13楼: Originally posted by Edstrayer at 2014-05-07 14:19:10
如果\mid f(x)-f'(x)\mid\leq 1
则-1\leq f'(x)-f(x)\leq 1
令f(x)=c(x)e^x，则f'(x)=(c(x)+c'(x))e^x
于是就有：
-1\leq c'(x)e^x\leq 1
-e^{-t}\leq c'(t)\leq e^{-t}
两边对t从0到x进行积分就得到：
e^{- ...

2楼: Originally posted by wurongjun at 2014-05-07 00:49:30
命题不成立!
比如f指数函数f(x)=e^x满足有连续导数
且f(x)=f'(x),所以f(x)-f'(x)=0;
但|f(x)|<1并不不成立!

3楼: Originally posted by jeato at 2014-05-07 07:48:51
()() 可是e^x是无界
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