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Dirac functions, inner products and $T \in \mathcal{L}(G)$

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.

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