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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

$$\langle x, y \rangle = \sum_{t \in G} x(t) \overline{y(t)}$$
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

$$x(t)= \langle x , \delta_t \rangle$$

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.