Consider $f(A) = \underset{x \in \mathcal{X}} \max x^\top A x$ where the set $\mathcal{X}$ represents the domain of $x$. If $\mathcal{X} = \{x : ||x||_2 \leq 1\}$ then the problem reduces to an eigenvalue problem. However, for $\mathcal{X} = \{x : ||x||_1 \leq 1\}$ there is no closed form solution.
For two positive definite matrices, $C \text{ and } D$ such that $\lambda_{k}(C) < \lambda_{k}(D)$ and $D-C$ is indefinite, how do I lower bound the difference $f(D) - f(C)$ when $\mathcal{X}$ is the $\ell_1-$norm set? Trivial upper bound is $\lambda_{\max}(D)$ and a trivial lower bound is $\lambda_{min}(D) - \lambda_{max}(C)$. The question is can I do better in the lower bound ?