# Lipschitz Condition and Differentiability

1. Feb 21, 2009

### MatthewSmith2

Let K>0 and a>0. The function f is said to satisfy the Lipschitz condition if
|f(x)-f(y)|<= K |x-y|a ..

I am given a problem where I must prove that f is differentiability if a>1.

I know I need to show that limx->c(f(x)-f(c))/ (x-c) exists. I am having quite a hard time. Any hints?

2. Feb 21, 2009

### Unknot

I hope I remember my analysis, but that's usually called Hölder condition. Lipschitz we generally reserve for case when a=1. if a>1, the function is not just differentiable, but constant.

Proving that might be easier.

3. Feb 22, 2009

### HallsofIvy

Staff Emeritus
I have often seen "Lipschitz of order a" for that. I notice that Planet Math gives both:
http://planetmath.org/encyclopedia/HolderContinuous.html [Broken]

MathewSmith2, look at
$$\lim_{x\rightarrow y}\frac{|f(x)- f(y)|}{|x-y|}= \lim_{x\rightarrow y}\frac{K|x-y|^a}{|x-y|}$$

How is that related to the derivative and what happens on the right when a> 1?

Yes, it is true that such a function is constant, but I think proving that involves proving the derivative is 0 which requires first proving that the derivative exists.

Last edited by a moderator: May 4, 2017
4. Apr 30, 2011

### AxiomOfChoice

I'm interested in this, too. The definition of locally Lipschitz is something like this: $f$ is locally Lipschitz at $x_0$ if there exists $M > 0$ and $\epsilon > 0$ such that

$$|f(x) - f(x_0)| \leq M|x-y| \quad \text{whenever} \quad |x-x_0| < \epsilon.$$

Doesn't differentiability at $x_0$ imply this? I mean, if $f$ isn't locally Lipschitz at $x_0$, it's not differentiable there, right?