Injectiveness of convolution operator in $\ell^2(G)$

Question

Let $G$ be an infinite countable group and consider the Hilbert space $\ell^2(G)$. Let us define the convolution product as: $$(\xi*\eta)(g)=\sum_{a\in G}\xi(a)\eta(a^{-1}g), ~\text{ for all }~ \xi, \eta \in \ell^2(G).$$ This is easy by Cauchy-Schwarz inequality that $\|\xi*\eta\|_\infty\le \|\xi\|_2\|\eta\|_2.$ Let $\xi \in \ell^2(G)$ be a fixed element. Let us now consider a map $\phi_\xi$ on $\ell^2(G)$ as $$\phi_\xi(\eta)=\xi*\eta ,~\forall~ \eta \in \ell^2(G).$$ Now note that, the map $\phi_\xi$ is bounded operator on $\ell^2(G)$ if $\xi*\eta\in \ell^2(G)$ for every $\eta \in \ell^2(G)$. Now stuck at: Is the map $~\xi \longmapsto \phi_\xi~$ injective?
Please help me to solve this. Any hint, comment or answers is highly appreciated. Thank you.

Answer 1

Note $(\xi * \delta_e)(g) = \xi(g)$ for all $g$ (where $e$ is the identity of $G$). So if $\xi * \delta_e = \tilde{\xi} * \delta_e$, which occurs if $\phi_\xi = \phi_{\tilde{\xi}}$, then $\xi = \tilde{\xi}$.