, A6, A7, A9, step 1.1, step 2.1, algebra] We prove $V^{n}d=T_{k_n}d$ for every $d\in D$. At $n=1$ this is the definition of $V$. Suppose the identity holds for $n$. Choose a bounded Borel step representative of $d$, since its rational half-open boxes are Borel. For each fixed $x\in [0,1]$, the integrand on $[0,1{]}^2$ $(s,t)⟼{\mathbf{1}}_{\{0\le t\le s\le x\}}(s-t{)}^{n-1}d(t)$ is product-measurable by [A5] and bounded; the product space has finite measure. It is therefore in $L^1$ by [A6], so Fubini changes the order of integration. If $t=x$ the inner integral is zero. If $t<x$, [A9], applied to the primitive $(s-t{)}^n\mathrm{/}n!$, gives ${\int }_t^x\frac{(r-t{)}^{n-1}}{(n-1)!}dr=\frac{(x-t{)}^n}{n!}\mathrm{.}$ Indeed $r\mapsto r-t$ has derivative $1$: the identity has derivative $1$ by the $n=1$ power case, while the constant $-t$ has zero difference quotient. The chain rule differentiates the shifted power, and the factorial recursion cancels its factor $n$. Consequently $V^{n+1}d(x)={\int }_0^x\frac{(x-t{)}^n}{n!}d(t)dt$ for almost every $x$. This proves the induction. Both $V^n$ and $T_{k_n}$ are bounded; since $D$ is dense by step 1.1, equality on $D$ extends to all $H$. Thus $\mathrm{\parallel }{V}^n\mathrm{\parallel }=\mathrm{\parallel }{T}_{k_n}\mathrm{\parallel }\le \frac{1}{(n-1)!}(n\ge 1)\mathrm{.}$

Exclude every nonzero scalar from $\sigma (V^2)$. [A2, A10, step 3.1, algebra] Fix $\lambda \mathrm{\ne }0$ and put $R_N:={\sum }_{m=0}^N{\lambda }^{-m-1}(V^2{)}^m\mathrm{.}$ For $m\ge 1$, step 3.1 gives $\mathrm{\parallel }{\lambda }^{-m-1}(V^2{)}^m\mathrm{\parallel }\le \frac{\mathrm{\mid }\lambda {\mathrm{\mid }}^{-m-1}}{(2m-1)!}\le \frac{\mathrm{\mid }\lambda {\mathrm{\mid }}^{-m-1}}{m!},$ because $(2m-1)!\ge m!$. The scalar majorant is summable by the exponential-series fact [A10]. Since $\mathcal{B}(H)$ is Banach [A10], its series criterion gives an operator-norm limit $R_{\lambda }={\lim }_N{R}_N$. Finite telescoping gives on both sides $(\lambda I-V^2)R_N=R_N(\lambda I-V^2)=I-{\lambda }^{-N-1}{V}^{2N+2}\mathrm{.}$ The remainder tends to zero by step 3.1 and the vanishing terms of the convergent scalar majorant. Submultiplicativity [A10] lets the products pass to the operator-norm limit, so $R_{\lambda }$ is a bounded two-sided inverse of $\lambda I-V^2$. Therefore every nonzero $\lambda$ lies in the resolvent set [A10], and $\sigma (V^2)\subseteq \{0\}\mathrm{.}$

Prove quasinilpotence, the trace and determinant conclusions, and the nonzero witness. [A1, A2, A4, A11, A12, step 1.1, step 1.3, step 3.1, step 4.1, algebra] By [A2], $H$ is a complex Hilbert space; it is separable by step 1.1 and $V^2$ is trace class by step 1.3. Step 4.1 puts its spectrum inside $\{0\}$, while [A12] makes the spectrum nonempty; hence $\sigma (V^2)=\{0\}$ and $V^2$ is quasinilpotent. The local quasinilpotent lemma [A11] applies, yielding $\mathrm{tr}(V^2)=0,D_{V^2}(z)=1(z\in \mathbb{C})\mathrm{.}$ This particular operator is not zero: $1\in H$, $\mathrm{\parallel }1{\mathrm{\parallel }}_2=1$ by [A4], and the same Newton–Leibniz calculation gives $V1(x)=x$ and $V^{2}1(x)=x^2\mathrm{/}2$, which is positive on $[1\mathrm{/}2,1]$ of positive measure [A4]. The formula includes the degenerate endpoint $x=0$ because that integral is zero; the closed triangle convention retains both endpoints, and no boundary point is discarded in Tonelli or Fubini. The base case $n=1$ is step 3.1. AC is the exact declared assumption [A1]: it supplies the Hilbert basis used in step 1.3 and supplies Countable Choice for the stated auxiliary results; the separability approximation selects only two approximants for a single tolerance. There is no one-dimensional branch: the intervals $I_n=(2^{-n},2^{-(n-1)})$, indexed by integers $n\ge 1$, lie in $[0,1]$, are pairwise disjoint and have measure $2^{-n}>0$ by [A4]; if a finite linear combination of their indicator classes is zero, restricting to each $I_n$ forces its coefficient to vanish. Thus $H$ is infinite-dimensional. The assertion is a conjunction, not an iff, so neither iff direction applies.

