If G is a countable group with neutral element e (and with the composition written multiplicatively).
$\ell^2(G)$ consist of functions $x: G \to \mathbb{C} $ such that $\sum_{t \in G} \vert x(t) \vert^2 < \infty$ and with inner product
\begin{equation} \langle x, y \rangle = \sum_{t \in G} x(t) \overline{y(t)} \end{equation} For $x,y \in \ell^2(G)$.
For each $t \in G$ let $\delta_t \in \ell^2(G)$ be given by $\delta_t (t) = 1$ and $\delta_t(s)=0$ whenever $s \neq t$. \
The set $\lbrace \delta_t \rbrace_{t \in G}$ is an orthonormal basis for $\ell^2 (G) $ and
\begin{equation} x(t)= \langle x , \delta_t \rangle \end{equation}
For $x \in \ell^2(G)$ and $t \in G$.
Let $T \in \mathcal{L}(G)$ be given and put $x = T \delta_e \in \ell^2 (G)$
(i) Show that $x(t)= \langle T \delta_s , \delta_{ts} \rangle$ for all $s,t \in G$
Idea: I believe that the orthonormality part is essential, as every element x in $\ell^2 (G) $ can thus be written as $\sum_{t \in G} \langle x, \delta_t \rangle \delta_t$, right? However, what I have tried so far gives me that
$x(t)=\langle x, \delta_t \rangle = \langle T \delta_e, \delta_t \rangle = \sum_{t \in G} T \delta_e (t) \overline{\delta_t (t)}$ However, I can't see how this will get me any further.
I assume by $\mathcal L(G)$ you mean the group von-Neumann algebra of $G$ (if this is stated somewhere in your question, I couldn't find it). In this case, fix $s,t\in G$, approximate $T$ by an element $T_0\in\mathbb CG$ in the weak operator topology, and show that we have $$(T_0\delta_e)(t)=\langle T_0\delta_s,\delta_{ts}\rangle$$ (The idea here being that the left- and right-regular representations of $G$ on $\ell^2(G)$ commute.)