These days I am not taking any subway since we are on holidays, but let's just keep the glorious name of the project and use it for every idea that I wanted to prove.
I actually don't know what to call to my results, and looked up the differences between Theorems, Propositions and Lemmas, since my results aren't nearly as heavy as the Theorems on research papers. But according to this answer on Math Exchange, we can call our results Theorems if those are the main characters of the text we are writing about. So here you have my Theorem:
If \(f:\mathbb{R}\longrightarrow\mathbb{R}\) is continuous and periodic with period \(p\in\mathbb{R}, p>0\) (meaning that \(f(a)=f(a+p), \forall a\in\mathbb{R}\)), and we denote \((a_n)_{n\geq 1}\) as a sequence, then one of the following holds:
Moreover, if (2) holds, then there exists an increasing sequence \((a_n)_{n\geq 1}\subseteq \mathbb{N}\) such that \((f(a+a_n))_{n\geq 1}\) converges to \(f(a)\), for whatever \(a\in\mathbb{R}\).
I initially came up with the Theorem (it was conjecture back then XD) with just an intuition that it should happen, since one can of course approximate the reals using rationals. Although (1) was easy to prove, (2) initially had a painful amount of details to care about and I was stuck for a long while, but it turned out well in another way.
All in all, it has been super interesting!
Suppose \(p\in\mathbb{Q}\), then \(p=\frac{q}{r}, q,r\in\mathbb{N}\). Now obviously \(b=rp=q\in\mathbb{N}\), and therefore \(f(a)=f(a+p)=f(a+rp)=f(a+b)\), \(\forall a\in \mathbb{R}\).
First we have to present the definitions:
A few theorems that come in handy:
And now let's start with a lemma:
Given \(\varepsilon > 0, \alpha \in \mathbb{R}\), there exist \(b,n\in\mathbb{N}\) such that \(|n\alpha-b|<\varepsilon\).
Proof (Lemma 2A). Here we apply the Dirichlet's Approximation Theorem. Since \(\lim_{N\rightarrow \infty}{\frac{1}{N}}=0\), given \(\varepsilon>0\) we have that \(\exists N_0 : \forall N \geq N_0, \frac{1}{N}\leq \varepsilon\). Let's plug \(N_0\) in Dirichlet's Appr. Theorem, and we have that \(\exists b,n\in\mathbb{N}\) such that \(|n\alpha - b| < \frac{1}{N_0}< \varepsilon\).
Now let us proceed with the final proof.
Suppose \(p\in\mathbb{R}\setminus \mathbb{Q}\). Let \(\varepsilon > 0\). By Heine-Borel Theorem, in \(\mathbb{R}\) one characterization of compactness is that
we have that, fixed \(a\in\mathbb{R}\), the interval \([a,a+p]\) is compact.
By Heine-Cantor Theorem, since \(f\) is continuous on \([a,a+p]\) compact in a metric space, then \(f\) is uniformly continuous. So, given \(\varepsilon > 0\), we have a uniform \(\delta > 0\) that is independent on the point \(x\) chosen.
Since \(f\) is continuous, we have a \(\delta=\delta(\varepsilon)>0\). Applying Lemma 2A, we have the pair \(b,n\in\mathbb{N}\) such that \(|np-b|<\delta\), here \(p\in\mathbb{R}\) is the period of \(f\). Dividing by \(n\) we have that \(|p-\frac{b}{n}|<\frac{\delta}{n}\). Now, we have that \(\forall x\in[a,a+p]\) we have that \(|f(x)-f(x+b)|=|f(x)-f(x+b-np)|\) (because \(f\) is \(p\)-periodic and \(n\in\mathbb{N}\)). Let's see that the points \(x\) and \(x+b-np\) are close enough with each other:
Which implies that \(|f(x)-f(x+b)|<\varepsilon\).
And the result can be easily extended to the entire real line since \(f\) repeats every \(p\). (If you don't like this sentence, there is already a lemma saying that if \(f\) is periodic and continuous in \(\mathbb{R}\), then \(f\) is uniformly continuous in \(\mathbb{R}\)).
Now we still have one sentence stating that if (2) holds then we can build a increasing sequence \((a_n)_n\subseteq\mathbb{N}\) such that \((f(a_n))_n\) converges to \(f(a)\), for \(a\in\mathbb{R}\). This is very easy after all that.
Let \((\varepsilon_n)_n\) be a sequence defined as \(\varepsilon_n = \frac{1}{n}\), for \(n \geq 1, n\in \mathbb{N}\). For each \(\varepsilon_n\) we have a number \(b_n\) that \(|f(a)-f(a+b_n)|<\varepsilon_n,\forall a\in\mathbb{R}\). Let \((b_n)_n\subseteq\mathbb{N}\) be that sequence. This sequence ensures that \((f(a+b_n))_n\) converges to \(f(a)\) by the definition of limit. Now, in order to make it increasing, we use something similar to Stalin sort, and the philosophy is to start with \(b_1\) and delete all the non-increasing successors. By Pigeonhole Principle there must be increasing successors, otherwise the sequence \((f(a+b_n))_n\) doesn't converge. You may observe that \((b_n)_n\) diverges towards \(\infty\), which is pretty interesting.
Initially I wanted to prove it through rational approximation, which is not much more far from what I eventually did. But I needed the Dirichlet's Approximation Theorem. I was heavily thinking about it without it and even almost proved the Dirichlet's Appr. Theorem, but I was too distracted by my Theorem, and did not come up with Pigeonhole Principle. Anyways, it has been an extremely good practise for Pigeonhole Theorem, Dirichlet's Appr. Theorem and many other math theorems!
Here you have one draft of the prior incorrect proof:
Proof: Two ingredients:
Given \(\varepsilon>0\), let's plug it in the definition of the continuity of \(f\) and we get an \(\delta_1=\delta(\varepsilon,x)\). Now let \(\delta_2=\frac{\delta_1}{s_1}\), and we select again \(q_2\in(p-\delta_2,p+\delta_2)\), \(q_2=\frac{r_2}{s_2}\). Let \(b=s_1q_2\). Let's see:
Now if we see that the distance between \(a\) and \(s_2(q_2-p)\) is smaller than \(\delta_1\), we would have it.
Therefore we have proven that, for a specific \(x\), we have a \(b\in\mathbb{N}\) such that \(|f(x)-f(x+b)|<\varepsilon\).
I have slightly modified to hide an error inside. Can you catch it? It is not about the non-uniform continuity.
Time flies by as you're not taking action.
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