Holder-Continuous Functions for a>1

[calculus-stud]

I'm new to analysis, so I'm still trying to grapple with the concepts... there is one that has been bugging me forever now ---

||f(x) - f(y)|| <= ||x -y||^a is the Holder-continuous equality. What happens if a becomes > 1, does that still remain Holder-continuous or is Holder-continuity valid only for 0<a<1?

 

[snipez90]

If a > 1 f should be constant. Try to prove f' = 0 using the definition of the derivative. Intuitively the growth condition on f(x) - f(y) (for a > 1) implies that f(x) - f(y) is much smaller than y-x when y-x is small.

So you still call it Holder continuity, but the name doesn't mean much. If a > 1 you have a constant function so there is not much to say. Thus the only interesting cases are for a less than or equal to 1. If a = 1, you have lipschitz continuity which has many consequences in say differential equations.

 

[calculus-stud]

So, using the definition of derivative ---
f'(x) = lim h->0 f (x+h) - f(x)/h, if we have h = y-x then as you said from the growth condition, then the numerator which is f(y) - f(x) is much smaller than the denominator y-x, so f' = 0 for y-x close to 0. Is that right?

Also, when I'm changing h to y-x, how will the limit change?

 

[snipez90]

Well you might want to show |f'| = 0 since then you will have absolute values in both the numerator and denominator of the difference quotient.

But you just repeated what I said instead of giving a proof, which is what you're expected to do in analysis. So try to show |f'| = 0 using the definition. It's really just one step, since all you can do is apply the given inequality.

Intuitively, replacing h with y-x shouldn't make a difference. If we fix x and let y approach x, h is of course just measuring how far y is from x; letting h approach 0 is the same as letting y move closer and closer to x. The formal way of demonstrating this is via the epsilon-delta definition of the limit. It's completely analogous to showing that continuity of some function g at a point b means g(x) -> g(b) as x -> b or equivalently g(b + h) -> g(b) as h -> 0.