F _ n (x) > can be deduced when f_0(x)>0; 0 (except the origin), f_n(x) is continuous and strictly increasing, so we might as well start with f_ 1

Assuming that f_n has a limit and that the limit and integral are interchangeable, we can solve f(x)=x/2 by squaring first and then taking the derivative. Of course, here can only guess the answer, not as a basis for reasoning.

1. consider f _ 1 (x) = a * x b, a>0, b & gt=0. The recurrence formulas of a _ n and b _ n in f _ n (x) = a_n * x {b_n} can be obtained by using the recurrence relation, and it is not difficult to find that lim a _ n = 1/2 and lim b _ n = 1, that is, the initial value conclusion for a * x b type is valid.

2. Riemannian integrable function is bounded, so | f _ 0 (x) |

3. It is difficult to find the lower bound of A * X B type, but it can be modified slightly.

For any d>0 (0, of course

G_ 1(x)=0 on [0, d] and g _1(x) = u (x-d)/(1-d) on [d, 1].

So 0

Substitute g_ 1(x) into the iterative format of f_n to generate the sequence g_n(x). Because g_ 1(x) has the form of a * (x-d) b, it is easy to verify G _ n (on [d, 1]) by translation. (x-d)/2, so liminff _ n (x) >; = (x-d)/2

Since d is arbitrary, liminff _ n (x) >: = x/2.