I would like to compute the concavity of the distance function in $\mathbb{R}^n$.
Let $ f(x) =- \Vert x \Vert $ in $\mathbb{R}^n$. Then $\nabla_xf=- \frac{x}{\Vert x \Vert}$. And $$-\operatorname{Hess}_x(f)=\frac{1}{\Vert x \Vert} E_n + \frac{1}{\Vert x \Vert^3} (x_i\cdot x_j)_{ij}.$$
The function f is concave if $\operatorname{Hess}f$ is negative definite. Therefore I want to compute the Determinant of the second matrix. Does anyone has sugestions how to do so?
I want to have a more subtle result than "just" concavity. I want to compute how concave the distance function is on $B_r(0)$ (depending on $r$) in the sense of $\lambda$-concavity. A function is $\lambda$ concave on a riem. Manifold if for every unit speed geodesic $\gamma$ between $x$ and $y$ the function $f\circ \gamma(t) - \lambda t^2$ is concave for some $\lambda \in \mathbb{R}$.
Reason behind computation:
For a nonnegatively curved space $M$ the distance function $\operatorname{dist}(x,\partial C)$ is concave for every convex subset $C$ on $C \setminus \partial C$. How much better does it get for balls in the eulidean plane? How much more ($\lambda$)-concave is $\operatorname{dist}(x,\partial B_r(0))=r- \Vert x \Vert$?
I get a slightly different result then you when I calculate the Hessian. Starting with
$$ \frac{\partial f}{\partial x_i} = -\frac{x_i}{\|x\|} $$
I get that
$$ \frac{\partial^2 f}{\partial x_i^2} = -\frac{\|x\|^2 - x_i^2}{\|x\|^3} $$
and
$$ \frac{\partial^2 f}{\partial x_i\partial x_j} = \frac{x_i x_j}{\|x\|^3} $$
Therefore, we should have that
$$ \text{Hess}_x(f) = -\frac{1}{\|x\|}\left(I - \bar{x}\bar{x}^T\right), $$
where $\bar{x} = x/\|x\|$. From here you can show that the Hessian is negative definite as follows. Let $y$ be any nonzero vector in $\mathbb{R}^n$, then
$$ y^T \text{Hess}_x(f) y = -\frac{1}{\|x\|}[y^T y - (y^T\bar{x})^2] $$
From the Cauchy-Schwarz inequality we have that
$$ |y^T\bar{x}|^2 \leq (y^Ty)\cdot(\bar{x}^T\bar{x}) = y^Ty $$
Thus $y^Ty - (y^T\bar{x})^2 \geq 0$ and hence $y^T \text{Hess}_x(f) y \leq 0$.