p*****d
发帖数: 237 | 1 A real-valued function f on [0,1] is said to be Holder
continuous of order α if there is a constant C such that
abs(f(x)-f(y)) ≤ C(abs(x-y)^α).
Define Norm(f)_α
=max(abs(f(x))+sup(abs(f(x)-f(y))/(abs(x-y))^α).
Show that for 0< α ≤ 1, the set of functions with
Norm(f)_α ≤ 1 is a compact subset of C[0,1].
Note: C[0,1] is the space of continuous functions on [0,1].
多谢多谢！ | b*****e
发帖数: 474 | 2 if Norm(f)_α ≤ 1, then f is Holder continuous of order α,
obviously f is continous, thus f belongs to C[0,1].
Compactness? for any f(x) > 1, Norm(f)_α > 1, thus not in
{ f | Norm(f)_α ≤ 1 }
【在 p*****d 的大作中提到】
: A real-valued function f on [0,1] is said to be Holder
: continuous of order α if there is a constant C such that
: abs(f(x)-f(y)) ≤ C(abs(x-y)^α).
: Define Norm(f)_α
: =max(abs(f(x))+sup(abs(f(x)-f(y))/(abs(x-y))^α).
: Show that for 0< α ≤ 1, the set of functions with
: Norm(f)_α ≤ 1 is a compact subset of C[0,1].
: Note: C[0,1] is the space of continuous functions on [0,1].
: 多谢多谢！
| d*z
发帖数: 150 | 3 for each n, we can select a m(n)>0, and a function list
F(n)={f(n,k),k=1,2,...},and the same time F(n) is a sub list
of F(n-1），
that in those x in {1/m(n),2/m(n),...,m(n)/m(n)},
Limit(k-->+00,f(m(n),k)(x) Exist.
Let F={f(1,1),f(2,2),f(3,3),...}, Then we can found that the
Limitation of the function list exist. So It is a compact
set. |
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