Radial primitive and five differentiated decay bounds

Put $E=\{x:\mathrm{\mid }x\mathrm{\mid }>R\}$ and $h_i(s,x)=s{x}^j{B}_{ji}(sx)$. For fixed $s\ge 1$, fixing the other two coordinates makes $x^{\mathrm{\ell }}\mapsto B_{ji}(sx)$ a composition of the corresponding one-variable coordinate slice of $B_{ji}$ with $x^{\mathrm{\ell }}\mapsto s{x}^{\mathrm{\ell }}$. The slice is $C^5$ and the inner map is affine, so [F8] applies successively through five coordinate derivatives, with one factor $s$ per chain-rule differentiation. The product rule also differentiates the factor $x^j$ at most once. Thus, for $\mathrm{\mid }\beta \mathrm{\mid }=k\le 5$, these rules express ${\mathrm{\partial }}_x^{\beta }{h}_i$ as a sum of terms of the forms $s^{k+1}{x}^j({\mathrm{\partial }}^{\beta }{B}_{ji})(sx)$ and, when $k\ge 1$, constant multiples of $s^k({\mathrm{\partial }}^{\beta -e_{\mathrm{\ell }}}{B}_{\mathrm{\ell }i})(sx)$ with ${\beta }_{\mathrm{\ell }}>0$. Thus, whenever $sr\ge R_1$, each sum is bounded by $C_k{r}^{-3-k}{s}^{-3}$. Let $K\subset E$ be nonempty and compact, let $a={\min }_K\mathrm{\mid }x\mathrm{\mid }>R$, and choose $S>1$ with $Sa\ge R_1$. On $K\times [1,S]$ the same derivatives are continuous and hence bounded. Enlarging a constant to absorb this finite interval gives $\mathrm{\mid }{\mathrm{\partial }}_x^{\beta }{h}_i(s,x)\mathrm{\mid }\le C_{K,\beta }{s}^{-3}$ for all $s\ge 1$ and $x\in K$. This is a local uniform majorant, not a bound uniform up to the inner boundary.

Since ${\int }_1^L{s}^{-3}ds=(1-L^{-2})\mathrm{/}2$, the majorant in step 1.1 is integrable. Comparison gives absolute convergence of every integral of ${\mathrm{\partial }}_x^{\beta }{h}_i$ for $\mathrm{\mid }\beta \mathrm{\mid }\le 5$. These integrals are jointly continuous in $x$: on a compact neighborhood $K$ the tails beyond $L$ are bounded uniformly by $C_{K,\beta }\mathrm{/}(2L^2)$, while the integral over $[1,L]$ is continuous by uniform continuity of its integrand on $K\times [1,L]$ and the estimate of an integral by interval length times the uniform bound. Choose an open rectangular neighborhood with compact closure $K\subset E$. For each $\mathrm{\mid }\beta \mathrm{\mid }\le 4$, fix the other two coordinates and apply [F4] to ${\mathrm{\partial }}_x^{\beta }{h}_i$ with each remaining coordinate as the real parameter and integration domain $(1,\mathrm{\infty })$. The integrand and its parameter derivative are continuous, every base slice is absolutely integrable, and step 1.1 supplies an integrable derivative bound on every compact parameter interval. The endpoint $s=1$ is regular and has no effect on the integral. Successively applying this argument gives ${\mathrm{\partial }}_x^{\beta }{A}_i(x)=-{\int }_1^{\mathrm{\infty }}{\mathrm{\partial }}_x^{\beta }{h}_i(s,x)ds$ for every $\mathrm{\mid }\beta \mathrm{\mid }\le 5$; their joint continuity proves $A\in C^5(E)$.

For $r\ge R_1$ the asymptotic estimate in step 1.1 holds for every $s\ge 1$, so $\mathrm{\mid }{\mathrm{\partial }}_x^{\beta }{A}_i(x)\mathrm{\mid }\le C_k{r}^{-3-k}{\int }_1^{\mathrm{\infty }}{s}^{-3}ds=(C_k\mathrm{/}2)r^{-3-k}$. Hence $A_i=O_5(r^{-3})$ in exactly the convention of [F1]. If the hypotheses hold at every finite order, repeat steps 1.1 and 2.1 at each such order, with constants and radii allowed to depend on the order; this yields smoothness of the same integral without an all-order conclusion from the finite-order hypothesis alone.

Closedness and antisymmetry give ${\mathrm{\partial }}_i{B}_{kj}-{\mathrm{\partial }}_j{B}_{ki}={\mathrm{\partial }}_k{B}_{ij}$. The first derivative formula from step 2.1 therefore yields ${\mathrm{\partial }}_i{A}_j-{\mathrm{\partial }}_j{A}_i=-{\int }_1^{\mathrm{\infty }}[2s{B}_{ij}(sx)+s^2{x}^k{\mathrm{\partial }}_k{B}_{ij}(sx)]ds$. For fixed $x\in E$, the ray ${\gamma }_x(s)=sx$ lies in $E$ on the open interval $(R\mathrm{/}r,\mathrm{\infty })$ containing $[1,L]$. Its derivative is the vector $x$. The $C^5$ coefficients of $B$ have continuous first partial derivatives, so [F9] makes them totally differentiable and gives $\frac{d}{ds}{B}_{ij}(sx)=x^k{\mathrm{\partial }}_k{B}_{ij}(sx)$. The product and power rules in [F8] now identify the integrand as the continuous derivative of the $C^1$ function $s\mapsto s^2{B}_{ij}(sx)$. On $[1,L]$, [F6] evaluates its integral as $L^2{B}_{ij}(Lx)-B_{ij}(x)$. The first term tends to zero by [F1], since it is $O(L^{-2}{r}^{-4})$ as $L\to \mathrm{\infty }$ for fixed $x$. Passing to the improper limit gives ${\mathrm{\partial }}_i{A}_j-{\mathrm{\partial }}_j{A}_i=B_{ij}(x)$, which is $dA=B$ by [F2]. Finally $x^i{A}_i=-{\int }_1^{\mathrm{\infty }}s{x}^i{x}^j{B}_{ji}(sx)ds=0$ by antisymmetry and convergence.