The first-order emitted-photon energy law: factorised kernel, normalisation, concentration at the transition energy and positivity on positive-measure windows

[Arbitrary-coefficient push-forward] Since $\mathfrak{h}$ is linear, the finite sum $M$ lies in it, with $\mathrm{\parallel }M{\mathrm{\parallel }}_2\le {\sum }_j\mathrm{\mid }{c}_j\mathrm{\mid }\mathrm{\parallel }{g}_j{\mathrm{\parallel }}_2<\mathrm{\infty }$. The common cutoff gives $M=0$ a.e. outside $\{0<\epsilon \le \mathrm{\Lambda }\}$. The weight $\mathrm{\parallel }M{\mathrm{\parallel }}^2$ is nonnegative and integrable, and the inverse image of a Borel set under continuous $\epsilon$ is Borel. Thus ${\nu }_b$ is nonnegative and finite, with ${\nu }_b(\mathbb{R})=\mathrm{\parallel }M{\mathrm{\parallel }}_2^2$ and the stated support. For disjoint Borel $B_i$, the inverse-image indicators of their finite unions increase to the indicator of their union; monotone convergence gives countable additivity. The identity $\int fd{\nu }_b=\int f(\epsilon (k))\mathrm{\parallel }M(k){\mathrm{\parallel }}^2{d}^{3}k$ holds first for indicators by definition, then for nonnegative simple functions by finite additivity and then for every nonnegative Borel $f$ by monotone convergence. It also holds for integrable real or complex functions by their positive/negative or real/imaginary decompositions. All conclusions include $M=0$ and do not import the source's special matrix coefficients.

[Exact zeros and their momentum pullback] Since $K_t(0)=t\mathrm{/}{\mathrm{\hslash }}^2>0$, its zero set is exactly $Z_t=\{2\pi \mathrm{\hslash }n\mathrm{/}t:n\in \mathbb{Z}\setminus \{0\}\}$ by [F2], and this is countable and one-dimensionally null by [F4]. The zeros of $K_t(E-\mathrm{\Delta })$ form $\mathrm{\Delta }+Z_t$. For each fixed energy $E$, ${\epsilon }^{-1}(\{E\})$ is empty if $E<0$, is the singleton $\{0\}$ if $E=0$, and is the sphere $\{\mathrm{\mid }k\mathrm{\mid }=E\mathrm{/}\alpha \}$ if $E>0$. The singleton is null by [F4]. For a positive radius $r_0$, apply the polar formula [F4] to the indicator of $\{\mathrm{\mid }k\mathrm{\mid }=r_0\}$: its integral is $\sigma (S^2){\int }_0^{\mathrm{\infty }}{\mathbf{1}}_{\{r_0\}}(r)r^{2}dr=0$ because the one-dimensional singleton is null. Thus every such energy level is null in ${\mathbb{R}}^3$, and the countable union ${\epsilon }^{-1}(\mathrm{\Delta }+Z_t)$ is null by countable subadditivity. Consequently $K_t(\epsilon (k)-\mathrm{\Delta })>0$ for a.e. momentum $k$. This establishes the exact needed pullback statement for arbitrary real $\mathrm{\Delta }$.

[Amplitude and factorization including zero cases] For $t\ge 0$ the time integral $I_t(E)={\int }_0^t{e}^{i(E-\mathrm{\Delta })s\mathrm{/}\mathrm{\hslash }}ds$ is continuous in $E$ and bounded by $t$. Therefore $A_b(k,t)=-(i{\eta }_0\mathrm{/}\mathrm{\hslash })I_t(\epsilon (k))M(k)$ is an $L^2$ vector with norm at most $\mathrm{\mid }{\eta }_0\mathrm{\mid }t\mathrm{\parallel }M{\mathrm{\parallel }}_2\mathrm{/}\mathrm{\hslash }$, and scalar multiplication preserves transversality, so $A_b(\cdot ,t)\in \mathfrak{h}$. At $t=0$ it is zero. For $t>0$, the kernel definition gives $\mathrm{\mid }{I}_t(E){\mathrm{\mid }}^2={\mathrm{\hslash }}^{2}t{K}_t(E-\mathrm{\Delta })$; hence $\mathrm{\parallel }{A}_b(k,t){\mathrm{\parallel }}^2={\eta }_0^{2}t{K}_t(\epsilon (k)-\mathrm{\Delta })\mathrm{\parallel }M(k){\mathrm{\parallel }}^2$ a.e. Integrating with step 1.1 proves the norm formula in (a). If $M=0$ or ${\eta }_0=0$, both sides are zero; no division by that norm was used.

