If $f(t,x)=A(t)x+b(t)$ is continuous, prove that $f$ is locally lipschitz in the second variable.
I suppose I have to prove that given $\delta>0$, if $\| x-x_0 \| < \delta$ then there is $M > 0$ such that $$\|f(t,x)-f(t,x_0)\| \le M \| x-x_0 \|$$
Theorem: There is $M$ such that $\| A(t) \| < M \| t \|$ (Spivak Calculus Manifolds, I. Ex.10)
Can I use that? The inequality $\| A(t) x + b(t) \| \le \| A(t) x \| + \| b(t) \|$ doesn't take me far, but can't the continuity make both terms smaller.
Or Picard's theorem for Lipschitz? I've avoided because I don't really understand how to use it.
I'm assuming that $t\mapsto A(t)\in M_{n\times n}({\mathbb R})$ is continuous. Given a point $(t_0,x_0)\in{\mathbb R}\times{\mathbb R}^n$ there is a neighborhood $U:=\>]t_0-h,t_0+h[\>$ of $t_0$ and a constant $L>0$ such that $\|A(t)\|\leq L$ for all $t\in U$. One then has for arbitrary $x$, $x'\in{\mathbb R}^n$, in particular for arbitrary $x$, $x'$ in a neighborhood $V$ of $x_0$, the estimate $$|f(t,x)-f(t,x')|=|A(t)x-A(t)x'|\leq L|x-x'|\ .$$ This shows that $f$ is Lipschitz with respect to $x$ in the neighborhood $U\times V$ of $(t_0,x_0)$.