Suppose $f\in L_2(\mathbb R)$ and $f$ is a continuous function on $\mathbb R$. Is $\displaystyle\sum_{k\in\mathbb{Z}}|f(k)|^2<\infty $?

Question

Suppose $f\in L_2(\mathbb R)$ and $f$ is a continuous function on $\mathbb R$. Is $\displaystyle\sum_{k\in\mathbb{Z}}|f(k)|^2<\infty $ ?

I am trying to prove the relation $\displaystyle\sum_{k\in\mathbb{Z}}|f(k)|^2<\int_{\mathbb R}|f(x)|^2dx$.

Is that relation true? I am struggling to prove this.

Please help me!

Answer 1

Let $t(x) = \max(0, 1-|x|)$, note that $t(0) = 1$ and $\|t\|^2 = {2 \over 3}$.

Note that $\| x \mapsto t(k^2x) \|^2 = { 1\over k^2} {2 \over 3}$.

Let $f(x) = \sum_{k \neq 0} t(k^2(x-k))$, note that $\|f\|^2 \le {2 \over 3} \sum_k {1 \over k^2}$ and $f(k) = 1$ (for $k \neq 1$), hence $\sum_k |f(k)|^2 $ is unbounded.

(I ignored $k=0$ so the support of each of the constituent functions in the sum has disjoint supports. Not necessary, but makes the relationship $g(f(x)) = \sum_{k \neq 0} g(t(k^2(x-k)))$ true for any function $g$.)