→ znmkhxrw :觀念就像樓上推文 02/18 01:02
7.
Assuem f(x) =/= 0, i.e. there exists a€[-1,1] s.t. f(a) =/=0 => f^2(a) > 0
Since f is continuous on [-1,1], f^2 is also conti. on [-1,1]
so by definition
take ε= 0.5*f^2(a) , there exists δε>0 ,s.t. when │x-a│< δε
│f^2(x) - f^2(a)│< 0.5*f^2(a)
we have │f^2(a)│ - │f^2(x)│< 0.5*f^2(a)
so │f^2(x)│= f^2(x) > 0.5*f^2(a) > 0 , for │x-a│< δε
1 a+δε a+δε
S f^2(x) dx >= S f^2(x) dx >= S 0.5*f^2(a) dx > 0.5*f^2(a)*2δε>0
-1 a-δε a-δε
矛盾
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