Soft-current angular norm and exact cutoff coherent model

On the open set $U=(0,R)\times (0,\pi \mathrm{/}2)\times (0,2\pi )\times (0,2\pi )$ define $T(r,\theta ,\varphi ,\psi )=(r\cos \theta \cos \varphi ,r\cos \theta \sin \varphi ,r\sin \theta \cos \psi ,r\sin \theta \sin \psi )\mathrm{.}$ Let $L={\mathbb{R}}^2\setminus \{(a,0):a\ge 0\}$, the plane slit along the nonnegative horizontal ray. For $(a,b)\in L$, put $\rho =\sqrt{a^2+b^2}>0$ and define ${\mathrm{Arg}}_L(a,b)$ to be $\arccos (a\mathrm{/}\rho )$ if $b>0$, $2\pi -\arccos (a\mathrm{/}\rho )$ if $b<0$, and $\pi$ if $b=0$ (then $a<0$). The inverse and branch identities [F8]–[F10] give $\cos {\mathrm{Arg}}_L=a\mathrm{/}\rho$ and $\sin {\mathrm{Arg}}_L=b\mathrm{/}\rho$ and make this the unique angle in $(0,2\pi )$. Each non-axis branch is $C^1$ by [F9]. Near the negative horizontal axis, the same angle equals $\pi -\arcsin (b\mathrm{/}\rho )$, so it is $C^1$ there too. Positive square roots are $C^1$: continuity follows from $\mathrm{\mid }\sqrt{u}-\sqrt{v}\mathrm{\mid }\le \sqrt{\mathrm{\mid }u-v\mathrm{\mid }}$, and the difference quotient is $1\mathrm{/}(\sqrt{u+h}+\sqrt{u})$, with continuous limit $1\mathrm{/}(2\sqrt{u})$ for $u>0$. Thus all radii in these formulas are $C^1$ by [F7]. Define the open set $V=\{x:\mathrm{\mid }x\mathrm{\mid }<R,(x_1,x_2)\in L,(x_3,x_4)\in L\}$. Write $A=\sqrt{x_1^2+x_2^2}>0$, $B=\sqrt{x_3^2+x_4^2}>0$. The inverse of $T$ on $V$ is explicitly $r=\sqrt{A^2+B^2}$, $\theta =\arccos (A\mathrm{/}r)\in (0,\pi \mathrm{/}2)$, $\varphi ={\mathrm{Arg}}_L(x_1,x_2)$, $\psi ={\mathrm{Arg}}_L(x_3,x_4)$. Indeed $\cos \theta =A\mathrm{/}r$, $\sin \theta =B\mathrm{/}r$ by [F8]–[F10], and substitution into $T$ gives $x$; the same branch identities recover every parameter of $U$. Both directions are $C^1$, so $T:U\to V$ is a $C^1$ diffeomorphism. Using [F6]–[F8], The four derivative columns are pairwise orthogonal, with squared lengths $1,r^2,r^2{\cos }^2\theta ,r^2{\sin }^2\theta$; hence $\mathrm{\mid }\det DT\mathrm{\mid }=r^3\sin \theta \cos \theta$. The complement of $V$ in the closed ball is contained in the union of the boundary sphere and the four coordinate hyperplanes: the omitted angular branch cuts lie in $x_2=0$ or $x_4=0$, and the omitted endpoint planes lie in $x_1=x_2=0$ or $x_3=x_4=0$. The coordinate hyperplanes are null by [F2]. The boundary sphere is null by [F4], applied to its indicator, since its polar radial integral is supported at the singleton $\{R\}$, which has one-dimensional Lebesgue measure zero. The origin is already in the coordinate hyperplanes.

