Does the n'th derivative of a smooth transition have to have more than n+1 roots for some n? I'm wondering if the space
$$S=\left\{f\in C^{\infty}([0,1])\ \ :\ \ f(0)=0\bigwedge f(1)=1\bigwedge\forall n\in\mathbb{N},\ f^{(n)}(0)=f^{(n)}(1)=0\right\}$$
contains an element $f_0$ such that $f_0^{(n)}$ has at most $n+1$ roots for all $n\in\mathbb{N}_0$.

Intuitively I'm somewhat inclined to think it's false. Do anyone know if this holds, or how I should go on about finding out?

If $f_0^{(n)}$ has $n+1$ roots, then by the mean value theorem $f_0^{(n+1)}$ has $n$ distinct roots in the interior of the support in addition to the boundary points being roots. Hence by induction $f_0^{(n)}$ must have $n+1$ roots for all $n\in\mathbb{N}$.

As to whether the lower bound is non-strict, I think maybe the way to go is to assume that $(a_i)_{i\in\mathbb{N}}$ is a sequence with $a_i \in S$ and $a_i^{(i)}$ having at most $i+1$ roots. Because if those properties imply that not all derivatives of $a_i$ remain bounded as $i\to\infty$, then no such sequence can have an element of $S$ as limit, and so the lower bound must be strict. 

I made some numerically experiments with linear programs which indicated the existence of such a sequence where also $\sup_x a_i^{(1)}(x)&lt;2$. So several derivatives must be considered simultaneously.
I would guess that the answer to your question is yes and my candidate for a function would be as follows. Let us work on the interval $[-1,1]$ and consider the bump function
$$ \psi(x) = \exp\left(\frac {-1}{1-x^2} \right)\,.$$
Then all derivatives at the endpoints vanish,
$$ \psi^{(n)}(-1) = \psi^{(n)}(1) = 0\,.$$
Now define
$$ f(x) = a \int_{-1}^x \psi(y) \,dy\,,$$
and choose $a \in \mathbb R$ such that $f(1) = 1$.

To prove your required properties one would need explicit expressions for the derivatives of $\psi$ which involve polynomials of high degrees. I believe that intuitively $\psi$ and its derivatives have no more roots than necessary, but I have not checked the details.
I would guess that the answer to your question is yes and my candidate for a function would be as follows. Let us work on the interval $[-1,1]$ and consider the bump function
$$ \psi(x) = \exp\left(\frac {-1}{1-x^2} \right)\,.$$
Then all derivatives at the endpoints vanish,
$$ \psi^{(n)}(-1) = \psi^{(n)}(1) = 0\,.$$
Now define
$$ f(x) = a \int_{-1}^x \psi(y) \,dy\,,$$
and choose $a \in \mathbb R$ such that $f(1) = 1$.

To prove your required properties one would need explicit expressions for the derivatives of $\psi$ which involve polynomials of high degrees. I believe that intuitively $\psi$ and its derivatives have no more roots than necessary, but I have not checked the details.