Suppose we have a matrix $\mathbf B \in \mathbb R^{n\times n}$. Is there an analytical solution to the following problem?
\begin{equation} \begin{aligned} &\max_{\mathbf x \in \mathbb R^n} && \|\mathbf B\mathbf x\|_{\infty} \\ &\ \ \text{s.t.} && \|\mathbf x\|_2\leq 1 \end{aligned} \end{equation}
I had a follow-up thought... By definition, $||\mathbf B\mathbf x||_\infty = \max_i \{ |\mathbf b_i^\text{T} \mathbf x|\}$, where $\mathbf b_i^\text{T}$ is the $i$th row of the matrix $\mathbf B$. Then, if $||\mathbf x||_2 \leq 1$, we have
\begin{equation} \begin{aligned} ||\mathbf B\mathbf x||_\infty &= \max_i \{ |\mathbf b_i^\text{T} \mathbf x|\} \\ &\leq \max_i \{ ||\mathbf b_i ||_2 ||\mathbf x||_2 \} \\ &\leq \max_i \{ ||\mathbf b_i ||_2 \}. \end{aligned} \end{equation}
Thus, $\max_i \{ ||\mathbf b_i ||_2 \}$ is an upper bound on the optimal value of the problem.
Let $j \equiv \arg \max_i \{ ||\mathbf b_i ||_2 \}$, and let $\mathbf y \equiv \mathbf b_j/||\mathbf b_j||_2$. Then, $\mathbf y$ is feasible for the optimization problem, because $||\mathbf y||_2 = 1$, and
\begin{equation} \begin{aligned} ||\mathbf B\mathbf y||_\infty &= \max_i \{|\mathbf b_i^\text{T} \mathbf y|\} \\ &\geq |\mathbf b_j^{\text T}\mathbf y| \\ & = ||\mathbf b_j||_2 \\ &= \max_i \{ ||\mathbf b_i ||_2 \}. \end{aligned} \end{equation}
It follows that $||\mathbf B\mathbf y||_\infty = \max_i \{ ||\mathbf b_i ||_2 \}$. Thus, $\max_i \{ ||\mathbf b_i ||_2 \}$ is the optimal value of the optimization problem, with $\mathbf y =\mathbf b_j/||\mathbf b_j||_2$ as the optimal solution.