What's the inverse of the gradient of a gradient?

Question

I am wondering if, for a given functio $f(\mathbf{x})$, there exists a tensor $\mathbf{A}$ such that $$\nabla \nabla f \cdot \mathbf{A} = \mathbf{I}$$ To make it clearer, in index notation (with summation convention). $$\frac{\partial^2 f}{\partial x_i \partial x_j} A_{jk} = \delta_{ik}$$ What conditions does $f$ have to satisfy for the inverse to exhist?

Answer 1

The "gradient of a gradient" $\nabla^2 f$ is the Hessian of F. In order for $A$ to exist, you are asking the Hessian to be invertible. This is the case for instance if $f$ is $\lambda$-convex with $\lambda>0$ (this is what Wikipedia calls strongly convex).