Question

已知函數f(x)=(x∈R)滿足下列條件：對任意的實x₁、x₂都有λ（x₁-x₂）²≤(x₁-x₂)［f(x₂)-f(x₂)］和|f(x₁)-f(x₂)|≤|x₁-x₂|,其中λ是大于0的常數.設實數a₀,a,b滿足f(a₀)=0和b=a-λf(a).

(1)證明λ≤1,并且不存在b₀≠a₀,使得f(b₀)=0;

(2)證明（b-a₀）²≤(1-λ²)(a-a₀)².

Answer 1

解析：(1)任取x₁,x₂∈R,x₁≠x₂，則由?λ(x₁-x₂)²≤(x₁-x₂)［f(x₁)-f(x₂)］                ①

和|f(x₁)-f(x₂)|≤|x₁-x₂|                                                                                       ②

可知λ(x₁-x₂)²≤(x₁-x₂)［f(x₁)-f(x₂)］≤|x₁-x₂|·|f(x₁)-f(x₂)|≤|x₁-x₂|²,從而λ≤1.

假設有b₀≠a₀，使得f(b₀)=0,則由①式知0＜λ(a₀-b₀)²≤(a₀-b₀)·［f(a₀)-f(b₀)］=0矛盾，故不存在b₀≠a₀,使得f(b₀)=0.

(2)由b=a-λf(a)

可知（b-a₀）²=［a-a₀-λf(a)］²=(a-a₀)²-2λ(a-a₀)f(a)+λ²［f(a)］²                      ③

由f(a₀)=0和①式知，（a-a₀）f(a)=(a-a₀)［f(a)-f(a₀)］≥λ(a-a₀)²                         ④

由f(a₀)=0和②式知，［f(a)］²=［f(a)-f(a₀)］²≤(a-a₀)²                                        ⑤

將④⑤代入③得（b-a₀）²≤(a-a₀)²-2λ²(a-a₀)²+λ²(a-a₀)²=(1-λ²)(a-a₀)².