Can the definition of almost periodic functions be simplified?

Question

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 $\mathbb{R}_+$.

Answer 1

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 $x_+=\max(x,0)$.

Answer 2

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.