I have to prove that a certain function $F(x): \mathbb{R}^m \rightarrow \mathbb{R}^n$ is continuously differentiable and its Jacobian $J(x)$ is Lipschitz continuous.
Are both criteria fulfilled if
$ \left\Vert J(y) - J(x) \right\Vert \leq L \left\Vert y- x \right\Vert \quad \forall x,y \in \mathbb{R}^m$
is fulfilled? And what does the Lipschitz continuity of the Jacobian tell me about the Lipschitz continuity of the function $F(x)$ itself?
Thank you!
Yes, if you prove that the Jacobian function $J:\mathbb{R}^m\to {\cal L}(\mathbb{R}^m,\mathbb{R}^n)$ is Lipschitz continuous, then you've also proven that it's continuous. (Lipchitz continuity $\implies$ continuity!) To say $F:\mathbb{R}^m\to \mathbb{R}^n$ is differentiable is to say that the map $J:\mathbb{R}^m\to {\cal L}(\mathbb{R}^m,\mathbb{R}^n)$ is well-defined. To say that $F$ is continuously differentiable is to say that the map $J$ is not only well-defined but also continuous. So, yes, if you prove that $J$ is Lipschitz continuous, then you've proven something stronger than is required to prove that $F$ is continuously differentiable.
The function $f:\mathbb{R}\to \mathbb{R}$ defined by the rule $f(x)=x^2$ has Lipchitz continuous derivative $f':\mathbb{R}\to\mathbb{R}$ (defined by the rule $f'(x)=2x$) but is not itself Lipschitz continuous.
I hope this helps!
