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Let $f\colon (0,\infty) \to \mathbb{R}$ a continuous function such that for every $\delta >0$ $$l = \lim_{n \to \infty} f(n\delta) $$ (the limit doesn't depend on $\delta$)

Is it possible to prove that $\lim_{x\to\infty}f(x)$ exist?

user90803
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  • When you write $\lim_{n\to\infty}$ do you mean the limit of a sequence (I.e. You only have $n$ being an integer)? – Dan Robertson Oct 09 '16 at 21:35
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    Relevant: http://math.stackexchange.com/questions/63870/a-classical-problem-about-limit-of-continuous-function-at-infinity-and-its-conne – Aloizio Macedo Oct 10 '16 at 01:41

1 Answers1

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Note: Partial answer only at the moment.

I'm assuming that the limits you assume are only taken over integers. Certainly your statement works the other way round (the continuous limit implies the limit of the series) and if the continuous limit converges then it must converge to $\ell$. Now let's expand your statement with the two definitions of the different types of limit in your question. You want to prove: $$\exists\ell\in\Bbb R\quad\forall\delta>0\quad \forall\epsilon>0\quad\exists N\in\Bbb N \quad\forall N<n\in\Bbb N \quad |f(n\delta)-\ell|<\epsilon $$ $$\Rightarrow$$ $$[\exists\ell\in\Bbb R]\quad\forall \epsilon>0 \quad\exists y>0 \quad\forall x>y\quad |f(x)-\ell|<\epsilon $$

My next step would be "Let $\epsilon>0$...". And then I would look at what I'd need to assume and try to prove it or spot a counter example.

Now there is a counter example if $f$ needn't be continuous:

Let $p_1<p_2<\cdots$ be prime.

Define $1<q_1<2<q_2<3<\dots$ as: $$q_i=\frac{i p_i+1}{p_i}$$

Let $f$ be zero everywhere except that $f(q_i)=1$ for infinitely many $i.$

Perhaps this can be modified to give us a continuous function that also works.

If $\delta$ could only take rational values then you could splice together higher-and-higher frequency sinusoids so that the sequence $(f(n\delta))$ would eventually become constant for any $\delta\in\Bbb Q$ but that is no good for $\delta\not\in\Bbb Q$.

Ok try this. We define $p_i$ and $q_i$ as in the first counterexample except we omit a few small primes as they won't give us sufficiently large denominators. Now define $i<q_i<r_i<i+1$ by: $$r_i=q_i+\frac1{p_i}$$

Now let $f$ be the simplest piece wise linear function which is zero for all values in $(0,1],\,[r_1,2],\,[r_2,3],\,\dots$ and is 1 only on $q_1,q_2,\dots$. If you draw this in a graph, it appears as a series of increasingly sharp spikes (we note that while $f$ is continuous, it is not uniformly continuous). I strongly suspect (although I have not proven it) that this is a counterexample. We not that for $\delta=\frac1{p_i}$ that the spikes soon become thin enough to fit in the gap between $n$ and $\frac{np_i+1}{p_i}$. Similar arguments can be made about any rational $\delta$. This leaves irrational $\delta$ to prove. This is probably the hardest case.

Dan Robertson
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