Choose . Since the limit of is zero for , we get some such that for . Consider a covering of consisting of intervals of length . Choose and show that by adding and substracting inside the absolute value.
Let Lip_m[a,b] be the set of all Lipschitz functions of order m in [a,b] that is all functions satisfying |f(x)-f(y)|<= K * |x-y|^m for a certain constant K and for all x ,y in [a,b].
Now define the following function:
A_m (f) = sup { |f(x) - f(y)|/|x-y|^m : x,y are in [a,b] and x is not equal to y}. sup represents supremum.
Now for z >0 and f in Lip_m[a,b] and t>0 define:
R_z (f,t) = sup { |f(x)- f(y)| / |x-y|^z : x,y are in [a,b], 0< |x- y| <= t}.
Now another definition:
The "small Lipschitz" space denoted by lip_m[a,b] is the following set:
lip_m[a,b] = { f in Lip_m[a,b] : limit ( R_m (f,t) when t-> +0) = 0 } where t->+0 means from the right.
If m = 1 show that lip_m [a,b] contains only the constant functions.
Can you please help?