b is contained in the set of all Lipschitz functions of order a.
2. Is the set of all Lipschitz functions of order b a closed subspace of those
of order a?
Definitie orde van Lipschitz continuïteit
A function f: [a,b] -> R is Lipschitz of order a if there exists a constant K such that |f(x) - f(y)| <= K |x-y|^a and for all x,y in [a,b].
Voor 1 heb ik:
\( |f(x) -f(y)|<K|x-y|^b =K |x-y|^{b-a}|x-y|^a\)
\( |x-y| \leq |x|+|y| \leq 2b\leq 2 \)
en omdat \(|b-a|\leq 1\)
geldt er dus \( |x-y|^{b-a} \leq 2 \)
Conclusie
\(|f(x)-f(y)| <2K|x-y| <C|x-y|^a \)
Is dit correct?
