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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 $g,h$?

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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.

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It's been awhile since I've worked with Lipschitz continuity, but from what I worked out it's a means of employing the triangle inequality and the continuity conditions on functions $g(x)$ and $h(x)$. Like you asserted in your post. $g(x) = x^{2}$ is locally Lipschitz continuous. In other words, $$||g(x)-g(y)|| \leq L_{g} ||x-y||$$ where the value $L_{g}$ will vary depending on what $x$ and $y$ are. And $h(x) = |x|$ is globally Lipschitz so $$||h(x) - h(y)|| \leq L_{h} ||x-y||$$ where $L_{h}$ is invariant upon your choices of $x$ or $y$. Now if we put those two together we see that $$||g(x)-g(y)|| \: + \: ||h(x)-h(y)|| \leq (L_{g} + L_{h}) ||x-y||$$ Since $f(x) = g(x) + h(x)$ we can see from the triangle inequality that $$||f(x) - f(y)|| = ||g(x) - g(y) + h(x) - h(y)|| \leq ||g(x) - g(y)|| + ||h(x) - h(y)||$$ And thus we can easily see that $$||f(x) - f(y)|| \leq (L_{g} + L_{h}) ||x - y||$$ Thus making it locally but not globally Lipschitz. A similar process can be done to show differentiability everywhere except $x=0$.

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