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Basically, my question is as given the headline: "Is the composition of Lipschitz continuous functions Lipschitz continuous in the limit to $\infty$"?

To elaborate, consider two functions $f_1$ and $f_2$ which are both Lipschitz continuous, say with constants $L_1$ and $L_2$. Then we can easily prove that the composition is also Lipschitz:

$$\|f_1(f_2(x)) - f_1(f_2(y)) \| \leq L_1 \|f_2(x) - f_2(y)\| \leq L_1L_2 \|x-y\|$$

However, the new Lipschitz constant is the product of the two previous. This made me think what if we extend this to an infinite number of compositions? What else do we have to require of the elements to ensure the constant being finite? I.e. For Lipschitz continuous functions $f_1,f_2,\ldots, f_n$ is $\lim_{n\to \infty} (f_1 \circ \cdots \circ f_n)(x)$ Lipschitz continuous with finite constant? Alternatively let $f^{n} = (f \circ f \circ \cdots \circ f)(x)$ where there are $n$ compositions. Is $\lim_{n\to \infty}f^{n}(x)$ Lipschitz.

I am a aware that if $f$ is a contraction, such a construction, will make (under some constraints) $f^{n}(x)$ converge to $f^{n}(y)$ (or vice versa). Though I am interested in weaker requirements, which allows me to bound the Lipschitz constant, basically for any $n$.

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The limit of compositions need not be Lipschitz continuous, or continuous at all. Consider the function $f(x)=x^2$ on the interval $[0, 1]$. It is Lipschitz continuous with $L=2$. The $n$th iterate of $f$ is $x^{2^n}$, and these converge to the discontinuous function $$ g(x) = \begin{cases} 1, \quad x = 1; \\ 0, \quad 0\le x <1 \end{cases} $$

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  • $\begingroup$ This was also what I sort of anticipated, but are there any requirements one can put on the base-function to ensure that won’t happen? $\endgroup$ – Nicky Mattsson Mar 17 '18 at 7:47

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