Positive Elements of $\mathbb{C}G$: as functionals versus as elements of the C*-algebra I might have thought about this problem a little longer but am quite confused so said I would put this question to the good people here...
Consider a finite group $G$ or rather the algebra of functions on $G$, $\mathbb{C}^G=F(G)$ with multiplication in $F(G)$ defined pointwise. We have an involution on $F(G)$:
$$f^*(g)=\overline{f(g)}.$$
If we take the supremum norm on $F(G)$:
$$\|f\|_\infty=\max_{g\in G}|f(g)|,$$
we have a C*-algebra. The positive elements of $F(G)$ are those of the form $f^*f$ and so those that have positive coefficients with respect to the basis $\{\delta_s:s\in G\}$ are the positive functions.
Denote by $\mathbb{C}G$ the linear functionals on $F(G)$. This set is spanned by elements dual to the $\{\delta^s:s\in G\}$:
$$\delta^s(\delta_t)=\delta_{t,s},$$
where the latter $\delta$ is the Kronecker delta. It is not difficult to show that the positive linear functionals of $\mathbb{C}(G)$ are those with positive coefficients with respect to the $\{\delta^s:s\in G\}$ basis.
So for example, if $G=\{\mathbb{Z}_3,\,\oplus_3\}$ then $\displaystyle \nu=\frac{\delta^1+\delta^2}{2}\in\mathbb{C}G$ is a positive linear functional.
However one can also give $\mathbb{C}G$ the structure of a C*-algebra. Extend by linearity the convolution product
$$(\delta^s,\delta^t)\mapsto \delta^{st};\qquad\delta^s\star\delta^t=\delta^{st},$$
to the whole of $\mathbb{C}G$. My understanding is that there are C*-norms on $\mathbb{C}G$ when equipped with the involution:
$$\nu^*(\delta_s)=\overline{\nu(\delta_{s^{-1}})}.$$
Now positivity in this C*-algebra is not the same as positivity as a linear functional. In particular it is not difficult to show that there is no $\mu\in\mathbb{C}\mathbb{Z}_3$ such that 
$$\mu^*\star \mu=\frac{\delta^1+\delta^2}{2}=:\nu.$$
Therefore, as an element of the C*-algebra $\mathbb{C}\mathbb{Z}_3$, $\nu$ is not positive.

Is there an easy way of recognising if an element of the C*-algebra
  $\mathbb{C}G$ is positive?

Random thoughts... is it necessary for $\nu(\delta_e)\neq0$, symmetry plays a role? $\nu(\delta_s)=\nu(\delta_{s^{-1}})$... have we any positive elements of the algebra which are not positive functionals. Alternatively I need to look at self-adjoint --- $\nu(\delta_s)=\overline{\nu(\delta_{s^{-1}})}$ --- and positive spectrum... can I find the spectrum of an element of $\mathbb{C}G$?
Thank you.
 A: Consider the space $\mathbb{C}G$ as a Hilbert space with orthonormal basis $e_g$.  $G$ acts on it by left multiplication which are isometries. 
$$L_g \colon h \mapsto g h$$
Hence the adjoint of the map $L_g$ is $L_{g}^{-1} = L_{g^{-1}}$.
Now $\mathbb{C}G$ acts on $\mathbb{C}G$ by left multiplication so we can look at 
$\mathbb{C}G$ as an algebra of linear operators on the Hilberts spaced $\mathbb{C}G$. This algebra is  closed under $*$, since
$$\left(\sum a_g g\right)^* = \sum \bar a_g g^{-1}$$
so it's a $*$ subalgebra of $End(\mathbb{C}G)$, hence a $C^{*}$ algebra. Now, the positivity is preserved under the imbeddings of $C^{*}$ algebras. Now, a basic fact about $C^{*}$ algebras is that   an element of $\mathbb{C}G$ is positive, if and only if its image in $End(\mathbb{C}G)$ is positive. So you consider the corresponding  $|G|$ dimensional matrix check whether that one is positive.
In general, an element $\sum a_g g$ will have the corresponding matrix 
\begin{eqnarray}
\sum a_g g &\mapsto& (x_{gh})_{g, h \in G}\\
x_{gh} &=& a_{g h^{-1}}
\end{eqnarray}
a so called group matrix. 
Consider an element of the group algebra $\mathbb{C} ( \mathbb{Z}/3)$
$$\alpha = a_0  e_0 + a_1 e_1 + a_2 e_2$$
with corresponding matrix 
$$
\left( \begin{array}{ccc} a_0 & a_2 & a_1 \\
a_1  & a_0 & a_2\\
a_2 & a_1 & a_0
\end{array} \right )
$$
The element $\alpha$ and corresponding matrix are selfadjoint if and only if $a_2 = \bar a_1$ and $a_0 \in \mathbb{R}$.
So take $\alpha = 1 \cdot e_1 + 1 \cdot e_2$. The corresponding matrix 
$$
\left( \begin{array}{ccc} 0 & 1 & 1 \\
1  & 0 & 1\\
1 & 1& 0
\end{array} \right )
$$
is not positive definite so neither is $\alpha$. 
One can do this in general for other groups. 
The idea: Map your $C^{*}$ algebra into a matrix algebra and test positivity there.
You can also map $\mathbb{C}G$ isomorphically to a product of matrix algebras ( representation of finite groups ) and test positivity on each component. 
Eg: for $G = \mathbb{Z}/n$, a self-adjoint matrix $(a_{m-n})$ is positive if and only if all the sums $\sum a_n \omega^{n}$ are positive, where $\omega$ is an $n-th$ root of $1$ ( these are the eigenvalues of this circular matrix). 
$\bf{Added:}$
Let's mention that every injective morphism of $C^{*}$ algebras preserves the norms. The way it goes with $\mathbb{C} G \subset End (\mathbb{C}G)$ that's clear  Now let's assume that some $\alpha$ from $\mathbb{C} G$ has a square root in $End (\mathbb{C}G)$. Then that square root is the limit of a sequence of form $r_{n+1} = \frac{1}{2} ( \alpha + \frac{r_n}{\alpha})$, $r_0 = 1$, so that will also be in $\mathbb{C}G$. 
