I have the exercise.
Let $ \{f_n\}_{n=1}^{\infty} $ be a series of functions on $I = [0,1]$ defined with $f_1(x) =x, f_{n+1}=\sqrt{f_{n}(x)+2}$
(1) Show $ \{f_n\}_{n=1}^{\infty} $ is bounded for each $x \in I$ and $f_n(x) \leq f_{n+1}(x)$
(2) $ \{f_n\}_{n=1}^{\infty} $ is uniformly convergent on $I$
(3) Find $\lim_{n \to \infty} \int_I f_{n}(x) dx$
I can solve only (1).
(1) We show $f_n(x) \leq 2$. $n=1$ is obvious. Suppose it is true for $n=k$. Then $f_{k+1}=\sqrt{f_{k}(x)+2} \leq \sqrt{4} =2$. Next, $f_{1}(x) \leq f_{2}(x)$ for each $x$ is obvious, since $1 \leq f_{2}(x)$ for all $x \in I$. For general $n$, this is shown with induction.
(2)(my approach) $f_n$ is pointwise convergent, so suppose convergent to $f$. Let $g_n = 2- f_n$. If I can show $f$ is continuous, then by using Dini's thorem to $g$ and get result.
(3)(my approach) From (2), $\int_I \lim_{n \to \infty} f_{n}(x) dx$ is essential.
Thank you in advance.
Let $f(x)=\lim f_n(x)$. Then $f(x)=\sqrt {f(x)+2}$ so $y^{2}-y-2=0$ where $y=f(x)$. Solving this quadratic equation we get $y=f(x)=2$ or $f(x) =-1$. But $f_n \geq 0$ for all $n$ so $f(x) \geq 0$. Hence $f(x)=2$ for all $x$. Also $\int f_n(x)dx=\int 2 dx=2$ by Monotone Convergnce Theorem or DCT or by 2). Your argument for 2) is fine.