Uniform convergence and composition

Question

Suppose we have a sequence of analytic functions $f_j:(1,M) \to (1,M)$ for some $M \in \mathbb{R}^+$. We are guaranteed that there is an $n \in \mathbb{N}$ such that its $n$'th iterate $$g_j(x) =f_j( f_j(...(n\,times)...(f_j(x)) = (f_j \circ f_j \circ ...(n\,times).. \circ f_j)(x)$$ converges uniformly, $g_j \to g$.

Does this guarantee that $f_j \to f$ uniformly?

If so how would one go about showing this? I think a proof by contradiction would work, if $f_j \not\to f$ uniformly then $g_j \not\to g$ uniformly. Is this sufficient enough of an argument?

Any help or suggestions would be greatly appreciated. Thanks.

Do you mean that $n$th iterate converges uniformly for every $n\ge 2$? — ajotatxe, Mar 20 '15 at 15:49
No sorry, I meant there exists an $n$ such that this is true and for all $k \in \mathbb{N}$ it is true for $n \cdot k$. — JmsNxn, Mar 20 '15 at 15:52
Oh sorry, should have mentioned that as well. They're bounded and take values in a bounded set. I edited the question to be more specific. — JmsNxn, Mar 20 '15 at 15:58
You can probably reduce this to proving that if $f(f(x))$ convergence uniformly then $f(x)$ converges uniformly. — abnry, Mar 20 '15 at 16:12

score 0 · Answer 1 · answered Mar 20 '15 at 16:02

0

Consider $f_n(x) = e^{n/z} = \exp(n/z)$ defined on $(1, M)$. Then

$$g_n(x) = f_n(f_n(x)) = \exp(n \exp(-n/z))$$

converges uniformly to 1, but $f_n$ doesn't converge at all!

answered Mar 20 '15 at 16:02

abnry

14,664

But $\lim_n f_n$ grows unbounded in this example. My sequence is necessarily bounded. – JmsNxn Mar 20 '15 at 16:10
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I see, you snuck that condition in! – abnry Mar 20 '15 at 16:11

Uniform convergence and composition

1 Answers1