Let $f(x)$ be an analytic function on $\mathbb{R}$. Suppose we have two points $z, z' \in \mathbb{R}$ and let $P_z^k(x)$ be the Taylor polynomial of degree $(k-1)$ around $z$. That is, $$ P_z^k(x) = \sum_{j=0}^{k-1} \frac{f^{(j)}(z)}{j!}(x-z)^j, $$ where $f^{(j)}$ is the $j$-th order derivative of $f$.
Suppose $\sup_{x} |f^{(j)}(x)| \le M$ for all $j=0,\cdots, k$. I want to show that $$ |D^{\alpha}(P_z^k - P_{z'}^k)(z)| \le M|z-z'|^{k-\alpha}, \quad \forall \alpha = 0,\cdots, k-1. $$ If $z=z'$, the statement is obvious. Thus we assume $z\ne z'$.
Here is my attempt. When $\alpha = k-1$, since $P_z^k$ is a polynomial of degree $k-1$, we have $$ D^{k-1}P_z^k(x) = f^{(k-1)}(z). $$ Therefore, $$ |D^{k-1}(P_z^k - P_{z'}^k)(z)| = |f^{(k-1)}(z) - f^{(k-1)}(z')| \le M|z-z'|, $$ where the last inequality holds since $f^{(k-1)}$ is Lipschitz $M$: ($\sup_{x} |f^{(k)}(x)| \le M$).
If $\alpha < k-1$, we have $$ |D^{\alpha}(P_z^k - P_{z'}^k)(z)| = \left|f^{(\alpha)}(z) - f^{(\alpha)}(z') - \sum_{j=1}^{k-1-\alpha}\frac{f^{(\alpha+j)}(z')}{j!}(z-z')^{j} \right|. $$ Since $|f^{(\alpha)}(z) - f^{(\alpha)}(z')| \le M|z-z'|$ and $\frac{|f^{(\alpha+j)}(z')|}{j!} \le \frac{M}{j!}$, we have $$ |D^{\alpha}(P_z^k - P_{z'}^k)(z)| \le M\left[|z-z'| + \sum_{j=1}^{k-1-\alpha} \frac{1}{j!}|z-z'|^{j}\right]. $$ However, I am not sure how to obtain $M|z-z'|^{k-\alpha}$ from there.
Any answers/comments/suggestions would be very appreciated.
I also would like to mention that this is a simpler version of the theorem (Theorem A) I am trying to understand.
I figured this out. Note that the Taylor expansion of $f^{(\alpha)}(x)$ at $z'$ is $$ f^{(\alpha)}(x) = f^{(\alpha)}(z') + \sum_{j=1}^{k-1-\alpha}\frac{f^{(\alpha+j)}(z')}{j!}(x-z')^j + \frac{f^{k-\alpha}(\bar{z})}{(k-\alpha)!}(x-z')^{k-\alpha}. $$ Thus, $$ |D^{\alpha}(P_z^k - P_{z'}^k)(z)| = \left|f^{(\alpha)}(z) - f^{(\alpha)}(z') - \sum_{j=1}^{k-1-\alpha}\frac{f^{(\alpha+j)}(z')}{j!}(z-z')^{j} \right| \le \frac{M}{(k-\alpha)!}|z-z'|^{k-\alpha}, $$ which completes the statement.