stability of two-sided sectional curvature bounds in Lorentzian geometry Suppose that $(M,g)$ is a Lorentzian manifold of signature $(-,+,\ldots,+)$. Given a two plane $\Pi=\textrm{Span}\{X,Y\}$ with $X,Y \in T_pM$, we say that $\Pi$ is non-degenerate if
$$ g(X,X)g(Y,Y)-g(X,Y)^2 \neq 0.$$
Moreover, given a non-degenerate two-plane we say that it is timelike or spacelike if the above quantity is negative or positive respectively. Finally, the sectional curvature $\textrm{Sec}(\Pi)$ for a non-degenerate two plane $\Pi$ as above is defined by
$$\textrm{Sec}(\Pi)=\frac{g(R(X,Y)X,Y)}{ g(X,X)g(Y,Y)-g(X,Y)^2 }.$$
Question. Suppose that given any non-degenerate space-like two plane in $(M,g)$ its sectional curvature is bounded from above by a fixed $K_1<0$ and that given any non-degenerate time-like two plane in $(M,g)$ its sectional curvature is bounded from below by a fixed $K_2>K_1$. Suppose that $\widetilde{g}$ is another Lorentzian metric on $M$ that is obtained by adding a sufficiently small smooth compactly supported tensor $h$ to $g$. Is it true that there exists constants $K_1'$ and $K_2'$ such that similar sectional curvature bounds hold for the perturbed metric $\widetilde g$ with the new constants $K_1'$ and $K_2'$ in place of $K_1$ and $K_2$?
 A: Your condition is somewhat restrictive, and might not be what you want. Let $Y$ be a future directed lightlike vector and let $X$ be a spacelike vector orthogonal to it. Consider $Y'=Y+tU$ where $U$ is a unit future directed timelike vector. Then for  $t<0$ with $\vert t\vert $ sufficiently small, $\textrm{Span}(X,Y')$ is spacelike, that is, $g(X,X)g(Y',Y')-g(X,Y')^2$ is positive actually going to zero for $t\to 0$. For $t>0$, $\textrm{Span}(X,Y')$ is timelike and $g(X,X)g(Y',Y')-g(X,Y')^2$ is negative actually going to zero for $t\to 0$. Thus if $g(R(X,Y)X,Y)> 0$ the 'spacelike' sectional curvature for $t\to 0-$ goes to plus infinity while in the 'timelike' sectional curvature for $t\to 0+$  goes to minus infinity, so  the two bounds cannot be satisfied. We conclude that $g(R(X,Y)X,Y)\le 0$ for a generic pair as above. Maybe from here one can obtain further constraints on the Riemann tensor.
A: To add on to Ettore Minguzzi's answer:

*

*If $g(R(X,Y)X,Y) < 0$ for all pairs $X,Y$ as he defined, then since this is an open condition stability is automatic for small smooth compactly supported perturbations.


*If $g(R(X,Y)X,Y) = 0$ for some pair, then stability can be violated.
This can be seen from taking $\tilde{g} = e^{2\phi} g$ a small conformal change, where $\phi\in C^\infty_0$ is small. We will choose further that $\phi$ to have a critical point at the base of $X,Y$
Since $\tilde{g}$ is conformal, $X,Y$ are still a pair with the same properties. Our goal is to try to make the corresponding curvature tensor
$ \tilde{g}(\tilde{R}(X,Y)X,Y) > 0$. From well-known formulas since we chose $\nabla\phi = 0$ at the point of interest, we have
$$ \tilde{g}(\tilde{R}(X,Y)X,Y) = (-\tilde{g}\odot \nabla^2\phi)(X,Y,X,Y)$$
where $\odot$ is the Kulkarni-Nomizu product.
The orthogonality properties of $X$ and $Y$ further implies (I'm too lazy to figure out the currect sign)
$$ \tilde{g}(\tilde{R}(X,Y)X,Y) = \pm\tilde{g}(X,X) \nabla^2_{Y,Y}\phi $$
So choosing $\phi$ suitably convex/concave in the direction of $Y$ you can make the quantity positive.
In particular, applying this construction to Minkowski space gives you a counterexample to stability (of Minguzzi's condition).

Returning to your actual question: it turns out that the case $g(R(X,Y)X,Y)=0$ is automatically ruled out for smooth manifolds.
Using Minguzzi's analysis, consider
$$ \rho(t) = g(R(X, Y+tU)X, Y+tU)$$
and
$$ \sigma(t) = g(X,X)g(Y+tU,Y+tU) - g(X, Y+tU)^2$$
Observe that $\sigma(t)$ vanishes at $t = 0$ to only first order.
The requirement that $\rho(t) / \sigma(t) > K_2$ when $t > 0$ and $\rho(t) / \sigma(t) < -K_1$ for $t < 0$ requires that $\rho$ is strictly negative when $t \neq 0$. So if $\rho(0) = 0$ this would make it a critical point and hence vanish to second order.
But this would mean that $\lim_{t\to 0} \rho(t)/\sigma(t) = 0$ which is a contradiction.
Therefore in the end, we find that you must be in the situation where stability holds.
