Monotonic version of Weierstrass approximation theorem Let $f\in\mathcal{C}^1([0,1])$ be an increasing function over $[0,1]$. 
Prove or disprove the existence of a sequence of real polynomials $\{p_n(x)\}_{n\in\mathbb{N}}$ with the properties:


*

*$p_n(x)$ is a non-decreasing function over $[0,1]$;

*the degree of $p_n$ is $n$;

*$\|f-p_n\|_{\infty}=\max_{x\in[0,1]}|f(x)-p_n(x)|=O\left(\frac{1}{n}\right)$.


This is a comment too long to fit in the usual format.
If you had asked for a bound in $\frac{1}{\sqrt{n}}$ rather
than $\frac{1}{n}$, then there would be
a nice answer using Bernstein polynomials (I find it especially nice
as the proof consists of a chain of inequalities, each using
a different part of mathematics). In this case everything 
behaves smoothly : one can compute an optimal bound for the error, and it is reached 
at rational values.
Recall that the $n-th$ Berstein approximation for $f$ is
$$
B_n(f)(x)=\sum_{k=0}^n f(\frac{k}{n})B_{n,k}(x), \
B_{n,k}(x)=\binom{n}{k}x^k(1-x)^{n-k} \tag{1}
$$
As explained in the linked MO answer, an elegant probabilistic argument shows that
$B_n(f)$ is nondecreasing if $f$ is. Let us put $M=\big|\big|\ f'\big|\big|_{\infty}$, so that
$f$ is $M$-Lipschitz. One can then show 
the surprisingly simple chain of inequalities, each one “optimal” in its way : for
any $x\in[\frac{i}{n},\frac{i+1}{n}]$, we have
$$
\big|B_n(f)(x)-f(x)\big| \leq 
2M\binom{n-1}{i}x^{i+1} (1-x)^{n-i} \tag{2}
$$
which is optimal because the inequality tends to equality when $f'$ tends to a 
staged function that equals $-M$ on $[0,x]$ and $+M$ on $[x,1]$. Next
$$
2M\binom{n-1}{i}x^{i+1} (1-x)^{n-i} 
\leq 
2M\binom{n-1}{i} \frac{(i+1)^{i+1} (n-i)^{n-i}}{(n+1)^{n+1}} \tag{3}
$$
which is optimal because it becomes an equality for $x=\frac{i+1}{n+1}$ ;
and 
$$
2M\binom{n-1}{i} \frac{(i+1)^{i+1} (n-i)^{n-i}}{(n+1)^{n+1}}
\leq
2M\binom{n-1}{\lfloor\frac{n+1}{2}\rfloor} 
\frac{(\lfloor\frac{n+3}{2}\rfloor)^{\lfloor\frac{n+3}{2}\rfloor} 
(\lceil\frac{n-1}{2}\rceil)^{\lceil\frac{n-1}{2}\rceil}}{(n+1)^{n+1}} \tag{4}
$$
which is optimal because it becomes an equality for 
$i=\lfloor\frac{n+1}{2}\rfloor$. Finally, Stirling's formula shows that the RHS above is asymptotically
equivalent to $\frac{1}{\sqrt{\pi n}}$.
