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My brain can't grasp the concept of Lipschitz continuity for some how... The definition is stated here: Lipschitz continuous

Now, can someone please run through an example and explain how we can determine whether a function is lipschitz continuous?

Let's look at the function $f(x,y)=x^2-x^3$ for instance... (or any other simple function). Is that lipschitz continuous, and why?

Sebastiano
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econmajorr
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  • this simplest functions would be to either a constant function or $f(x)=x$ from $\mathbb{R}$ to $\mathbb{R}$ – C Squared May 04 '21 at 21:48
  • also, do you mean $f(x,y)=x^2-y^3$ or $f(x,y)=y^2-x^3$ ? It would help if you specified the domain of your function as well, since that will play a role in determining if a function is Lipschitz or not – C Squared May 04 '21 at 21:52
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    no I actually, mean what I write: $x^2-x^3$. This is due to fact that in numerical analysis it is often a function a function og two dimention. For instance: $y'=f(y,x)=y$ where y is a funtion of x itself. Just an example – econmajorr May 04 '21 at 21:56
  • okay, just making sure. thanks – C Squared May 04 '21 at 21:58

2 Answers2

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Common examples of Lipschitz functions are continuously differentiable functions with certain domains. If $f \in C^1([a, b], \mathbb{R})$, then $f$ is Lipschitz continuous since for $x, y \in [a, b]$ by the mean value theorem, $$|f(x) - f(y)| \leq \sup_{[a, b]}|f'| \cdot |x - y|.$$ $\sup_{[a, b]}|f'|$ exists since $f'$ is continuous on the compact set $[a, b]$. Also if $\Omega \subset \mathbb{R}^n$, $f \in C^1(\Omega, \mathbb{R}^m)$, and $K \subset \Omega$ is compact and convex (actually the convexity of $K$ is not necessary), then $f|_{K}$ is Lipschitz (a proof is given here by zhw $C^1$ function on compact set is Lipschitz). So these are some commonly encountered Lipschitz functions.

Mason
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    how does this answer op’s question? – C Squared May 04 '21 at 21:51
  • @CSquared the OP asked "Let's look at ... any other simple function). Is that lipschitz continuous, and why?" This answer looked at a lot of functions and showed why they are Lipschitz continuous – miracle173 May 05 '21 at 06:21
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To address your request, let's try to determine whether $f:\mathbb{R}^2\to\mathbb{R}$ given by $f(x,y)=x^2-x^3$ is Lipschitz or not. I often find that proving a function is not Lipschitz is harder than proving that it is Lipschitz.

I claim that $f$ is not Lipschitz. Indeed let $K\geq 0$. We need to produce $(x,y)$ and $(z,w)$ in $\mathbb{R}^2$ such that $$|f(x,y)-f(z,w)|>K\sqrt{(x-y)^2+(z-w)^2} $$

Choose $(z,w)=(0,0)$ and $(x,y)=(\sqrt{K+1}+1,0)$. Notice that,

$$|x||x-1|-K=(\sqrt{K+1}+1)(\sqrt{K+1})-K=1+\sqrt{K+1}>0$$ which implies $$|x^3-x^2|-K|x|>0$$ and finally $$|x^3-x^2|>K|x|$$ hence $f$ is not Lipschitz.

If anything is unclear, please ask in the comments.

C Squared
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