# Show it converges uniformly

• Nov 15th 2009, 11:18 AM
thaopanda
Show it converges uniformly
Suppose that g : R $\rightarrow$ R is such that there exists r $\in$ (0,1) with the property that |g'(x)| $\leq$ r for every x $\in$ R. Let $F_{0}$ : R $\rightarrow$ R be a bounded function and define recursively $F_{n}(x)$ := $g(F_{n-1}(x))$, for each n $\geq$ 1. Show that { $F_{n}$} $_{n \in N}$ converges uniformly on R.
• Nov 15th 2009, 11:45 AM
redsoxfan325
Quote:

Originally Posted by thaopanda
Suppose that g : R $\rightarrow$ R is such that there exists r $\in$ (0,1) with the property that |g'(x)| $\leq$ r for every x $\in$ R. Let $F_{0}$ : R $\rightarrow$ R be a bounded function and define recursively $F_{n}(x)$ := $g(F_{n-1}(x))$, for each n $\geq$ 1. Show that { $F_{n}$} $_{n \in N}$ converges uniformly on R.

This is similar to the contraction mapping principle.

$\frac{|g(x)-g(y)|}{|x-y|}=|g'(c)|\leq r\implies |g(x)-g(y)|\leq r|x-y|$

Have you learned the CMP yet?
• Nov 15th 2009, 12:21 PM
thaopanda
I don't think I've learned that yet...
• Nov 15th 2009, 12:59 PM
redsoxfan325
Quote:

Originally Posted by redsoxfan325
This is similar to the contraction mapping principle.

$\frac{|g(x)-g(y)|}{|x-y|}=|g'(c)|\leq r\implies |g(x)-g(y)|\leq r|x-y|$

Have you learned the CMP yet?

$|F_{n+1}-F_n|=|g(F_n)-g(F_{n-1})|\leq r|F_n-F_{n-1}|=$ $r|g(F_{n-1})-g(F_{n-2})|\leq r^2|F_{n-1}-F_{n-2}|\leq...\leq r^n|F_1-F_0|$ (You might want to prove this using induction.)

Let $d=|F_1-F_0|$. For $m>n$, by the triangle inequality,

$|F_m-F_n|\leq |F_m-F_{m-1}|+|F_{m-1}+F_{m-2}|+...+|F_{n+1}-F_n|\leq r^{m-1}d+r^{m-2}d+...+r^nd=$ $dr^n\sum_{k=0}^{m-n-1}r^k\leq dr^n\sum_{k=0}^{\infty}r^k=\frac{dr^n}{1-r}$ (You might also want to prove this by induction.)

Since $r<1$ (and $d$ is bounded), $\forall \epsilon>0$, $\exists N$ such that $n>N$ implies $r^n<\frac{\epsilon(1-r)}{d}$. So for all $\epsilon>0$, $m>n>N$ implies

$|F_m-F_n|\leq\frac{dr^n}{1-r}<\frac{d}{1-r}\cdot\frac{\epsilon(1-r)}{d}=\epsilon$

so $\{F_n\}$ converges uniformly by the Cauchy criterion.

Note: The above proof I gave is basically the proof of the CMP.