Sequence of Contractive Maps Converges? Suppose I have a sequence of contractive maps with contraction constants $(1-a_i)$, where $0 < a_i < 1$. If the sum of the $a_i$ diverges, how can I show that the sequence $x_{i+1} = f_i(x_i)$ converges? Assume we are in a compact complete metric space.
 A: The principal trick suggested by these assumptions is that
$$
\prod_{i=1}^\infty(1-a_i)\le e^{-\sum_{i=1}^\infty a_i}
$$
converges to zero.

You can combine sufficient numbers of maps $f_i$ so that their concatenation $g_j$ has a contraction constant smaller some pre-selected $\varepsilon>0$. I.e., without loss of generality one can assume that $a_i\ge 1-ε$.
Now each map $f_i$ has a fixed point $z_i$ and if $R$ is the radius of the compact $X$, then the image of $f_i$ is contained in $B(z_i,2εR)$. Assume $ε$ is so small that even $2εR$ is very small. Then the sequence $x_i$ will jump from close to $z_i$ to close to $z_{i+1}$. While there must be limit points of the sequence, there does not need to be convergence. 

Example: On $X=[0,1]$ alternate the maps $f_{2m}(x)=εx$ and $f_{2m+1}(x)=1-εx$. $f_1(f_0(x))=1-ε^2x$ has the fixed point $1/(1+ε^2)$, $f_0(f_1(x))= ε-ε^2x$ has the fixed point $ε/(1+ε^2)$, $a_0=a_1=1-ε$, so the series of the $a_i$ diverges.

Thinking about it again, there was a reason this seemed familiar. This setup readily applies to the situation of a chaos game for the construction resp. approximation of an IFS fractal. There you generate a random sequence out of a set of contrative maps and get a sequence of points that accumulates to give the shape of Julia sets, Barnsley ferns or Sierpinski triangles.


