maximin of difference between projections I am wondering if the following problem is well studied. Given two matrices $A, B \in R^{n\times d}$, solve
$$\max_{x:\|x\|\leq 1}\min_{y:\|y\|\leq 1} \|Ax - By\|^2,$$
where $x, y \in R^d$. Assume $n\geq d$, and $A$, $B$ are full column-rank.
I have derived something, but I wonder if it's already been studied somewhere?
 A: I have a partial solution and am glad to see more discussions.
First, let's solve the inner minimization problem for any given $x$,
$$y^\ast(x)\triangleq \arg\min_{\|y\|\leq 1} \|Ax-By\|^2 = (B^\top B + \lambda I)^{-1}B^\top Ax,$$
where $\lambda=0$ if $(B^\top B)^{-1}B^\top Ax \leq 1$. Otherwise $\lambda$ is such that $(B^\top B + \lambda I)^{-1}B^\top Ax=1$. In general, we can write $\lambda=\lambda(x)$ to emphasize that the $\lambda$ value depends on $x$. This is not hard. A detailed derivation can be found at Least Squares with Euclidean ($ {L}_{2} $) Norm Constraint
Now substitute $y^\ast(x)$ to the outer maximization problem, solve $$\max_{\|x\|\leq 1} \|Ax-B(B^\top B + \lambda I)^{-1}B^\top Ax\|^2 \equiv \max\|[I-B(B^\top B + \lambda I)^{-1}B^\top] Ax\|^2.$$
If $\lambda$ were a constant, the maximizer is simply the leading right singular vector of $[I-B(B^\top B + \lambda I)^{-1}B^\top] A$.
However in reality, $\lambda$ also depends on $x$. So I'm not sure what we can do with this. Could anyone give more ideas?
