3
$\begingroup$

One of the questions in my complex analysis book (Stein's text) is the following:

Prove that if $f$ is holomorphic in the unit disc, bounded, and not identically zero, and $z_1,z_2,\ldots,z_n,\ldots$ are its zeros, $(|z_k|<1)$, then $$ \sum_{n=1}^\infty(1-|z_n|)<\infty. $$

I proved this just fine using Jensen's formula, but I am still not able to think of an example for such a function. Obviously it will have to have infinitely many zeros, otherwise we're only adding finitely many terms and the problem becomes trivial. Since there are infinitely many, the limit point(s) has to be on the boundary (otherwise the function is identically zero). At one point, I think someone suggested a function like $\sin(\pi/z)$ but this isn't bounded (indeed, this has lots of problems around $0$).

Does anyone have an example of such a function?

$\endgroup$
6
$\begingroup$

The simplest examples are probably infinite Blaschke products:

$$f(z) = \prod_{n=1}^\infty \frac{z-z_n}{1-\bar z_n z}.$$

You can check that this product converges when your condition (which is often called the Blaschke condition) is satsified.

$\endgroup$
3
$\begingroup$

Here's an example which is somewhat close to the thoughts you had. $$f(z)=\exp\left(\frac{z-1}{z+1}\right)-e^{-1}$$ Here $\zeta=(z-1)/(z+1)$ maps the disk onto the left half-plane; the function $\zeta \mapsto e^\zeta-e^{-1 }$ vanishes at the points $-1+2\pi i n$ for $n\in\mathbb Z$, and satisfies $| e^\zeta -e^{-1}|\le 1+e^{-1}$.

$\endgroup$
  • $\begingroup$ Is $f$ bounded on the line segment $(-1,-1/2)$? $\endgroup$ – AD. May 3 '13 at 5:15
  • $\begingroup$ @AD. I think so. The fractional linear transformation takes values in the left halfplane, where the exponential function is bounded by $1$. $\endgroup$ – 75064 May 3 '13 at 5:16
  • $\begingroup$ Of course you are right. But the function needs to be fixed, since the image of -$1/2+2n\pi i$ under the map $\zeta\mapsto 1-2e^\zeta$ is $1-2e^{-1/2}\ne0$. Perhaps the easiest fix is to look at just $\exp(\zeta)-1$ at the points $2n\pi i$. $\endgroup$ – AD. May 4 '13 at 5:24
  • $\begingroup$ @AD. Thanks, I fixed the function. I could not use $2n\pi i$ because these points are on the boundary of the halfplane, and the question asks for zeros inside of the domain. $\endgroup$ – 75064 May 4 '13 at 5:31
  • $\begingroup$ Looks better +1 :) $\endgroup$ – AD. May 4 '13 at 14:34
2
$\begingroup$

Clayton, another way to think of it is to transfer this question to the upper half-plane and use a Weierstrass product. Try putting zeroes at $z_n=in^2$ and then you shouldn't need any correction factors to get uniform convergence on compacts.

$\endgroup$

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.