The originally problem is as follows:
Give an example for two different probability generating functions that coincide at countably many points $x_i ∈ (0, 1)$, $i ∈ \mathbb N$.
In other words, find $\{a_n\}_{n \in \mathbb N},\{b_n\}_{n \in \mathbb N}$, two different sequences of non negative real numbers which sum to $1$, but such that: $$\sum_{n \in \mathbb N} a_nx^n = \sum_{n \in \mathbb N} b_nx^n $$ For countably infinite many points $x\in (0,1)$.
My thoughts are as follows, if we let the LHS be $f(x)$ and the RHS be $g(x)$ then $f,g$ are holomorphic with radius of convergence $1$, and hence so is $f-g$. If the zero set of $f-g$ (i.e. where $f=g$) has a limit point in the interior of the unit disc, then $f=g$ identically, so we must have that the zero set of $f-g$ (in the unit interval) is a sequence approaching $1$. Furthermore, let $t_n$ be the sequence of zeros of $f-g$, we may cite the Blaschke condition to rule out sequences such as the harmonic series (i.e. $t_n = 1-\frac 1n$ since we need the sum of $1-t_n$ to be finite).
Also, suppose $h(x): = \sum_{n \in \mathbb N} c_nx^n$ is analytic such that the sum of its coefficients is absolutely convergent, and such that $h(x_i) = 0$ for some sequence $x_i$ tending to 1, then we can normalize $c_n$ such that $\sum_{n \in \mathbb N}c_n^+ = \sum_{n \in \mathbb N}c_n^-=1$ (the superscript $+$ denotes $\min(\cdot,0)$, similarly for $-$), then letting $f=\sum_{n \in \mathbb N}c_n^+x^n, g=\sum_{n \in \mathbb N}c_n^-x^n$ gives us the desired coefficients.
To this end, I think $h(x)=\sin\left(\frac \pi {\ln 2}\ln(1-x)\right)$ works but I have no proof of this (the coefficients are hard to compute explicitly).
My question is as follows:
- I know that the product of Blaschke factors (which satisfy the Blaschke condition) defines a bounded holomorphic function on the unit disc, is it true that such a function also has absolutely convergent coefficients?
- can the coefficients of $h(x)=\sin\left(\frac \pi {\ln 2}\ln(1-x)\right)$ be explicitly found/proven to be absolutely convergent in sum or is it just too much effort? If so is there a example with simpler example that's easier to work with?
- If such an example cannot be written down exactly. Is it possible to at least prove the existence of it? Or is there no such example possible?
Context: This problem was originally part of homework of a graduate probability class which was not meant to be handed in. I deleted the question originally because this particular problem was made into extra credit for the first person who solved it. I am now undeleting it since I think this is a fairly interesting problem and the extra credit has been claimed, so I think it might be good to have a record of this solution for future reference. The idea of the following solution is from my classmate, the explicit example is my original work.
