How translation plays a role to create a pointwise convergence in $\ell^{1}(\mathbb Z)$?

Question

Suppose that $f\in \ell^{1}(\mathbb Z)$ with $\|f\|_{\ell ^1}=1$ (where $f:\mathbb Z \to \mathbb C$ and $\|f\|_{\ell^{1}}= \sum_{m\in \mathbb Z} |f(m)|$.) Let $\{x_{n}\} \subset \mathbb Z$ such that $|x_{n}| \to \infty$ as $n\to \infty.$

Define $g^n= \tau_{x_n}f,$ where $\tau_{x_n}f(m)= f(m-x_{n}), m\in \mathbb Z$

My Question: Can we say $g^n$ converges to $0$ pointwise as $n\to \infty$? In other words, can we say $g^{n}(m)\to 0$ for all $m\in \mathbb Z$ as $n\to \infty$?

Answer 1

Yes. We know that $$ \lim_{|n|\to\infty}f(n)=0. $$ Now, for fixed $m$ we have $|m-x_n|\to\infty$ as $|n|\to\infty$, thus $$ \lim_{|n|\to\infty}g^n(m)=\lim_{|n|\to\infty}f(m-x_n)=0. $$