Let $f\in\mathcal{C}^1([0,1])$ be an increasing positive function over $[0,1]$.
In this answer, it is shown that there exists a squence of $n$th degree real polynomials $p_n(x)$ that approximate $f$:
- $p_n(x)$ is a non-decreasing function over $[0,1]$
- $\|f-p_n\|_{\infty}=\max_{x\in[0,1]}|f(x)-p_n(x)|=O\left(\frac{1}{\sqrt{n}}\right)$
I'll call a polynomial nice if $p_n(x)=\sum_{i=0}^n a_i x^{i-1}$ with $a_i \geq 0$ (This is the only distinction).
Question: Can we approximate any increasing function over $[0,1]$ using a sequence of nice polynomials? If so, does the $O(\frac{1}{\sqrt{n}})$ bound still go through? As far as I see, the construction in the answer above is not necessarily nice, unless I am missing something.
It is not possible. Any polynomial with non-negative coefficients is convex on $\mathbb R^+$, and a sequence of convex functions cannot approximate a non-convex function with arbitrary precision.
This is obvious enough intuitively, but to formalize it consider a non-convex function $f(x)$. There must exist points $a,b,t$ with $0 \le a \lt t \lt b \le 1$ such that $f(t) \gt \lambda f(a) + \mu f(b)$ where $\lambda = (b-t)/(b-a)$, $\mu = (t-a)/(b-a)$, otherwise the function would be convex. Let $\varepsilon = - \lambda f(a) + f(t) - \mu f(b) \gt 0$.
Consider any polynomial $p(x)$ with positive coefficients, then $p''(x) \ge 0$ on $(0,1)$ so the polynomial is convex on $(0,1)$ and therefore $p(t) \le \lambda p(a) + \mu p(b)$. It follows that:
$$ \begin{align} 0 \color{red}{+ \varepsilon} &\le \lambda p(a) - p(t) + \mu p(b) \color{red}{+ \varepsilon} \\ &= \lambda\left(p(a)-f(a)\right) - \left(p(t) - f(t)\right) + \mu \left(p(b)-f(b)\right) \\ &\le \lambda \big|p(a)-f(a)\big| + \big|p(t) - f(t)\big| + \mu \big|p(b)-f(b)\big| \end{align} $$
The latter inequality means the weighted sum of the approximation errors at points $a, t, b$ can never be smaller than $\varepsilon$, so the approximations can not be made arbitrarily precise.