Ex. 1.3.10 in Tao's Additive Combinatorics

Question

For $k \geq 2$ and $B \subset \mathbb N$ set $r_{k,B}(n) = |\{ (x_1,\dots,x_k) \in B^k: \sum x_i = n\}|$. The problem at hand is to show that it can not happen that for some $m \geq 1$ that $r_{k,B}(n) = m$ for all $n >> 0$. There is also a hint in Tao's book: Consider the function $f(z) = \sum_{n \in B} z^n$, holomorphic for $|z| <1$. Now assuming that $r_{k,B}(n) = m$ for all $n >> 0$ we get that $p(z) = f^k(z)(z-1)$ is a polynomial. How does this give a contradiction? If we knew that $f$ could be extended past the unit circle around $1$ we could consider $1$ to be a pole and then get a contradiction as $p$ has a simple pole at $1$. But I do not see why this has to be the case. Can someone enligthen me?

Answer 1

So we have $f^k(z) = p(z)/ (z-1)$. Hence $f^k$ can be extended as a holomorphic function in the deleted neighborhood of $1$. Now $p(1) \ne 0$, as that would have made $f(z)$ bounded near $1$. So, in particular there is a punctured disk of certain radius around $1$ such that the extended $ f^k(z)$ doesn't vanish on that disk. So, we can define $\widetilde{f}$ on that disk which analyticaly extends $f$ in a deleted neighborhood of $1$. So there is a Laurent series of $\widetilde{f}$ around $1$, which has to be a pole. If that pole has order $m$, then the pole of $f^k(z)$ has order $km$, but $f^k(z)$ had a simple pole, so $km = 1$, contradiction for $k \ge 2$.

Answer 2

No, I do not think that this proof works. It is simply not clear that $\tilde{f}$ can be defined in a whole small disk near 1 (the puntured disk is not simply connected, so maybe you can not take the $k$th root. Look at a particular example: By taking the square root of the function $1/(1-z)$ defined in the unit disk (this can be done as the unit disk is simply connected and this function has no zeros there) we find a holomorphic function $g$ defined in the unit disk such that $g(z)^2 = 1/(1-z)$ and there is no contradiction.

The point is of course that the function $f$ given by the set $B$ has only coefficients 0 and 1 when expanded into a power series around 0, so the contradiction has to come from this direction.