[Normalization when M is nonzero] The defining time integral has modulus at most $t$, so $0\le K_t(x)\le t\mathrm{/}{\mathrm{\hslash }}^2$ everywhere. Thus $d_t:=\int K_t(E-\mathrm{\Delta })d{\nu }_b(E)\le (t\mathrm{/}{\mathrm{\hslash }}^2)\mathrm{\parallel }M{\mathrm{\parallel }}_2^2<\mathrm{\infty }$. If $M\mathrm{\ne }0$, the set on which $\mathrm{\parallel }M(k){\mathrm{\parallel }}^2>0$ has positive Lebesgue measure. By step 1.2, the product $K_t(\epsilon (k)-\mathrm{\Delta })\mathrm{\parallel }M(k){\mathrm{\parallel }}^2$ is positive a.e. on that set, and its integral is positive by [F5]. Step 1.1 makes this integral exactly $d_t$. Hence $B\mapsto d_t^{-1}{\int }_B{K}_t(E-\mathrm{\Delta })d{\nu }_b(E)$ is a Borel probability measure, and its support is in $(0,\mathrm{\Lambda }]$ because ${\nu }_b$ has that support. This proves (b). For ${\eta }_0=0$, $A_b=0$ but this formal normalized kernel law is still defined whenever $M\mathrm{\ne }0$; it is not asserted to arise by normalizing the zero amplitude. If $M=0$ then $d_t=0$ and no normalized law is asserted.

[Concentration for every bounded continuous complex test] Assume (c), and let $f$ be bounded continuous. The density equality and translation from [F5] give $\int fd{P}_t=(\int K_t(x)f(\mathrm{\Delta }+x)J_b(\mathrm{\Delta }+x)dx)\mathrm{/}(\int K_t(x)J_b(\mathrm{\Delta }+x)dx)$; both integrals are finite because $K_t$ is bounded and $J_b,f{J}_b\in L^1$. The function $f{J}_b$ is continuous at $\mathrm{\Delta }$, so [F2] applied to its real and imaginary parts makes the numerator tend to $(2\pi \mathrm{/}\mathrm{\hslash })f(\mathrm{\Delta })J_b(\mathrm{\Delta })$. Applied to $J_b$ it makes the denominator tend to $(2\pi \mathrm{/}\mathrm{\hslash })J_b(\mathrm{\Delta })>0$. The denominator is also strictly positive at every time by step 3.1. The ratio therefore tends to $f(\mathrm{\Delta })$, proving (c), with no restriction that the test or coefficients be real.

[Positive-measure windows] Under (d), the numerator ${\int }_B{K}_t(E-\mathrm{\Delta })J_b(E)dE$ is nonnegative and is strictly positive a.e. on the positive-measure Borel set $B$: $J_b>0$ a.e. there, and the kernel vanishes only on the countable energy set $\mathrm{\Delta }+Z_t$ from step 1.2. By [F5] its integral is positive, whereas step 3.1 gives a finite positive denominator. Thus $P_t(B)>0$ for every $t>0$, proving (d).

[Correctly shifted tail test] Under (c), fix $\delta >0$ and put ${\phi }_{\delta }(x)=\max \{0,\min \{1,2(\mathrm{\mid }x\mathrm{\mid }-\delta \mathrm{/}2)\mathrm{/}\delta \}\}$. This is continuous, lies between zero and one, vanishes on $\mathrm{\mid }x\mathrm{\mid }\le \delta \mathrm{/}2$, and equals one on $\mathrm{\mid }x\mathrm{\mid }\ge \delta$. Use the bounded continuous function $f_{\delta }(E)={\phi }_{\delta }(E-\mathrm{\Delta })$, so $f_{\delta }(\mathrm{\Delta })={\phi }_{\delta }(0)=0$. Pointwise ${\mathbf{1}}_{\{\mathrm{\mid }E-\mathrm{\Delta }\mathrm{\mid }\ge \delta \}}\le f_{\delta }(E)$, whence $0\le P_t(\{\mathrm{\mid }E-\mathrm{\Delta }\mathrm{\mid }\ge \delta \})\le \int f_{\delta }d{P}_t\to f_{\delta }(\mathrm{\Delta })=0$ by step 4.1. This proves (e) with the required energy shift.

[Assembly and scope] Step 2.1 proves (a) including every zero case; step 3.1 proves (b) only for $M\mathrm{\ne }0$; steps 4.1–5.1 give the full concentration, positivity and tail claims. The arbitrary-vector push-forward was established locally in step 1.1, and the scalar kernel/approximate-identity clauses of the source were used unchanged. No dynamics, physical transition probability, rate, detector identification or normalization of a zero physical amplitude was asserted.