I know that if we have a Lipschitz gradient
$$\|\nabla f(x) - \nabla f(y)\|\leq L\|x-y\|,\, \forall x,y, $$ we can say that $\nabla^2f\preceq LI.$ I have a problem where difference of gradient is bounded as below $$\|\nabla f(x) - \nabla f(y)\|\leq L\|x-y\|^n,\ \forall x,y, $$
where $n$ is an integer greater than or equal to 2. Is there some theorem which says about bounds on $\nabla^2f$ in the form $\nabla^2f\preceq LI \|x-y\|$. I am trying to bound $\nabla^2f$, but I am not sure how to do it. Any direction in this question would be really helpful. Here is example of $f(X) = \|A - XYY^TX^T\|_F^2$ where the gradient of $f$ follows above bound where $A,X,Y \in R^{n \times n}$
If $n>1$ then the inequality implies that $\nabla f$ is constant, and $\nabla^2f$ is zero: divide by $\|x-y\|$ and let $x\to y$.