A quasinilpotent trace-class operator has zero trace

Let $z\mathrm{\ne }0$ and set $\lambda =-1\mathrm{/}z$. The spectral hypothesis gives $\lambda \mathrm{\notin }\sigma (Q)$; [A2] and [A3] put $Q$ in the bounded-operator spectrum setting. By [A4, L1], $\lambda I-Q$ has a bounded inverse, and $I+zQ=-z(\lambda I-Q)$ is boundedly invertible. The local zero criterion [A5] gives $D_Q(z)\mathrm{\ne }0$.

At $z=0$, [L2] gives $D_Q(0)=1$. Thus [A5] and step 1.1 make $D_Q$ zero-free on $\mathbb{C}$, while [A6] makes it entire.

Put $f=D_Q$. The quotient $f^{\mathrm{\prime }}\mathrm{/}f$ is entire because $f$ is entire and zero-free, using the local power-series and quotient rules (The sum of a complex power series is analytic throughout its open disc of convergence, A complex power-series sum has complex derivatives of every order, obtained by repeated termwise differentiation, Complex analytic functions are closed under finite linear combinations, products, quotients with nonzero denominator, and composition, A complex function is holomorphic if and only if it is analytic). Cauchy's theorem on the convex plane gives a primitive $h$ of $f^{\mathrm{\prime }}\mathrm{/}f$; subtract a constant so that $h(0)=0$ (Cauchy's theorem on a convex complex domain, For a continuous function on a complex domain, endpoint independence, zero closed-contour integrals, and existence of a primitive are equivalent). The product and chain rules, together with $(e^{-h}{)}^{\mathrm{\prime }}=-h^{\mathrm{\prime }}{e}^{-h}$, show $(f{e}^{-h}{)}^{\mathrm{\prime }}=0$ (Linearity, product, reciprocal, and quotient rules for complex derivatives, The chain rule for complex derivatives, The complex exponential is entire and its complex derivative is itself). Hence $f{e}^{-h}=f(0)=1$ and $f=e^h$ (A holomorphic function with zero derivative on a domain is constant, $\exp (z+w)=\exp z\exp w$, and the complex exponential extends the real exponential).

For every $z$, [A8] with $\epsilon =1$ gives $\mathrm{\mid }f(z)\mathrm{\mid }\le C_1{e}^{\mathrm{\mid }z\mathrm{\mid }}$ for some $C_1\ge 1$, since $f(0)=1$. The exponential modulus formula [A12] gives $e^{\mathrm{Re}h(z)}\le C_1{e}^{\mathrm{\mid }z\mathrm{\mid }}$; [A10, A11] then imply $\mathrm{Re}h(z)\le A+\mathrm{\mid }z\mathrm{\mid }$, where $A=\log C_1\ge 0$.

We prove the required disk estimate. Fix $w\mathrm{\ne }0$ and $0<t<1$, let $R=\mathrm{\mid }w\mathrm{\mid }\mathrm{/}t$ and $K=A+R+1$. For $\mathrm{\mid }\zeta \mathrm{\mid }<1$, set $u=h(R\zeta )$ and $G(\zeta )=u\mathrm{/}(2K-u)$. The bound from step 4.1 gives $\mathrm{Re}u\le A+R$, so $K-\mathrm{Re}u\ge 1$ and $\mathrm{\mid }2K-u{\mathrm{\mid }}^2-\mathrm{\mid }u{\mathrm{\mid }}^2=4K(K-\mathrm{Re}u)>0.$ Thus $G$ is holomorphic, $G(0)=0$, and $\mathrm{\mid }G\mathrm{\mid }<1$. The local power series of $G$ shows that $G(\zeta )\mathrm{/}\zeta$ extends holomorphically through zero. On every circle $\mathrm{\mid }\zeta \mathrm{\mid }=r<1$, the maximum-modulus principle bounds this quotient by $1\mathrm{/}r$; letting $r\uparrow 1$ gives $\mathrm{\mid }G(\zeta )\mathrm{\mid }\le \mathrm{\mid }\zeta \mathrm{\mid }$ (The sum of a complex power series is analytic throughout its open disc of convergence, Boundary maximum modulus principle on a bounded domain). Since $\mathrm{\mid }G\mathrm{\mid }=\mathrm{\mid }u\mathrm{\mid }\mathrm{/}\mathrm{\mid }2K-u\mathrm{\mid }$, this implies $\mathrm{\mid }u\mathrm{\mid }\le \mathrm{\mid }\zeta \mathrm{\mid }(2K+\mathrm{\mid }u\mathrm{\mid })$ and hence $\mathrm{\mid }u\mathrm{\mid }\le 2K\mathrm{\mid }\zeta \mathrm{\mid }\mathrm{/}(1-\mathrm{\mid }\zeta \mathrm{\mid })$. Take $\zeta =tw\mathrm{/}\mathrm{\mid }w\mathrm{\mid }$, so $R\zeta =w$ and $\mathrm{\mid }\zeta \mathrm{\mid }=t$. Therefore $\mathrm{\mid }h(w)\mathrm{\mid }\le \frac{2Kt}{1-t}=\frac{2t(A+1)+2\mathrm{\mid }w\mathrm{\mid }}{1-t}\mathrm{.}$ Letting $t\downarrow 0$ gives $\mathrm{\mid }h(w)\mathrm{\mid }\le 2\mathrm{\mid }w\mathrm{\mid }$; it also holds at $w=0$ because $h(0)=0$.

