Question

設(shè)y=f(x)是定義在［-1，1］上的函數(shù)，且滿足:

(i)f(-1)=f(1)=0;

(ii)對(duì)任意的u、v∈［-1,1］都有|f(u)-f(v)|≤|u-v|.

(1)證明對(duì)x∈［-1,1］都有x-1≤f(x)≤1-x;

(2)證明對(duì)任意的u、v∈［-1,1］都有|f(u)-f(v)|≤1.

Answer 1

證明:(1)令u=x,v=1得

|f(x)-f(1)|≤|1-x|,

∵f(1)=0,∴|f(x)|≤1-x.

∴x-1≤f(x)≤1-x.

(2)當(dāng)|u-v|≤1時(shí),

|f(u)-f(v)|≤|u-v|≤1;

    當(dāng)|u-v|＞1時(shí),

    則u、v異號(hào).不妨設(shè)u＜0,v＞0,

    則v-u＞1.

∴|f(u)-f(v)|=|f(u)-f(-1)+f(1)-f(v)|≤|f(u)-f(-1)|+|f(1)-f(v)|≤|u+1|+|1-v|=u+1+1-v=2-(v-u)＜1.

    綜上,|f(u)-f(v)|≤1.