Derivative of the quadrant modification

Put $A(s,u)=\mathcal{G}(s,u)-\mathcal{G}(u,s)$, so $A(u,s)=-A(s,u)$. For a matrix unit $e_{ij}$, the coordinate $c_{ij,kl}(T)=(T(e_{ij}){)}_{kl}$ satisfies $\mathrm{\mid }{c}_{ij,kl}(T)\mathrm{\mid }\le \mathrm{\parallel }T\mathrm{\parallel }$, because $\mathrm{\parallel }{e}_{ij}{\mathrm{\parallel }}_{\mathrm{s}\mathrm{p}}=1$ and every matrix entry is bounded by the spectral norm. Conversely, let $H_{ij,kl}(X)=X_{ij}{e}_{kl}$. Then $\mathrm{\parallel }{H}_{ij,kl}\mathrm{\parallel }=1$ and $T={\sum }_{ij,kl}{c}_{ij,kl}(T)H_{ij,kl}$, so $\mathrm{\parallel }T\mathrm{\parallel }\le {\sum }_{ij,kl}\mathrm{\mid }{c}_{ij,kl}(T)\mathrm{\mid }$. There are $N=d^4$ such coordinates. The coordinate-swap map is measurable on the product sigma-algebra (inverse images of rectangles are rectangles), hence $A$, its coordinates and its norm are product-measurable. Apply [F7] to $q\in L^1(\mathbb{R})$ with the affine maps $s\mapsto s-v$ and $s\mapsto v-s$. Both transformed integrals equal $\int q=\mathrm{\Gamma }\mathrm{/}2$; thus for every $v$, ${\int }_{\mathbb{R}}\mathrm{\parallel }A(s,v)\mathrm{\parallel }ds\le {\int }_{\mathbb{R}}(q(s-v)+q(v-s))ds=\mathrm{\Gamma }$. The same bound holds for the other section orientation by these same two substitutions.

By [F5], the intervals $[-m,m]$ of finite measure cover $\mathbb{R}$, so both Lebesgue spaces are sigma-finite. For a fixed $c\in \mathbb{R}$, the sets $\mathbb{R}\times \{c\}$ and $\{c\}\times \mathbb{R}$ are product-measurable. Tonelli [F2] applied to their indicators, integrating first in the singleton coordinate, gives product measure zero: each inner integral is $\lambda (\{c\})=0$ by [F5] and [F6], and the outer integral of the zero function is zero by [F4]. This avoids multiplying zero by infinite measure. By [F6], the integral of $\mathrm{\parallel }A\mathrm{\parallel }$ over either line is zero; the coordinate bound in step 1.1 and [F4] then show that every coordinate integral over either line is zero. The same holds over finite unions of these lines by splitting into disjoint measurable pieces. Also [F5] and [F6] give ${\int }_{J}1d\lambda =b-a$ for every interval $J$ with finite endpoints $a<b$, independently of its endpoint convention; [F4] supplies the corresponding constant-integrand formulas.

Put $f(r)=q(r)+q(-r)$. By [F7] and scalar linearity [F4], $f$ is measurable and integrable with $\int f=\mathrm{\Gamma }$. Applying [F7] to the integrable function $\mathrm{\mid }r\mathrm{\mid }q(r){\mathbf{1}}_{\{r<0\}}$ with reflection, and splitting its first moment by [F4], gives ${\int }_0^{\mathrm{\infty }}rf(r)dr={\int }_{\mathbb{R}}\mathrm{\mid }r\mathrm{\mid }q(r)dr\le \mathrm{\Gamma }\tau$. Fix $t$ and set $Q_t=\{(s,u):s>t,u<t\}$. For each $s>t$, apply [F7] with $u\mapsto s-u$ to the integrable function $f(r){\mathbf{1}}_{\{r>s-t\}}$ to obtain ${\int }_{-\mathrm{\infty }}^{t}f(s-u)du={\int }_{s-t}^{\mathrm{\infty }}f(r)dr$. Since $\mathrm{\parallel }A(s,u)\mathrm{\parallel }\le f(s-u)$, Tonelli [F2] applied to the product-measurable function ${\mathbf{1}}_{\{t<s<t+r\}}f(r)$ yields ${\int }_t^{\mathrm{\infty }}ds{\int }_{-\mathrm{\infty }}^{t}du\mathrm{\parallel }A(s,u)\mathrm{\parallel }\le {\int }_0^{\mathrm{\infty }}dr{\int }_t^{t+r}dsf(r)={\int }_0^{\mathrm{\infty }}rf(r)dr\le \mathrm{\Gamma }\tau$. The last equality uses the constant-integrand formula of step 2.1. Apply [F2] also to ${\mathbf{1}}_{Q_t}\mathrm{\parallel }A\mathrm{\parallel }$, which is product-measurable by step 1.1. Its product integral is finite by the preceding iterated bound. Each coordinate of ${\mathbf{1}}_{Q_t}A$ is therefore absolutely product-integrable. Its actual inner section integrals are finite at every parameter by step 1.1 and measurable by applying [F2] to the positive and negative real and imaginary coordinate parts. Thus [F3] identifies their product and iterated integrals, and the definition of $M$ in [F1] gives $M(t)={\int }_{Q_t}Ad(\lambda \times \lambda )$ coordinatewise for every $t$. This derives the needed quadrant identity locally.

