Show that function series is bounded

Question

Let $h$ be a continuous function on $\mathbb{R}$ such that $h(x) = o(|x|^{-\alpha})$ at $\infty$ for some $\alpha > 1$. Let also $(b_k)_{k\in\mathbb{Z}}$ an increasing sequence such that, $$\lim_{k\to\infty} b_k - b_{k-1} > 0.$$

I consider the function $f$ on $\mathbb{R}$ defined as, $$\forall~ u\in\mathbb{R}, ~f(u) = \sum_{k=0}^{+\infty} h(u - b_k).$$

For any $u\in\mathbb{R}$ it is clear that $f(u) < \infty$. But how to prove that $f$ is bounded? Put differently, $\displaystyle \lim_{u\to\infty} f(u) < \infty$?

For example, the proof is straightforward if $(b_k)$ is an arithmetic sequence: $b_k = a_0k$, where $a_0 > 0$. One can prove that $f$ is modulo $a_0$: $f(u + a_0) = f(u).$ This implies that $f$ is necessarily bounded.

Answer 1

Without loss of generality, I redefine $f$ as, $$f(u) = \sum_{r=0}^{\infty}h(u-a_r),$$ where $(a_r)_r$ is positive on $\mathbb{N}$.

I will prove that $f$ is bounded (this proof can be generalized when $f$ is defined as in initial post).

$h(x) =o\left(|x|^{-\alpha}\right) \implies h(x) =o\left((|x|+1)^{-\alpha}\right) \implies |h(x)| =o\left((|x|+1)^{-\alpha}\right)$. Thus, $\exists~x_0\in\mathbb{R}_+~/~\forall~x<-x_0 \text{ or } x>x_0,~|h(x)| < \left(|x| + 1\right)^{-\alpha} $. In addition, $\exists~M\geq1~/~\forall ~x\in\mathbb{R},~|h(x)| \leq M\left(|x| + 1\right)^{-\alpha} $. For example, one can choose any $\displaystyle M \geq \max\left\{1, \left(\max_{x\in[-x_0,x_0]}h(x)\right)\left(|x_0| + 1\right)^{\alpha}\right\}$.

It follows that, $\forall ~u\in\mathbb{R}$, $r\in\mathbb{N}$, \begin{equation}|h(u-a_r)| \leq M\left(|u-a_r|+1\right)^{-\alpha}. \end{equation}

Let $f^*$ be the real-valued function defined as $\displaystyle f^*(u) = \sum_{r = 0}^{\infty} \left(|u-a_r|+1\right)^{-\alpha}, ~\forall ~u\in\mathbb{R}$.

$\displaystyle \lim_{r\to\infty} a_{r+1} - a_{r} > 0 \implies \exists~r_u \in \mathbb{N} \text{ and } a > 0$ such that $\forall~r\geq r_u$, $a_{r+1}-a_r \geq a$. As $(a_r)_r$ grows to $\infty$, $\forall~u\in\mathbb{R}$, it is possible to choose $r_u$ sufficiently large such that $a_{r_u} > u$. It follows that, $\forall~r\geq r_u$,

\begin{align*} &a_r \geq (r-r_u)a + a_{r_u},\\ &|u-a_r| = a_r-u+1 \geq (r-r_u)a + a_{r_u} - u \geq 0, \\ &\left(|u-a_r| + 1\right)^{-\alpha} \leq \left((r-r_u)a + a_{r_u} - u\right)^{-\alpha}. \end{align*}

Therefore, $\displaystyle f^*(u) < \infty$, for any $u\in\mathbb{R}$.

To prove that $f$ is bounded, it is sufficient to prove that $f^*$ is also bounded. As $f^*$ is a continuous function, it is also sufficient to prove that $\displaystyle \lim_{u\to -\infty} f^*(u) < \infty$ and $\displaystyle \lim_{u\to +\infty} f^*(u) < \infty$.

If $\displaystyle u\leq0$, then $\left(|u-a_r|+1\right)^{-\alpha} = \left(a_r - u +1\right)^{-\alpha} \leq \left(a_r +1\right)^{-\alpha}.$

Thus $\forall ~u\leq 0, ~f^*(u) \leq f^*(0)$. As $f^*$ is a positive function, then $\displaystyle \lim_{u\to -\infty} f^*(u) < \infty$.

