An entrance exam problem relating to sequences and limits 
this is a problem from the entrance exam of the University of Tokyo and unfortunately, the official doesn't offer solutions.

1) Using mathematical induction. Assume $f_{n}(x)=c_{n} x^{a_{n}}$ holds for $n$, and by $f_{n+1}(x)=p \int_{0}^{x}\left(f_{n}(t)\right)^{1 / q} \mathrm{d} t$ we can get $f_{n+1}(x)$. Then comparing the coeffient and the exponent will show that it holds for $n+1$ too.
2) 3) 4) 5) 6) I've no idea. I tried to calc the derivatives of $g_n$, but I don't know what to do next.
I can get $a_{n+1}$ from the recursive formula, but the form of it is kinda complex which makes it hard to get $c_{n+1}$. So I guess the rest questions could be done without knowing what actually $a_n$ and $c_n$ are. But I don't know how to do that.

Also, $1 / p+1 / q=1$ seems a frequent condition, how is it usually used in solutions?
 A: To get you started, the main notion to know to handle the first two questions is that of an arithmetico-geometric sequence (in the French context which differs from the English one).
Then, you will easily notice how $a_n$ is an arithmetico-geometric sequence.
That gives you both:


*

*an explicit formula for $a_n = \frac{1-q^{-n+1}}{1-q^{-1}}=\sum_{k=0}^{n-2}q^{-k} = p(1-q^{-n+1})$

*its limit $\lim_{n\to \infty} a_n=\frac{1}{1-q^{-1}}=p$
For 2):
$$
g'_n(x_n)=a_nx_n^{a_n-1}-px_n^{p-1}=0 \Leftrightarrow x_n=\left( \frac{a_n}{p} \right)^{\frac{1}{p-a_n}}
$$
which holds because $\lim_{n\to \infty} a_n=p$  and $(a_n)_{n\geq 0}$ is increasing so $p-a_n> 0$ thus $x_n$ exists and is unique.
It is a maximum because of the sign of the difference of $x\mapsto a_nx^{a_n-1}$ and $x\mapsto px_n^{p-1}$ (try to get this result by yourself).
For 3):
Since $y\mapsto x^y=e^{y\ln x}$ is continuous over $\mathbb{R}$ whatever $x\in \mathbb{R}^{+*}$, $\lim_{n\to \infty}x^{a_n}=x^{\lim_{n\to \infty}a_n}=x^p$. Thus, $\lim_{n\to \infty}g_n(x)=0$ for $x\in ( 0,1 ]$. For $x=0$, the result is trivial.
