Does $g\in C([0,1],[0,1])=A$ exist such that $\{g^n ,n\in\mathbb N\}$ is dense in $A$ provided with the uniform norm?
with $g^2=g \circ g $
If we can find $g$ then $F$ a closed of $A$, $id \in F$ with $g\circ F\subset F$, we have $F=A$?
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1$\begingroup$ Which part of this is the question? The existence of $g$? The existence of $g$ and $F$? The existence of $g$ such that every $F$ behaves as indicated? You highlight one thing, but then appear to ask another question. $\endgroup$– LSpiceCommented Mar 21, 2022 at 14:06
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$\begingroup$ g is the grail, if it exists. $\endgroup$– DattierCommented Mar 21, 2022 at 14:08
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$\begingroup$ Right, but which part is the question? You make an assertion about $F$, but follow it by a question mark. Are you asserting that, if $g$ exists, then every such $F$ must be all of $A$, or are you asking whether that is the case (given the existence of $g$)? $\endgroup$– LSpiceCommented Mar 21, 2022 at 14:13
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$\begingroup$ "Does.......... .? " $\endgroup$– DattierCommented Mar 21, 2022 at 14:14
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4$\begingroup$ Brouwer fixed point theorem. $\endgroup$– bathalf15320Commented Mar 21, 2022 at 14:14
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1 Answer
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The answer is no. If there was, then for some $N$ we would have $\|g^N\|\leq 1/2$, in order for $g^N$ to be within distance $1/2$ of the constant $0$ function. This means that the range of $g^N$ is contained in $[0,1/2]$. But then the same holds for $g^n$ for all $n\geq N$, since $g^n=g^N\circ g^{n-N}$. In particular $\{g^n\}$ cannot be dense in $A$.
Essentially the same argument shows that if some constant function is a limit point of the sequence $g^n$, then it is in fact the limit of this sequence.
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$\begingroup$ Show the answer of @bathalf15320. $\endgroup$– DattierCommented Mar 21, 2022 at 14:26
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4$\begingroup$ What do you mean? They have not posted an answer. $\endgroup$– WojowuCommented Mar 21, 2022 at 14:27
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$\begingroup$ You answer is correct but I can t choose it, bathalf is the first. $\endgroup$– DattierCommented Mar 21, 2022 at 14:28
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1$\begingroup$ With this in mind, I think the question that question, which could more appropriately be called the "grail in functional analysis", is whether the set $\mathrm{span} \{g^n ; n \in \mathbb{N}\}$ is dense in $A$. $\endgroup$ Commented Mar 21, 2022 at 14:58
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9$\begingroup$ @Dattier If someone makes a comment but chooses not to post an answer, that should not prevent you from accepting someone else's answer. $\endgroup$ Commented Mar 21, 2022 at 17:20