Does a smooth "transition function" with bounded derivatives exist? Does there exist a function $f: \mathbb{R} \to \mathbb{R}$ having the following properties?


*

*$f(x) = 0$ for all $x \le 0$.

*$f(x) = 1$ for all $x \ge 1$.

*For $0 < x < 1$, $f$ is strictly increasing.

*$f$ is everywhere $C^\infty$.

*The sequence of $L^\infty$ norms $\langle \left\lVert f \right\rVert_\infty, \left\lVert f' \right\rVert_\infty, \left\lVert f'' \right\rVert_\infty, \dots \rangle$ is bounded.


If we impose only the first four conditions, there is a well-known answer: for $0 < x < 1$, define $f(x)$ by
$$ f(x) = \frac{e^{-1/x}}{e^{-1/x} + e^{-1/(1-x)}} = \frac{1}{1 + e^{1/x - 1/(1-x)}} $$
However, the derivatives of this function appear to grow quite rapidly. (I'm not sure how to verify this, but it seems at least exponential to me.)
If such a function does not exist, what is the smallest order of asymptotic growth that the sequence $\langle \left\lVert f \right\rVert_\infty, \left\lVert f' \right\rVert_\infty, \left\lVert f'' \right\rVert_\infty, \dots \rangle$ can have?
 A: Suppose $f$ satisfies your first four hypotheses.  By the mean value theorem, there is some $x_1\in (0,1)$ such that $f'(x_1)=1$.  There is then some $x_2\in(0,x_1)$ such that $f''(x_2)=1/x_1$, and then there is some $x_3\in(0,x_2)$ such that $f'''(x_3)=1/x_1x_2>1/x_1^2$, and so on.  By induction, we see that we have points $x_n$ such that $f^{(n)}(x_n) \ge 1/x_1^{n-1}$ for all $n \ge 1$.  Thus the norms $\|f^{(n)}\|_\infty$ must grow at least exponentially in $n$.
By a more careful argument, you can show that in fact the derivatives must grow factorially.  More precisely, if $\|f^{(n)}\|_\infty=M$, we get that for any $x$, $|f^{(n-1)}(x)|\leq M|x|$, and then we get $|f^{(n-2)}(x)|\leq M|x|^2/2$, and so on, up to $|f(x)|\leq M|x|^n/n!$.  It follows that if there exists a constant $c>0$ such that $\|f^{(n)}\|_\infty<c^nn!$ for all $n$, then $f(x)=0$ for $|x|<1/c$.  But then replacing $f(x)$ with $g(x)=f(x\pm 1/c)$, we get that $f(x)=0$ for all $|x|<2/c$, and iterating this we get that $f$ vanishes everywhere.  So the norms $\|f^{(n)}\|_\infty$ must grow faster than $c^nn!$ for all $c>0$.
A: If the sequence of infinity norms $\{\|f^{(n)}\|_{\infty}\}_{n = 0}^{\infty}$ is bounded, i.e. there exists a constant $M$ such that for all $n \in \mathbb{N}_0$, $|f^{(n)}(x)| \le M$ for all $x \in \mathbb{R}$, then by this theorem $f$ is analytic. 
However, if an analytic function is $0$ on any interval (such as $(-\infty,0)$), then its Taylor series about any point in that interval is identically $0$. Then, since $f$ is analytic, the Taylor series converges to $f$ everywhere, i.e. $f \equiv 0$.
Therefore, no such function exists.
