Question

设f(x)=ax²+bx+c,当|x|≤1时,总有|f(x)|≤1,求证:|f(2)|≤8.

Answer 1

证明:∵0∈［-1,1］,∴|f(0)|≤1,即|c|≤1.?

又f(1)=a+b+c,f(-1)=a-b+c,?

∴|2b|=|f(1)-f(-1)|≤|f(1)|+|f(-1)|≤2.?

∴|b|≤1.?

又|2a|=|f(1)+f(-1)-2c|?

≤|f(1)|+|f(-1)|+2|c|?

≤2+2=4,?

∴|a|≤2.?

从而|f(2)|=|4a+2b+c|?

=|(a+b+c)+3a+b|?

≤|f(1)|+3|a|+|b|?

≤1+6+1=8.