When reading on rigidity I become interested in the following question. Let $\mathbb{H}$ be the hyperbolic space of dimension at least 3, with metric $g_0$ and upper half space model. There is no deformation of hyperbolic metric for finite-volume hyperbolic manifolds. But how much can we deform the metric to a negatively curved one? Using local coordinate, let $p$ be a point and $R$ positive. Is it possible to perturb the metric on the ball $B_R(p)$ so that
- the new smooth metric $g$ on the ball has curvature $\leq -1$, but not identically $-1$,
- the new metric is smooth in the in the upper half space (so it is smooth in the interior of the ball, up to and across the boundary of the ball)?
If we use polar coordinates and write $g(r, \theta)= exp(2f(r))g_0(r,\theta)$, then we obtain the differential inequality $f''(r) \geq exp(2f(r))-1$ and such perturbation of the metric is equivalent to a nonzero smooth solution $f(r)$ such that $f(r) \leq 0$ for $r \in [0, R]$. For $r>R$, $f(r)=0$ in order to match the original hyperbolic metric outside the ball. Using Wolfram, a nontrivial solution seems to exist but it only provides a graph near $R$. For the metric to be smooth it also requires the function to behave like constant near $r=0$. Are there known methods to demonstrate existence rigorously? Such functions are not analytic, so I have abandon power series. Searches online also didn't grant me perturbation which preserves the curvature upper bound. Thank you very much!