I was reading a paper and the autor uses the following lemma of complex analysis

Let $R_{1},R_{2},\rho \in \mathbb{R}$ with $R_{2}>R_{1}>0$ and $R_{2}\geq \rho >0$. Let $f$ be a holomorphic function in a open set that contains the closed disc $\{|z|\leq R_{2}\}$. If $f$ is not identically zero in $\{|z|\leq R_{2}\}$, then the number of zeros of $f$ in $\{|z|\leq \rho\}$, $n(f,\rho)$ satisfies $$n(f,\rho) \log \left( \frac{R_{2}^{2}-R_{1}\rho}{R_{2}(R_{1}+\rho)} \right)\leq \log \left( \frac{||f||_{R_{2}}}{||f||_{R_{1}}} \right).$$

The proof is the following

First of all, notice that the set of zeros in $\{|z|\leq \rho\}$ can not be infinite, because if this happen, then there will be a accumulation point and then, as $f$ is holomorphic, $f \equiv 0$. So, we have finite zeros in $\{|z|\leq \rho\}$. Let us call them $z_{1},\ldots,z_{t}$. The function $$g(z):=f(z) \prod_{j=1}^{t} \frac{R_{2}^{2}-z \overline{z}_{j}}{R_{2}(z-z_{j})}$$ is holomorphic in an open set that contains $\{|z| \leq R_{2}\}$. By the maximum modulus principle, $||g||_{R_{1}}\leq ||g||_{R_{2}}$. Also, notice that if $|z|=R_{2}$ then $$\frac{|R_{2}^{2}-z\overline{z}_{j}|}{|R_{2}(z-z_{j})}| = \frac{|z\overline{z}-z\overline{z}_{j}|}{|R_{2}(z-z_{j})|} =1,$$ so, $||g||_{R_{2}}=||f||_{R_{2}}$. Also, using the triangle inequality we obtain that if $|z_{j}| \leq \rho$ and $|z|=R_{1}$ $$\frac{|R_{2}^{2}-z\overline{z}_{j}|}{|R_{2}(z-z_{j})|} \geq \frac{|R_{2}^{2}-R_{1}\rho|}{|R_{2}(z-z_{j})|} \geq \frac{|R_{2}^{2}-R_{1}\rho|}{|R_{2}(R_{1}+\rho)|}=\frac{R_{2}^{2}-R_{1}\rho}{R_{2}(R_{1}+\rho)},$$ and then, $$||g||_{R_{1}}\geq ||f||_{R_{1}} \left( \frac{R_{2}^{2}-R_{1}\rho}{R_{2}(R_{1}+\rho)} \right)^{t}.$$ Finally, $$||f||_{R_{2}}=||g||_{R_{2}}\geq ||g||_{R_{1}}\geq ||f||_{R_{1}} \left( \frac{R_{2}^{2}-R_{1}\rho}{R_{2}(R_{1}+\rho)} \right)^{t}.$$

Then the author says (without proof) that one can improve the inequality

$$\frac{|R_{2}^{2}-z\overline{z}_{j}|}{|R_{2}(z-z_{j})|} \geq \frac{R_{2}^{2}-R_{1}\rho}{R_{2}(R_{1}+\rho)}$$

to the following:

$$\frac{|R_{2}^{2}-z\overline{z}_{j}|}{|R_{2}(z-z_{j})|} \geq \frac{R_{2}^{2}+R_{1}\rho}{R_{2}(R_{1}+\rho)}.$$

I know this seems trivial, but I can't prove it. I tried the usual tricks of triangle inequality but I only obtained bad results.


1 Answer 1


By scaling (replacing $z,z_j$ by $\frac{z}{R_2}, \frac{z_j}{R_2}$) it seems that you want to prove the inequality:

$z\ne w, |z|=r, |w|=\rho <1$ implies $\frac{|1-z\bar w|}{|z-w|}\ge \frac{1+r\rho}{r+\rho}$.

If we let $z=re^{i\alpha}, w=\rho e^{i\beta}$ and we square, this translates to:

$\frac{1+(r\rho)^2-2r \rho cos(\alpha-\beta)}{r^2+\rho^2-2r \rho cos(\alpha-\beta)}\ge (\frac{1+r\rho}{r+\rho})^2$

But now we notice the standard inequality $1+(r\rho)^2 >r^2+\rho^2$ so renaming the terms by $b=1+(r\rho)^2, c=r^2+\rho^2,a= 2r \rho>0, b>c, b,c \ge a, \theta =\alpha-\beta$ we need to find the minimum on $\theta$ of :

$\frac{b-acos(\theta)}{c-a cos(\theta)}=1+\frac{b-c}{c-a cos(\theta)}$.

But $c-a cos(\theta) \le c+a$, so

$\frac{1+(r\rho)^2-2r \rho cos(\alpha-\beta)}{r^2+\rho^2-2r \rho cos(\alpha-\beta)}=1+\frac{b-c}{c-a cos(\theta)} \ge 1 + \frac{b-c}{c+a}=\frac{b+a}{c+a}=(\frac{1+r\rho}{r+\rho})^2$

and we are done!


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .