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Let $x \in \mathbb{R}$ and $n \in \mathbb{N}$. Let $f(x)$ be continous over the whole domain of $a<x<b$. Let the composition of functions $f^{(n)}(x) =f(f(...f(x)))$. Let $g(x)$ defined by

$$g(x)= \lim_{n \to \infty}f^{n}(x)$$

Is $g(x)$ continous? How can I proof that?

Erivelton Geraldo Nepomuceno
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2 Answers2

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$g$ does not need to be continous; in fact $g$ is not necessarily well-defined. For a counterexample, take $a=-b=1$ and $f(x)=-x$; then $f^n(x)$ is alternatively $\pm x$ hence $\lim_{n \to \infty} f^n(x)$ does not exist for nonzero $x$.

user133281
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No. $g(x)$ need not be continuous even if it does exists, which as reuanis points out might not be the case.

Take $f: [0,1] \to \mathbb R$, $f(x) = x^2$

Then you might quickly realize that $f^n$ tends pointwise to

$$g(x) = \begin{cases}0 & x, \in [0,1) \\ 1, & x = 1\end{cases}$$

which exhibits a jump at $x = 1$.

Squid
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  • This is defined on a closed interval, though. – Andrea Mori Mar 05 '14 at 14:00
  • I don't see how it's relevant, but it could be slightly modified to an open interval if that's more desirable. I could define $f(x)$ to be $f(x) = x^2$ on $(0,1]$and to be $f \equiv 1$ for positive $x$ up to some open limit, say in $(1,2)$ totaling $(0,2)$. The same general behavior is observed, in fact we get the same limit function. – Squid Mar 05 '14 at 14:04