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Problem Statement: Let $f_{v}(x)=\sin (v\pi x)$. Show that $\lim \inf_{v\rightarrow\infty} f_{v}(x)=-1$ and $\lim \sup_{v\rightarrow\infty} f_{v}(x)=1$ whenever $x$ is irrational. [Hint: If $x$ is irrational, then every arc of the circle $s^{2}+t^{2}=1$ contains $(\cos(v\pi x), \sin(v\pi x))$ for infinitely many $v$.]

I would like to verify if my proof of this problem is correct, given the hint. I am a little confused with my conclusion that there is some sequence that converges to $(0,-1)$, because the sequence must be of the form $\sin (v\pi x)$, and the behavior of the sequence is dependent on our choice of $x$. But wouldn't many sequences of this form continue iterating through the circle?

"Proof."

Let $x\in\mathbb{I}=\mathbb{R}\setminus \mathbb{Q}$. Then consider the equation $$s^{2}+t^{2}=1.$$ Since $x\in \mathbb{I}$, then any arc of the circle contains $(s,t)=(\cos(v\pi x), \sin(v\pi x))$ for infinitely many $v$. Since this is true for any arbitrarily small arc of the circle, then it must be that $(\cos(v\pi x), \sin(v\pi x))$ is dense in the circle, and in particular, $(0,-1)$ is a limit point of $(\cos(v\pi x), \sin(v\pi x))$. Thus, $\exists x\in \mathbb{I}$ so that

$$\lim_{v\rightarrow \infty} f_{v}(x)=\lim_{v\rightarrow \infty} \sin (v\pi x)=-1.$$ Therefore, $$\lim \inf_{v\rightarrow \infty} f_{v}(x)=\lim_{v\rightarrow \infty} (\inf_{m\geq v} f_{v}(x))=\lim_{v\rightarrow \infty} (\inf_{u\geq v} \sin (v\pi x))=-1.$$

And if this is correct, then the $\lim\sup$ case would be analogous? I appreciate any advice!

1 Answers 1

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Yes, $\{(\cos(\nu\pi x), \sin(\nu\pi x))\}_{\nu \in \Bbb N}$ has a subsequence that converges to $(0,-1)$. In fact, it has a subsequence converging to every point of the circle. And yes, that follows from the hint, and it shows what the $\liminf \sin(\nu\pi x) = -1$.

But there is something that I believe you have misunderstood: the density of $\{(\cos(\nu\pi x), \sin(\nu\pi x))\}_{\nu \in \Bbb N}$ was a hint as to which direction you should go. Not a given result that you can just assume. I.e., you also need to prove that this sequence is dense in the unit circle.

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    Thank you for your response! A question about showing that the closure is the unit circle: I wanted to take point $(s_{0}, t_{0})$ which lies on the circle, take an open set around this point and show that the open set intersects the sequence $((\cos v\pi x), \sin(v\pi x))$. But I am not sure what an open set looks like on the circle. I assume, an open arc, but then we would have the open set look something like $(s_{0}-\epsilon,s_{0}+\epsilon)\times$ (some open set dependent on $t_{0}$, $s_{0}$, and $\epsilon$) where any point in this set varies by replacing $\epsilon$ with a smaller number? – 2017-02-25
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    I have that $(s_{0}-\epsilon_{0}, \pm \sqrt{t_{0}^{2}+2s_{0}\epsilon_{0}-\epsilon_{0}^{2}})$ and $(s_{0}+\epsilon_{0}, \pm \sqrt{t_{0}^{2}-2s_{0}\epsilon_{0}-\epsilon_{0}^{2}})$ would lie on the circle and be in this open set, for $\epsilon_{0}<\epsilon$. – 2017-02-25
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    @yung_Pabs - An open set on the circle is just the intersection of the circle and an open set in $\Bbb R^2$. So yes, open arcs of the circle are open sets. But rather than prove this on the unit circle, you may want to prove instead that $vx\operatorname{ mod }2$ is dense in $[0,2]$, then use the continuity and bijectiveness of the map $(\cos \pi t, \sin \pi t)$ to show that any dense set in $[0,2)$ is mapped to a dense set in the circle. – 2017-02-25
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    Ah yes, that is very helpful. Thank you! – 2017-02-25