0
$\begingroup$

For a function $f$ from the Schwartz space $\mathcal{S}(\mathbb{R})$ do we have that
$$\sum\limits_{k\in\mathbb{Z}}\int_{\mathbb{R}}|f(x+k)f'(x)|dx$$ converges?

$\endgroup$
1
  • $\begingroup$ Should the summation in the title really be over $\mathbb{R}$? $\endgroup$
    – tj_
    Aug 6, 2019 at 21:38

1 Answer 1

4
$\begingroup$

Trivially, you have the pointwise bound

$$\sum_{k\in\mathbb{Z}} |f(x+k)f'(x)| \leq |f'(x)|\sup_{y\in\mathbb{R}} \sum_{k\in\mathbb{Z}} |f(y+k)|$$

By translation invariance, $$\sup_{y\in\mathbb{R}}\sum_{k\in\mathbb{Z}} |f(y+k)| \leq \sup_{y\in [0,1)} \sum_{k\in\mathbb{Z}} |f(y+k)|,$$ and since $f$ is Schwartz by assumption, $$\sup_{y\in [0,1)} \sum_{k\in\mathbb{Z}} |f(y+k)| \lesssim \sum_{k\in\mathbb{Z}} (1+|k|)^{-2} <\infty.$$ Since $f$ is Schwartz, $f'\in L^1(\mathbb{R}))$. So by monotone convergence theorem, it follows that $$\sum_{k\in\mathbb{Z}} \int_{\mathbb{R}} |f(x+k)f'(x)|dx <\infty.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .