Zeros of the local Fredholm determinant

If $T=0$, including when $H=\{0\}$, choose $F=T$ and $E=\{0\}$ in [A11]. Then $D_T\equiv 1$ and $I+zT=I$ has its identity as bounded inverse for every $z$; there is no nonzero eigenvalue to consider. Hence assume $H\mathrm{\ne }\{0\}$ and $T\mathrm{\ne }0$.

For the multiplicity claim, let $\lambda \mathrm{\ne }0$ be an eigenvalue of $T$. By [A2] and [A3], $H$ is a complex Banach space and $T$ is compact, so the Riesz–Schauder suppliers apply. The value $\lambda$ lies in $\sigma (T)$. By [A5], $S=\{\mu \in \sigma (T):\mathrm{\mid }\mu \mathrm{\mid }\ge \mathrm{\mid }\lambda \mathrm{\mid }\mathrm{/}2\}$ is finite; every spectral point within distance $\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{/}2$ of $\lambda$ lies in $S$. If $S\setminus \{\lambda \}$ is empty, take $\delta =\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{/}4$; otherwise take $\delta$ positive and less than both $\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{/}2$ and the finitely many distances $\mathrm{\mid }\lambda -\mu \mathrm{\mid }$ for $\mu \in S\setminus \{\lambda \}$. Then $\delta$ isolates $\lambda$, so $E=\{\lambda \}$ is clopen in $\sigma (T)$ and [A1, A6] define its Riesz projection $P$.

For every trace-class $U$ and $c\in \mathbb{C}$, [A10] and [A14] give $D_{cU}(1)={\sum }_{n\ge 0}\mathrm{tr}({\mathrm{\Lambda }}^n(cU))=D_U(c),$ because the degree-zero term is $1$ on both sides and the degree-$n$ terms agree for each $n\ge 1$; this includes $c=0$.

Let $U$ be trace class and suppose $I+wU$ has bounded inverse $R$. Set $V:=-wUR$, which is trace class by [A12]. The right-inverse equation $(I+wU)R=I$ gives $wU+V+wUV=wU-wUR-w^2{U}^{2}R=wU-wU(I+wU)R=0$. Apply [A13] to $wU,V$ and use $D_0(1)=1$ from [A11] to get $1=D_{wU}(1)D_V(1)=D_U(w)D_V(1)$ by step 1.3. Thus $D_U(w)\mathrm{\ne }0$.

Put $M=\mathrm{ran}P$ and $N=\ker P$. By [A7], $H=M\oplus N$ is a bounded topological direct sum and both summands are $T$-invariant. By [A9], $M=G_{\lambda }(T)$ and $d:=\dim M=m_{\mathrm{a}\mathrm{l}\mathrm{g}}(\lambda ;T)<\mathrm{\infty }$; since $\lambda$ is an eigenvalue, $d\ge 1$. The restriction $J:=(T{\mathrm{\mid }}_M)-\lambda I_M$ is nilpotent because $M=\ker (T-\lambda I{)}^{m_0}$ for a stabilized exponent. Both $TP$ and $T(I-P)$ are trace class by [A12].

Set $A=zTP$ and $B=zT(I-P)$. Since $PT=TP$, $AB=z^{2}TPT(I-P)=z^2{T}^{2}P(I-P)=0$, while $A+B=zT$. Apply [A13] and then step 1.3 to obtain, for every $z\in \mathbb{C}$, $D_T(z)=D_{zT}(1)=D_{zTP}(1)D_{zT(I-P)}(1)=D_{TP}(z)D_{T(I-P)}(z)\mathrm{.}$

The operator $TP$ has finite rank and range in $M$, and $M$ is invariant; since $P{\mathrm{\mid }}_M=I_M$, [A11] gives $D_{TP}(z)={\det }_M(I_M+z(T{\mathrm{\mid }}_M))$. A Jordan-string basis from [A16] makes the matrix of the nilpotent $J$ triangular with zero diagonal; hence the matrix of $I_M+z(T{\mathrm{\mid }}_M)$ is triangular with diagonal entries $1+z\lambda$. By [A17] and [A18] its operator determinant is this matrix determinant, and [A19] gives $D_{TP}(z)=(1+z\lambda {)}^d\mathrm{.}$

Put $z_0=-1\mathrm{/}\lambda$ and $Q(z):=D_{T(I-P)}(z)$. On $M$, $T(I-P)$ is zero; on $N$, it is $T{\mathrm{\mid }}_N$. If $N\mathrm{\ne }\{0\}$, [A8] excludes $\lambda$ from $\sigma (T{\mathrm{\mid }}_N)$, so $I_N+z_{0}T{\mathrm{\mid }}_N=(\lambda I_N-T{\mathrm{\mid }}_N)\mathrm{/}\lambda$ has a bounded inverse. Its direct-sum inverse with $I_M$ is bounded because the projections $P$ and $I-P$ are bounded. If $N=\{0\}$, the complementary operator is just the identity on $M$. Thus $I+z_{0}T(I-P)$ is boundedly invertible on $H$. Since $T(I-P)$ is trace class by step 2.2, step 2.1 yields $Q(z_0)\mathrm{\ne }0$.

By steps 3.1 and 3.2, $D_T(z)={\lambda }^d(z-z_0{)}^{d}Q(z)\mathrm{.}$ The function $Q$ is entire by [A10], and step 3.3 gives $Q(z_0)\mathrm{\ne }0$; since $\lambda \mathrm{\ne }0$, ${\lambda }^{d}Q$ is holomorphic and nonzero at $z_0$. Thus [A20] shows that $D_T$ has a zero of order $d=m_{\mathrm{a}\mathrm{l}\mathrm{g}}(\lambda ;T)$ at $z_0$.

If $z\mathrm{\ne }0$ and $I+zT$ is not boundedly invertible, set $\lambda =-1\mathrm{/}z$. Then $I+zT=z(T-\lambda I)$, so $\lambda \in \sigma (T)$; by [A2] and [A3], T is a compact operator on a complex Banach space, and [A4] makes $\lambda$ an eigenvalue. Step 4.1 applies and gives $D_T(z)=0$. The case $z=0$ is invertible and has $D_T(0)=1$ by [A10].

If $I+zT$ is boundedly invertible, step 2.1 with $U=T$ gives $D_T(z)\mathrm{\ne }0$; together with step 5.1 this proves both directions of the stated iff. For every nonzero spectral value, [A4] supplies an eigenvalue, so step 4.1 proves the multiplicity claim on the full nonzero spectrum.

The zero operator and zero Hilbert space are covered by step 1.1; on $H=\mathbb{C}$ with $T=tI$, the determinant is $1+zt$, so for $t\mathrm{\ne }0$ its zero is simple and $m_{\mathrm{a}\mathrm{l}\mathrm{g}}(t;T)=1$, while $t=0$ has no zero and every $I+zT$ is invertible. An empty nonzero spectrum yields no noninvertible $I+zT$ by step 5.1. There is no interval parameter or one-sided endpoint. The exact assumption is AC [A1]; it is used through the Riesz–Schauder, Riesz-projection, and algebraic-multiplicity suppliers, and AC supplies the AC${}_{\omega }$ required by the local determinant and trace-class suppliers through [A15]. The finite Jordan-string argument makes no additional choice. Both iff directions were proved in step 6.1; the multiplicity assertion is a factorization claim, not an iff.