1
$\begingroup$

Suppose $f$ is a bounded holomorphic function on the open unit disk $D$. Show that $$\left(1-\left|z\right|\right)\left|f'\left(z\right)\right|\leq\sup_{w\in D}\left|f\left(w\right)\right|$$ for all $ z\in D$.

The first thing that occurs to me is the maximal principle. But it didn't work. I don't know how to find connection between right side and left side of the inequality.

$\endgroup$
2
$\begingroup$

The simplest proof of the inequality you ask about is as in @Kelener's answer; apply the Cauchy Integral Formula on the circle $|w-z|=r$ for $0<r<1-|z|$. (Or one could apply the inequalities often known as "Cauchy's Estimates", which is really the same thing, since those inequalities follow from CIF.)

Let's say $M=\sup_{|w|<1}|f(w)|$.

In fact a stronger inequality is true: $$(1-|z|^2)|f'(z)|\le M.$$(This is stronger since $1-|z|^2=(1+|z|)(1-|z|)\ge1-|z|$.) This follows from CIF on the circle $|w|=r$, $|z|<r<1$. We have $$f'(z)=\frac{1}{2\pi i}\int_{|w|=r}\frac{f(w)}{(w-z)^2}\,dw.$$Parametrizing that circle as $w=re^{it}$ shows that $$|f'(z)|\le \frac{Mr}{2\pi}\int_0^{2\pi}\frac{1}{|re^{it}-z|^2}\,dt\le \frac{M}{2\pi}\int_0^{2\pi}\frac{1}{|re^{it}-z|^2}\,dt,$$and now we need to calculate that integral. A fun way to do that is using Fourier series.

Note first that $|re^{it}-z|=|r-e^{-it}z|$. Since $|e^{-it}z|<r$ we can apply the formula for the sum of a geometric series: $$\frac{1}{r-e^{-it}z}=\frac1r\sum_{n=0}^\infty\left(\frac{e^{-it}z}{r}\right)^n=\frac1r\sum_{n=0}^\infty\left(\frac{z}{r}\right)^ne^{-int}.$$So Parseval shows that $$\frac1{2\pi}\int_0^{2\pi}\frac1{|r-e^{-it}z|^2}=\frac1{r^2}\sum_{n=0}^\infty\frac{|z|^{2n}}{r^{2n}}= \frac{1}{r^2}\frac1{1-\left(\frac{|z|}{r}\right)^2}=\frac{1}{r^2-|z|^2}.$$ Now let $r\to1$ and insert the result above: $$|f'(z)|\le\frac{M}{1-|z|^2}.$$


The Last Word: In fact the best possible bound on $|f'(z)|$ in terms of $M$ and $|f(z)|$ is $$|f'(z)|\le\frac1M\frac{M^2-|f(z)|^2}{1-|z|^2}.$$Hint: First show wlog $M=1$. Now compose $f$ on the left and on the right by an appropriate linear-fractional transformation and apply the Schwarz Lemma. (Martin R is right when he says I should mention that this is the Schwarz-Pick theorem.)

$\endgroup$
2
$\begingroup$

I tkink that your $f^{\prime}(t)$ is $f^{\prime}(z)$. If true:

Let $z\in D$, and $0<r<1-|z|$. Let $\gamma$ the circle centered at $z$, with radius $r$. Then $\gamma$ is included in $D$. We have $$f^{\prime}(z)=\frac{1}{2i\pi}\int_{\gamma}\frac{f(t)}{(t-z)^2}dt$$ Hence with $M=\sup_{|z|< 1}|f(z)|$, we have $$ |f^{\prime}(z)|\leq \frac{1}{2\pi}\frac{M2\pi r}{r^2}=\frac{M}{r}$$ Now let $r\to 1-|z|$.

$\endgroup$
2
$\begingroup$

Let's use a consequence of Cauchy's integral formula. The following holds.

