Does there exist a set $F$ of monotone continuous functions $f \colon [0,1] \to [0,1]$ with the following properties?


1. For each $f \in F$ there exists $x \in [0,1]$ such that $f(x)=1$.
2. There exist $0<m<M<\infty$ such that for all $x \in [0,1)$ and $f \in F$, if $f(x)<1$ then $f$ is $C^2$ at $x$ with
$$ m \ < \ f'(x), \! f''(x) \ < \ M. $$
3. There exists $c \in (0,1)$ such that for all $x \in [0,1)$ and $f \in F$ with $f(x)<1$,
$$ x < c \ \Longleftrightarrow \ f'(x)<1. $$
4. The semigroup $S\_F := \{f\_j \circ \ldots \circ f\_1 \, : \, j \geq 1, \, f\_1,\ldots,f\_j \in F \}$ contains functions $g\_1,g\_2,g\_3,\ldots$ converging uniformly to the identity function.



Can the identity function be approximated by compositions of a "uniformly monotone-and-convex" set of functions?
No such class $F$ exists, and in fact we do not even need condition 3.




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First of all, note that every function $f$ in $F$ is non-decreasing and $f(1) = 1$.


Consider the composition $g = f\_n \circ \ldots \circ f\_1$. For every $x \in (0, 1)$, unless $g(x) = 1$, we have
$$ \log g'(x) = \sum\_{i = 1}^n \log f\_i'(x\_i(x)) , $$
where $x\_i(x) = f\_{i-1} \circ \ldots \circ f\_1(x)$. Note that $x\_i' \geqslant 0$, $f\_i' \geqslant 0$, $f\_i'' \geqslant 0$, and $x\_1(x) = x$. Therefore,
$$ \begin{aligned} (\log g'(x))' & = \sum\_{i = 1}^n (\log f\_i'(x\_i(x)))' \\ & = \sum\_{i = 1}^n \frac{f\_i''(x\_i(x)) x\_i'(x)}{f\_i'(x\_i(x))} \\ & \geqslant \frac{f\_1''(x\_1(x)) x\_1'(x)}{f\_1'(x)} \geqslant \frac{m}{M} \, . \end{aligned} $$
It follows that if $a < x$ and $g(x) < 1$, then
$$ \log \frac{g'(x)}{g'(a)} \geqslant \frac{m (x - a)}{M} \, ,$$
and thus
$$ \frac{g(b) - g(a)}{g'(a)} \geqslant \int\_a^b \exp\biggl(\frac{m (x - a)}{M}\biggr) dx \tag{1} $$
whenever $a < b$ and $g(b) < 1$. Similarly, if $x < b$ and $g(b) < 1$, then
$$ \log \frac{g'(x)}{g'(b)} \leqslant -\frac{m (b - x)}{M} \, ,$$
and consequently
$$ \frac{g(b) - g(a)}{g'(b)} \leqslant \int\_a^b \exp\biggl(-\frac{m (b - x)}{M}\biggr) dx \tag{2} $$
whenever $a < b$ and $g(b) < 1$.


We proceed by contradiction. Fix a sufficiently small $\varepsilon \in (0, \tfrac13)$ (to be specified later), and suppose that $g$ is $\varepsilon$-close to the identity function. Note that $g(\tfrac23) \leqslant \tfrac23 + \varepsilon < 1$. Set $a = \tfrac13$ and $b = \tfrac23$ in (1), and $a = 0$ and $b = \tfrac13$ in (2) to find that
$$ \frac{g(\tfrac23) - g(\tfrac13)}{g'(\tfrac13)} \geqslant \int\_{\frac13}^{\frac23} \exp\biggl(\frac{m (x - \tfrac13)}{M}\biggr) dx =: A $$
and
$$ \frac{g(\tfrac13) - g(0)}{g'(\tfrac13)} \leqslant \int\_0^{\frac13} \exp\biggl(-\frac{m (\tfrac13 - x)}{M}\biggr) dx =: B , $$
where $A > \tfrac13$ and $B < \tfrac13$ depend only on $m$ and $M$. It follows that
$$ \frac{g(\tfrac13) - g(0)}{g(\tfrac23) - g(\tfrac13)} \leqslant \frac{B}{A} \, , $$
which, in turn, implies that
$$ \frac{\tfrac13 - 2 \varepsilon}{\tfrac13 + 2 \varepsilon} \leqslant \frac{B}{A} \, . $$
If, however, $\varepsilon$ was chosen sufficiently small, the above inequality is false, and the proof is complete.


