I am working on existence and uniqueness of nonlinear systems and one of the prerequisites is the notion of Lipschitz continuity. I am a little bit confused on how we use the definitions to actually show that a function is Lipschitz continuous or not.
First, I'll provide the definitions that I am aware of:
A function $f$ is locally Lipschitz on an open and connected subset $D\subset\mathbb R^n$ if each point of $D$ has a neighborhood $D_0$ such that
$$||f(x)-f(y)||\leq L||x-y||\qquad (1)$$
for all point in $D_0$ with some Lipschitz constant $L_0$. We say that $f$ is globally Lipschitz if $f$ satisfies condition $(1)$ for all points in a set $W$ with the same constant $L$.
Now, a simple example is to consider $f(x)=x^2+|x|$. One way to determine if $f$ is Lipschitz continuous is to write $f=g+h$ where $g(x)=x^2$ and $h(x)=|x|$.
Now $g$ is locally Lipschitz, thus continuous, and continuously differentiable but not globally Lipscitz. In a similar manner, $h$ is not continuously differentiable, but locally and globally Lipschitz and thus continuous. From these arguments, one may conclude that $f$ is not continuously differentiable, but it is locally Lipschitz and continuous.
Why is that? Why are we allowed to argue Lipschitz continuity of $f$ by studying the functions $g,h$?
Best Answer
Lemma: Locally Lipschitz functions form a vector space, closed under addition and scalar multiplication.
Proof: Let $f$ and $g$ be locally Lipschitz. Choose a point $z$. There is a neighborhood $D_f$ of $z$ such that $|f(x)-f(y)|\le L_f|x-y|$ for all $x,y\in D_f$, and a neighborhood $D_g$ such that $|g(x)-g(y)|\le L_g|x-y|$ for all $x,y\in D_g$. Then, on the neighborhood $D_f\cap D_g$, $|af(x)-af(y)|\le aL_f|x-y|$ and $|(f(x)+g(x))-(f(y)+g(y))|\le (L_f+L_g)|x-y|$.
This can be done everywhere. Every point has a neighborhood on which $af$ and $f+g$ are Lipschitz, so the definition holds and $af$ and $f+g$ are locally Lipschitz. Done.
So then, we can break things down. We wrote $f=g+h$, and showed $g$ and $h$ were locally Lipschitz. By the lemma, $g+h=f$ is locally Lipschitz as well.