① For any x∈[0, 1], there is always f (x) ≥ 0; ②f( 1)= 1； ③ If x 1≥0, x2≥0, x 1+x2≤ 1, then F(X1+x2) ≥ f (x1)+f (x2) holds.

(1) If f(x) is known as "friendship function", find the value of f(0).

(2) Is the function g (x) = (2 x)- 1 a "friendship function" in the interval [0, 1]? And give reasons.

(3) It is known that f(x) is a "friendship function", assuming that x0∈[0, 1] exists, then f(x0)∈[0, 1] and f[f(x0)]=x0, which proves that f (.

Answer:

Let x 1 = 0 and x2 = 0, then f (0); =0, so f(x)=0.

G(x) satisfies the first condition, because when x >; =0, f(x)>=0, and the second condition is also satisfied. When x= 1, f(x)= 1, and the third condition is satisfied, (2 (x1+x2))-(2x1)-(2x2)+. F(x 1)+f(x2), which holds.

The reduction to absurdity, assuming that f (x0) = x 1, f (x 1) = x0, and x 1 is not equal to x0, then we might as well put x1> X0, let t=x 1-x0, t>0, f(t)>0, then f (x1) = f (x0+t) > = f (x0)+f (t) > x1. X0, which contradicts the original conditions, so the hypothesis is not established, so the original verification is established and proved.