Hessian of the Euclidean norm ($D^2|x|=|x{|}^{-1}(I-\hat{x}{\hat{x}}^T)$)

For a nonzero vector $x\in {\mathbb{R}}^n$, put $r=|x|$ and $\hat{x}=x/r$. The gradient of the Euclidean norm is $\nabla r=\hat{x}$. Differentiate its components to obtain the Hessian matrixThe factor $I-\hat{x}{\hat{x}}^T$ is the orthogonal projection matrix onto the orthogonal complement of the radial direction. Thus the Hessian matrix has eigenvalue zero radially and $1/r$ on each tangential direction, explaining the convexity of the Euclidean norm. Taking the trace gives $\Delta r=(n-1)/r$ away from the origin. The Euclidean norm is not differentiable at the origin: its restriction to any line is a nonzero multiple of the absolute value.