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A bounded continuous function $f : \mathbb{R} \to \mathbb{C}$ is almost periodic if for every $\epsilon>0$, there exists some $L>0$, such that every interval of $\mathbb{R}$ with length $\ge L$ contains some real number $T$ such that $\Vert f(\cdot+T)-f \Vert_\infty \le \epsilon$.

An equivalent definition is the relative compactness of the set of all functions $f(\cdot+T)$ with $T$ varying in $\mathbb{R}$ in the space of all continuous bounded functions endowed with $\Vert \cdot \Vert_\infty$.

It is necessary to have $$\liminf_{T \to \infty} \Vert f(\cdot+T)-f \Vert_\infty = 0.$$ Is it also sufficient?

My thought: the triangle inequality and the invariance of the norm $\Vert \cdot \Vert_\infty$ under translations show that the function $T \mapsto \Vert f(\cdot+T)-f \Vert_\infty$ is sub-additive on $\mathbb{R}_+$.

Christophe Leuridan
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  • What norm is this? –  Jan 18 '24 at 08:44
  • Usual $\Vert \cdot \Vert_\infty$ norm on the space of all continuous bounded functions. I edited the post. – Christophe Leuridan Jan 18 '24 at 09:02
  • It might help if you'd link a pdf where your quasiperiodic is defined. –  Jan 20 '24 at 09:42
  • I recall the two equivalent definitions in my post. – Christophe Leuridan Jan 20 '24 at 09:53
  • Isn't the "necessary" part obvious? I might be missing something. –  Jan 20 '24 at 10:12
  • The sufficient part sounds hard at first glance (if it's true), you'd need show that assuming there exists some shifts that there have to be linearly many shifts. (I might be misunderstanding something.) –  Jan 20 '24 at 10:21
  • For the sufficient part: If you get one shift and then shift again by that shift best you can do is $2\epsilon$. This obviously can't go on since you are guaranteed to have some bigger shift that works so there has to be some cancellation... Hmm... –  Jan 20 '24 at 10:43
  • Oh my bad, I've read "Is it" instead of "It is" on the "necessary" part. Nice question, +1! –  Jan 20 '24 at 11:07
  • I am thinking about a counterexample, namely, when the function $f(x)$ is like $\sum_{n=1}^\infty \frac 1{n^2}\left(1+\sin\frac x{2^n}\right)$. – Alex Ravsky Jan 20 '24 at 12:19
  • Adding some more tags to this might help get it seen. – Alexander Gruber Jan 20 '24 at 12:20
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    @AlexRavsky A uniform limit of generalized trigonometric polynomials is almost periodic. – Christophe Leuridan Jan 20 '24 at 12:30
  • Someone post a solution to this please else I can't sleep tonight! –  Jan 20 '24 at 15:42
  • @ChristopheLeuridan Could you show some non trivial properties of a function satisfying your definition? –  Jan 20 '24 at 15:46

2 Answers2

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The answer is "No". Just take a hump at $0$ and place copies of it on the line along a finite arithmetic progression with some big difference slowly scaling it down to $0$. Then take the whole resulting picture and place copies of it on the line along a (longer) finite arithmetic progression with much bigger difference even more slowly scaling it down to $0$. And so on. You'll get a function with your properties that is identically $0$ on arbitrarily long intervals, which is, clearly, not almost-periodic.

By eulersgroupie's request here is a formula. Let $\psi$ be your favorite bump on $[-1,1]$. Choose a sequence of numbers $T_j$ going up really fast. Now put $$ f(x)=\sum_{k=(k_1,\dots)}\prod_{j\ge 1}\left(1-\tfrac{|k_j|}{j}\right)_+\psi(x+\sum_j k_jT_j) $$ where the first summation is taken over sequences $k$ having only finitely many integer non-zero entries and $x_+=\max(x,0)$.

