Matrix-valued function

Question

I have a problem about matrix-valued function. Given a function $f:\mathbb{R}^k \rightarrow {\cal M}_{k \times k}$ of class $C^1$, where ${\cal M}_{k \times k}$ is the set of all $k \times k$ matrices. Define a function $g$ by \begin{equation} g(x)=K\{K^Tf(x)K\}^{-1}Kf(x). \end{equation} Suppose $x_0 \in \mathbb{R^k}$ satisfies $g(x_0)x_0=x_0$. I'd like to obtain \begin{equation} \|[g(x)-g(x_0)]x_0\|\leq M\|x-x_0\|^2. \end{equation} for some real number $M>0$. However, what I got is only \begin{equation} \|[g(x)-g(x_0)]x_0\|\leq M\|x-x_0\|. \end{equation} due to the continuity of $g$. Does anyone have suggestion about this? Thanks...

Answer 1

As $g(x)-g(x_0)$ will contain a nonzero first-order term in general, what you want to achieve is not always possible. For example, consider \begin{align*} K&=\pmatrix{1&0&0\\ 0&1&1\\ 0&0&-1}, \quad f\pmatrix{u\\ v\\ w} = \pmatrix{\cos u&-\sin u&0\\ \sin u&\cos u&0\\ 0&0&1},\\ g(x)&=K(K^\top f(x)K)^{-1}Kf(x)\\ &= f(x)^{-1}K^{-\top}Kf(x)\\ &=\pmatrix{\cos u&\sin u&0\\ -\sin u&\cos u&0\\ 0&0&1} \pmatrix{1&0&0\\ 0&1&1\\ 0&1&2} \pmatrix{\cos u&-\sin u&0\\ \sin u&\cos u&0\\ 0&0&1}\\ &=\pmatrix{1&0&\sin u\\ 0&1&\cos u\\ \sin u&\cos u&2}. \end{align*} Let $x_0=(2\pi,0,0)^\top$. Then $g(x_0)x_0=x_0$. Let $u$ be small and $x=(2\pi+u,0,0)^\top$. Then \begin{align*} \left[g(x)-g(x_0)\right]x_0 =(g(x)-I)x_0 =\pmatrix{0\\ 0\\ 2\pi\sin u} =\pmatrix{0\\ 0\\ 2\pi u} + O(u^2) \end{align*} and it is not true that $\|(g(x)-g(x_0))x_0\|=O(\|x-x_0\|^2)$.