Before revision:  For each x\in \mathbb{R^m such that \nabla f(x)}  = 0, let \delta (x) = \delta _j, where j is the smallest number so that \nabla ^2f(x)+\delta _j||\nabla f(x)||^{1+\alphaId is invertible.
After revision:  For each x\in \mathbb{R^m such that \nabla f(x)} = 0, let \delta (x) = Id is invertible.