Question

一道定积分的不等式证明题

设Pn(x)为n次多项式，求证：

∫(a,b)|Pn'(x)|dx<=2n*Max|Pn(x)|,x∈[a,b]

Answer 1

设a≤x1<x2<...<xk≤b为Pn'(x)在[a,b]上的k个零点，k≤n-1<n
∫(a,b)|Pn'(x)|dx=|∫(a,x1)Pn'(x)dx|+|∫(x1,x2)Pn'(x)dx|+...+|∫(xk,b)Pn'(x)dx|
=|Pn(x1)-Pn(a)|+|Pn(x2)-Pn(x1)|+...+|Pn(xk)-Pn(b)|
≤|Pn(x1)|+|Pn(a)|+|Pn(x2)|+|Pn(x1)|+...+|Pn(xk)|+|Pn(b)|
≤Max|Pn(x)|+Max|Pn(x)|+...+Max|Pn(x)|=2k*Max|Pn(x)|
≤2n*Max|Pn(x)|