Uniform convergence of a recursive increasing function Let $\phi:[0,+\infty)\rightarrow\mathbb{R}$ an increasing, continous function such that $\frac{1}{2}\leq\phi(x)\leq1$ for all $x$.
Let $f_0:[0,+\infty)\rightarrow\mathbb{R}$ an increasing function (not neccesarily continous) such that $f_0(0)\geq0$. For any $x\in[0,+\infty)$ define $f_n$ as a recursive function such that:
$$f_{n+1}(x) = \phi(x)f_n(x)+\frac{1}{ \phi(x)f_n(x)},n\in\mathbb{N}$$
Prove that $(f_n)_{n\in\mathbb{N}}$ uniformly converges in any closed interval $[0,R]$.
I tried to prove this sequence follows the Cauchy condition to get an instant uniform convergence. From what I got that
$$|f_m(x)-f_n(x)|\leq|f_{m-n}(x)-f_0(x)|+2\sum^{n}_{k = 0}|\frac{1}{f_{m-k}(x)}+\frac{1}{f_{n-k}(x)}|$$
Obviously this approach didn't seem to work. I tried to prove $f_n$ is monotone to use other theorems I have on monotone functions, getting no positive results. Which should be my approach for this problem?
 A: Some main elements for the answer. First I suppose that $\phi$ is increasing and stricty less than $1$. If you allow to have $\phi(x)=1$ the sequence $((f_n(x))$ diverges to $+ \infty$.
For $1/2 \le \phi < 1$, consider the sequence $(f_n)$ defined by the following recurrence relation:
$$f_{n+1} = \phi f_n + \frac{1}{\phi f_n} = g_\phi(f_n) \text{ for } n \in \mathbb{N}$$ where $g_\phi$ is defined as follow:
$$g_\phi:
\left|\begin{array}{lrl}
[0, +\infty & \longrightarrow & \mathbb{R} \\
y & \longmapsto & \phi y + \frac{1}{\phi y}\end{array} \right.$$
You should be able to prove following properties about $g_\phi$:


*

*$g_\phi$ is decreasing on $(0,1/\phi)$, has a minimum at $y = 1/\phi$ (with $g_\phi(1/\phi)=2$) and is increasing on $(1/\phi,+\infty)$

*$g_\phi$ is crossing the line $y=x$ at a single point $l_\phi=\frac{1}{\sqrt{(1-\phi)\phi}}$

*For $y \in (0,\frac{1}{\phi}\sqrt{\frac{1-\phi}{\phi}})$: $g_\phi(y) > l_\phi > y$.

*For $y \in (\frac{1}{\phi}\sqrt{\frac{1-\phi}{\phi}},\frac{1}{\sqrt{(1-\phi)\phi}})$: $y < g_\phi(y) < l_\phi$.

*For $y \in (\frac{1}{\sqrt{(1-\phi)\phi}},+ \infty)$: $y > g_\phi(y) > l_\phi$.

*For $y \ge 2$: $\phi \ge g^\prime_\phi(y) =\phi - \frac{1}{\phi y^2} \ge 0$

*$g_\phi(l_\phi)=l_\phi=g_\phi(\frac{1}{\phi}\sqrt{\frac{1-\phi}{\phi}})$.

*$g$ is convex.


From points above, it follows that for $x > 0$:


*

*The sequence $(f_n(x))$ is monotonic from $n=1$ and for $n \ge 2$ $f_n(x) \ge g_\phi(2)$

*The sequence $(f_n(x))$ is strictly increasing when $f_1(x) < \frac{1}{\sqrt{(1-\phi(x))\phi(x)}}$

*The sequence $(f_n(x))$ is strictly decreasing when $f_1(x) > \frac{1}{\sqrt{(1-\phi(x))\phi(x)}}$

*The sequence $(f_n(x))$ is constant when $f_1(x) = \frac{1}{\sqrt{(1-\phi(x))\phi(x)}}$

*The sequence $(f_n(x))$ converges to $\frac{1}{\sqrt{(1-\phi(x))\phi(x)}}$.


when $\phi(x) < 1$, while $(f_n(x))$ diverges to $+ \infty$ when $\phi(x)=1$.
You can then invoke the second theorem from Dini (sorry this is in French and the English version of Wikipedia is not including this second theorem from Dini) to say that $(f_n)$ is uniformly convergent to $\frac{1}{\sqrt{(1-\phi(x))\phi(x)}}$ on all closed segments. I have to check precisely this last point however!
However we can get rid of Dini second theorem.
By the mean value theorem, for $y \in (0,+ \infty)$ there exists $c_y \in (y,l_{\phi(x)})$ such that:
$$g_{\phi(x)}(y)-g_{\phi(x)}(l_{\phi(x)})=g_{\phi(x)}(y)- l_{\phi(x)}=g_{\phi(x)}^\prime(c_y)(y-l_{\phi(x)})$$
applying the equality to $y=f_n(x)$ you get
$$f_{n+1}(x) - l_{\phi(x)}=g_{\phi(x)}^\prime(c_y)(f_n(x)-l_{\phi(x)})$$
and for $n \ge 2$ you obtain:
$$ \vert f_{n+1}(x) - l_{\phi(x)} \vert \le \phi(x) \vert f_n(x)-l_{\phi(x)} \vert$$.
Now take any closed interval $[a,R]$ with $ 0 < a < R$. As $\phi(x)$ is supposed to be increasing and strictly less than $1$ you have $0 \le \phi(x) \le \phi(R) < 1$ for $x \in [a,R]$ hence
$$ \vert f_{n}(x) - l_{\phi(x)} \vert \le (\phi(R))^{n-2} \vert f_2(x)-l_{\phi(x)} \vert$$
for $n \ge 2$.
Proving that $(f_n(x))$ converges uniformly to the function $\frac{1}{\sqrt{(1-\phi(x))\phi(x)}}$ on any compact $[a,R]$.
