解: a =(x1,y1), b =(x2,y2),则λ a +(1-λ) b =(λx1+(1-λ)x2, λy1+(1-λ)y2} 对于①,f[λ a +(1-λ) b ]=λx1+(1-λ)x2-λy1-(1-λ)y2=λ(x1-y1)+(1-λ)(x2-y2) 而λf( a )+(1-λ)f( b )=λ(x1-y1)+(1-λ)(x2-y2)满足性质P 对于②f2(λa+(1-λb))=[λx1+(1-λ)x2]2+[λy1+(1-λ)y2],λf2(a)+(1-λ)f2(b)=λ(x12+y1)+(1-λ)(x22+y2) ∴f2(λa+(1-λb))≠λf2(a)+(1-λ)f2(b),∴映射f2不具备性质P. 对于③f[λ a +(1-λ) b ]=λx1+(1-λ)x2+λy1+(1-λ)y2+1=λ(x1+y1)+(1-λ)(x2+y2)+1 而λf( a )+(1-λ)f( b )=λ(x1+y1+1)+(1-λ)(x2+y2+1)═λ(x1+y1)+(1-λ)(x2+y2)+1 满足性质p 故答案为:①③