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Let $f$ be an analytic function in the unit disk with $|f(z)|\leq 1$. $f(\alpha_1)=f(\alpha_2)=0$ with $|\alpha_j|<1$ and $\alpha_1\not=\alpha_2$. Prove: $$|f(0)|\leq |\alpha_1\alpha_2|$$

When I see this I remember the Schwarz–Pick Theorem which has these exact conditions on $f$. However, applying this to $f$ and the pairs $(z_1,z_2)=(0,\alpha_1)$ and $(z_1,z_2)=(0,\alpha_2)$ yields only that $|f(0)|\leq \sqrt{|\alpha_1 \alpha_2|}$. So I tried to adapt the proof of Schwarz-Pick's Theorem to this particular case. Consider the Mobius transformations for $k=1,2$:

$$M_k(z)=\frac{z-\alpha_k}{z\overline{\alpha_k}-1}$$

$M_k$ maps the unit disk to the unit disk and so does it's inverse, so we may consider

$$f\circ M_k^{-1}: D \rightarrow D$$

Indeed, if $f$ attained the value $1$ in $D$, it would be constant by the Maximum Modulus Principle, as it attains the value $0$, $f$ would be identically equal to zero, which is absurd with it satisfying $|f(z)|=1$ for some point. Thus $f\circ M^{-1}_k:D\rightarrow D$ indeed.

We also know that $f \circ M_k^{-1} (0)=f(\alpha_k)=0$ and so the $f\circ M_k^{-1}(z)/z$ is analytic (as is easily seen by expanding the taylor series). Also one shows that $|f\circ M_k^{-1}(z)/z|\leq 1/r$ for $z\in B(0,r)\subset D$. Taking the limit as $r$ tends to one yields $f\circ M_k^{-1}(z)\leq |z|$. This yields taking $z$ to be $M_k(z)$:

$$|f(z)|\leq |M_k(z)|=\left|\frac{z-\alpha_k}{z\overline{\alpha_k}-1}\right|$$

Thus $|f(0)|\leq |\alpha_k|$ and again $|f(0)|\leq \sqrt{|\alpha_1\alpha_2|}$. Not that surprising that this is exactly the result by using Schwarz-Pick's Theorem directly. But this also means I do not know how to improve the bound. Indeed, $|\alpha_1\alpha_2|\leq \sqrt{|\alpha_1\alpha_2|}$ to our sadness so some improvement is needed.

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    $\begingroup$ fairly sure this has been asked before here - in a nutshell take the Blaschke product $B$ with zeroes at both $\alpha_{1,2}$ and note that $f/B$ is analytic in the unit disc and bounded by $1$ still since $|B|=1$ on the unit circle and maximum modulus applies; hence $|f(z)| \le |B(z)|, |z|<1$ and $B(0)=\alpha_1\alpha_2$ gives you the required result $\endgroup$
    – Conrad
    Apr 2 at 21:31
  • $\begingroup$ I will try to learn about this Blaschke product tomorrow (I have never heard of it). This must be in any standard complex analysis book right? $\endgroup$
    – Kadmos
    Apr 2 at 21:56
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    $\begingroup$ yes - it is just the product (pointwise so $M_1(z)M_2(z)$) of the two automorphisms with roots at $\alpha_1$ and $\alpha_2$ you wrote above); more generally a finite Blaschke product is a product of finitely many such automorphisms (some of which can be equal); these play for the unit disc the role that polynomials play for the plane since with them you can take out finitely many zeroes in the unit disc without changing the maximum value of the function since $|B|=1$ on the circle $\endgroup$
    – Conrad
    Apr 2 at 22:10
  • $\begingroup$ What a nice trick. Thank you so much. :). I would have never thought of dividing it by $B(z)$, but I guess it makes sense, because $f(z)$ has the exact regularity needed to "compensate" for the zeros of $B$. $\endgroup$
    – Kadmos
    Apr 2 at 22:29
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    $\begingroup$ No you do not need - maximum modulus applies for $f/B$ - in other words since $\limsup_{|z|\to 1}|f/B| \le 1$ (as otherwise you find $|f(z_n)/B(z_n)| \to c>1$ but for $z_n$ close enough to the circle $|B(z_n)|<(1+c)/2$ so $|f(z_n)|>1$ which is not possible) it follows that if $|f/B|$ has supremum larger than $1$ it must be attained inside a smaller disc so by continuity it is a maximum $f/B$ constant $\endgroup$
    – Conrad
    Apr 3 at 3:14

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Using Conrad's trick:

$$ B(z)=\prod_{k=1}^2\frac{\alpha_k-z}{1-\overline{\alpha_k}z}$$

With this, we have that $B(z)$ is analytic in a disk slightly bigger than the unit circle because $|1-\alpha_k z|\geq 1-|\alpha_k z|>0$ if $|z|<\frac{1}{|\alpha_k|}$. Also, $B(z)$ has only two zeros, namely at $\alpha_k$ for $k=1,2$. At these points $Q(z)=\frac{f(z)}{B(z)}$ is analytic because $f(z)$ has a zero of order greater or equal to one in $\alpha_k$ which makes up for these zeros of $B(z)$. It is clearly analytic elsewhere. For $\max\{|\alpha_1|,|\alpha_2|\}<r<1$ we have that $Q(z)$ is analytic and continuos in $B(0,r)$:

$$\sup_{z\in B(0,r)}\frac{|f(z)|}{|B(z)|}\leq \sup_{ \partial B(0,r)}\frac{|f(z)|}{\left|\prod_{k=1}^2\frac{\alpha_k-z}{1-\overline{\alpha_k}z}\right|}$$

Notice $\max\{|\alpha_1|,|\alpha_2|\}<r$, then we are in every write to consider:

$$\sup_{z\in B(0,r)}\frac{|f(z)|}{|B(z)|}\leq \sup_{ \partial B(0,r)}\frac{1}{\left|\prod_{k=1}^2\frac{\alpha_k-z}{1-\overline{\alpha_k}z}\right|}=\prod_{k=1}^2\left|\frac{\alpha_k-re^{i\theta_r}}{1-r\overline{\alpha_k}e^{i\theta_r}}\right|=\prod_{k=1}^2\left|\frac{\alpha_k-re^{i\theta_r}}{e^{-i\theta_r}-r\overline{\alpha_k}}\right|$$

Hence, because $\max{|\alpha_1|,|\alpha_k|}<r<1$ is arbitrary we may take the limit as both sides tend to one and we have that for any $z$ with $|z|<1$, $|f(z)|\leq|B(z)|$. In particular:

$$|f(0)|\leq\left|\prod_{k=1}^2\frac{\alpha_k-0}{1-\overline{\alpha_k}0}\right|= |\alpha_1\alpha_2|$$

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