已知二次函数f(x)满足|f(0)|≤1,|f(1)|≤1,|f(-1)|≤1,求证当|x|≤1...
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设f(x)=ax²+bx+c,则
f(0)=c;
f(1)=a+b+c;
f(-1)=a-b+c;
于是解得
a = 1/2·f(1) + 1/2·f(-1) - f(0)
b = 1/2·f(1) - 1/2·f(-1)
c = f(0)
因此|f(x)|=|[1/2·f(1) + 1/2·f(-1) - f(0)]·x²+[1/2·f(1) - 1/2·f(-1)]·x+f(0)|
=|(1/2·x²+1/2·x)·f(1)+(1/2·x²-1/2·x)·f(-1)+(-x²+1)·f(0)|
≤|1/2·x²+1/2·x|·|f(1)|+|1/2·x²-1/2·x|·|f(-1)|+|-x²+1|·|f(0)|
≤1/2·|x|·|x+1|+1/2·|x|·|x-1|+|1-x²|
=1/2·|x|·(|x+1|+|x-1|)+1-x²
=|x|+1-x²
=-(|x|-1/2)²+5/4
≤5/4
于是，原命题得证。
