$\int_{\mathbb R}|f(x)|^{2} dx <\infty \implies \sum_{m\in \mathbb Z}\int_{m-\beta}^{m+\beta}|f(x)|^{2} dx <\infty$?

Question

Let $f\in L^{2}(\mathbb R),$ that is, $\int_{\mathbb R}|f(x)|^{2} dx <\infty,$ and $\beta>0.$

My Question: Is it true that that: $\sum_{m\in \mathbb Z}\int_{m-\beta}^{m+\beta}|f(x)|^{2} dx <\infty.$ If yes, how to justify it?

Answer 1

$\int_{m - \beta}^{m + \beta} |f(x)|^2 \,dx$ is the same as $\int_{-\infty}^{\infty} \chi_{[m - \beta, m + \beta]}(x)|f(x)|^2 \,dx$, where $\chi_A(x)$ means the characteristic function of $A$. So $$\sum_{m \in {\mathbb Z}}\int_{m - \beta}^{m + \beta} |f(x)|^2 \,dx = \sum_{m \in {\mathbb Z}}\int_{-\infty}^{\infty} \chi_{[m - \beta, m + \beta]}(x)|f(x)|^2 \,dx$$ $$= \int_{-\infty}^{\infty} (\sum_{m \in {\mathbb Z}}\chi_{[m - \beta, m + \beta]}(x))|f(x)|^2 \,dx$$ Since $|\sum_{m \in {\mathbb Z}}\chi_{[m - \beta, m + \beta]}(x)| \leq 2\beta$ the above is finite whenever $\int_{-\infty}^{\infty} |f(x)|^2\,dx$ is finite.

Answer 2

If $\beta \leq 1/2$, then $$\sum_{m \in \mathbb{Z}} \int_{m - \beta}^{m + \beta} |f(x)|^2 dx \leq \sum_{m \in \mathbb{Z}} \int_{m - 1/2}^{m + 1/2} |f(x)|^2 dx = \int_{\mathbb{R}} |f(x)|^2 dx < \infty$$

If $\beta > 1/2$, then let $K = \lceil \beta \rceil$. Since the interval $[m - \beta, m + \beta]$ is contained in $\bigcup_{k = -K}^{K} [m + k - 1/2, m + k + 1/2]$, it follows that $$\int_{m - \beta}^{m + \beta}|f(x)|^2 dx \leq \sum_{k = -K}^{K} \int_{m + k - 1/2}^{m + k + 1/2} |f(x)|^2 dx$$ and so $$\begin{aligned} \sum_{m \in \mathbb{Z}} \int_{m - \beta}^{m + \beta}|f(x)|^2 dx & \leq \sum_{m \in \mathbb{Z}} \sum_{k = -K}^{K} \int_{m + k - 1/2}^{m + k + 1/2} |f(x)|^2 dx \\ &= \sum_{k = -K}^{K} \sum_{m \in \mathbb{Z}} \int_{m + k - 1/2}^{m + k + 1/2} |f(x)|^2 dx \\ &= \sum_{k = -K}^{K} \int_{\mathbb{R}}|f(x)|^2 dx \\ &= (2K+1) \int_{\mathbb{R}}|f(x)|^2 dx \\ &< \infty \\ \end{aligned}$$