Lipschitz-constant gradient implies bounded eigenvalues on Hessian
This is not true as stated. For example, the function $f(x)=x|x|$ on the real line has Lipschitz gradient, but is not twice differentiable. Also, the function $f(x)=-x^4$ satisfies $f''\le LI$ with $L=0$, but its gradient is not Lipschitz continuous.
The two properties are equivalent for functions that are convex and twice differentiable. For such functions, $\nabla^2 f$ is a positive semidefinite matrix, so its norm is its largest eigenvalue. Hence, $$\nabla^2 f \preceq LI \iff \|\nabla^2 f\|\le L \iff \|\nabla f(x)-\nabla f(y)\|\le L\|x-y\|$$ where the last equivalence is based on the mean value theorem.