用定义验证下列各集合是凸集： (1)S={(x1 x2)|x1+2x2≥1 x1—x2≥1)； (2

正确答案：(1)对集合S中任意两点及每个数λ∈[01有由题设有 [λx₁⁽¹⁾+(1一λ)x₁⁽²⁾+2[λx₂⁽¹⁾+(1一λ)x₂⁽²⁾ =λ(x₁⁽¹⁾+2x₂⁽¹⁾)+(1一λ)(x₁⁽²⁾+2x₂⁽²⁾)≥λ+(1一λ)=1 [λx₁⁽¹⁾+(1一λ)x₁⁽²⁾一[λx₂⁽¹⁾+(1一λ)x₂⁽²⁾ =λ(x₁⁽¹⁾一x₂⁽¹⁾)+(1一λ)(x₁⁽²⁾一x₂⁽²⁾)≥λ+(1一λ)=1因此λx⁽¹⁾+(1-λ)x⁽²⁾∈S故S是凸集．由题设有 λx₂⁽¹⁾+(1一λ)x₂⁽²⁾≥λ|x₁⁽¹⁾+(1一λ)|x₁⁽²⁾|≥|λx₁⁽¹⁾+(1一λ)x₁⁽²⁾|因此λx⁽¹⁾+(1一λ)x⁽²⁾∈S故S是凸集．由题设有 [λx₁⁽¹⁾+(1一λ)x₁⁽²⁾+[λx₂⁽¹⁾+(1一λ)x₂⁽²⁾² =λ²x₁⁽¹⁾²+2λ(1一λ)x₁⁽¹⁾x₁⁽²⁾+(1一λ)²x₁⁽²⁾²+λ²x₂⁽¹⁾²+2λ(1一λ)x₂⁽¹⁾x₂⁽²⁾ +(1一λ)²x₂⁽²⁾²=λ²[x₁⁽¹⁾²+x₂⁽¹⁾²+(1一λ)²[x₁⁽²⁾²+x₂⁽²⁾²+λ(1一λ)[2x₁⁽¹⁾x₁⁽²⁾ +2x₂⁽¹⁾x₂⁽²⁾≤10λ²+10(1一λ)²+λ(1一λ)[x₁⁽¹⁾²+x₁⁽²⁾²+x₂⁽²⁾²+x₂⁽²⁾² ≤10λ²+10(1一λ)²+20λ(1一λ)=10因此λx⁽¹⁾+(1-λ)x⁽²⁾∈S故S是凸集．
(1)对集合S中任意两点及每个数λ∈[0,1,有由题设,有[λx1(1)+(1一λ)x1(2)+2[λx2(1)+(1一λ)x2(2)=λ(x1(1)+2x2(1))+(1一λ)(x1(2)+2x2(2))≥λ+(1一λ)=1,[λx1(1)+(1一λ)x1(2)一[λx2(1)+(1一λ)x2(2)=λ(x1(1)一x2(1))+(1一λ)(x1(2)一x2(2))≥λ+(1一λ)=1,因此,λx(1)+(1-λ)x(2)∈S,故S是凸集．由题设,有λx2(1)+(1一λ)x2(2)≥λ|x1(1)+(1一λ)|x1(2)|≥|λx1(1)+(1一λ)x1(2)|,因此λx(1)+(1一λ)x(2)∈S,故S是凸集．由题设,有[λx1(1)+(1一λ)x1(2)+[λx2(1)+(1一λ)x2(2)2=λ2x1(1)2+2λ(1一λ)x1(1)x1(2)+(1一λ)2x1(2)2+λ2x2(1)2+2λ(1一λ)x2(1)x2(2)+(1一λ)2x2(2)2=λ2[x1(1)2+x2(1)2+(1一λ)2[x1(2)2+x2(2)2+λ(1一λ)[2x1(1)x1(2)+2x2(1)x2(2)≤10λ2+10(1一λ)2+λ(1一λ)[x1(1)2+x1(2)2+x2(2)2+x2(2)2≤10λ2+10(1一λ)2+20λ(1一λ)=10,因此λx(1)+(1-λ)x(2)∈S,故S是凸集．