Lorentzian metrics are stable under small compactly supported symmetric perturbations

(Positive and negative directions.) Let $B$ be the symmetric form with matrix $D+E$, $D=\mathrm{diag}(-1,1,1,1)$, in a basis $e_0,e_1,e_2,e_3$ and suppose $\mathrm{\mid }{E}_{ij}\mathrm{\mid }<1\mathrm{/}8$ for all $i,j$. For $w={\sum }_{i=1}^3{w}_i{e}_i\mathrm{\ne }0$, writing $\mathrm{\mid }w{\mathrm{\mid }}^2=\sum w_i^2$ and using $(\mathrm{\mid }{w}_1\mathrm{\mid }+\mathrm{\mid }{w}_2\mathrm{\mid }+\mathrm{\mid }{w}_3\mathrm{\mid }{)}^2\le 3\mathrm{\mid }w{\mathrm{\mid }}^2$, $B(w,w)=\mathrm{\mid }w{\mathrm{\mid }}^2+E(w,w)\ge \mathrm{\mid }w{\mathrm{\mid }}^2-{\max }_{i,j}\mathrm{\mid }{E}_{ij}\mathrm{\mid }({\sum }_{i=1}^3\mathrm{\mid }{w}_i\mathrm{\mid }{)}^2\ge \mathrm{\mid }w{\mathrm{\mid }}^2-3\cdot \frac{1}{8}\mathrm{\mid }w{\mathrm{\mid }}^2>0.$ Also $B(e_0,e_0)=-1+E_{00}<0$.

(Inertia from the two directions.) Let $(w_1,w_2,w_3)$ be a $B$-orthonormal basis of $W=\mathrm{span}(e_1,e_2,e_3)$, which exists by [L3] because $B$ is positive definite there by step 1.1; write $u:=e_0$ and ${\beta }_i:=B(u,w_i)$. In the basis $(u,w_1,w_2,w_3)$ the matrix of $B$ is $\left(\begin{array}{cc}B(u,u)& {\beta }^{\mathsf{T}}\\ \beta & I_3\end{array}\right)$, and the congruence with the invertible lower triangular matrix $\left(\begin{array}{cc}1& 0\\ -\beta & I_3\end{array}\right)$ transforms it into $\mathrm{diag}(B(u,u)-{\beta }^{\mathsf{T}}\beta ,I_3)$: expanding the product gives $B(u,w_j)-{\beta }_j=0$ in the off-diagonal block and $I_3$ in the lower block. Since ${\beta }^{\mathsf{T}}\beta \ge 0$ and $B(u,u)<0$, the first diagonal entry is strictly negative; by Sylvester's law [L2] the inertia of $B$ is $(3,1,0)$.

(Pointwise openness.) Every symmetric form whose matrix in the basis $e$ is within maximum-entry distance $1\mathrm{/}8$ of $D$ has inertia $(3,1,0)$ by steps 1.1 and 2.1. Given any symmetric form $G$ of inertia $(3,1,0)$ on $V$, [L1] supplies a basis in which its matrix is $D$; applying the previous sentence in that basis shows the set of such matrices is open around the matrix of $G$ in the maximum-entry norm of that basis. Changing bases multiplies matrices by congruences $P^{\mathsf{T}}(\cdot )P$ with $P$ invertible, a linear isomorphism of the space of symmetric matrices onto itself; hence openness holds in every basis and, at any $p\in M$, around $g_p$ in the space of symmetric forms at $p$.

(Uniform openness on the support.) If $K=\mathrm{\varnothing }$, this uniform statement is vacuous; assume $K\mathrm{\ne }\mathrm{\varnothing }$. For each point of the compact set $K=\mathrm{supp}h$, choose a coordinate ball whose closure is compact and lies inside its chart domain; such balls exist by taking a smaller Euclidean ball with closure inside the chart image. Compactness gives finitely many such balls $V_1,\dots ,V_m$ covering $K$. Write $U_j$ for the chart domain containing $V_j$ and $K_j:=\overline{V_j}\subseteq U_j$ for its compact closure. Let $G_j(q)$ be the matrix of $g_q$ in that chart, continuous on $K_j$. I claim there is ${\delta }_j>0$ such that every symmetric matrix $B$ with ${\max }_{i,j}\mathrm{\mid }{B}_{ij}-G_j(q{)}_{ij}\mathrm{\mid }<{\delta }_j$ for some $q\in K_j$ has inertia $(3,1,0)$. Otherwise there are $q_n\in K_j$ and matrices $B_n$ with $\max \mathrm{\mid }{B}_n-G_j(q_n)\mathrm{\mid }<1\mathrm{/}n$ and $B_n$ of inertia different from $(3,1,0)$; passing to a subsequence with $q_n\to q\in K_j$ (compactness, [L4]), continuity gives $B_n\to G_j(q)$, which is a matrix of inertia $(3,1,0)$; but step 3.1 makes a whole maximum-entry neighborhood of $G_j(q)$ consist of such matrices, so $B_n$ has inertia $(3,1,0)$ for large $n$, a contradiction.

(Conclusion.) If $h=0$, equivalently $K=\mathrm{supp}h=\mathrm{\varnothing }$, then $g+\lambda h=g$ is a smooth Lorentzian metric with the same time orientation for every $\lambda$, so the conclusion holds with any ${\lambda }_0>0$; assume henceforth $K\mathrm{\ne }\mathrm{\varnothing }$. Let $\delta :={\min }_j{\delta }_j>0$, let $M_h:={\max }_j{\max }_{q\in K_j}{\max }_{a,b}\mathrm{\mid }{h}_{ab}^{(j)}(q)\mathrm{\mid }$ where $h^{(j)}$ denotes components in chart $j$, a finite number because each $K_j$ is compact and $h$ is smooth (Smooth covariant 2-tensor fields on a smooth manifold). For $\mathrm{\mid }\lambda \mathrm{\mid }<{\lambda }_0:=\delta \mathrm{/}(1+M_h)$ and $q\in K$, the matrix of $(g+\lambda h{)}_q$ differs from that of $g_q$ by $\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\mid }{h}_{ij}(q)\mathrm{\mid }<\delta$, so it has inertia $(3,1,0)$ by step 4.1; off $K$ the field equals $g$. Hence $g+\lambda h$ is a smooth symmetric nondegenerate field of signature $(-,+,+,+)$ on $M$ for every such $\lambda$. Since $N$ is nowhere zero and $g(N,N)<0$, the same smallness argument applied to the continuous function $q\mapsto g_q(N_q,N_q)$ on $M$ (which is negative and, on the support of $h$, bounded away from $0$; off the support it is unchanged) shows that $N$ remains timelike for $g+\lambda h$ after shrinking ${\lambda }_0$ if necessary.