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on Voronovskaja-type asymptotic formula for Bézier–Bernstein operators with shape parameter

The statement

Let $I = [0,1]$ and fix a real parameter $\alpha > 0$ with $\alpha \neq 1$ . For each $n \in \mathbb{N}$ , the Bernstein basis polynomial is defined by

p_{n,j}(x) = \binom{n}{j} x^j (1-x)^{n-j}

for $j = 0, 1, \ldots, n$ , and the cumulative tail sum is

J_{n,k}(x) = \sum_{j=k}^{n} p_{n,j}(x)

for $k = 0, 1, \ldots, n$ , with $J_{n,n+1}(x) = 0$ . The Bézier-type Bernstein operator $B_{n,\alpha}$ acts on a function $f: I \to \mathbb{R}$ by

(B_{n,\alpha} f)(x) = \sum_{k=0}^{n} f\!\left(\frac{k}{n}\right) \Bigl( J_{n,k}(x)^{\alpha} - J_{n,k+1}(x)^{\alpha} \Bigr).

Determine the smallest natural number $m$ and an explicit formula $\Lambda(f,x)$ such that for every function $f: \mathbb{R} \to \mathbb{R}$ and every point $x \in I$ , if $f$ is $C^m$ on $I$ (that is, $m$ -times continuously differentiable on $I$ with all derivatives up to order $m$ continuous on $I$ ), then

\lim_{n \to \infty} \sqrt{n}\, \bigl( (B_{n,\alpha} f)(x) - f(x) \bigr) = \Lambda(f,x).

Equivalently, determine the pair $(m, \Lambda)$ where $m \in \mathbb{N}$ and $\Lambda: (\mathbb{R} \to \mathbb{R}) \times I \to \mathbb{R}$ is a functional, such that the sequence $\sqrt{n}\bigl( B_{n,\alpha} f(x) - f(x) \bigr)$ converges to $\Lambda(f,x)$ for all $C^m$ functions $f$ and all $x \in I$ .

What gets accepted

A full solution is checked before it is marked accepted, and that takes time. Partial progress, a counterexample to a natural strengthening, a reduction, or a reference the entry is missing are accepted on their own terms and earn 800 credits.

Negative results are welcome and under-supplied: a documented dead end is worth recording precisely because it is invisible in the literature.