# Supremum of holomorphic function on the unit disk

Let $f\colon B \to \mathbb{C}$ be holomorphic where $B$ is the unit disk in the complex plane centered at the origin. For $0<r<1$, let $M_r=\sup\{|f(z)|:|z|=r\}$.

Show that $|f(0)|≤M_r$ $\forall r \in (0,1)$.

Now suppose further that $|f(0)|≥|f(z)| \forall z\in B$. Show that $|f(z)|$ is constant on $B$,

For the first bit I use Cauchy's integral formula and got: $$\left| \frac{1}{2\pi i} \int_{|z|=r}^{} \frac{f(z)}{z} \right| = \left|f(0)\right|\leq \frac{1}{2\pi} \int_{|z|=r}^{} \left|\frac{f(z)}{z} \right|$$ Now I'm not sure how to implement the $M_r$

Need help with the second part...

• The following is proper typesetting: $M_r=\sup\{|f(z)|:|z|=r\}$. I edited accordingly. ${}\qquad{}$ Commented Dec 4, 2015 at 1:31
• Thanks is it ok to use | | or is $\left | \right |$ better? Commented Dec 4, 2015 at 1:33
• For something like $|1+f(z)|$ I just write |1+f(z)|. For something like $\displaystyle \left| 1 + \frac{f(z)} z\right|$ I use \left 1 + \frac{f(z)} z\right| because it you omit \left and \right then what you see is $\displaystyle | 1 + \frac{f(z)} z |$. ${}\qquad{}$ Commented Dec 4, 2015 at 2:00

For the first part, you can conclude using the "ML-inequality" (you may know it under a different name):

$$|f(0)| = \Big| \frac{1}{2\pi i} \int_{|z|=r} \frac{f(z)}{z}\,dz \Big| \le \frac{1}{2\pi} \cdot 2\pi r \cdot \sup_{|z|=r} \Big|\frac{f(z)}{z} \Big| \le r \cdot \frac{M_r}{r} = M_r.$$

• Thanks I know it as the estimation lemma. For the second part I have $0≤|f(z)|≤M_r$. Not sure what to do now. Commented Dec 3, 2015 at 23:50
• I also need help on this, Could you give a hint for the second part Commented Dec 6, 2015 at 12:04

$$\underbrace{\,\,\left| \frac{1}{2\pi i} \int_{|z|=r} \frac{f(z)}{z} \, dz \right| = |f(0)|\leq \frac{1}{2\pi} \int_{|z|=r} \left|\frac{f(z)}{z} \, dz \right|\,\,}_{\huge \text{?}\vphantom{\displaystyle\int}}$$ Instead of the above, I'd write $$|f(0)| = \left| \frac{1}{2\pi i} \int_{|z|=r} \frac{f(z)}{z} \, dz \right| \leq \frac{1}{2\pi} \int_{|z|=r} \left|\frac{f(z)}{z} \, dz \right|.$$ The point is we know $\text{“}=\text{''}$ is true because of Cauchy's formula, and we know $\text{“}\le\text{''}$ is true for a different reason, and we initially put $\text{“}=\text{''}$ and $\text{“}\le\text{''}$ only where we already know they're true rather than where we're trying to prove they're true. Then we can say we've proved $$|f(0)| \leq \frac{1}{2\pi} \int_{|z|=r} \left|\frac{f(z)}{z} \, dz \right|.$$ We can go on to say $$\frac{1}{2\pi} \int_{|z|=r} \left|\frac{f(z)}{z} \, dz \right| = \frac 1 {2\pi r} \int_{|z|=r} |f(z)| \, |dz|.$$ What does it mean to write $|dz|$ instead of $dz$ (which you omitted)? It means we're integrating with respect to arc length.

That last integral is the average value of $|f(z)|$ on the circle of radius $r$ centered at $0$. So what we're trying to show is that the average value is less than or equal to the largest value. We can say $$\int_{|z|=r} |f(z)| \, |dz| \le \int_{|z|=r} M_r \, |dz| = 2\pi r M_r.$$

Notice that it would be grave error to write $\displaystyle\int_{|z|=r} M_r \,dz$ instead of $\displaystyle\int_{|z|=r} M_r \, |dz|$, since that integral (with $dz$ instead of $|dz|$) evaluates to $0$ rather than to $2\pi rM_r \vphantom{\displaystyle\int}$.

• Sorry I forgot the dz. What about the second part. I have that |f(z)| is bounded Commented Dec 4, 2015 at 1:55
• Do I need Loiuvilles theorem or something else? Commented Dec 4, 2015 at 2:15
• The value of a harmonic function, including any holomorphic function, at the center of a disk is the average of its values on the boundary of the disk. That will do it. But I'm not sure I want to write a proof of that from scratch right now. Commented Dec 4, 2015 at 2:39
• I also need help with the second part could you please expand on what you mean Commented Dec 6, 2015 at 16:04