Lipschitz-constant gradient implies bounded eigenvalues on Hessian

Question

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 $.)

Answer 1

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.

Answer 2

Implication from gradient to Hessian holds true for a twice differentiable function. From the definition of the Hessian of a twice differentiable function $f(\mathbf{x})$, we know that for any vector $\mathbf{v}\in\mathcal{R}^n$

\begin{align} \nabla^2f(\mathbf{x})\mathbf{v}&=\lim_{h\to0}\frac{\nabla f(\mathbf{x}+h\mathbf{v})-\nabla f(\mathbf{x})}{h}\\ \implies ||\nabla^2f(\mathbf{x})\mathbf{v}||&\leq\lim_{h\to0}\frac{||\nabla f(\mathbf{x}+h\mathbf{v})-\nabla f(\mathbf{x})||}{|h|}\\ \implies ||\nabla^2f(\mathbf{x})\mathbf{v}||&\leq\lim_{h\to0}L\frac{|h|||\mathbf{v}||}{|h|}\\ \implies ||\nabla^2f(\mathbf{x})\mathbf{v}||&\leq L||\mathbf{v}|| \end{align}

Since this is true for any $\mathbf{v}$, it is also true for the eigenvectors for matrix $\nabla^2f(\mathbf{x})$. If $\mathbf{v}$ is such an eigenvector \begin{align} ||\nabla^2f(\mathbf{x})\mathbf{v}||&=||\lambda\mathbf{v}||\leq L ||\mathbf{v}||\\ \implies |\lambda|\leq L \end{align}

So all eigenvalues are upper bounded by $L$