TITLE: Does a smooth "transition function" with bounded derivatives exist?
QUESTION [20 upvotes]: 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?

REPLY [12 votes]: 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.