Can the identity function be approximated by compositions of a "uniformly monotone-and-convex" set of functions?

Question

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.

Answer 1

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.