Suppose I have a real unitary matrix $U$ and a unit vector $\mathbf{x}, \|\mathbf{x}\|\_2 = 1$. What is the solution to the following problem?


$$
\widehat{\mathbf{x}} = \arg\max\_{\mathbf{x}, ~\|\mathbf{x}\|\_2 = 1}\|U\mathbf{x}\|\_1.
$$



Maximal norm-1 projection
One solution is to take $x$ to be proportional to the column sums of $U$. Indeed, we have that $$\widehat{x} = \arg \max\_{||y||\_{\infty} = 1, ||x||\_2 = 1} y^T U x $$ which suggests we should pick $x$ proportional to $U^T \widehat{y}$ where $$\widehat{y} = \arg \max ||U^T y||\_2 $$ where the maximum is taken over $y$ with $\infty$-norm equal to $1$. But this is maximizing a convex function over a convex set, so the maximum occurs at the corners of the cube $[-1,1]^n$. Moreover, because $U$ is unitary, its easy to see it actually doesn't matter which corner we pick. So we might as well pick $y = (1,1,..,1)^T$, leading to the claim in the first sentence of this answer. Picking other corners gives other solutions. 


