schwarz lemma exercies

mokong · New Member Offline Joined: 23 Mar 2004 Posts: 4

f(z) is analytic on and within the unit circle. Also, |f(z)| < 1 for |z| <= 1 and f(0) = 0. Show that f(z) < |z| for |z|<= 1.

I know this can be done by the max mod principle, but I'm following a way which doesn't seem to use it.

After showing f(z)/z is analytic, I am to express [f(z_0)/z_0]^n by the Cauchy integral formula, so that maybe later I would come with |f(z_0)/z_0|^n <= 1. I still can't get any expression via the Cauchy integral formula that does this. Maybe someone has an idea to do this?

Any help would be greatly appreciated! Thanks..

**Posted:** Sun Sep 12, 2004 4:56 pm

Kent Merryfield

I'm not quite sure why you're trying to avoid the maximum modulus principle, since that principle is so naturally suited to the proof of this theorem. I need to guess what you mean by "[f(z_0)/z_0]^n".

Maybe this is what you're looking for: Suppose that $|h(\zeta)|<1$ for $|\zeta|= 1$ and note that for $\zeta$ on the boundary circle, $\frac1{|\zeta-z|}\le\frac1{1-|z|}$ . Put this into the Cauchy Integral Formula for $h$ to get that $|h(z)|<\frac1{|2\pi i|}\int_{|\zeta|=1}\frac{|h(\zeta)||d\zeta|}{|\zeta-z|} < \frac1{1-|z|}$ . Now let $h(z) = (g(z))^n$ where $g(z)=\frac{f(z)}z$ . Since this satisfies the boundary estimate no matter what $n$ is, we get that $|g(z)|^n<\frac1{1-|z|}$ $\forall n$ . This estimate can only be true independent of $n$ if $|g(z)|\le 1$ . That's our proof. It also seems to be an extra-hard way to prove the maximum modulus principle in the first place.

mokong · New Member Offline Joined: 23 Mar 2004 Posts: 4

wow! that was great kent. thanks a lot! Very Happy

just some minor questions..

how did you get the inequality $|{z_1 - z_2}|\ge\,||z_1| - z_2|$ , and did you mean $|h(z)| \le \frac{1}{|2 \pi i|} \displaystyle\int_{|\zeta|=1}\frac{|h(\zeta)||d\zeta|}{|\zeta - z|}$

i cant use the max mod becaue it wasnt proved in the book.. its a math physics text, unfortunately. Sad

again, thanks a lot!

**Posted:** Sat Sep 18, 2004 3:35 am

Kent Merryfield

I've just realized a mistake in what I wrote: that shouldn't be $\frac1{|\zeta-z|}\le\frac1{|1-z|}$ ; instead it should be $\frac1{|\zeta-z|}\le\frac1{1-|z|}$ . I'll go back and edit this to fix it.

As to where that inequality involving $|h(z)|$ and the integral: that simply is the Cauchy Integral Formula, followed by applying the theorem (think of it as an analogue of the triangle inequality) that the absolute value of an integral is less than or equal to the integral of the absolute value. (Have you seen the proof of that for complex valued functions?)

I still don't particularly like this proof, becuase I would prefer to first prove, then use, the Maximum Modulus Principle. I see Max Mod as a more elementary and more accessible theorem than Schwarz, so why not prove it first?

**Posted:** Sat Sep 18, 2004 4:09 pm

mokong · New Member Offline Joined: 23 Mar 2004 Posts: 4

thanks for clearing that up. Smile

we just did it via max mod in class this morning, and yup, it was more straightforward and much easier.

thanks for helping out! it was nice knowing two proofs of the same theorem.

**Posted:** Sat Sep 18, 2004 9:08 pm