That is, +[g (0)] 2 = g (0),

Therefore, g(0)=0 or 1,

Assuming g(0)=0, then:

When x 1=x2= 1, [f (1)] 2+[g (1)] 2 = g (0),

That is, 1+[g (1)] 2 = 0, which is impossible.

So only g(0)= 1 holds.

②: when x 1=x2=- 1, there is f (-1) * f (-1)+g (-1) * g =

That is g (-1) = 0;

When x 1= 1 and x2=- 1, there is f (1) * f (-1)+g (1) g (-1).

That is, g(2)=0.

So g(2) can't be equal to 2.

③: Because it can be [f (x)] 2+[g (x)] 2 = g (0) = 1,

Therefore, [f (x)] 2+[g (x)] 2 = 1 holds.

④: Because [f (x)] 2+[g (x)] 2 = 1, [f (x)] 2 ≤1;

[g(x)]^2≤ 1，

(When the natural number n>2 points) so | [f (x)] n| ≤ [f (x)] 2;

|[g(x)]^n|≤[g(x)]^2，

Therefore | [f (x)] n| +| [g (x)] n|≤ [f (x)] 2+[g (x)] 2 =1,

That is, [f (x)] n+[g (x)] n ≤| [f (x)] n |+| [g (x)] n |≤1,

When x=0, [f (0)] n +| [g (0)] n = 1,

Therefore, the maximum value of [f (0)] n +| [g (0)] n is 1.

To sum up, choose D.