Apply [F1] to $T$ and the nonnegative Borel function $x\mapsto \phi (\mathrm{\mid }x\mathrm{\mid })$ on $V$. The null-complement calculation in step 1.1 shows that its integral over $V$ equals its integral over the closed ball, even when $\phi$ takes the value $+\mathrm{\infty }$ on a radius whose preimage in the omitted set is null. Therefore the ball integral equals ${\int }_U\phi (r)r^3\sin \theta \cos \theta drd\theta d\varphi d\psi \mathrm{.}$

The integrand in step 2.1 is nonnegative and product-measurable, so [F3] separates the four integrations. The angular factors are ${\int }_0^{\pi \mathrm{/}2}\sin \theta \cos \theta d\theta =1\mathrm{/}2$ and ${\int }_0^{2\pi }1d\varphi ={\int }_0^{2\pi }1d\psi =2\pi$; the first equality follows from [F5] with antiderivative ${\sin }^2\theta \mathrm{/}2$, whose derivative is $\sin \theta \cos \theta$ by [F6]–[F7]. The angular integrands are continuous and bounded on their compact intervals, so [F11] identifies these calculus evaluations with the Lebesgue integrals in [F3]. Their product is $2{\pi }^2$, which proves the formula in the Statement. Endpoints do not change any Lebesgue integral, and all equalities are valid as extended nonnegative integrals.

The integrand $r\mapsto r^3\mathrm{/}(r^2+{\mu }^2)$ is nonnegative and continuous on $[0,\mathrm{\Lambda }]$. Apply [F2], then divide by $(2\pi {)}^4$, to obtain $I_{\mathrm{\Lambda }}=\frac{1}{8{\pi }^2}{\int }_0^{\mathrm{\Lambda }}\frac{r^3}{r^2+{\mu }^2}dr\mathrm{.}$ Polynomial division gives $r^3\mathrm{/}(r^2+{\mu }^2)=r-{\mu }^{2}r\mathrm{/}(r^2+{\mu }^2)$. By [F4], an antiderivative is $G(r)=r^2\mathrm{/}2-({\mu }^2\mathrm{/}2)\log (r^2+{\mu }^2)$ on this interval. Its derivative is continuous; its compact-interval integrands are bounded and Riemann integrable, so [F9] identifies their Lebesgue integrals with the calculus integrals. Thus [F3] evaluates the integral as $G(\mathrm{\Lambda })-G(0)=\frac{1}{2}({\mathrm{\Lambda }}^2-{\mu }^2\log (1+{\mathrm{\Lambda }}^2\mathrm{/}{\mu }^2))\mathrm{.}$ Multiplying by $1\mathrm{/}(8{\pi }^2)$ gives the claimed expression.

By [F6], $-\exp (-x)$ has derivative $\exp (-x)$; [F3] on $[0,R]$ gives ${\int }_0^R\exp (-x)dx=1-\exp (-R)$. The positive series [F5] gives $\exp (R)\ge 1+R$, so $\exp (-R)\to 0$ and $M_0=1$. For each integer $n\ge 0$, [F5] with $y=x\mathrm{/}2$ gives $x^n\exp (-x)\le 2^{n}n!\exp (-x\mathrm{/}2)$. The majorant has finite integral $2^{n+1}n!$ by [F3] and [F6]; the positive truncated integrals are increasing and bounded by this finite integral, hence have a finite improper limit. All the integrands are continuous on every compact interval; [F9] identifies these convergent nonnegative improper integrals with their half-line Lebesgue integrals. Thus $M_n={\int }_0^{\mathrm{\infty }}{x}^n\exp (-x)dx$ is finite. The same bound and $\exp (x\mathrm{/}2)\ge 1+x\mathrm{/}2$ show $x^n\exp (-x)\to 0$. For $n\ge 1$ and $R>0$, take $u(x)=x^n$, $v(x)=-\exp (-x)$ on $[0,R]$. They are continuously differentiable with continuous, hence integrable, derivatives. Thus [F7] gives ${\int }_0^R{x}^n\exp (-x)dx=-R^n\exp (-R)+n{\int }_0^R{x}^{n-1}\exp (-x)dx\mathrm{.}$ The endpoint at zero vanishes since $n\ge 1$. Taking $R\to \mathrm{\infty }$ is justified by the finite positive improper integrals and the vanishing boundary term just proved. It gives $M_n=n{M}_{n-1}$, and induction from $M_0=1$ gives $M_n=n!$. Finally $0\le \exp (-x)\mathrm{/}(1+\lambda x)\le \exp (-x)$ for $\lambda \ge 0$, so $F(\lambda )$ is finite.

