A question on discrete group

Question

There is a quotation below:

For a discrete group $\Gamma$ , $f\in l^{\infty}(\Gamma)$ and $s, t\in \Gamma$. We let $s.f \in l^{\infty}(\Gamma)$ be the function $s.f(t)=f(s^{-1}t)$.

My question is

If $f\in l^{\infty}(\Gamma)$, then what is $f(t)$ here? I mean $f(t)=?$

Answer 1

You are probably used to think of $\ell^\infty(\Gamma)$ as the set of bounded sequences indexed by $\Gamma$. That's nothing but $$\{f:\Gamma\to\mathbb C:\ \sup\{|f(t)|:\ t\in \Gamma\}<\infty\}. $$

If you think about it $f_t$ is nothing but notation for $f(t)$.

Answer 2

In your setting, $f$ is just a bounded function on the group with values in real or complex numbers. $f(t)$ is its value at $t$.