Let $(x_n)_n$ be a bounded sequence. Let $a,b >0$ s.t. $\frac{a}{b}$ is an irrationnel. The sequences $(e^{iax_n})_n$ and $(e^{ibx_n})_n$ converge. Show that $(x_n)_n$ converges.
Starting with the fact that $(e^{iax_n})_n$ and $(e^{ibx_n})_n$ converge, we get by definition that:
Suppose that $(e^{iax_n})_n$ converges to $l_1$, then $\forall \epsilon_1 >0 \ \exists N_1>0$ s.t. $\forall n > N_1$: $|e^{iax_n}-l_1|< \epsilon_1$.
Suppose that $(e^{ibx_n})_n$ converges to $l_2$, then $\forall \epsilon_2 >0 \ \exists N_2>0$ s.t. $\forall n > N_2$: $|e^{ibx_n}-l_2|< \epsilon_2$.
From here I don't see how to show that $(x_n)_n$ is convergente.
Best Answer
Let $x$ and $y$ be two limit points of $(x_n)$. Then, because the sequences $(e^{iax_n})$ and $(e^{ibx_n})$ both converge, you get $$e^{iax} = e^{iay} \quad \text{and} \quad e^{ibx} = e^{iby}$$
so $$ax = ay + 2k\pi \quad \text{and} \quad bx=by + 2l\pi$$
for some integers $k$ and $l$.
If $x \neq y$, you get $$a= \frac{2k\pi}{x-y} \quad \text{and} \quad b= \frac{2l\pi}{x-y} $$
which contradicts the fact that $a/b$ must be irrationnal. So $x=y$.
So you have a bounded sequence with only one limit point : it is necessarily a convergent sequence.