Let $f_{0}$ in $ \mathbb R $ be an increasing and continuous function. Define $f_{n}$ as
$f_{n+1}=\arctan\left(x+f_{n}(x)\right)$ for $x$ in $\mathbb R$, $n \geq 1$.
Show that $(f_{n})$ converges uniformly in each bounded interval $[a,b]$.
Solution:
I can show recursively that $(f_{n})$ is monotone and increasing in $\mathbb R$. Considering that we are in a bounded interval, is that enough to assert that this function converges uniformly?
How may I write this in a formal statement?
Thanks
$f_{n}$ is increasing for each $x\in [a,b]$, so the pointwise limit, $f$, exists.
Then $\left | f_{n+1}\left ( x \right )-f(x)) \right |= \left | \tan^{-1}\left ( x+f_{n}\left ( x \right ) \right )-\tan^{-1}\left ( x+f\left ( x \right ) \right ) \right |\leq \frac{1}{1+c^{2}}\left ( f\left ( x \right ) -f_{n}\left ( x \right )\right )$, for $c\in (x+f(x),x+f_{n}(x))$.
Now, if we restrict $x$ to $[a,b]$, then $c$ is contained in a bounded interval, so we may choose $r>0$ s.t $\frac{1}{1+c^{2}}< r< 1$ for all $x\in [a,b]$. This means that
$\left | f_{n+1} -f\right |< r\left | f_{n} -f\right |$ and therefore that $\left | f_{n+1} -f\right |<r^{n}\left | f-f_{0} \right |$.
But $\left | f(x)-f_{0}\left ( x \right ) \right |$ is bounded by some $M$, so we have finally
$\left | f_{n+1} -f\right |<r^{n}M$, so if $\epsilon >0$ is given then this is less than $\epsilon $ as soon as $n$ is large enough.