p*****n
发帖数: 758 | 1 show that there exists a nonnegative continuous function on [0,1] such that
f(0)=f(1)=0 and for a.e. x, f'(x) exists and is strictly positive. | Q***5
发帖数: 994 | 2 I think you can make this even more striking: f(0)=1, f(1)=0, and yet f'(x)>
0 a.e.
First, we construct a weaker example: f(0)=1, f(1) = 0, and f'(x)=0 a.e.
Consider the construction of Cantor set.
(1) On [1/3 2/3], define f = 1/2;
(2) On [1/9 2/9] f = 1/2*(f(0)+ f(1/3)) and [7/9 8/9] f= 1/2(f(/3)+f(1))
....., and so on.
It can be shown that such a f is continuous, and f'(x) = 0, except on the
set of all the end points -- which is of measure 0.
Now, to get an example of f'(x)>0 a.e., we only hav
【在 p*****n 的大作中提到】
: show that there exists a nonnegative continuous function on [0,1] such that
: f(0)=f(1)=0 and for a.e. x, f'(x) exists and is strictly positive.
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