Classical transverse Fourier amplitudes

In the whole-space componentwise convention, the absolute-tail estimate in the Definition makes the integer-radius integrals of each real or imaginary component a real Cauchy sequence. Its convergence follows from [F8], and the same uniform tail estimate gives that limit for all real radii tending to infinity. This argument applies separately to each of the finitely many components and introduces no choice principle. Since $K$ is compact and excludes $0$, it has positive distance from the origin. Thus extending $a$ by zero across a neighborhood of $0$ gives a smooth compactly supported map on all of ${\mathbb{R}}^3$. For any compact time interval $J$, each function $f_t(k):=a(k)\exp (-ic\mathrm{\mid }k\mathrm{\mid }t)$ and each of its $t$-derivatives is smooth and supported in $K$; all its derivatives are uniformly bounded on $K\times J$. Every integral in the Definition is therefore an ordinary Riemann integral of a continuous compactly supported function. Differentiating in $x$ or $t$ under the integral is justified because the corresponding difference quotients converge uniformly on $K$ for each fixed $(x,t)$: the phase derivatives in $(x,t)$ are bounded because $k$ ranges over $K$ and $t$ lies in $J$, while the complex exponential has unit modulus. The one-variable mean-value estimate for the exponential is therefore uniform in $k$. The absolute-integral estimate in [F5] then bounds the difference between the integrals by the volume of a containing cube times the uniform error, which tends to zero. Iterating this argument for the resulting derivatives proves existence and continuity of every mixed space-time derivative.

For multi-indices $\alpha ,\beta$, differentiating in $x$ multiplies the integrand by $(ik{)}^{\beta }$. To control a factor $x_j$, integrate the $k_j$-derivative of the product of the compactly supported factor and $\exp (ik\cdot x)$ over a containing cube. The one-dimensional FTC in [F5] gives zero boundary terms because the factor vanishes near both endpoints; Riemann-Fubini in [F5] integrates that identity over the other two coordinates. Repeating this argument transfers every monomial $x^{\alpha }$ to derivatives in $k$, leaving an integral of ${\mathrm{\partial }}_k^{\alpha }((ik{)}^{\beta }{f}_t(k))$ times a unit-modulus exponential, up to the fixed factor of modulus one produced by integration by parts. Its absolute value is uniformly bounded in $x$ and locally uniformly in $t$. The same argument applies after any $t$-derivative, since it only adds the smooth factor $(-ic\mathrm{\mid }k\mathrm{\mid }{)}^q$ on $K$. By [F1], $A_{+}$, $A$, $E$, and $B$ are therefore Schwartz in $x$ at each $t$, with the claimed local bounds.

Expanding ${\prod }_{j=1}^3(1+\mathrm{\mid }{x}_j\mathrm{\mid }{)}^2$ gives finitely many absolute monomials, each controlled by a Schwartz seminorm in [F1]. Hence each Schwartz component $h$ is bounded by $C_h{\prod }_j(1+\mathrm{\mid }{x}_j\mathrm{\mid }{)}^{-2}$. Applying this bound to $E$ and $B$ gives a constant $C\ge 0$, depending on the fixed $t$, such that $0\le u(x,t)\le CG(x)$, where $G(x):={\prod }_{j=1}^3(1+\mathrm{\mid }{x}_j\mathrm{\mid }{)}^{-4}$; nonnegativity follows from [F4]. For $R>0$, the ordinary one-dimensional Riemann integral and [F5] give ${\int }_{-R}^R(1+\mathrm{\mid }s\mathrm{\mid }{)}^{-4}ds=2{\int }_0^R(1+s{)}^{-4}ds=\frac{2}{3}(1-(1+R{)}^{-3})\le \frac{2}{3}\mathrm{.}$ Riemann-Fubini in [F5] therefore gives ${\int }_{Q_R}G{d}^{3}x={(\frac{2}{3}(1-(1+R{)}^{-3}))}^3\le (2\mathrm{/}3{)}^3$, and so $0\le W_R(t):={\int }_{Q_R}u(x,t)d^{3}x\le C(2\mathrm{/}3{)}^3$ for every $R>0$. If $R^{\mathrm{\prime }}>R$, cutting $Q_{R^{\mathrm{\prime }}}$ at the six coordinate planes $x_j=\pm R$ decomposes it into finitely many rectangles, including $Q_R$; finite-slice additivity and nonnegativity of the remaining cell integrals in [F5] give $W_R(t)\le W_{R^{\mathrm{\prime }}}(t)$. Thus $\{W_R(t):R>0\}$ is nonempty and bounded above. Let $W(t)$ be its supremum, which exists by [F8]. For every $\epsilon >0$, the least-upper-bound property gives some $R_0>0$ with $W(t)-\epsilon <W_{R_0}(t)$; for all $R\ge R_0$, monotonicity gives $W(t)-\epsilon <W_R(t)\le W(t)$. Hence ${\lim }_{R\to \mathrm{\infty }}{W}_R(t)=W(t)<\mathrm{\infty }$, which is the whole-space energy defined above.