The quotient $q(z)=h(z)\mathrm{/}z$ extends to an entire function by the local power series of $h$, with $q(0)=h^{\mathrm{\prime }}(0)$. Step 5.1 gives $\mathrm{\mid }q(z)\mathrm{\mid }\le 2$ for $z\mathrm{\ne }0$, so continuity bounds it at zero as well. Liouville's theorem (Liouville's theorem: every bounded entire function is constant) makes $q$ constant, hence $h(z)=az$ with $a=h^{\mathrm{\prime }}(0)=f^{\mathrm{\prime }}(0)\mathrm{/}f(0)=D_Q^{\mathrm{\prime }}(0)\mathrm{/}{D}_Q(0)$. Thus $D_Q(z)=e^{az}$.

By [L2] and $D_Q(0)=1$, the coefficient in step 6.1 is $a=D_Q^{\mathrm{\prime }}(0)=\mathrm{tr}(Q)$.

Suppose for contradiction that $a\mathrm{\ne }0$, and choose $\epsilon =\mathrm{\mid }a\mathrm{\mid }\mathrm{/}2>0$. The bound [A8] has $C_{\epsilon }\ge 1$ by evaluation at zero and [L2]. Set $t=2(\log C_{\epsilon }+1)\mathrm{/}\mathrm{\mid }a\mathrm{\mid }>0$ and $z_t=t\bar{a}\mathrm{/}\mathrm{\mid }a\mathrm{\mid }$. By [A13, L4], $\mathrm{\mid }{z}_t\mathrm{\mid }=t$ and $a{z}_t=t\mathrm{\mid }a\mathrm{\mid }$. Using step 6.1, [A12] and the [A8] bound gives $e^{t\mathrm{\mid }a\mathrm{\mid }}=\mathrm{\mid }{D}_Q(z_t)\mathrm{\mid }\le C_{\epsilon }{e}^{\epsilon t}$. Apply [A10, A11, L5] to take logarithms: $t\mathrm{\mid }a\mathrm{\mid }\le \log C_{\epsilon }+\epsilon t$, hence $t\mathrm{\mid }a\mathrm{\mid }\mathrm{/}2\le \log C_{\epsilon }$. But the definition of $t$ makes the left side $\log C_{\epsilon }+1$, a contradiction. Thus $a=0$.

If $Q=0$, [A7] with $E=\{0\}$ gives $D_Q\equiv 1$, and [L2] then gives $\mathrm{tr}(Q)=D_Q^{\mathrm{\prime }}(0)=0$; this includes $H=\{0\}$. On $H=\mathbb{C}$ with $Q=qI$, the spectral condition forces $q=0$. Indeed, if $q\mathrm{\ne }0$, then $Q1=q1$ and $qI-Q=0$ is not invertible, so [A4, L1] give $q\in \sigma (Q)$. For $q=0$, [A7] with $E=H$ calculates $D_Q(z)=1+zq=1$, and [L2] gives trace zero. If $\sigma (Q)=\mathrm{\varnothing }$, step 1.1 still applies for every nonzero $z$ and no spectral enumeration is used. There is no endpoint parameter. The exact assumption is AC [A1]; it supplies AC${}_{\omega }$ through [A14] for the trace-class and determinant inputs. The scalar argument uses no further choice. The claim is an implication, not an equivalence, so both iff directions are inapplicable.

Step 6.1 with $a=0$ gives $D_Q(z)=1$ for every $z$; step 7.1 gives $\mathrm{tr}(Q)=0$. This proves both conclusions. [step 6.1, step 7.1, step 7.2] \qed