(1)
f(0)=2f(0)-1, f(0)=1
只要证对在x附近,f(x)>f(0)
1=f(0.5)+f(-0.5)-1 ,f(0.5)=2
f(0.5)=2f(0.25)-1,f(0.25)=1.5
...
f(1/2^n)>1,对任意n成立
又因为f在0附近连续,所以只要x足够靠近0,f(x)>1
证毕
(2)例子y=x+1
(1)
f(0)=2f(0)-1, f(0)=1
只要证对在x附近,f(x)>f(0)
1=f(0.5)+f(-0.5)-1 ,f(0.5)=2
f(0.5)=2f(0.25)-1,f(0.25)=1.5
...
f(1/2^n)>1,对任意n成立
又因为f在0附近连续,所以只要x足够靠近0,f(x)>1
证毕
(2)例子y=x+1