Open-problem statements, with their sources attached.

Let $C>0$. There exists $\epsilon>0$ such that if $n$ is sufficiently large the following holds.

For any $x_1,\ldots,x_n\in [-1,1]$ there exist $y_1,\ldots,y_n\in [-1,1]$ such that, if $P$ is a polynomial of degree $m<(1+\epsilon)n$ with $P(x_i)=y_i$ for at least $(1-\epsilon)n$ many $1\leq i\leq n$, then $\max_{x\in [-1,1]}\lvert P(x)\rvert >C.$

Lower bound for $c(p)$ for $1 < p \le \infty$, improving the known value at $p = 2$ or $p = \infty$.

Upper bound for $c(p)$ for $1 < p \le \infty$, improving the best-known value at $p = \infty$.

Assume for $n>1$, $f:\mathbb{R}^n\to\mathbb{R}^n$ is a bijection, where $\mathbb{R}^n$ is equipped with the standard topology. Does the connectedness of (the induced power set map) $f$ imply that of $f^{-1}$?

Let $f : \mathbb R^n \to \mathbb R, n \geq 2$ be a $C^1$ function. Does the equality $\sup_{x \in \mathbb R^n}f(x) = \sup_{x\in \mathbb R^n} f(x+\nabla f(x))$ hold when both suprema are finite?

Conjecture: Voronovskaja-type formula for Bézier-Bernstein operators with shape parameter $\alpha > 0$, $\alpha \neq 1$.

The source asks for sufficiently smooth functions. This concrete version uses ContDiffOn ℝ 2 f I as a readable baseline regularity assumption; since the domain is the compact interval $[0,1]$, this also explains why no separate boundedness assumption is included here. The variants below record the unknown smoothness threshold more explicitly.

Variant of the Bézier-Bernstein Voronovskaja problem which treats "sufficiently smooth" as an eventual condition in the smoothness order $m$: for all sufficiently large finite $m$, every $C^m$ function on $[0,1]$ should have the asserted asymptotic formula.

Existence-only version of the eventual-smoothness variant. This separates the first part of the source problem, proving that the scaled sequence has some limit, from the stronger task of finding an explicit expression for that limit.

Variant of the Bézier-Bernstein Voronovskaja problem with the required smoothness order itself left as an answer. Replacing (answer(sorry) : ℕ × ((ℝ → ℝ) → ℝ → ℝ)) by a concrete value lets one state the conjecture for a chosen regularity threshold.