$U \subset \mathbb{C}$ open$, f: U \to \mathbb{C}$ holomorphic, then $f$ can be differentiated any number of times. Furthermore, we have:
$$f^{(n)}(z_0) = \frac{n!}{2\pi i} \int\limits_{\partial B} \frac{f(\xi)}{(\xi - z_0)^{n + 1}} d\xi$$ where $B \subset U$ is a disk and $\partial B$ is the boundary.

Okay. You are interested in $|f'(z)|$, so let's use the formula:
$$f'(z) = \frac{1!}{2\pi i} \int\limits_{\partial B} \frac{f(\xi)}{(\xi - z)^2} d\xi$$ We must now guess a good disk so that we obtain what is needed. Let's use $D_r(z)$ which is a disk centered at $z$ with radius $r = 1 - |z|$. For a given $z$ the radius is the distance of $z$ to $1$. You see that our disk is a subset of the unit disk $D$. You also already see, at the boundary of this disk $r = 1 - |z|$ is pretty close to the denominator $\xi - z$ and matches the leftmost factor of your result $1 - |z|$.

We now estimate the size of the integral by the length multiplied with the biggest element. The length is $2\pi (1 - |z|)$. $$f'(z) \le \frac{1!}{2\pi i} (2\pi \cdot (1 - |z|)) \sup_{\xi \in D_r(z)} \frac{f(\xi)}{(\xi - z)^2} = \frac{1}{i} \cdot (1 - |z|) \sup_{\xi \in D_r(z)} \frac{f(\xi)}{(\xi - z)^2}$$ Next we use $|\cdot|$ as you're interested in $|f'(z)|$ and by that get rid of the $i$. $$|f'(z)| \le \left|\frac{1}{i}\right| \cdot \left|1 - |z|\right| \sup_{\xi \in D_r(z)} \left|\frac{f(\xi)}{(\xi - z)^2}\right| = \left|1 - |z|\right| \sup_{\xi \in D_r(z)} \left|\frac{f(\xi)}{(\xi - z)^2}\right|$$

We know that $\frac{1}{|\xi - z|} \le \frac{1}{1 - |z|}$ thus we simplify. $$|f'(z)| \le \left|1 - |z|\right| \sup_{\xi \in D_r(z)} \left|\frac{f(\xi)}{(\xi - z)^2}\right| \le \left|1 - |z|\right| \sup_{\xi \in D_r(z)} \left|\frac{f(\xi)}{(1 - |z|)^2}\right|$$

Next, we see that the denominator now is independent of $\xi$, we pull it out. We also know that $|1 - |z|| = 1 - |z|$ because of $1 - |z| \ge 0$ since |z| is bounded by $1$. $$|f'(z)| \le \left|1 - |z|\right| \sup_{\xi \in D_r(z)} \left|\frac{f(\xi)}{(1 - |z|)^2}\right| = \left|1 - |z|\right| \cdot \frac{1}{(1 - |z|)^2}\sup_{\xi \in D_r(z)} \left|f(\xi)\right| = \frac{1}{1 - |z|} \sup_{\xi \in D_r(z)} \left|f(\xi)\right|$$

Right factor to the left and we receive: $$(1 - |z|) |f'(z)| \le \sup_{\xi \in D_r(z)} \left|f(\xi)\right|$$

Almost finished, we need $D$ and not $D_r(z)$. We already noticed $D_r(z) \subseteq D$. Thus, we simply can expand the expression to $D$ and would only receive a bigger value.

$$(1 - |z|) |f'(z)| \le \sup_{\xi \in D_r(z)} \left|f(\xi)\right| \le \sup_{\xi \in D} \left|f(\xi)\right|$$

And we are finished, hope you understood everything :)

$\endgroup$
  • $\begingroup$ The fact that you use Cauchy's integral formula doesn't mean that this should also be how you tag the question. I'd rather say that the question should be given the extra tag "holomorphic-functions" - but clearly not the one that you have suggested. $\endgroup$ – Alex M. Jul 10 '16 at 12:34
  • $\begingroup$ As it can be solved using this formula I thought it would be a good idea for other users that search for problems related to the formula. Then they will also see this problem which may help since it is related. Your suggested tag also sounds good. $\endgroup$ – Zabuza Jul 10 '16 at 13:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.