# Why is the convergence set of this series (in two variables) a closed set?

I am reading Salem's "A remarkable class of algebraic integers: proof of a conjecture of Vijayaraghavan" in which Salem proves that the set of Pisot-Vijayaraghavan (PV) numbers is closed. My question regards a specific step in the proof (and doesn't have to do with PV numbers specifically).

In the proof of the first theorem of section 3, Salem states that the set of points in $$R := \{(x,y) \in \mathbb{R}^2: 1 \leq x \leq b, a \leq y \leq b\}$$ such that \begin{align*} \sum_{n=0}^\infty \sin^2(\pi xy^n) \leq \pi^2 \left( \frac{2b + 1}{a-1} \right)^2 =: C && (*) \end{align*} is a closed set. Why is this true? I have never encountered convergence sets like this, nor could I find something similar online, so I figured that this would be a good example to have on here. What follows is my attempt.

Let $$(x_i,y_i)$$ be a sequence in $$R$$ converging to $$(x,y) \in R$$ and further let $$(x_i,y_i)$$ satisfy equation $$(*)$$. Then we need to show that $$(x,y)$$ also satisfies equation $$(*)$$. To this end, let's try to bound $$|\sin^2(\pi (x-s)(y-t)^n) - \sin^2(\pi xy^n)| = |G_n(s,t) - G_n(0,0)|$$ for any $$s$$ and $$t$$, where $$G_n(s,t) := \sin^2(\pi (x-s)(y-t)^n)$$. Compute $$\triangledown G_n = -2 \pi \sin(\pi (x-s)(y-t)^n) \cos(\pi (x-s)(y-t)^n)(y-t)^{n-1} \cdot \langle y-t, n(x-s) \rangle$$ to see that $$|\triangledown G_n| \leq 2 \pi |y-t|^{n-1} \cdot \sqrt{(y-t)^2 + n^2(x-s)^2}$$.

Now, let $$(x_i,y_i)$$ be closer than some distance $$\delta$$ to $$(x,y)$$. Also let $$C$$ denote the straight line from $$(x_i,y_i) =: (x-s_i,y-t_i)$$ to $$(x,y)$$, and let r denote the straight-line parametrization of $$C$$. Then I want to be able to compute \begin{align*} |G_n(s_i,t_i) - G_n(0,0)| &= \left| \int_C \triangledown G_n(\textbf{r}) \cdot ~d\textbf{r} \right| \\ &\leq \int_C |\triangledown G_n| \cdot |d\textbf{r}| \\ &\leq \int_{r=0}^\delta 2\pi |y-t|^{n-1} \cdot \sqrt{(y-t)^2 + n^2(x-s)^2} ~dr & \text{by earlier computation} \end{align*} to get a nice bound $$|G_n(s_i,t_i) - G_n(0,0)|$$ but it doesn't work out to something useful. Ideally, I want to bound each of these so that I can obtain $$\left| \sum_{n=0}^\infty \sin(\pi x_i y_i^n) - \sum_{n=0}^\infty \sin(\pi xy^n) \right| \leq \sum_{n=0}^\infty |G_n(s_i,t_i) - G_n(0,0)| \leq \epsilon$$ for any $$\epsilon$$.

How else might one prove closed-ness? Does there exist some more general theory which makes the closed-ness assertion true?

• I think you should investigate whether the RHS of $(*)$ is a continuous function of $x$ and $y$. If it is than the solutions of $(*)$ will certainly comprise a closed set. Commented Apr 16, 2020 at 22:41
• Do you mean the LHS? And that strategy makes sense, but it seems equivalent to what I've attempted: showing that for any $\epsilon$, a sufficiently small perturbation in $x$ and $y$ changes the infinite sum by at most $\epsilon$. Commented Apr 16, 2020 at 23:56
• I did mean the LHS. Sorry for the typo. Commented Apr 17, 2020 at 0:15