Let $f(\lambda)$ have a continuous $r$-th derivative on the closed unit interval.
A well-known result says that $f$ can be written as— $$f(\lambda) = R_{f,r}(\lambda, x_0) + f(x_0) + \sum_{i=1}^{r} (\lambda-x_0)^i f^{(i)}(x_0)/(i!).$$
Thus, $f$ equals the Taylor polynomial of degree $r$ at $x_0$ plus the Lagrange remainder, $R_{f,r}(\lambda, x_0)$.
A result found in Piţul, P. "Evaluation of the Approximation Order by Positive Linear Operators", 2007, bounds the remainder in terms of its modulus of continuity: $$|R_{f,r}(\lambda, x_0)| \le \frac{|\lambda-x_0|^r}{r!}\tilde\omega(f^{(r)}, |\lambda-x_0|/(r+1)),$$ where $\tilde\omega(f, h)$ is a concave modulus of continuity of $f$, that is, a continuous, concave, and nondecreasing function satisfying $|f(x)-f(y)|\le\tilde\omega(|x-y|)$ whenever $x$ and $y$ are in $f$'s domain.
Thus, for example, if $f$'s $r$-th derivative is $\alpha$-Hölder continuous with Hölder constant $L$ and $0\lt\alpha\le 1$ then— $$|R_{f,r}(\lambda, x_0)| \le \frac{|\lambda-x_0|^r}{r!}L \left(\frac{|\lambda-x_0|}{r+1}\right)^{\alpha}. \tag{1}$$
This question is about bounding the Lagrange remainder in terms of the Zygmund class. In this question, a continuous function $f$ is in the Zygmund class if there is a constant $D>0$ such that— $$|f(x) + f(y) - 2f((x+y)/2)| \le D\epsilon, $$ for every $\epsilon>0$, whenever $x$ and $y$ are points in $f$'s domain such that $|x-y|\le\epsilon$.
The Zygmund class cannot be described in terms of $\tilde\omega(f,h)$ described above. However, a function in the Zygmund class is Hölder continuous for $\alpha$ arbitrarily close to 1, so that if $f$'s $r$-th derivative is in the Zygmund class, perhaps $(1)$ holds for $\alpha=1$ and $L$ equal to $D$ or another constant, but I haven't found a proof of this.
Therefore:
- If $f$'s $r$-th derivative is in the Zygmund class, does $(1)$ hold true for $\alpha=1$ and $L$ equal to $D$ or another constant?
- A second-order modulus of continuity $\omega_2(f, h)$ is a function such that $|f(x) + f(y) - 2f((x+y)/2)| \le \omega_2(|x-y|/2)$ whenever $x$ and $y$ are in the domain of $f$. Are there results that bound the Lagrange remainder in terms of a concave second-order modulus of continuity?