The set of all sigmoid functions with slope $1$ at the origin and $\lim\limits_{x→+∞}f(x)=1$ I was reading about sigmoid functions and especially the ones with slope $1$ at the the origin.
For some data I think a sigmoid function with slope $1$ at the origin and upper bound $1$ may be an exact fit. Now I want to try to loop through the set of sigmoid functions.
However I need to know all possible sigmoid functions with slope $1$ at the origin and upper bound $1$, and so far I have just seen a few examples that happen to suffice. Are there more functions with these properties?
In short, is it possible to construct a set that contains all S-shaped functions that have slope $1$ at the origin and an upper bound of $1$? 
 A: $\def\d{\mathrm{d}}$For any sigmoid function $f$, if $f'$ is discontinuous at some point $x_0$, then Darboux's theorem implies that at least one of the left limit and right limit of $f'$ at $x_0$ does not exist (see Wiki). Thus the set of sigmoid functions below are sufficient for practical use.
Define $\mathscr{S}$ to be the set of all functions\begin{align*}
f:&& \mathbb{R} &\longrightarrow \mathbb{R}\\
&& x &\longmapsto \int_0^x g(t) \,\d t
\end{align*}
where $g \in C(\mathbb{R}),$ $0 \leqslant g \leqslant 1,$ $g(0) = 1,$ $\displaystyle \int_{-∞}^0 g(t) \,\d t < + ∞$, $\displaystyle \int_0^∞ g(t) \,\d t = 1$, $c \in \mathbb{R}$. For any $f \in \mathscr{S}$, $f' = g$ implies $f$ being increasing and $0 \leqslant f' \leqslant 1$, $f(0) = 0$, $f'(0) = 1$, and$$
\lim_{x → +∞} f(x) = \int_0^∞ g(t) \,\d t = 1,\\
|f(x)| \leqslant \int_{-∞}^∞ g(t) \,\d t < +∞. \quad \forall x \in \mathbb{R}
$$
thus $f$ satisfies the given conditions.

Now define $\mathscr{T}$ to be the set of all functions $h: \mathbb{R} → \mathbb{R}$ satisfiying:
*

*$h \in C^1(\mathbb{R})$,

*$h(x) = x\ (\forall x \leqslant 0)$, $h(x) → +∞\ (x → +∞)$,

*$h'(0) = 1$, $0 \leqslant h' \leqslant 1$.

It is easy to see that $(\mathscr{T}, \circ)$ is a semigroup, where $\circ$ means function composition. For any $f \in \mathscr{S}$ and $h \in \mathscr{T}$, since$$
(f \circ h)'(x) = f'(h(x)) h'(x) = g(h(x)) h'(x), \quad \forall x \in \mathbb{R}
$$
then $f \circ h \in \mathscr{S}$. Note that for any $x \geqslant 0$,$$
h(x) \leqslant x \Longrightarrow f(h(x)) \leqslant f(x),
$$
thus $f \circ h$ is a sigmoid function the graph of which is below that of $f$.
A: It is certainly possible to describe the set of all such functions. However it is hopeless to try and loop through the set of all such functions or even to describe it with finitely many parameters over which one could try to optimize. 
The basic underlying issue is that any such function could be perturbed by a $C^\infty$ function that vanishes to sufficiently high order at the origin and also vanishes at $\pm \infty$. And the set of such perturbations is infinite dimensional in a very strong sense (it contains an ideal of differentiable functions that is not finitely generated).    