fedja
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  • Why does this function satisfy the liminf condition? –  Jan 20 '24 at 19:20
  • @eulersgroupie Shift by the differences of the corresponding progressions. – fedja Jan 20 '24 at 19:23
  • Let's say you shift by the second progression, what guarantees that the humps from the first progression get matched well? –  Jan 20 '24 at 19:26
  • @eulersgroupie The whole picture gets shifted by the second progression. Just take paper and draw the first two iterations :-) – fedja Jan 20 '24 at 19:28
  • @eulersgroupie If it is still hard to comprehend what's going on, I can write a formula, but IMHO, the picture is clearer :) – fedja Jan 20 '24 at 19:33
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    Well I have to admit I do not see your idea as of right now. A formula would clarify things. –  Jan 20 '24 at 19:35
  • @eulersgroupie Enjoy. – fedja Jan 20 '24 at 19:46
  • Thanks! It's a nice formula :) And it works :) Is way more clear than your text! –  Jan 20 '24 at 20:41
  • @eulersgroupie It is an exact translation of my text. But yeah, some people prefer pictures and some prefer formulas. We just belong to opposite categories :-) – fedja Jan 20 '24 at 20:44
  • No, I prefer words but I didn't get your words. –  Jan 20 '24 at 20:44
  • @eulersgroupie Not words. Pictures. You were supposed to draw according to the instructions :-). But nevermind; the message went through and that is all that matters. – fedja Jan 20 '24 at 20:46
  • You provided a text not a picture and I didn't understand your text. That's why we have formulas :) –  Jan 20 '24 at 20:56
  • @eulersgroupie Good point. OK, maybe I'll add a picture too, but not now :-) – fedja Jan 20 '24 at 21:25
  • Btw I did understand your text AFTER you posted your pretty formula :) –  Jan 21 '24 at 09:10
  • @fedja Thank you for your answer. It looks interesting. Let me think about it. – Christophe Leuridan Jan 21 '24 at 21:08
  • @ChristopheLeuridan Notice that only one term in the sum is nonzero for a specific $x$. The finiteness of the arithmetic progressions is encoded in the brackets with the $+$ and the decay is encoded in the product. So around a big progression $T_j$ we have all the smaller finite progressions and if we shift by $T_j$ all the smaller ones get shifted as well and the difference is between $0$ and $1/j$ times the hump at $x$ (maximal difference appears right on the finite progression $T_j$ and the difference is smaller on it's clones coming from the bigger finite progressions). –  Jan 22 '24 at 09:06
  • With all the smaller finite progessions I mean the clones of them (around each smaller finite progression there is again each even smaller finite progression around them...(until you reach the smallest one)). Kinda awkward in words, thinking about it now I think fedja has done a great job in explaining his idea in words. –  Jan 22 '24 at 09:38
  • Thank fedja and user1279061 for your explanations. I have validates fedja's answer. And I give more details in a answer which formalizes her arguments. n – Christophe Leuridan Jan 22 '24 at 20:15
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Reformulation of an example of the answer given by fedja.

Start from $f_0$ given $f_0(x)=(1-|x|)_+$, with compact support $[-1,1]$.

The function $f_1 := \displaystyle{\sum_{k=-1}^1} \Big(1-\Big|\frac{k}{2}\Big|\Big)_+ f_0(\cdot-2k)$ has compact support $[-3,3]$.

The function $f_2 = \displaystyle{\sum_{k=-2}^2} \Big(1-\Big|\frac{k}{3}\Big|\Big)_+ f_1(\cdot-6k)$, has compact support $[-15,15]$.

Assume that $f_n$ has support $[-P_n,P_n]$ where $P_n = \prod_{k=0}^n(2k+1)$.

Then the function $f_{n+1} = \displaystyle{\sum_{k=-n-1}^{n+1}} \Big(1-\Big|\frac{k}{n+2}\Big|\Big)_+ f_{n-1}(\cdot-2P_nk)$ has support $[-P_{n+1},P_{n+1}]$.

Since $f_{n+1}$ and $f_n$ coincide on $[-P_n,P_n]$, and since these intervals are larger and larger, the sequence $(f_n)$ converges uniformly on compacts sets to some continuous function $f$.

By recursion we check that $||f_n||_\infty = f_n(0) = 1$ for all $n$. Thus $||f||_\infty \le 1$.

Using that $k \mapsto (1-|k|/(n+2))_+$ is Lipschitz with ration $1/(n+2)$, we derive that $$||f_{n+1}(\cdot+P_n)-f_{n+1}||_\infty \le 1/(n+2)||f_n||_\infty = 1/(n+2).$$ More generally $||f_m(\cdot+P_n)-f_m||_\infty \le 1/(n+2)$ for all $m \ge n+1$, because $f_m$ is a sum of functions with disjoint supports (up to the boundaries) which are shifts of $f_{n+1}$. Hence $||f(\cdot+P_n)-f||_\infty \le 1/(n+2)$.

Yet, $f$ is not almost periodic. For example, the set of all real numbers $x$ such that $|f(x)| \ge 1/2$ contains no point in arbitrary long intervals.

Christophe Leuridan
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