Substituting the formula from step 1.1, the subtracted quantity equals $\frac{{\mu }^2}{16{\pi }^2}\log \left(\frac{{\mu }^2}{{\mu }_R^2}\frac{{\mathrm{\Lambda }}^2}{{\mathrm{\Lambda }}^2+{\mu }^2}\right)\mathrm{.}$ As $\mathrm{\Lambda }\to \mathrm{\infty }$, the second factor inside the logarithm tends to $1$; the limiting logarithm is defined because ${\mu }^2\mathrm{/}{\mu }_R^2>0$. Continuity of $\log$ on $(0,\mathrm{\infty })$ proves the stated finite limit.

For every $t\ge 0$, finite multiplication of a geometric sum gives $\frac{1}{1+t}={\sum }_{n=0}^N(-t{)}^n+(-1{)}^{N+1}\frac{t^{N+1}}{1+t}\mathrm{.}$ Put $t=\lambda x$, multiply by $\exp (-x)$, and integrate. Step 1.2 evaluates each polynomial moment as $n!$, while the remainder is bounded by ${\lambda }^{N+1}{M}_{N+1}={\lambda }^{N+1}(N+1)!$. This gives the exact finite identity and its stated bound, including $\lambda =0$.

For each fixed $\lambda >0$, the absolute terms $n!{\lambda }^n$ in the formal real power series have successive ratio $(n+1)\lambda$, eventually greater than $2$. They therefore do not tend to zero, so the series does not converge at any positive $\lambda$. By [F10], no $r>0$ has absolute convergence at every $\mathrm{\mid }\lambda \mathrm{\mid }<r$: choosing $\lambda =r\mathrm{/}2$ contradicts the divergence just proved. The only admissible radius is $r=0$, so the supremum in [F10] is zero. Separately, the bare positive coefficient grows quadratically because $\frac{I_{\mathrm{\Lambda }}}{{\mathrm{\Lambda }}^2}=\frac{1}{16{\pi }^2}(1-\frac{{\mu }^2}{{\mathrm{\Lambda }}^2}\log (1+{\mathrm{\Lambda }}^2\mathrm{/}{\mu }^2))⟶\frac{1}{16{\pi }^2}\mathrm{.}$ For the last limit, with $y=1+{\mathrm{\Lambda }}^2\mathrm{/}{\mu }^2$, [F4] gives $\log y={\int }_1^{y}du\mathrm{/}u\le {\int }_1^{y}du\mathrm{/}\sqrt{u}\le 2\sqrt{y}$, so the logarithmic term divided by ${\mathrm{\Lambda }}^2$ tends to zero.

Steps 1.1 and 2.1 give the exact cutoff integral and selected finite subtraction limit; steps 1.2 and 2.2 prove the finite factorial remainder; step 2.3 proves that the formal power series diverges for every $\lambda >0$. These conclusions concern only the two displayed mathematical integrals and do not assert a full QED renormalization theorem or a continuum limit.

For every leg and $n\in S^2$, $1-v_i\cdot n\ge 1-\mathrm{\mid }{v}_i\mathrm{\mid }>0$. Since the leg set is finite, each summand of $K(n)$ is continuous and bounded on $S^2$. Moreover $h(K(n),(1,n))={\sum }_i{\eta }_i{Q}_i\frac{1-v_i\cdot n}{1-v_i\cdot n}={\sum }_i{\eta }_i{Q}_i=0.$ Thus $K^0(n)=n\cdot \mathbf{K}(n)$ and $h(K,K)=(K^0{)}^2-\mathrm{\mid }\mathbf{K}{\mathrm{\mid }}^2=-\mathrm{\mid }{P}_T(n)\mathbf{K}(n){\mathrm{\mid }}^2$. Consequently the angular integrand is nonnegative, bounded and measurable; the coefficient $C$ is finite and nonnegative.

