TT plane waves with arbitrary bounded profiles solve the linearized vacuum equations, are in harmonic gauge, and give Lorentzian small-amplitude metrics with a quadratic Einstein-tensor residual

(Lorentzian signature.) In the inertial chart the components of $g(\lambda )=\eta +\lambda h$ are $g_{00}=-1$, $g_{0i}=0$, $g_{ij}={\delta }_{ij}+\lambda h_{ij}$. Since $h$ has nonzero components only at positions $11,22,12,21$ with $\mathrm{\mid }{h}_{ij}\mathrm{\mid }\le M$, for every $v\in {\mathbb{R}}^3\setminus \{0\}$ $\mathrm{\mid }{v}^i{v}^j{h}_{ij}\mathrm{\mid }\le M((v^1{)}^2+(v^2{)}^2)+2M\mathrm{\mid }{v}^1{v}^2\mathrm{\mid }\le 2M((v^1{)}^2+(v^2{)}^2)\le 2M\mathrm{\mid }v{\mathrm{\mid }}^2$ by [F4], so $v^i{v}^j({\delta }_{ij}+\lambda h_{ij})\ge (1-2M\mathrm{\mid }\lambda \mathrm{\mid })\mathrm{\mid }v{\mathrm{\mid }}^2>0$ when $\mathrm{\mid }\lambda \mathrm{\mid }<1\mathrm{/}(2M)$. The matrix of $g(\lambda )$ is therefore the orthogonal direct sum of the negative entry $-1$ and a positive definite $3\times 3$ matrix; diagonalizing the latter gives three positive diagonal entries, so the inertia is $(3,1,0)$ and $g(\lambda )$ is a Lorentz form at every point by [F3]. The components are smooth by [F6] because $h$ has smooth components, so $g(\lambda )$ is a smooth Lorentzian metric; $g(\lambda )({\mathrm{\partial }}_0,{\mathrm{\partial }}_0)=-1<0$ exhibits a smooth everywhere timelike vector field, which orients the metric by [F3], and ${\mathrm{\partial }}_0$ is future-directed since $g({\mathrm{\partial }}_0,{\mathrm{\partial }}_0)<0$ with the flat time orientation. This proves (a).

(Gauge, trace and wave equation.) For the TT profile $F$, the trace vanishes: ${\eta }^{cd}{h}_{cd}=h_{11}+h_{22}=A_{+}-A_{+}=0$, and $h_{00}=h_{01}=h_{02}=h_{03}=h_{33}=0$; hence $\bar{h}=h$. By [F1], ${\mathrm{\partial }}^a{h}_{ab}={\eta }^{ac}{\mathrm{\partial }}_c{h}_{ab}=F_{ab}^{\mathrm{\prime }}(u)n^a$; for each $b$ this is $F_{0b}^{\mathrm{\prime }}(u)n^0+F_{3b}^{\mathrm{\prime }}(u)n^3$, with $n^0=-1$ and $n^3=-1$ by step 1.1 of The phase coordinate u = x^0 - x^3 is null and profile functions are flat-harmonic and constant along its transport curves, so it equals $-F_{0b}^{\mathrm{\prime }}(u)-F_{3b}^{\mathrm{\prime }}(u)=0$ because the profile satisfies $F_{a0}=F_{a3}=0$ for every $a$. Hence ${\mathrm{\partial }}^a{\bar{h}}_{ab}=0$: harmonic gauge. Also $\mathrm{\square }{\bar{h}}_{ab}=\mathrm{\square }{h}_{ab}=0$ for every $a,b$ by [F1]. By [F2] the first-order Einstein coefficient in harmonic gauge is $G_{ab}^{(1)}(h)=-\frac{1}{2}\mathrm{\square }{\bar{h}}_{ab}=0$; this is the linearized vacuum equation of Plane-wave perturbations of the flat linearization: planar profiles, wave covectors, amplitudes, TT form and the linearized field equations clause (e). The transversality for the wave covector is the vanishing of ${\eta }^{ac}{k}_c{h}_{ab}$ for every $b$: for the wave covector $k_c=\omega {\eta }_{cd}{T}^d$ with $T={\mathrm{\partial }}_0+{\mathrm{\partial }}_3$ of The phase coordinate u = x^0 - x^3 is null and profile functions are flat-harmonic and constant along its transport curves clause (a) this equals $\omega (h_{0b}+h_{3b})=0$ because the TT profile has $F_{0b}=F_{3b}=0$. This proves (b).

(Quadratic residual.) Fix the nonempty compact $K$ and $L>0$. Let $b_j$ be the supremum over $K$ of the maximum absolute value of the order-$j$ component derivatives of $h$, for $j=0,1,2$; these numbers are finite by [F4]. The quantity in clause (c) is $M_K=b_0+L^{-1}{b}_1+L^{-2}{b}_2$, while the supplier's control is $B_K(h)=b_0+L{b}_1+L^2{b}_2$. Set $q_L:=\max \{1,L^2,L^4\}$. Termwise comparison gives $B_K(h)\le q_L{M}_K$ for every $L>0$. Take ${\delta }_0,C$ from [F5] for this nonempty $K$ and $L$, and choose ${\lambda }_0>0$ with ${\lambda }_0{B}_K(h)\le {\delta }_0$ and, when $M>0$, ${\lambda }_0<1\mathrm{/}(2M)$; if $B_K(h)=0$, its constraint is automatic. For $\mathrm{\mid }\lambda \mathrm{\mid }\le {\lambda }_0$, put $A=\lambda h$, so $B_K(A)=\mathrm{\mid }\lambda \mathrm{\mid }{B}_K(h)\le {\delta }_0$. The metric is Lorentzian on all of ${\mathbb{R}}^4$ by step 1.1. Applying the residual estimate of [F5] on $K$ gives $\mathrm{\mid }{G}_{ab}(\eta +\lambda h)(p)-G_{ab}^{(1)}(\lambda h)(p)\mathrm{\mid }\le C{\lambda }^2{B}_K(h{)}^2\mathrm{/}{L}^2\le C{q}_L^2{\lambda }^2{M}_K^2\mathrm{/}{L}^2$. By linearity and step 2.1, $G_{ab}^{(1)}(\lambda h)=\lambda G_{ab}^{(1)}(h)=0$. Thus clause (c) holds with $C_K:=C{q}_L^2<\mathrm{\infty }$, also when $M_K=0$.

(Conclusion.) Steps 1.1, 2.1 and 3.1 prove clauses (a)-(c). Only the explicit components of the TT planar perturbation, the chain rule of [F1], the harmonic-gauge reduction of [F2] and the quadratic-remainder estimate of [F5] are used; no field equation is assumed, no source or matter model occurs, and no choice principle beyond the inherited countable choice of the cited suppliers is used.