A follow-up question concerning giving a estimation of a derivative This is a followed-up question of Proving the following integral formula. I highly think the conclusion will help us in proving this estimate.
Consider an analytic $f$ on $D(0;R)$ where $r<R$, we try to prove
$$|f^{(n)}(w)| \le \frac{2n!}{(R-r)^n} (A(R,f)-\Re{f(w)}),$$
where $A(R,f) $ stands for $\max_{|z| \le R} \Re{f(z)}$
My initial idea is, of course,  related to Cauchy Estimate, where we have $|f^{(n)}(w)| \le \frac{n!}{r^n} sup|f(z)|$ for $z$ on the boundary of $D(w,\rho)$, and I can proceed no further from here. A real problem for me is we cannot interchange $\max_{|z| \le R} \Re{f(z)}$, and I don't know which property to use to deal with the annoying maximum. (conjugate, Maximum Principle,....)
 A: We can start with the integral formula from Proving the following integral formula  on the circle $|z-w| = \rho$ with $0 < \rho < R-r$:
$$
f^{(n)}(w) = \frac{n!}{ \pi \rho^n} \int_0^{2 \pi} \operatorname{Re}\left(f( w+ \rho e^{i\theta}) \right)  e^{-in\theta} \, d\theta \\
= \frac{n!}{ \pi \rho^n} \int_0^{2 \pi} \left(\operatorname{Re}\left(f( w+ \rho e^{i\theta}) \right) -A(R, f)\right) e^{-in\theta} \, d\theta + 
A(R, f)\frac{n!}{ \pi \rho^n} \int_0^{2 \pi}  e^{-in\theta} \, d\theta \, .
$$
The second integral is zero, and
$$
\operatorname{Re}\left(f( w+ \rho e^{i\theta}) \right) -A(R, f) \le 0
$$
because of the maximum principle for the harmonic function $\operatorname{Re}f$. If follows that
$$
|f^{(n)}(w)| \le \frac{n!}{ \pi \rho^n} \int_0^{2 \pi} \left(A(R, f) - \operatorname{Re}\left(f( w+ \rho e^{i\theta}) \right) \right) \, d\theta \\
= \frac{n!}{ \pi \rho^n} \bigl( 2 \pi A(R, f) - 2 \pi \operatorname{Re}f(w)\bigr) \\
=\frac{2n!}{ \pi \rho^n} \bigl( A(R, f) -  \operatorname{Re}f(w)\bigr)
$$
using the mean-value formula for  the harmonic function $\operatorname{Re}f$. The desired inequality follows now by taking the limit $\rho \to R-r$.
Remark: It is assumed that $n \ge 1$.
