Question

f(x)=ax+b,当x绝对值≤1时,都有f（x)的绝对值≤1,求证b的绝对值≤1,a的绝对值≤1

Answer 1

由题意,x=0时|f(0)|=|b|≤1

又x=1时,|f(1)|=|a+b|≤1,即-1≤a+b≤1 (1)

而x=-1时,|f(-1)|=|-a+b|≤1,则|a-b|≤1,即有-1≤a-b≤1 (2)

将(1)(2)两式相加即得：-2≤2a≤2,即-1≤a≤1,也即|a|≤1,证毕.