Improving regularity of the boundary of a convex set in Riemannian manifolds

Question

Let $X$ be a geodesically complete Riemannian manifold (we may assume that $X$ is simply connected and negatively curved, although I don't think it matters). Given a closed, convex subset $K \subset X$, there is a result by Rolf Walter that the $\epsilon$-neighbourhood of $K$ (denoted $K_{\epsilon}$) has $C^{1,1}$-regular boundary, i.e. the topological boundary of $K_{\epsilon}$ is a $C^{1,1}$-submanifold of $X$. I was wondering if the regularity can be improved if we allow for more deformations of the convex set. Specifically, I would like to know the following:

${\bf Question:}$ Given a closed, convex subset $K \subset X$ and $\epsilon > 0$, does there exist a convex closed $K'$ such that $K \subset K' \subset K_{\epsilon}$, and $\partial K'$ is a $C^2$-submanifold of $X$?}

Considering that Walter's result goes back to the 70s (Rolf Walter, "Some analytical properties of geodesically convex sets"), this seems like a question whose answer should be known, but I'm struggling to find anything. All methods I could come up with to deform $K$ so that its boundary becomes more smooth while keeping the set convex rely on the second fundamental form, which requires that we make the boundary $C^2$ first.

I should empathize that I do not want to assume that $K$ is compact. (I'm fine assuming that there is a cocompact action by isometries on $K$ though.)

Answer 1

The answer is "yes" we used a similar argument in our "An optimal lower curvature...". Let me sketch the proof.

I will assume that curvature is negative. The argument will use the existence of cocompact isometric action.

Note that the function $f=\mathrm{dist}^2_K$ is strongly convex in $K_{2\cdot\varepsilon}\backslash K_\varepsilon$.

Let us smooth $f$, applying the argument of Green and Wu [Theorem 2(a) in "$C^∞$ convex functions..."]. Here we have to use the cocompact action; otherwise there might be problems.

Take $K'$ to be set bounded by a level set of smoothed $f$ that lies in $K_{2\cdot\varepsilon}\backslash K_\varepsilon$. This way we get $K'$ with $C^\infty$-smooth boundary.