Fix $a<b$ and put $I=(a,b)$. Use the sigma-finiteness and interval-integral formulas from step 2.1. Apply [F2] to the product-measurable function ${\mathbf{1}}_I(v)\mathrm{\parallel }A(s,v)\mathrm{\parallel }$; step 1.1 and monotonicity give ${\int }_{\mathbb{R}\times I}\mathrm{\parallel }A(s,v)\mathrm{\parallel }d(\lambda \times \lambda )\le \mathrm{\Gamma }(b-a)<\mathrm{\infty }$. Thus each coordinate of $A$ restricted to this strip, its measurable subsets, and the square $I\times I$ has integrable absolute value. The quadrant restrictions are also absolutely integrable by step 3.1. Apply [F3] coordinatewise on these regions whenever changing integration order. Every section needed here is integrable by step 1.1; [F2] applied to the positive and negative parts of its real and imaginary coordinates gives measurable actual section integrals. Thus the almost-everywhere convention in [F3] does not affect the outer integrals or substitute for any fixed-time marginal.

Write $Q_t=\{(s,u):s>t,u<t\}$ and ${\int }_{D}A={\int }_{D}Ad(\lambda \times \lambda )$, interpreted coordinatewise. Use the locally derived product-integral identity in step 3.1. Subtract the absolutely convergent quadrant integrals and split their indicators using [F4]. Up to null boundary lines, the gained region is $(b,\mathrm{\infty })\times I$ and the lost region is $I\times (-\mathrm{\infty },a)$. Consequently $M(b)-M(a)={\int }_{I}dv{\int }_b^{\mathrm{\infty }}A(s,v)ds-{\int }_{I}dv{\int }_{-\mathrm{\infty }}^{a}A(v,u)du={\int }_{I}dv{\int }_{\mathbb{R}\setminus I}A(s,v)ds$, where antisymmetry changes the second minus sign to plus. The square integral $S={\int }_{I}dv{\int }_{I}A(s,v)ds$ equals $-S$ by [F3] and antisymmetry, so $S=0$. Adding it gives $M(b)-M(a)={\int }_a^{b}dv{\int }_{\mathbb{R}}A(s,v)ds$. Every split and change of order is among the integrable restrictions established in step 4.1; step 2.1 supplies the zero product measures and zero coordinate integrals of all boundary lines.

By the definitions in [F1] and scalar linearity [F4], ${\int }_{\mathbb{R}}A(s,v)ds=L(v)-B(v)$ for every $v$. Put $F=L-B$. For each $t$ and $\epsilon >0$, choose a continuity radius for each of $L,B$ with tolerance $\epsilon \mathrm{/}2$ and take their minimum. The triangle inequality gives $\mathrm{\parallel }F(v)-F(t)\mathrm{\parallel }\le \mathrm{\parallel }L(v)-L(t)\mathrm{\parallel }+\mathrm{\parallel }B(v)-B(t)\mathrm{\parallel }<\epsilon$ within this radius, so $F$ is continuous in the sense defined in the Statement. Step 5.1 gives $M(b)-M(a)={\int }_a^{b}F(v)dv$ for every $a<b$. Therefore for every nonzero real $h$, $(M(t+h)-M(t))\mathrm{/}h-F(t)=h^{-1}{\int }_t^{t+h}(F(v)-F(t))dv$, with ${\int }_t^{t+h}=-{\int }_{t+h}^t$ when $h<0$. All these integrals are coordinate integrals of integrable functions; the interval-integral formula in step 2.1 gives $hF(t)$ for the constant term, with the stated orientation when $h<0$.

Fix $t$ and $\epsilon >0$. Continuity of $F$ gives $\delta >0$ such that $\mathrm{\parallel }F(v)-F(t)\mathrm{\parallel }<\epsilon \mathrm{/}N$ whenever $\mathrm{\mid }v-t\mathrm{\mid }<\delta$. If $0<\mathrm{\mid }h\mathrm{\mid }<\delta$, each coordinate of the last expression in step 6.1 has modulus at most $\mathrm{\mid }h{\mathrm{\mid }}^{-1}{\int }_{\min (t,t+h)}^{\max (t,t+h)}\mathrm{\mid }{c}_{ij,kl}(F(v)-F(t))\mathrm{\mid }dv\le \epsilon \mathrm{/}N$ by [F4], the coordinate bound in step 1.1 and the interval-integral formula in step 2.1. Summing the $N$ coordinate estimates gives $\mathrm{\parallel }(M(t+h)-M(t))\mathrm{/}h-F(t)\mathrm{\parallel }\le \epsilon$. This proves the two-sided derivative $\dot{M}(t)=F(t)=L(t)-B(t)$ at every $t$; it is continuous since $F$ is continuous.