Is a differentiable matrix function Lipschitz Continuous?

Question

Let $L_{\infty}(\mathbb{R^{n \times n}})$ be the space of $n \times n$ matrices $A$ such that $\| A\|_{\infty} = \max_i \sum_{j=1}^n |A_{ij}| < \infty$. Is it true that if $f:L_\infty (\mathbb{R^{n \times n}})\to L_\infty(\mathbb{R^{n \times n}})$ has an everywhere bounded bounded derivative $\|f'(A)\|_{\infty} < \lambda$ then $f$ is Lipschitz-continuous in the sense that $$\|f(A)-f(B)\|_{\infty} \leq \lambda\|A-B\|_{\infty}$$

Answer 1

Yes, this can be reduced to the scalar case. Let $\phi(t)=(1-t)A+tB$ be a parameterization of the line segment from $A$ to $B$. Also let $\psi$ be a unit-norm functional such that $$\psi(f(A)-f(B)) = \|f(A)-f(B)\|$$ Then $\psi\circ f \circ \phi$ is a real function of a real variable $t\in [0,1]$, and its derivative (by the chain rule) is bounded by $$\lambda \|A-B\|$$ where the second factor comes from $\phi'$. It follows that $$ \|\psi\circ f \circ \phi(1) - \psi\circ f \circ \phi(0)\| \le \lambda \|A-B\| $$ which is the desired Lipschitz property.