If a function $f$ is analytic in a disk $\Delta(z_0,r)$ and if $m$ is a constant such that $|f(z)-f(z_0)|\leq m$ for all $z \in \Delta(z_0,r)$, then $|f'(z_0)|\leq m/r$ and $|f(z)-f(z_0)|\leq(m/r)|z-z_0|$ for all $z \in \Delta(z_0,r)$.

This is my revised attempt at a proof:

Let $g(z)=\frac{f(z)-f(z_0)}{z-z_0}$ for $z \in \Delta(z_0,r)\setminus\{z_0\}.$ Let $0< \rho < r$. Then we have the following:

$$\left| g(z) \right|=\max\limits_{|z|=\rho}\left|g(z)\right|=\max\limits_{|z|=\rho}\left|\frac{f(z)-f(z_0)}{z-z_0}\right|\leq \frac{m}{r}.$$

I believe the above more clearly establishes the inequality.

To establish the estimate since $\Delta(z_0,r)$ is a disk that means any two points $z_1, z_2$ have a line entirely in $\Delta(z_0,r).$ Let $\gamma(t)=z_1-t(z_2-z_1)$. Then $|f(z_2)-f(z_1)|=\left|\int_{0}^{1}f(\gamma(t))\gamma'(t)\,dt\right|\leq \int_{0}^{1}\left|f(\gamma(t)) \right|\left| \gamma'(t)\right|\,dt=\frac{m}{r}\int_{0}^{1}\left|f(\gamma(t))\right|\,dt$

What I was attempting to do was follow the reasoning from this question I had asked earlier. I am still unable to establish the inequality.

  1. You haven't defined $g$. I assume that $$g(\zeta) = \frac{f(z_0+\zeta)-f(z_0)}{\zeta}$$ extended by $g(0)=f'(z_0)$ to become a holomorphic function in $\Delta(0,r)$.
  2. I do not understand your manipulations with $g$, especially with the mysterious "$z\in \Delta$" at the end. What is $z$ here? What is $\Delta$? Neither is defined.
  3. If (as is likely) $\Delta(0,r)$ means an open disk, then you can't talk about the values of $g(\zeta)$ when $|\zeta|=r$. Instead, having fixed $\zeta$, consider some circle of radius $|\zeta|<\rho<r$, and apply the maximum principle there: $$ |g(\zeta)|\leq \max_{|\zeta|=\rho} |g(\zeta)| \le \frac{m}{\rho}$$ Then let $\rho\to r$ to obtain the desired inequality. With $\zeta=0$, it gives $|f'(z_0)|\le m/r$. For general $\zeta\in \Delta(0,r)$ it gives $|f(z)-f(z_0)|\le (m/r)|z-z_0|$
  4. I do not see the logic of "follows from the fact that the disk is a convex set." It's not like you can integrate the estimate for $f'(z_0)$ along a segment: the point $z_0$ is fixed all along, it's the center of the disk.
  • $\begingroup$ I cleaned up my question a lot and hopefully cleared up some stuff. I am still having problems establishing the estimate at the end of the problem. $\endgroup$ – emka Feb 17 '14 at 4:04
  • $\begingroup$ @Mark You should give up the idea of obtaining the estimate on $|f(z)-f(z_0)|$ by integrating the derivative. This is just not how Schwarz lemma works: you don't have a suitable derivative estimate to integrate. Instead use the maximum principle for $g$. the bound $|g|\le m/r$ is exactly the estimate on $|f(z)-f(z_0)|$ that you are trying to prove. Also, $|g(z_0)|\le m/r$ is the estimate on $f'(z_0)$ that you need. $\endgroup$ – user127096 Feb 17 '14 at 4:10
  • $\begingroup$ Are you saying I just get it by multiplying both sides by $|z-z_0|$. $\endgroup$ – emka Feb 17 '14 at 4:31
  • $\begingroup$ @Mark Yes, that's how you should get it. $\endgroup$ – user127096 Feb 17 '14 at 4:36

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