Show that $f(x,t)=\begin{bmatrix}1+x_1 \\ x_2^2 \end{bmatrix} $ satisfies a Lipschitz condition when $x$ lies in any bounded domain D (e.g. $|x|<M$ where M is a constant), but cannot satisfy a Lipschitz condition for all $x$.
Define Lipschitz as there exists a constant L such that $|f(x,t)-f(y,t)| \leq L|x-y|$ and $|x|=\sum^n_{i=1}|x_i|$
So far I arranged the problem using the defined function such that
$|x_1-y_1|+|(x_2^2-y_2^2)| \leq L(|x_1-y_1|+|x_2-y_2|)$
I suppose it could further be simplified to $|(x_2+y_2)(x_2-y_2)| \leq (L-1)|x_1-y_1|+L|x_2-y_2|$
But I'm not sure that gets me anywhere. Do I need to parameterize the vector somehow? If so I am not as sure how to do that without a given bound.
My book does give a lemma after the Lipschitz condition that L can be found by partial derivatives of the system (this is a class on systems of differential equations) but it requires an nxn system and this is not such.
Thanks for any help.
Best Answer
When $f:\>\Omega\to{\mathbb R}^m$ is a $C^1$-map defined in a convex region $\Omega\subset{\mathbb R}^n$ then for any two points $a$, $b\in\Omega$ one has $$|f(b)-f(a)|\leq M|b-a|\ ,\tag{1}$$ where $$M:=\sup_{x\in[a,b]}\|df(x)\|\tag{2}$$ and $[a,b]$ denotes the segment connecting $a$ with $b$. The inequality $(1)$ is a vectorial form of the mean value theorem of differential calculus. The norms $\|df(x)\|$ appearing in $(2)$ can be estimated in terms of the matrix elements of $df(x)$ as follows: $$\|df(x)\|\leq\sqrt{\sum_{i, k}\left|{\partial f_i\over\partial x_k}(x)\right|^2}\ .$$ From this one can conclude that an $f\in C^1(\Omega)$ is locally Lipschitz on $\Omega$ and Lipschitz on any compact subset $K\subset\Omega$.
The $f$ in your example has a partial derivative which is unbounded when $\Omega={\mathbb R}^2$; therefore it is to be expected that there is no globally valid Lipschitz constant. In fact $${|f_2(0,x_2+h)-f_2(0,x_2)|\over h}=2x_2+h\qquad(h>0)$$ gets arbitrarily large as $x_2\to\infty$. (The variable $t$ does not enter the game.)