Consider the set of all trigonometrical polynomials of the form $P(x)=\sum_{j=1}^n a_n e^{ix\cdot\xi_n}$, where $\xi_n\in\mathbb{R}^d$. A function is said to be almost periodic in the sense of Bohr if it is a uniform limit of trigonometrical polynomials. Define mean value of a function $f\in L^1_{loc}(\mathbb{R}^d)$ to be the number
$$\mathcal{M}(f)=\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T f(x)\,dx.$$ Then, a function $u$ is said to be almost periodic in the sense of Besicovitch if there is a sequence of trigonometrical polynomials $P_n$ such that $\mathcal{M}(|u-P_n|^2)\to 0$ as $n\to\infty$.
My question is whether the definition of mean value depends on the sequence $T\to \infty$. Can it happen that for an almost periodic function, the definition of mean value gives different values for different sequences $T\to\infty$?
The reason for asking this question is that when proving convergences, can one take a particular sequence $T_n\to\infty$, or does one have to prove convergence for all sequences going to $\infty$?
Best Answer
One can prove that for any (Bohr) almost periodic function $f$ the limit $$ \mathcal{M}(f) = \lim_{T\rightarrow \infty} \frac{1}{2T} \int_{-T}^T f(x)dx $$ always exists. Thus, if you want to compute the actual value of $\mathcal{M}(f)$ you may take any sequence $(T_n)_{n\geq 1}\subseteq \mathbb{R}$ such that $\lim_{n\rightarrow \infty} T_n = \infty$ and you will get $$ \mathcal{M}(f) = \lim_{n\rightarrow \infty} \frac{1}{2T_n} \int_{-T_n}^{T_n} f(x)dx.$$
Not directly related what you have asked, but I like this fact and would like to share it: You can even prove as well that for any almost periodic function $f$ $$ \mathcal{M}(f) = \lim_{T\rightarrow \infty} \frac{1}{T} \int_0^T f(x)dx. $$