I am working on an optimization problem where the function $f$ is assumed to have Lipschitz-continuous gradients and Hessian \begin{align} \| \nabla^2 f(x) - \nabla^2f(y) \| \leq L_1 \| x -y \|, \quad \forall x,y \\ \| \nabla f(x) - \nabla f(y) \| \leq L_2 \| x -y \|, \quad \forall x,y, \end{align} but where $\nabla^2 f(x)$ is not necessarily positive-semidefinite.
Can we conclude from this that $\text{diag}(\nabla^2 f(x))$ is also Lipschitz (with some other Lipschitz-constant)?
I already found this post Lipschitz Hessian implies Lipschitz Hessian diagonal?, but it assumes positive-semidefiniteness of $\nabla^2 f(x)$ and I was wondering if we can say something similar without this assumption?