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

By [A1], AC supplies the Countable Choice hypothesis needed in [A3] and [A5]. The finite-rank $F$ in the statement is therefore trace class, so the arbitrary-Hilbert determinant theorem applies.

Fix $u,v\in H$, put $\alpha :=⟨u,v⟩$, and define $F(x):=⟨x,v⟩u$. By [A2], $F$ is linear; [A4] gives $\mathrm{\parallel }F(x)\mathrm{\parallel }=\mathrm{\mid }⟨x,v⟩\mathrm{\mid }\mathrm{\parallel }u\mathrm{\parallel }\le \mathrm{\parallel }v\mathrm{\parallel }\mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }$, so it is bounded. Its range lies in $\mathrm{span}\{u\}$, hence it has rank at most one. Also $F(u)=\alpha u$.

Let $E$ be any finite-dimensional subspace containing $\mathrm{ran}F$. For every $x\in E$, $F(x)\in \mathrm{ran}F\subseteq E$, so $E$ is $F$-invariant. By [A5], $D_H(I+zF)={\det }_E(I_E+z(F{\mathrm{\mid }}_E))$ for every $z\in \mathbb{C}$. If $F=0$, this says both sides are $1$ because $I_E$ has determinant $1$; this includes $E=\{0\}$, where the convention is also explicitly in [A5].

If $u=0$, then $F=0$, $\alpha =0$, and $E=\{0\}$ is finite-dimensional and contains the range. Step 2.1 gives $D_H(I+zF)=1=1+z\alpha$ for all $z$; this also covers $H=\{0\}$.

Suppose $u\mathrm{\ne }0$ and set $E=\mathrm{span}\{u\}$. By [A6], $E$ is the set of scalar multiples of $u$. The one-element list $(u)$ is independent: its only coefficient relation is $\lambda u=0$, which forces $\lambda =0$. More generally, an injective finite list into $\{u\}$ has at most one entry, since any two entries would both equal $u$; the empty list is independent, so $\{u\}$ is independent under [A6]. It spans $E$, hence it is a basis and $(u)$ is an ordered basis; thus $E$ has dimension one. The range of $F$ is contained in $E$, and $E$ is invariant because $F(u)=\alpha u$. Relative to $(u)$, [A7] gives the matrix $[\alpha ]$ for $F{\mathrm{\mid }}_E$ and $[1+z\alpha ]$ for $I_E+z(F{\mathrm{\mid }}_E)$. By [A8]–[A9], its ordinary determinant is $1+z\alpha$; step 2.1 therefore yields $D_H(I+zF)=1+z⟨u,v⟩$. If $v=0$ then this is the zero operator and the value is $1$; if $v\mathrm{\ne }0$ but $\alpha =0$, then $F(v)=⟨v,v⟩u\mathrm{\ne }0$ and, for every $x$, $F^2(x)=⟨F(x),v⟩u=⟨x,v⟩⟨u,v⟩u=0$. Thus the nonzero rank-one operator is nilpotent and its one-dimensional restriction is zero, so the determinant is again $1$.

At $z=0$ both sides of the finite-rank identity and rank-one formula equal $1$. The zero-rank case, the zero vector and zero Hilbert space, and the nilpotent rank-one case are covered in steps 2.1, 3.1, and 3.2; nonzero $\alpha$ gives the scalar factor $1+z\alpha$ in step 3.2. There is no interval endpoint parameter. The only assumption is AC [A1]; no choices are made in the finite-dimensional rank-one calculation, and both conclusions are equalities rather than iff claims. [A1, A5, A6, A8, A9, step 2.1, step 3.1, step 3.2] \qed