A function f: (a,b) ->R is said to be Holder continuous of order alpha>0, if there exists M>0 such that |f(x)-f(y)|<=M|x-y|^alpha for all x,y elements of (a,b).
a) Prove that if alpha>0 and f is Holder continuous of order alpha, then f is uniformly continuous on (a,b).
b) Prove that if alpha>1 and f is Holder continuous of order alpha, then f is constant on (a,b).
c) Prove that if f is differentiable and Holder continuous of order alpha=1, then f'(x) is a bound function on (a,b).
a) Take any E>0 and choose delta=(E/M)^(1/alpha)>0. Then for all x,y elements of (a,b) such that |x-y|<delta, we have |f(x)-f(y)|<=M|x-y|^alpha<M*delta^alpha=E. Is that all I need to prove part (a)?
I don't know where to begin for (b) and (c). Any help?
Dec 3rd 2012, 02:13 PM
Re: Holder Continuity
Hint for b): under these assumptions, f is differentiable and its derivative equals 0.