A matrix function is Lipschitz

Question

Let $f:\mathbb{R}^{n\times m}\to\mathbb{R}$ be defined as
$$
f(A) = \min_{v\in S}\|ABv\|_2,
$$
where $\|\cdot\|_2$ is the $\ell_2$-norm, $B\in\mathbb{R}^{m\times m}$ is fixed and
$$
S = \{v\in\mathbb{R}^m:\|Bv\|_2=1,\|v\|_2\leq c\},
$$
for some fixed $c>0$. I want to prove that $f$ is 1-Lipschitz, meaning that
$$
|f(A)-f(A')|\leq \|A-A'\|_{\infty},\quad\forall\,A,A'\in\mathbb{R}^{n\times m},
$$
where $\|\cdot\|_{\infty}$ is the operator norm (greatest singular value). What I know is that
$$
f(A)\leq \|A\|_{\infty},
$$
since $\|Bv\|_2=1$ for all $v\in S$. Also, $\|\cdot\|_{\infty}$ is 1-Lipschitz. But I don't know were to go from here. It could be the case that the norm in the Lipschitz property is not $\|\cdot\|_{\infty}$ but $\|\cdot\|_{F}$, the Frobenius norm.

Any ideas?

Answer 1

Note that $$ \|ABv\|_2\le\|A'Bv\|_2+\|(A-A')Bv\|_2\le\|A'Bv\|_2+\|A-A'\|\cdot\|Bv\|_2, $$ with the operator norm computed with respect to the $\ell_2$ norm. This gives $$ f(A)\le f(A')+\|A-A'\|. $$ Interchanging the roles of $A$ and $A'$ finally gives $$ |f(A)-f(A')|\le\|A-A'\|. $$

Finally note that:

$f$ is Lipschitz taking any norm on the domain since all norms on $\mathbb{R}^{n\times m}$ are equivalent.