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Consider the closed convex subset $\mathcal{F} = \{f \in C[0,1] : 0 \leq f \leq 1, f(0)=0, f(1)=1\}$. Consider the polynomial class $\mathcal{P} = \{p \text{ is a polynomial} : p(0)=0, p(1)=1, 0 \leq p \leq 1\}$. Is $\mathcal{P}$ dense in $\mathcal{F}$ in the sup norm?

Is anything known about Weierstrass theorem generalized as above to handle constraints? Or is there a counterexample?

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Indeed, $\mathcal{P}$ is dense in $\mathcal{F}$. The Bernstein polynomials approximating $f\in\mathcal{F}$ belong to $\mathcal{P}$, see Theorem 11.68 in:

http://www.pitt.edu/~hajlasz/Teaching/Math1530Fall2018/selection.pdf

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  • $\begingroup$ The answer is fitting, but perhaps does not fully address the $0\leq p\leq1$ constraint. So I would add that the reason that $p$ lies within the same bounds as $f$ can be most clearly seen via De Casteljau's construction of the corresponding polynomial approximation (convex hull property of Bezier curves). $\endgroup$
    – Iiro Ullin
    Commented May 18, 2021 at 23:50
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    $\begingroup$ @IiroUllin It does. If $0\leq f\leq 1$, then the Bernstein polynomials satisfy the same estimate. $p\geq 0$ is obvious, because all terms in the definition of $p$ are nonnegative and then replacing $f(k/n)$ by the upper bound of $1$ you get the binomial formula for $((1-x)+x)^n$ which equals $1$ which is the upper bound for $p$. If I find time, I will add details. $\endgroup$
    – Piotr Hajlasz
    Commented May 19, 2021 at 0:25
  • $\begingroup$ You are totally right of course, I am just commenting that it's not necessarily totally obvious for someone who's asking this question in the first place. $\endgroup$
    – Iiro Ullin
    Commented May 19, 2021 at 2:23

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