證 不妨設(shè)0≤x1<x2≤1,則(1)若x2-x1≤,則|f(x2)-f(x1)|<|x1-x2|=x2-x1≤.(2)若x2-x1>,則由f(0)=f(1),得|f(x2)-f(x1)|=|f(x2)-f(1)+f(0)-f(x1)|≤|f(x2)-f(1)|+|f(0)-f(x1)|<|x2-1|+|0-x1|=1-x2+x1<. (本文共 183 字 ) ." href="javascript:void(0)"> [閱讀本文] >>
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 證 不妨設(shè)0≤x1<x2≤1,則(1)若x2-x1≤,則|f(x2)-f(x1)|<|x1-x2|=x2-x1≤.(2)若x2-x1>,則由f(0)=f(1),得|f(x2)-f(x1)|=|f(x2)-f(1)+f(0)-f(x1)|≤|f(x2)-f(1)|+|f(0)-f(x1)|<|x2-1|+|0-x1|=1-x2+x1<. (本文共 183 字 ) ." href="javascript:void(0)"> [閱讀本文] >>