Lipschitz Condition and Differentiability

[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 lim_x->c(f(x)-f(c))/ (x-c) exists. I am having quite a hard time. Any hints?

 

[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.

 

[HallsofIvy]

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

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.

 

[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 &gt; 0 and \epsilon &gt; 0 such that

<br /> |f(x) - f(x_0)| \leq M|x-y| \quad \text{whenever} \quad |x-x_0| &lt; \epsilon.<br />

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?