Does $C^{k,s-k}$ function with lipschitz lower order derivatives give a certain bound on the Taylor remainder?

Question

Let $\Omega \subseteq \mathbb{R}$ be open (not necessarily an interval). Let $ s > 0$ and $k \in \mathbb{N}_0$ be such that $s \in (k, k+1]$. Suppose that $f \colon \Omega \to \mathbb{R}$ is an element of $C^{k, s-k}(\overline{\Omega})$ such that $$ \frac{d^j f}{dx^j} \in C^{0,1}(\overline{\Omega}) $$ for $j < k$. (So, also, $f \in C^{0,1}(\overline{\Omega})$).

Does it follow that there exists $M > 0$ such that for all $x, y \in \Omega$ we have $$ f(y) = \sum_{ j = 0 }^{ k } \frac{d^j f}{dx^j}(x) \cdot (y-x)^j + R(x,y), $$ with $R(x,y)$ satisfying $$ | R(x,y) | \leq M |x-y|^{s} ?$$

I know that such a result would not be possible without lipschitz bounds on lower order derivatives since we could take, for example, $\Omega = (- \infty, 0) \cup (0,\infty)$ and $f = \chi_{ (0, \infty) }$.

Answer 1

Your desired conclusion is the hypothesis of (a version of) the Whitney extension theorem: it implies that there exists $F \in C^{k, s-k}(\mathbb R)$ such that $F^{(j)}|_{\bar{\Omega}} = f^{(j)}$ for all $j$. It doesn't follow merely from the fact that $f$ and its derivatives are Lipschitz on $\bar \Omega$. For example, let $s=2$, and let $$\Omega = \bigcup\limits_{n=1}^\infty(n-1, n-a_n),$$ $$f(x) = c_n, f'(x) = 0 \text{ for } x \in [n-1, n - a_n], $$ where $a_n = 2^{-n}$, $c_1 = 1$, and $c_n = c_{n-1} + 2^{-n}$ for $n > 1$. Then $\text{Lip}(f) = 1$, $\text{Lip}(f') = 0$, but $$\frac{f(n) - f(n - a_n) - a_n f'(n - a_n)}{a_n^2} = 2^n.$$