Let $G$ be a discrete group and consider $l^1(G)$. I know that $l^1(G)$ is an unital Banach $^*$algebra with multiplication as convolution. I am trying to apply the Gelfand–Naimark–Segal construction with the positive functional $\tau(f) := f(e)$(positive meaning that $\tau(f^* * f) \geq 0$, which I have shown). I was wondering if my reasoning is correct so far, define the the function $$\langle f, g \rangle_\tau := \tau(g^* * f)$$Since $\tau$ is a positive functional $\langle \cdot, \cdot \rangle$ is a semi-inner product. Now consider the set $N_\tau = \{f \in l^1(G): \tau(f^* * f) = \langle f, f \rangle_\tau = 0\}$, this is a left ideal. Meaning $l^1(G)/N_\tau$ is an algebra. However notice that $N_\tau$ is a singleton set that only consists of the zero function in $l^1(G)$. So there is isomorphism between $l^1(G)$ and $l^1(G)/N_\tau$. Also our semi - norm defined above is also a norm. Meaning we can think of $l^1(G)$ as $l^2(G)$ under this norm. Which $l^2(G)$ is complete so our GNS hilbert space would be $l^2(G)$. I am a little effy on my reasoning(most of it makes sense) due to my lack of abstract algebra understanding.
Also I am trying to determine the GNS representation using left regular representation of $l^1(G)$ on $l^2(G)$ i.e. $\pi:l^1(G) \rightarrow B(l^2(G))$ with $\pi(f)(g) = f * g$. This is kinda confusing.
Let's carefully analyze this example without the GNS machine that only applies to $C^*$-algebras in general. Over a certain subspace of $C(G)$, the convolution is defined by
$$ (fg)(x)=\sum_{yz=x}f(y)g(z) $$
It's easy to see that $fg$ is well-defined for $f,g\in l^1(G)$ with $\|fg\|_1\le\|f\|_1\|g\|_1$. It's also well-defined for $f,g\in l^2(G)$ by Cauchy-Schwarz, but the problem is $fg$ is not necessarily in $l^2$ any more.
I assume the $*$-operation is defined by $f^*(x)=\overline{f(x^{-1})}$. Then $$\tau(g^*f)=(g^*f)(e)=\sum_x g^*(x^{-1})f(x)=\sum_x f(x)\overline{g(x)}$$
is just the natural inner product, which makes it look more obvious that $N_{\tau}$ is trivial.
As $l^1(G)$ is dense in $l^2(G)$, we can complete it according to the inner product to get $l^2(G)$, that is, we can not directly "think $l^1$ as $l^2$ under this norm": the two norms are different and have different completions. Nonetheless the left regular representation does give an injective $*$-algebra homomorphism $l^1(G)$ to $B(l^2(G))$: $$\langle xg, f\rangle =\phi((xg)^*f)=\phi(g^*(x^*f))=\langle x^*f, g\rangle$$ so the $*$-operation is preserved.
Here is a crucial property of $C^*$-algebras that are not enjoyed by merely Banach ones: An injective $*$-algebra homomorphism must be an isometric embedding. Therefore if $l^1(G)$ happens to be $C^*$, we can conclude the map from $l^1(G)$ to $B(l^2(G))$ preserves the norm, and vice versa: If the representation $l^1(G)\rightarrow B(l^2(G))$ is isometric, then $l^1(G)$ must be a $C^*$-algebra as a complete $*$-subalgebra of $B(\mathcal H)$.
In general, when $G$ is a locally compact Hausdorff group, we may consider the space of compactly supported subalgebra $C_{\mathcal c}(G)$ and there are many different ways to equip it with norms and make a $C^*$ or even von Neumann algebra after completion. The difference between them is subtle and these algebras still constitute many active research topics.
When $G$ is commutative, even though $L^1(G)$ is only Banach, it still has the Gelfand transform, but is not isomorphic to the space of all continuous functions on its Gelfand spectrum. In particular when $G=\mathbb R$, this is related to Fourier analysis for which the $L^1$-theory and $L^2$-theory are significantly different.