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> 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?
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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 $F=A$?



The grail of functional analysis?
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.


