Let $f:[0,1]\mapsto(0, +\infty)$ be a continuous function. Define by

$$ B_k(f,t):=\sum_{j=0}^k f(j/k) {k \choose j} t^{j} (1-t)^{k-j}, \quad t \in [0,1] $$

the associated Bernstein polynomial. For $k \geq 3$ denote

$$ \Delta^{(2)}f(j/k)=f((j+2)/k)-2f((j+1)/k)+f(j/k), \quad j=0,1,\ldots,k-2. $$

Assume that $f\in C^2[0,1]$, the class of twice continuously differentiable functions on the unit interval $[0,1]$. Then, the second derivative

$$ B^{(2)}_k(f,t)=k(k-1)\sum_{j=0}^{k-2}\Delta^{(2)}f(j/k) {k-2 \choose j}t^j (1-t)^{k-2-j}, \quad t \in [0,1] $$

converges uniformly to $f^{(2)}$, the second derivative of $f$, as $k \to +\infty$. My question is the following: assume now that $f \in C^1[0,1]$ and that $f^{(1)}$, the first derivative of $f$, is differentiable on $(0,1)$ with derivative $f^{(2)}$ possibly satisfying $\lim_{t\to 0}f^{(2)}(t)=\lim_{t\to 1}f^{(2)}(t)=+\infty$. Then, for every $\epsilon>0$ is it possible to find some $\delta \in (0,1)$ and some $k^*$ such that $\sup_{\delta \leq t \leq 1-\delta}|f^{(2)}(t)-B^{(2)}_k(f,t)|\leq \epsilon$ for every $k \geq k^*$? If so, I guess that such a pair $(\delta, k^*)$ can not be in the form $(1/k^*, k^{*})$, correct?


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