By [F10] at $p=2$, the real-valued scalar space $L^2({\mathbb{R}}^3;\mathbb{R})$ has a countable dense set $D$. It is nonempty because this scalar $L^2$ space is nonempty and $D$ is dense. By [F12], $D\times D$ is at most countable; [F14] supplies a surjection from $\mathbb{N}$ onto this nonempty product. Composing with $(d,e)\mapsto d+ie$ and applying [F14] again shows that $D_{\mathbb{C}}:=\{d+ie:d,e\in D\}$ is at most countable. For every complex scalar $u\in L^2$, its real and imaginary parts $u_R,u_I$ belong to real $L^2$, and $\mathrm{\parallel }u-(d+ie){\mathrm{\parallel }}_2^2=\mathrm{\parallel }{u}_R-d{\mathrm{\parallel }}_2^2+\mathrm{\parallel }{u}_I-e{\mathrm{\parallel }}_2^2$; density of $D$ therefore makes $D_{\mathbb{C}}$ dense in complex scalar $L^2$. By two applications of [F12], $D_{\mathbb{C}}^3$ is countable and dense in $L^2({\mathbb{R}}^3;{\mathbb{C}}^3)$, since each of three components can be approximated with squared error below ${\epsilon }^2\mathrm{/}3$. For $n\in S^2$ and $z\in {\mathbb{C}}^3$, put $a={\sum }_j{n}_j{z}_j$. Using [F13] and the reality of $n_j$, direct expansion gives $\mathrm{\parallel }{P}_T(n)z{\mathrm{\parallel }}^2={\sum }_j\mathrm{\mid }{z}_j-n_{j}a{\mathrm{\mid }}^2={\sum }_j\mathrm{\mid }{z}_j{\mathrm{\mid }}^2-\mathrm{\mid }a{\mathrm{\mid }}^2\le \mathrm{\parallel }z{\mathrm{\parallel }}^2$, and $(I-n{n}^T{)}^2=I-n{n}^T$. Thus $M_{P_T}$ is a bounded idempotent on the ambient vector-valued $L^2$ space. Its image of $D_{\mathbb{C}}^3$ is countable and dense in its range: for $z=M_{P_T}z$ choose $w\in D_{\mathbb{C}}^3$ close to $z$, and $\mathrm{\parallel }{M}_{P_T}w-z{\mathrm{\parallel }}_2\le \mathrm{\parallel }w-z{\mathrm{\parallel }}_2$.

By [F6] and [F7], the $d\mathrm{\Omega }$ used in the statement is the finite Euclidean surface measure, and [F8] gives $d^{3}q\mathrm{/}((2\pi {)}^{3}2\omega )=\omega d\omega d\mathrm{\Omega }\mathrm{/}(2(2\pi {)}^3)$. Multiplying by $\mathrm{\mid }{h}_{\lambda }(\mathbf{q}){\mathrm{\mid }}^2=g^2\mathrm{\mid }{K}_{\perp }(n){\mathrm{\mid }}^2\mathrm{/}{\omega }^2$ and integrating over $\lambda <\omega <\mathrm{\Lambda }$ gives $\mathrm{\parallel }{h}_{\lambda }{\mathrm{\parallel }}^2=\frac{g^2}{2(2\pi {)}^3}{\int }_{S^2}\mathrm{\mid }{K}_{\perp }{\mathrm{\mid }}^{2}d\mathrm{\Omega }{\int }_{\lambda }^{\mathrm{\Lambda }}\frac{d\omega }{\omega }=C\log (\mathrm{\Lambda }\mathrm{/}\lambda )\mathrm{.}$ The substitution $\omega =\lambda u$ and [F15] give the displayed logarithm. The cutoffs are strictly positive and finite, so the norm is finite. The endpoint set $\{\lambda ,\mathrm{\Lambda }\}$ is null by [F11], so the strict cutoff convention gives the same integral as including either endpoint.

