设f(x)=ax^2+bx+c(a,b,c属于R),已知|f(-1)|≤1,|f(0)|≤1,|f(1)|≤1,求证：当-1≤x≤1时,|f(x)|≤5/4

Question

设f(x)=ax^2+bx+c(a,b,c属于R),已知|f(-1)|≤1,|f(0)|≤1,|f(1)|≤1,求证：当-1≤x≤1时,|f(x)|≤5/4

Answer 1

证明:
依题意有
{f(0)=c
{f(-1)=a-b+c
{f(1)=a+b+c
解此方程组得
{a=1/2*[f(1)+f(-1)]-f(0)
{b=1/2*[f(1)-f(-1)]
{c=f(0)
∴|f(x)|=|[1/2*(f(1)+f(-1))-f(0)]x^2+1/2*[f(1)-f(-1)]x+f(0)|
=|1/2*(x^2+x)f(1)+(1-x^2)f(0)+1/2*(x^2-x)f(-1)|
≤1/2*|x|*|x+1|*|f(1)|+|1-x^2|*|f(0)|+1/2*|x|*|x-1|*|f(-1)|
≤1/2*|x|(1+x)+1-x^2+1/2*|x|(1-x)
=-x^2+|x|+1
=-(|x|-1/2)^2+5/4
≤5/4.
证毕.
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