推 ic6413 :Math 還是比 Rudin 厲害XD 11/16 19:05
※ 引述《Rudin5566 (魯丁.嗚嗚溜溜)》之銘言:
: If 1.f is continuous on [a b]
: 2.F(x)=supf([a x])
: Prove that F is continuous on [a b]
Since f is continuous on compact set [a,b], f is uniformly continuous on [a,b].
Given ε>0 there is a δ>0 such that
|f(x)-f(y)|<ε whenever 0<|x-y|<δ.
WLOG assume y>x.
For 0<|y-x|<δ, sup f([a,y]) = sup{ sup f([a,x]) , sup f([x,y]) }.
If sup f([a,y]) = sup f([a,x]), then F(y)-F(x)=0<ε.
If sup f([a,y]) = sup f([x,y]). Since f is continuous on [x,y], then
there exists y' in [x,y] such that sup f([x,y]) = f(y').
ε > f(y') - f(x) ≧ f(y') - sup f([a,x]) = F(y) - F(x) ≧ 0 > -ε
Thus |F(y)-F(x)|<ε whenever 0<|x-y|<δ □
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※ 編輯: Math 來自: 118.168.166.97 (11/16 04:08)