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The Weierstrass product has the form $W(z)=\prod E_{N(a)}(z/a)$, where the product is over the set of the desired zeros $a$, and the integers $N(a)$ can be chosen freely; they only need to be large en …

A more abstract argument is also possible: $f$ satisfies $p(z,f(z))=0$, and let's for convenience assume that $p$ is irreducible (but the argument works in general). We have two meromorphic maps on th …

This is false. We can take any function $g\not\equiv 0$ with $\lim_{z\to 0} g(z)=0$ such as $g(z)=z$ and then $f(z)=B(z)g(z)$, where $B$ is a Blaschke product with zeros $z_n$ (the linked page discuss …

Yes, this works, with the understanding that we may have to renormalize the products as $\prod (1-z/z_n)e^{z/z_n}$, as is usually done anyway. [Or you could only consider $\prod (1-z/z_n)$ with the un …

$S$ is the set of all entire zero-free functions $F$ with $F(0)=F'(0)=1$. To approximate such an $F$, just cut off its Taylor series $F(z)=1+z+\sum_{n\ge 2} a_n z^n$ at high enough degree $N_1=f(1)$. …

This is false. Consider a polynomial whose roots are uniformly distributed on the circle of radius $r<1$. If we now send $n\to\infty$, then we obtain the elliptic integral $$ \lim_{n\to\infty} \frac{1 …

We can assume that $w=0$ (replace $p$ by $q=p-w$). Then $p=c\prod (z-a_j)$ with $|a_j|\ge 1$. Since $$ \frac{p'}{p} =\sum_{j=1}^n \frac{1}{z-a_j} , $$ we want to show that $$ 1 - \frac{1-e^{i\psi}}{n} …

This will not work if $L^1$ refers to area measure: the function $f(w,z)=1/z$ on (let's say) $|w|,|z|<1$, $z\not= 0$ is a counterexample.

Yes, for Question 1. We can divide through by one of the functions on the left-hand side and also assume $F=G$, and then the question becomes: Can $1+|h(z)|^2$ be the square of the absolute value of …

This has an elementary proof, using the formula for the radius of convergence and Cauchy's estimates. It suffices to treat the case $R=1$. We then know that the coefficients satisfy $|a_n|\lesssim (1+ …

These zeros can be literally anywhere, by the result you quote that for any discrete set $\{z_n\}$ and any values $a_n$, there are entire functions with $f(z_n)=a_n$.

A function $f\ge 0$ on the circle is the absolute value $f=|F|$ of an $F\in H^p$, $F\not\equiv 0$, if and only if $f\in L^p$ and $\log f\in L^1$, and obviously you can make such functions as oscillato …

This is a typical walk-to-the-library problem. I used Boas, but probably other standard books would have worked too. Boas proves the following results: (1) if $f$ is of order $1$, then $\limsup m(r)M …

A concrete example is given by $f(z)=\cos{\sqrt{z}}$. Then $$ g_f(z) = \frac{|\sin\sqrt{z}|}{2|\sqrt{z}|(1+|\cos^2{\sqrt{z}}|)} . $$ Obviously, this is small for large $|z|$ if $\sin{\sqrt{z}}$ is not …

As I already hinted at in my comment, your formula is based on a miscalculation (the only alternative is that $f_1$ can actually be continued past $|z|=1$, which even without closer analysis looks hig …