Lipschitz modulus of quadratic loss function

Question

I'm trying to figure out the Lipschitz modulus of quadratic function $\ell(\xi)=\xi^{\top} Q \xi+2 q^{\top} \xi + q_0$ with positive definite $Q$. According to the definition of Lipschitz modulus $\operatorname{Lip}(\ell)=\sup \left\{\|z\|_{*}: \ell^{*}(z)<\infty\right\}$, I need to find the supremum of the dual norm in the domain of the conjugate function of the quadratic, which is $\frac{1}{2}(z-q)^{\top}Q^{-1}(z-q) - q_0$. Since $z$ could be arbitrary vector, the supremum of dual norm is $\infty$. Am I right? Any hint would be appreciated.

Answer 1

A quadratic function with positive definite matrix is strongly convex and for that reason, has unbounded Lipschitz modulus. First, notice that since $Q\in\mathbb{S}_{++}^n\Rightarrow \lambda_{\min}(Q)\geq m>0$, hence your quadratic function is strongly convex. Since it is $\sigma=\lambda_{\min}(Q)$ - strongly-convex, it implies that $$ f(\mathbf{y})- f(\mathbf{x})\geq \nabla f(\mathbf{x})^T(\mathbf{y}-\mathbf{x})+\frac{\sigma}{2}\|\mathbf{y}-\mathbf{x}\|^2\overset{\textbf{C.S.}}{\geq}\|\mathbf{y}-\mathbf{x}\|(\frac{\sigma}{2}\|\mathbf{y}-\mathbf{x}\|-\|\nabla f(\mathbf{x})\|) $$ where the RHS goes to infinity since the coercive function $\|x-y\|\to\infty$ . This means that $|f(\mathbf{x})- f(\mathbf{y})|$ is unbounded, as you correctly suspected.