Let $k_0\in\mathbb{N}^*$ such that $\forall~r\geq k_0$, $a_{r+1}-a_r \geq a$, for some $a > 0$.

For $u$ sufficiently large, $\exists ~k_u\in\mathbb{N}$, where $k_u > k_0$ and $\forall~r \leq k_u - 1$, $u > a_r$, and $\forall~r \geq k_u$, $u \leq a_r$. For $u$ sufficiently large $f^*(u)$ can be decomposed as, \begin{align*} f^*(u) &= \sum_{r = 0}^{k_0 - 1}\left(|u-a_r|+1\right)^{-\alpha} + \sum_{r = k_0}^{k_u - 1}\left(|u-a_r|+1\right)^{-\alpha} + \sum_{r = k_u}^{\infty}\left(|u-a_r|+1\right)^{-\alpha}\\ f^*(u) &\leq k_0 + \sum_{r = k_0}^{k_u - 1}\left(u-a_r+1\right)^{-\alpha} + \sum_{r = k_u}^{\infty}\left(a_r-u+1\right)^{-\alpha}\\ f^*(u) &\leq k_0 + \sum_{r = k_0}^{k_u - 1}\left(a_{k_u - 1}-a_r+1\right)^{-\alpha} + \sum_{r = k_u}^{\infty}\left(a_r-a_{k_u}+1\right)^{-\alpha} \end{align*}

If $k_0 \leq r \leq k_u -1$, then $a_{k_u - 1} - a_r \geq a(k_u - 1 - r)$, because $a_{r+1}-a_r \geq a$. Thus, $(a_{k_u - 1} - a_r + 1)^{-\alpha} \leq \left(a(k_u - 1 - r) + 1\right)^{-\alpha}.$

Analogously, if $k_u \leq r$, then $a_{r} - a_{k_u} \geq a(r - k_u)$. Thus, $(a_{r} - a_{k_u} + 1)^{-\alpha} \leq \left(a(r - k_u) + 1\right)^{-\alpha}.$

For $u$ sufficiently large,

\begin{align*} f^*(u) &\leq k_0 + \sum_{r = k_0}^{k_u - 1}\left(a(k_u - 1 - r) + 1\right)^{-\alpha} + \sum_{r = k_u}^{\infty}\left(a(r - k_u) + 1\right)^{-\alpha}\\ f^*(u) &\leq k_0 + \sum_{r = 0}^{k_u - k_0 - 1}\left(ar + 1\right)^{-\alpha} + \sum_{r = 0}^{\infty}\left(ar + 1\right)^{-\alpha}\\ f^*(u) &\leq k_0 + 2\sum_{r = 0}^{\infty}\left(ar + 1\right)^{-\alpha} \end{align*}

The quantity $\displaystyle k_0 + 2\sum_{r = 0}^{\infty}\left(ar + 1\right)^{-\alpha}$ does not depend on $u$ and is finite. Hence, $\displaystyle \lim_{u\to +\infty} f^*(u) < \infty$.

As a result $f$ is bounded.

Answer 2

Your way to use periodicity to prove boundedness is smart, but I couldn't generalize. My idea is to bound the series by the improper integral $\int_{-\infty}^{+\infty}|h(\alpha x)|dx$. Let $\alpha:=\liminf_{k\to \infty} b_k-b_{k-1} $ be $>0$.

Suppose for one moment that we have $b_k-b_{k-1} \ge \alpha$ for all $k\in \mathbb{Z}$, then I claim: $$\sum_{k=1}^\infty |h(u-b_k)|\le\int_{-\infty}^{+\infty}|h(\alpha x)|dx$$ so you can bound $f$ by $2\int_{-\infty}^{+\infty}|h(\alpha x)|dx+M$ where $M:=\sup h$.

If $b_k-b_{k-1} \ge \alpha$ is true only for $k\ge k_0$ then you can bound the first terms by $M\cdot k_0$ and repeat the previous step.