Lipschitz-constant gradient implies bounded eigenvalues on Hessian I've read in a few places that if we have a Lipschitz gradient

$$\|\nabla f(x) - \nabla f(y)\|\leq L\|x-y\|,\, \forall x,y, $$
we can equivalently say $\nabla^2f\preceq LI.$ But I'm having a hard time showing this. (Equivalently, I want to show $z^T \nabla^2f(x)z\leq z^TLIz=Lz^Tz,\forall\, x,z $.)
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.
It is equivalent for $||\bigtriangledown ^2f(x)||_2 \leq L$ where $||\bigtriangledown ^2f(x)||_2$ means the maximum singular value.