Even radial coefficients give a C² Cartesian center

For $r>0$, the chain rule and product rule in [F1] give ${\mathrm{\partial }}_{i}F=f^{\mathrm{\prime }}(r)\frac{x_i}{r},{\mathrm{\partial }}_j{\mathrm{\partial }}_{i}F=(f^{\mathrm{\prime }\mathrm{\prime }}(r)-\frac{f^{\mathrm{\prime }}(r)}{r})\frac{x_i{x}_j}{r^2}+\frac{f^{\mathrm{\prime }}(r)}{r}{\delta }_{ij}\mathrm{.}$ Evenness implies $f^{\mathrm{\prime }}(0)=0$. By [F2], $f^{\mathrm{\prime }}(r)\mathrm{/}r\to f^{\mathrm{\prime }\mathrm{\prime }}(0)$ and $f(r)=f(0)+\frac{1}{2}{f}^{\mathrm{\prime }\mathrm{\prime }}(0)r^2+o(r^2)$ as $r\to 0$. Thus $F$ is differentiable at $0$ with gradient $0$ and Hessian $f^{\mathrm{\prime }\mathrm{\prime }}(0)I_3$. Since the displayed first and second derivatives tend to those values as $x\to 0$, they extend continuously to $0$, proving $F\in C^2$.

For $r>0$, ${\mathrm{\partial }}_k{H}_{ij}={\delta }_{ik}{x}_{j}h(r)+{\delta }_{jk}{x}_{i}h(r)+x_i{x}_j{h}^{\mathrm{\prime }}(r)\frac{x_k}{r}\mathrm{.}$ Evenness gives $h^{\mathrm{\prime }}(0)=0$, so [F2] yields $h^{\mathrm{\prime }}(r)\mathrm{/}r\to h^{\mathrm{\prime }\mathrm{\prime }}(0)$ and $h^{\mathrm{\prime }\mathrm{\prime }}(r)-h^{\mathrm{\prime }}(r)\mathrm{/}r\to 0$. Differentiating the displayed formula once more, the terms without a factor $x_i$ or $x_j$ are $({\delta }_{ik}{\delta }_{j\mathrm{\ell }}+{\delta }_{jk}{\delta }_{i\mathrm{\ell }})h(r)$; every remaining term is a sum of terms containing either one factor $x_i$ or $x_j$ times $h^{\mathrm{\prime }}(r)x_{\mathrm{\ell }}\mathrm{/}r$, or two such factors times $h^{\mathrm{\prime }\mathrm{\prime }}(r)-h^{\mathrm{\prime }}(r)\mathrm{/}r$ or $h^{\mathrm{\prime }}(r)\mathrm{/}r$. Since $h^{\mathrm{\prime }}$ is $O(r)$ and $h^{\mathrm{\prime }\mathrm{\prime }}$, $h^{\mathrm{\prime }}\mathrm{/}r$ are bounded near $0$, these remaining terms tend to $0$. The Peano expansion in [F2] gives $H_{ij}(x)={\delta }_{ij}+h(0)x_i{x}_j+o(r^2)$, hence derivative $0$ and second derivative $({\delta }_{ik}{\delta }_{j\mathrm{\ell }}+{\delta }_{i\mathrm{\ell }}{\delta }_{jk})h(0)$ at $0$. Thus the first and second derivatives extend continuously and $H_{ij}$ is $C^2$.

For $r>0$, write $d{r}^2=(x_{i}d{x}_i{)}^2\mathrm{/}{r}^2$ and use the polar identity $d{r}^2+r^2\gamma ={\sum }_{i}d{x}_i^2$ from [F3]. Therefore $b^{-1}d{r}^2+r^2\gamma ={\sum }_{i}d{x}_i^2+\frac{b^{-1}-1}{r^2}(x_{i}d{x}_i{)}^2={\sum }_{i,j}({\delta }_{ij}+x_i{x}_j\frac{q(r)}{b(r)})d{x}_{i}d{x}_j,$ because $b=1-r^{2}q$. The quotient $q\mathrm{/}b$ is even $C^2$ since $b>0$, so step 1.2 proves the spatial extension. The time coefficient $-e^{2\varphi (r)}$ extends $C^2$ by step 1.1 whenever $\varphi$ is even $C^2$. This proves the metric clause.