Put $a=\mathrm{\hslash }c>0$ and $T(k)=ak$. For $h\in L^2({\mathbb{R}}^3,d^{3}q\mathrm{/}[(2\pi {)}^{3}2\mathrm{\mid }q\mathrm{\mid }];{\mathbb{C}}^3)$, define $G_h(q)=\mathrm{\mid }h(q){\mathrm{\mid }}^2\mathrm{/}[(2\pi {)}^{3}2\mathrm{\mid }q\mathrm{\mid }]$ for $q\mathrm{\ne }0$ and $G_h(0)=0$; the weighted $L^2$ hypothesis gives $G_h\in L^1({\mathbb{R}}^3,d^{3}q)$. The density of the energy measure is positive and finite off $q=0$, so every energy-measure null set is Lebesgue null there; the singleton $\{0\}$ is also Lebesgue null. By [F9], $T^{-1}$ preserves Lebesgue-null sets, so composition by $T$ is well-defined on these $L^2$ equivalence classes. Define $Uh(0)=0$ and, for $k\mathrm{\ne }0$, $(Uh)(k)=\frac{a}{\sqrt{(2\pi {)}^{3}2\mathrm{\mid }k\mathrm{\mid }}}h(ak)\mathrm{.}$ The change-of-variables formula [F9] for the $L^1$ function $G_h$ gives $\mathrm{\parallel }Uh{\mathrm{\parallel }}_{L^2(d^{3}k)}^2=a^3{\int }_{{\mathbb{R}}^3}{G}_h(ak)d^{3}k={\int }_{{\mathbb{R}}^3}{G}_h(q)d^{3}q=\mathrm{\parallel }h{\mathrm{\parallel }}^2\mathrm{.}$ Conversely, for $f\in L^2(d^{3}k;{\mathbb{C}}^3)$ define $Vf(0)=0$ and $Vf(q)=\sqrt{(2\pi {)}^{3}2\mathrm{\mid }q\mathrm{\mid }}{a}^{-3\mathrm{/}2}f(q\mathrm{/}a)$ for $q\mathrm{\ne }0$. Applying [F9] to $\mathrm{\mid }f{\mathrm{\mid }}^2$ and the inverse dilation shows $\mathrm{\parallel }Vf\mathrm{\parallel }=\mathrm{\parallel }f\mathrm{\parallel }$; direct substitution gives $UVf=f$ and $VUh=h$ almost everywhere. Thus $U$ is a unitary on the ambient spaces. Since $q=ak$ and the scalar multipliers do not change the transverse condition, with null sets preserved under $T$ and $T^{-1}$ the equivalence $q\cdot h(q)=0$ a.e. iff $k\cdot (Uh)(k)=0$ a.e. follows. Therefore $U$ restricts to a unitary ${\mathcal{H}}_{E,T}\to {\mathcal{H}}_T$; in particular the energy-coordinate transverse subspace is a Hilbert space because [F2] makes ${\mathcal{H}}_T$ a closed Hilbert space. Substitution of $h_{\lambda }$ gives $(U{h}_{\lambda })(k)={\mathbf{1}}_{\{\lambda \mathrm{/}(\mathrm{\hslash }c)<\mathrm{\mid }k\mathrm{\mid }<\mathrm{\Lambda }\mathrm{/}(\mathrm{\hslash }c)\}}\mathrm{\mid }k{\mathrm{\mid }}^{-3\mathrm{/}2}\frac{g{K}_{\perp }(k\mathrm{/}\mathrm{\mid }k\mathrm{\mid })}{\sqrt{2(2\pi {)}^3}},$ whose angular coefficient has squared integral $C$ by the definition of $C$.