The square of the Volterra operator has zero trace

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.