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Let $f:[0,1]\to [0,1]$ be continuous. Let—

$$B_n(f)(x)=\sum_{k=0}^n f(k/n) {n \choose k} x^k (1-x)^{n-k},$$

be the Bernstein polynomial of $f$ of degree $n$.

This question relates to the difference between two Bernstein polynomials of the same function $f(x)$, namely the question of finding a simple and explicit function $\phi(f, n)$ such that— $$|B_{2n}(f)(x) - B_{n}(f)(x)| \le \phi(f, n) \text{ whenever } 0 \le x \le 1. \tag{1}$$

I have not been able to find results answering this question. However:

  • Butzer (1953) showed a very similar result on a family of operators, one of which is $2 B_{2n}(f) - B_n(f)$, but without explicit error bounds.
  • For the polynomials $2x(1-x)$ and $2x^2(1-x)$, the left-hand side of $(1)$ appears to converge to 0 at the rate $O(1/n^2)$. I suspect this is the case whenever $f(x)$ has a continuous second derivative.

Motivation

Answers to this question may help solve another question of mine, on finding bounds on the expected value of a hypergeometric random variable.

Questions

Let $M(f, r)$ equal the maximum absolute value of $f$ and its derivatives up to the $r$-th derivative.

  1. Given that $f$ has a continuous second derivative, can $\phi(f, n)$ equal $CM(f, 2)/n^2$ and/or $C/n^2$? In either case, find an upper bound for $C$.

  2. Given that $f$ is continuous, what is an explicit upper bound for $\phi(f, n)$, perhaps given the modulus of continuity of $f$?

  3. Does the same bound for $C$ as in question 1 hold if, instead—

    • $f$ has a Lipschitz-continuous derivative, and
    • $M(f, r)$ equals the maximum absolute value of $f$ and the Lipschitz constants of $f$ and its derivative?
  4. Does the same bound for $C$ as in question 1 hold if, instead—

    • $f'$ is in the Zygmund class, that is, for some constant $D\ge 0$ it has the property $|f'(x) + f'(y) - 2f'((x+y)/2)| \le D\epsilon$ for every $\epsilon>0$, whenever $x$ and $y$ are in $[0, 1]$ and $|x-y|\le\epsilon$, and
    • $M(f, r) = \max(D, \max |f|, \max |f'|)$?

It would be nice if the answers to questions 1 and 4 can easily carry over to the problem of finding an upper bound for $|B_{n+1}(f)(x) - B_{n}(f)(x)|$.

References

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1 Answer 1

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Some comments. The estimate $\|B_{2n} f-B_n f\|_\infty \leq c \omega (f, n^{-1/2})$, ($\omega$ is the modulus of continuity of $f$) follows from $\|f-B_n f\|_\infty \leq c \omega (f, n^{-1/2})$. Both estimates can be improved to $\frac C n$ if $f \in C^1([0,1]$. However, since $$ n(f-B_nf) \to -\frac{x(1-x)}{2}f''$$ an estimate $\|f-B_n f\|_\infty =o(1/n)$ gives $f(x)=ax+b$ (see Chapter 1 of the book of Lorentz, Bernstein polynomials).

If $\|B_{2n} f-B_n f\|_\infty \leq Cn^{-\alpha}$, then $\|B_{2^{k+1}n} f-B_{2^k n} f\|_\infty \leq C2^{-\alpha k}n^{-\alpha}$ and summinng over $k$, $\|f-B_nf \|_\infty \leq C n^{-\alpha}$ and $\alpha \leq 1$ unless $f$ is linear.

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    $\begingroup$ Thus, the answers to 1, 3, and 4 are "no". I thought I found a rate of $O(1/n^2)$, but I had written the numerical code for it wrong. $\endgroup$
    – Peter O.
    Commented Aug 29, 2022 at 14:08
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    $\begingroup$ Indeed, $|B_{2n}(f) - B_n(f)| \le 2M/n$ whenever $|B_n(f) - f| \le M/n$ (Ditzian and May, A Saturation Result for Combinations of Bernstein Polynomials, Tôhoku Math. J. 28 [1976]), answering question 2. $\endgroup$
    – Peter O.
    Commented Aug 29, 2022 at 14:22

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