We wish to define $f,g:[0,1] \to [0,1]$ with all positive coefficients such that $f>g$ for infinitely many points and $g>f$ for infinitely many points, and such that $f(1)=g(1)=1$. To do so, we recursively define two sequences $\{(n_k,m_k)\}_{k \in \mathbb N}$ and $\{t_k\}_{k \in \mathbb N}$, we also now define $f_n(x): = \sum_{i=0}^n2^{-i-1}x^{n_i}$ and $g_n(x): = \sum_{i=0}^n2^{-i-1}x^{m_i}$. Let $t_0=0=n_0 = m_0-1$, then $f_0 = \frac 12$ and $g_0 = \frac t2$ and $f_0(t_0) = \frac 12 > 0 = g_0(t_0)$. We intend to make these defined such that $f_N(t_k)>g_N(t_k)$ for $k<N$ even and $g_N(t_k)>f_N(t_k)$ for $k<N$ odd, and the $n_k$s are the powers of $t$ in the power series expansion of $f$ with nonzero coefficient $2^{-k-1}$, and $m_k$ for $g$.\\ Now suppose $f_k,g_k,t_k,n_k,m_k$ are defined, we split into two cases, if $k$ even, then $\epsilon:=f_k(t_k)-g_k(t_k)>0$, and if odd, we have $g_k(t_k)-f_k(t_k)>0$ in which case we repeat the even construction, replacing $f \to g$. So we fix some $m_{k+1}$ such that $2^{-k-1}t_k^{m_{k+1}} < \epsilon$, so that we have: \begin{align*} f_k(t_k)>g_k(t_k)+2^{-k-1}t_k^{m_{k+1}} & \geq g_k(t_k)+2^{-k-2}t_k^{m_{k+1}}(\sum_{i \in \mathbb N}2^{-i}) \\ & \geq g_k(t_k)+2^{-k-2}t_k^{m_{k+1}}(\sum_{i \in \mathbb N}2^{-i}t_k^i)\\ & \geq g_k(t_k)+2^{-k-2}t_k^{m_{k+1}}+2^{-k-3}t_k^{m_{k+1}+1}+ \dots \end{align*} Hence we have: $$f_k(t_k)>g_k(t_k)+2^{-k-1}t_k^{m_{k+1}} > g_k(t_k)+2^{-k-2}t_k^{m_{k+1}}=g_{k+1}(t_k) $$ Now, if we evaluate at $1$: $$f_k(1) = \sum_{i=0}^k2^{-i-1} <\sum_{i=0}^k2^{-i-1}+2^{-k-2} = g_k(1)+2^{-k-2} = g_{k+1}(1)$$ Hence there exists $t_k<t_{k+1}<1$ such that $g_{k+1}(t_{k+1}) > f_k(t_{k+1})$. Now we repeat the process at $t_{k+1}$, choose $2^{-k-1}t_{k+1}^{n_{k+1}} < g_{k+1}(t_{k+1})-f_k(t_{k+1})$ so that we have: \begin{align*} g_{k+1}(t_{k+1})>f_k(t_{k+1})+2^{-k-1}t_{k+1}^{n_{k+1}} & \geq f_k(t_{k+1})+2^{-k-2}t_{k+1}^{n_{k+1}}(\sum_{i \in \mathbb N}2^{-i}) \\ & \geq f_k(t_{k+1})+2^{-k-2}t_{k+1}^{n_{k+1}}(\sum_{i \in \mathbb N}2^{-i}t_{k+1}^i)\\ & \geq f_k(t_{k+1})+2^{-k-2}t_{k+1}^{n_{k+1}}+2^{-k-3}t_{k+1}^{n_{k+1}+1}+ \dots \end{align*} Hence we have: $$g_{k+1}(t_{k+1}) > f_k(t_{k+1})+2^{-k-2}t_{k+1}^{n_{k+1}}=f_{k+1}(t_{k+1})$$ And we also have: $$g_{k+1}(1) = \sum_{i=0}^{k+1} 2^{-i-1}=f_{k+1}(1)$$ We define $f,g$ as the limits of $f_n,g_n$ respectively, both have coefficients which are absolutely convergent to $1$ since the coefficients are just dyadics. \\ Claim that $f(t_k) > g(t_k)$ for all even $k$ and the inequality reverses for all odd $k$. Indeed, fix $k$ even, by construction we have: $$f_k(t_k) > g_k(t_k) + 2^{-k-1}t_k^{m+1} \geq g_k(t_k)+2^{-k-2}t_k^{m_{k+1}} +2^{-k-3}t_k^{m_{k+2}}+2^{-k-3}t_k^{m_{k+2}}+ \dots=g(t_k)$$ And also: $$f_k(t_k) < f_k(t_k) + 2^{-k-2}t_k^{n_k+1} + \dots=f(t_k)$$ The proof for the odd case follows similarly by the exact same computation (but switch $f \to g$). Now we may conclude by IVT that there exists $\{s_n\}_{n \in \mathbb N}$ with $s_n \in (t_n, t_{n+1})$ with $f(s_n) = g(s_n)$ for all $n \in \mathbb N$. \\ In summary, we constructed: $$f: = \sum_{i \in \mathbb N} 2^{-i-1}t^{n_i}, g: = \sum_{i \in \mathbb N} 2^{-i-1}t^{m_i}$$ which clearly satisfy: $$f(1)=g(1) = \sum_{i \in \mathbb N} 2 ^{-i-1} = 1 $$ And which are equal on a countably infinite set $\{s_n\}_{n \in \mathbb N}$.