Let us consider the set $\mathcal{F}$ of strictly increasing continuous functions from $[0;1]$ on $[0,1]$ that cancel in $0$ and are equal to $1$ in $1$. So, if $f\in \mathcal{F}$ one has $f(0)=0$ and $f(1)=1$.
The function $h_{a}(x)=x+ax(1-x)$ is associated to each number $a\in[-1,1]$.
One can see that for all $a\in[-1,1]$, $h_{a}\in \mathcal{F}$.
We consider now
$$\mathcal{A}_{n}=\left\{\varphi_{a^{n}}=h_{a_{n}}\circ\ldots\circ h_{a_{1}},{\rm with}\;a^{n}=(a_{1},\ldots a_{n})\in[-1,1]^{n}\right\}$$
the set of polynomials obtained by composing the $n$ fonctions $h_{a_{1}},\ldots,h_{a_{n}} $.
Clearly, for all $n$ we have $\mathcal{A}_{n}\subset \mathcal{F}.$
Now here is my question :
Consider $f\in \mathcal{F}$, is it possible to find for large enough $n$ a $\varphi_{a^{n}}\in \mathcal{A}_{n}$ that approaches $f$ uniformly? In other terms, is the set $\bigcup\mathcal{A}_{n}$ dense in $(\mathcal{F},\| \|_{\infty})$ ?
Dear Didier,
Obviously $\mathcal {F}$ is convex. Therefore, $\bigcup \ mathcal{A} _ {n}$ must also be convex. Suppose that is the case. Since any cumulative distribution on [0,1] can be uniformly approximated by a beta distribution mixture (1), it should be sufficient to show that $\bigcup \ mathcal {A} _ {n}$ contains the beta distributions.
Hope the reference is at least useful. Best regards.
(1) Ghosal, S. (2001) Convergence rates for density estimation with Bernstein polynomials, Ann. Statist. 29, 1264-1280.
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