= Solution
\b[Use the intended strong convexity bound] $\gamma^\top D^2\varphi(x)\gamma\geq\lambda|\gamma|^2$. 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.
Set
$$
\boxed{\varphi_\varepsilon=\varphi+\varepsilon\zeta,\qquad T_\varepsilon=\nabla\varphi_\varepsilon.}
$$
This function is $C^2$, and its <Hessian matrix> satisfies
$$
\begin{aligned}
\gamma^\top D^2\varphi_\varepsilon(x)\gamma
&=\gamma^\top D^2\varphi(x)\gamma+\varepsilon\gamma^\top D^2\zeta(x)\gamma\\
&\geq\bigl(\lambda-\varepsilon\|D^2\zeta(x)\|_{\mathrm{op}}\bigr)|\gamma|^2\geq0.
\end{aligned}
$$
Part (i) therefore proves that $\varphi_\varepsilon$ 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.
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