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"I need a continuous almost periodic function $f(x)$ such that $\lim_{x\to\infty}f(x)$ exists. But this function should not be constant, which is a trivial example."

Definition of almost periodic function: http://mathworld.wolfram.com/AlmostPeriodicFunction.html

We take the standard metric on $\mathbb{R}$, i.e., $d(x,y)=|x-y|$.

Examples of almost periodic functions: (but $\lim_{t\to\infty}f(t)$ does not exist) $$f(t)=\frac{\cos t}{2+\cos\sqrt2t}\ ,\quad f(t)=\sin2\pi t+ \sin2\pi t\sqrt2 ~.$$

On the other hand, page 69, paragraph 4 of the article http://projecteuclid.org/download/pdf_1/euclid.pjm/1102812425 says the following:

"Once Bohr established his fundamental theorem, he was able to show that any continuous almost periodic function is the limit of a uniformly convergent sequence of trigonometric polynomials. This is the main result of his second paper. The converse of this result was also true."

Since we know that for every non-constant periodic function $g(x)$ (or trigonometric function), $g(x)$ does not exist as $x$ tends to infinity, can we conclude (from uniform convergence) that almost periodic functions also have the same property?

Thanks your help.

Max
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The definition quickly implies that a nonconstant almost periodic function $f$ cannot have a limit at infinity. Indeed, pick $a,b$ such that $f(a)\ne f(b)$. Fix $\epsilon>0$ such that $$|f(a)-f(b)|>3\epsilon\tag{1}$$ Let $\ell=\ell(\epsilon) $ be as in the definition. Then for every integer $n$ there exists $\tau_n\in [n, n+\ell]$ such that $$|f(a+\tau_n)-f(a)|<\epsilon,\qquad |f(b+\tau_n)-f(b)|<\epsilon\tag{2}$$ It follows from (1) and (2) that $$|f(a+\tau_n)-f(b+\tau_n)|>\epsilon \tag{3}$$ Since $a+\tau_n\to\infty$ and $b+\tau_n \to\infty$ as $n\to\infty$, it follows that $\lim_{x\to\infty }f(x)$ does not exist.

user127110
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