If $A$ is a bounded matrix-valued function, is the map $x \mapsto A(x)x$ Lipschitz?

Question

Let $A \colon \mathbb{R}^{m} \to \mathbb{R}^{m\times m}$ be a given matrix-valued function, and suppose that all its entries are smooth bounded functions $\mathbb{R}^{m} \to \mathbb{R}$. Does this imply that the map

$$ \begin{align} F \colon \mathbb{R}^{m} &\to \mathbb{R}^{m}\\ x &\mapsto A(x)x \end{align} $$ is globally Lipschitz?

Answer 1

No. Take $m = 1$ and $A(x) = \sin(x)$ so that $F(x) = \sin(x) \cdot x$. If you assume that $F$ is globally Lipschitz, then there exists $M > 0$ so that

$$ |F(x) - F(y)| \leq M|x - y| $$

for all $x,y \in \mathbb{R}$. Taking $x = 2\pi k + \varepsilon$ and $y = 2\pi k$ for $\varepsilon \neq 0$ and $k \in \mathbb{N}$ we get

$$ |F(x) - F(y)| = |(\varepsilon + 2\pi k) \sin (\varepsilon)| \leq M |\varepsilon| $$

which implies that

$$ \left| \frac{\sin(\varepsilon)}{\varepsilon} \right| \leq \frac{M}{|\varepsilon + 2\pi k|}. $$

Taking $\varepsilon \to 0$ we get

$$ 1 \leq \frac{M}{2\pi k} \iff 2\pi k \leq M $$

for all $k \in \mathbb{N}$, a contradiction.