Put $\gamma :=(2\pi {)}^{-3\mathrm{/}2}$. Direct differentiation gives $\mathrm{div}A=2\gamma \mathrm{Re}\int i(k\cdot a(k))\exp (i(k\cdot x-c\mathrm{\mid }k\mathrm{\mid }t))d^{3}k=0$. The identity $\phi =0$ and [F2] show that $(\phi ,A)$ represents exactly the displayed $(E,B)$ and is in Coulomb gauge.

Put $q(k,x,t):=\exp (i(k\cdot x-c\mathrm{\mid }k\mathrm{\mid }t))$ and $\gamma :=(2\pi {)}^{-3\mathrm{/}2}$. By [F3] and real-linearity of $\mathrm{Re}$, $E=2\gamma \mathrm{Re}\int ic\mathrm{\mid }k\mathrm{\mid }a(k)q{d}^{3}k$ and $B=2\gamma \mathrm{Re}\int i(k\times a(k))q{d}^{3}k$. Differentiating each integrand using [F3], [F7], and step 1.1 gives $\mathrm{div}E=2\gamma \mathrm{Re}\int -c\mathrm{\mid }k\mathrm{\mid }(k\cdot a(k))q{d}^{3}k=0$, $\mathrm{div}B=2\gamma \mathrm{Re}\int -k\cdot (k\times a(k))q{d}^{3}k=0$, $\mathrm{curl}E+{\mathrm{\partial }}_{t}B=2\gamma \mathrm{Re}\int [-c\mathrm{\mid }k\mathrm{\mid }k\times a(k)+c\mathrm{\mid }k\mathrm{\mid }k\times a(k)]q{d}^{3}k=0$, and $\mathrm{curl}B-c^{-2}{\mathrm{\partial }}_{t}E=2\gamma \mathrm{Re}\int [-k\times (k\times a(k))-\mathrm{\mid }k{\mathrm{\mid }}^{2}a(k)]q{d}^{3}k=0$. The last equality uses the cross-product identity in [F7] and $k\cdot a(k)=0$. These are the source-free equations in [A1], proving that the integrated fields satisfy the adopted smooth Maxwell branch.

The exponent $k\cdot x-c\mathrm{\mid }k\mathrm{\mid }t$ is dimensionless. Since $d^{3}k$ has units ${\mathrm{m}}^{-3}$, the assigned $[a]=\mathrm{T}{\mathrm{m}}^4$ gives $[A]=\mathrm{T}\mathrm{m}$; then $E=-{\mathrm{\partial }}_{t}A$ has units $\mathrm{T}\mathrm{m}{\mathrm{s}}^{-1}=\mathrm{V}{\mathrm{m}}^{-1}$, and $B=\mathrm{curl}A$ has units $\mathrm{T}$. These agree with [F6].