 Solution

Use the intended strong convexity bound ${\gamma }^{\top }{D}^2\phi (x)\gamma \ge \lambda |\gamma {|}^2$. The original PDF omits $|\gamma {|}^2$ on the right. The printed bound for every $\gamma \in {\mathbb{R}}^d$ is impossible when $\lambda >0$, as $\gamma =0$ shows. Equivalently, the intended bound is stated for unit vectors. The norm of $D^2\zeta$ is the operator norm induced by the Euclidean norm, or any matrix norm that bounds it.

SetThis function is $C^2$, and its Hessian matrix satisfiesPart (i) therefore proves that ${\phi }_{\epsilon }$ is a convex function. The perturbation can use up the entire positive lower bound defining a strongly convex function; only convexity, rather than strict positivity of the Hessian matrix, is asserted.